4.5 Isomorphisms and change of basis

A basis of length $𝑛$ turns $𝑉$ into $𝐹^{𝑛}$ : every vector gets a column of coordinates, and every linear map gets a matrix. But a vector space has many bases, and nothing in $𝑉$ prefers one of them. This section studies that freedom. First we make precise when two vector spaces differ only in the names of their vectors; for finite-dimensional spaces this happens exactly when their dimensions agree. Then we work out how coordinates and matrices change with the basis. This takes one formula, and it leads to similarity: two square matrices are similar exactly when they are matrices of one operator in two bases. Whatever all matrices of an operator have in common, such as their rank and their trace, is then a property of the operator itself, and the freedom to pick the basis becomes a proof technique.

Throughout, $𝐹$ is a field.

4.5.1Isomorphisms

The coordinate map $𝐯 \mapsto {[𝐯]}_{ℬ︀}$ of a basis $ℬ︀$ of length $𝑛$ is a linear bijection $𝑉 \to 𝐹^{𝑛}$ (Proposition 1.6.19, Proposition 1.6.16). Through it, $𝑉$ and $𝐹^{𝑛}$ have the same arithmetic: adding two vectors adds their coordinate columns, and scaling a vector scales its column. So $𝑉$ is $𝐹^{𝑛}$ with different names for the vectors. Linear bijections are exactly the maps that rename vectors in this way.

Definition 4.5.1 (Isomorphism). An isomorphism is an invertible linear map

𝑇 \in L (𝑉, 𝑊)

. If there is an isomorphism from

𝑉

𝑊

, then

𝑉

and

𝑊

are isomorphic, and we write

𝑉 ≅ 𝑊

By Proposition 2.7.7, a linear map is an isomorphism if and only if it is bijective, since the inverse function of a bijective linear map is automatically linear. So to show that $𝑇$ is an isomorphism, we may either show that it is injective and surjective, or exhibit a linear two-sided inverse.

Example 4.5.2 (Isomorphisms).

(a)For every basis $ℬ︀$ of $𝑉$ of length $𝑛 \geq 1$ , the coordinate map $𝑉 \to 𝐹^{𝑛}$ is an isomorphism, with inverse $(𝑎_{1}, \dots, 𝑎_{𝑛}) \mapsto$ $𝑎_{1} 𝐛_{1} +$ $\dots +$ $𝑎_{𝑛} 𝐛_{𝑛}$ (Example 2.5.14(c)). For $𝑉 = {𝐹 [𝑥]}_{\leq 𝑛}$ and the basis $1, 𝑥, \dots, 𝑥^{𝑛}$ , it is the map $𝑎_{0} +$ $𝑎_{1} 𝑥 +$ $\dots +$ $𝑎_{𝑛} 𝑥^{𝑛} \mapsto$ $(𝑎_{0}, \dots, 𝑎_{𝑛})$ , so ${𝐹 [𝑥]}_{\leq 𝑛} ≅ 𝐹^{𝑛 + 1}$ .
(b)Stacking the columns of a matrix, $𝐴 \mapsto$ $(𝑎_{11}, \dots, 𝑎_{𝑚 1}, 𝑎_{12}, \dots, 𝑎_{𝑚 2}, \dots, 𝑎_{𝑚 𝑛})$ , is a linear bijection $𝐹^{𝑚 \times 𝑛} \to 𝐹^{𝑚 𝑛}$ : both spaces use entrywise operations, and unstacking is an inverse. So $𝐹^{2 \times 3} ≅ 𝐹^{6}$ .
(c)The map $(𝑥, 𝑦) \mapsto (𝑥, 𝑦, 0)$ is an isomorphism from $𝐹^{2}$ onto the plane $𝑃 =$ ${(𝑥, 𝑦, 0) | 𝑥, 𝑦 \in 𝐹}$ of $𝐹^{3}$ , with inverse $(𝑥, 𝑦, 0) \mapsto (𝑥, 𝑦)$ . This is the sense in which $𝑃$ is “a copy of $𝐹^{2}$ ” (Exercise 4.1.41(f)), even though $𝐹^{2}$ is not a subset of $𝐹^{3}$ .
(d)The map $L (𝐹, 𝑉) \to 𝑉$ , $𝑇 \mapsto 𝑇 (1)$ , is an isomorphism. It is linear because the operations on $L (𝐹, 𝑉)$ are pointwise. Its inverse sends $𝐯$ to the map $𝑎 \mapsto 𝑎 𝐯$ , since every $𝑇 \in L (𝐹, 𝑉)$ satisfies $𝑇 (𝑎) = 𝑎 𝑇 (1)$ .

Non-example 4.5.3 (Bijections that are not isomorphisms).

(a)Complex conjugation $𝑧 \mapsto 𝑧$ is a bijection $ℂ \to ℂ$ that respects addition and multiplication by real numbers, so it is an isomorphism of vector spaces over $ℝ$ . It is not linear over $ℂ$ : $𝑖 \cdot 1 = - 𝑖$ , while $𝑖 \cdot 1 = 𝑖$ . Whether a map is an isomorphism depends on the field, as dimension does (Remark 4.1.6).
(b)The bijection $ℝ \to ℝ$ , $𝑥 \mapsto 𝑥^{3}$ , is not additive, since ${(1 + 1)}^{3} = 8 \neq 2 = 1^{3} + 1^{3}$ . A bijection between vector spaces renames the vectors, but unless it respects the operations, the arithmetic does not survive the renaming.

Proposition 4.5.4 (Isomorphism is an equivalence relation). For all vector spaces $𝑈$ , $𝑉$ , and $𝑊$ over $𝐹$ :

(a) $𝑉 ≅ 𝑉$ ;
(b)if $𝑉 ≅ 𝑊$ , then $𝑊 ≅ 𝑉$ ;
(c)if $𝑈 ≅ 𝑉$ and $𝑉 ≅ 𝑊$ , then $𝑈 ≅ 𝑊$ .

Proof.routine

Uses. Definition 4.5.1 (Isomorphism), Proposition 2.5.17 (Algebra of inverses), Example 2.5.14 (Invertible maps and matrices)

Hint 1. Each part asks for one isomorphism. Which maps are available?

Hint 2. Use

{id}_{𝑉}

, the inverse of a given isomorphism, and the composite of two given isomorphisms, and quote Proposition 2.5.17.

Sketch.

{id}_{𝑉}

is invertible. If

𝑇 : 𝑉 \to 𝑊

is invertible, so is

𝑇^{- 1} : 𝑊 \to 𝑉

. If

𝑆 : 𝑈 \to 𝑉

and

𝑇 : 𝑉 \to 𝑊

are invertible, so is

𝑇 𝑆 : 𝑈 \to 𝑊

(a)

{id}_{𝑉}

is its own inverse (Example 2.5.14(a)). (b) If

𝑇 \in L (𝑉, 𝑊)

is an isomorphism, then

𝑇^{- 1} \in L (𝑊, 𝑉)

is invertible by Proposition 2.5.17(a). (c) If

𝑆 \in L (𝑈, 𝑉)

and

𝑇 \in L (𝑉, 𝑊)

are isomorphisms, then

𝑇 𝑆 \in L (𝑈, 𝑊)

is invertible by Proposition 2.5.17(b), with inverse

𝑆^{- 1} 𝑇^{- 1}

. □

So the vector spaces over $𝐹$ fall into isomorphism classes (Section A.3). Within a class, every statement that is phrased in terms of the vector space operations transfers from one space to another. Here are the statements we use most.

Proposition 4.5.5 (Isomorphisms preserve linear structure). Let $𝑇 \in L (𝑉, 𝑊)$ be an isomorphism.

(a)A list $ℒ︀ = 𝐯_{1}, \dots, 𝐯_{𝑘}$ in $𝑉$ is linearly independent if and only if $𝑇 (ℒ︀) =$ $𝑇 (𝐯_{1}),$ $\dots,$ $𝑇 (𝐯_{𝑘})$ is. It spans $𝑉$ if and only if $𝑇 (ℒ︀)$ spans $𝑊$ , and it is a basis of $𝑉$ if and only if $𝑇 (ℒ︀)$ is a basis of $𝑊$ .
(b)For every subspace $𝑈$ of $𝑉$ , $im (𝑇 ↾_{𝑈})$ is a subspace of $𝑊$ , and $𝑇 ↾_{𝑈}$ is an isomorphism from $𝑈$ onto $im (𝑇 ↾_{𝑈})$ .
(c) $𝑉$ is finite-dimensional if and only if $𝑊$ is, and in that case $dim 𝑉 = dim 𝑊$ . More generally, $dim im (𝑇 ↾_{𝑈}) =$ $dim 𝑈$ for every finite-dimensional subspace $𝑈$ of $𝑉$ .

Proof.routine

Uses. Theorem 2.6.11 (Injectivity is decided at $𝟎$ ), Proposition 2.5.17 (Algebra of inverses), Proposition 4.2.3 (Properties of restriction), Proposition 2.6.8 (Kernels and images are subspaces), Proposition 2.7.7 (Bijective linear maps are invertible), Definition 4.1.5 (Dimension)

Techniques. Fix an arbitrary element and chase it (Technique 1.3.13)

Hint 1. Linearity gives

𝑐_{1} 𝑇 (𝐯_{1}) +

\dots +

𝑐_{𝑘} 𝑇 (𝐯_{𝑘}) =

𝑇 (𝑐_{1} 𝐯_{1} + \dots + 𝑐_{𝑘} 𝐯_{𝑘})

. For the “only if” directions use that

𝑇

is injective or surjective; for the “if” directions apply the same argument to

𝑇^{- 1}

Hint 2. For (b), a restriction of an injective map is injective, and every map is onto its own image. For (c), apply (a) to a basis.

Sketch. (a) A relation

\sum 𝑐_{𝑖} 𝑇 (𝐯_{𝑖}) = 𝟎

says

𝑇 (\sum 𝑐_{𝑖} 𝐯_{𝑖}) = 𝟎

, so

\sum 𝑐_{𝑖} 𝐯_{𝑖} = 𝟎

by injectivity. If

ℒ︀

spans, every

𝐰 = 𝑇 (𝑇^{- 1} 𝐰)

𝑇

of a combination of

ℒ︀

. The converses are the same statements for

𝑇^{- 1}

. (b)

𝑇 ↾_{𝑈}

is linear and injective, and bijective onto its image. (c) A finite spanning list or basis of

𝑉

is carried to one of

𝑊

of the same length, and back by

𝑇^{- 1}

; for

𝑈

, apply this to the isomorphism of (b).

Since $𝑇$ is linear, for all scalars $𝑐_{1}, \dots, 𝑐_{𝑘}$ ,

𝑐_{1} 𝑇 (𝐯_{1}) + \dots + 𝑐_{𝑘} 𝑇 (𝐯_{𝑘}) = 𝑇 (𝑐_{1} 𝐯_{1} + \dots + 𝑐_{𝑘} 𝐯_{𝑘}) .

(4.5.1)

(a) Suppose $ℒ︀$ is independent, and $\sum_{𝑖} 𝑐_{𝑖} 𝑇 (𝐯_{𝑖}) =$ $𝟎$ . By (4.5.1), $\sum_{𝑖} 𝑐_{𝑖} 𝐯_{𝑖} \in ker 𝑇$ , and $ker 𝑇 = {𝟎}$ because $𝑇$ is injective (Theorem 2.6.11). So $\sum_{𝑖} 𝑐_{𝑖} 𝐯_{𝑖} = 𝟎$ , and all $𝑐_{𝑖} = 0$ . Suppose $ℒ︀$ spans $𝑉$ , and let $𝐰 \in 𝑊$ . Then $𝑇^{- 1} (𝐰) =$ $\sum_{𝑖} 𝑐_{𝑖} 𝐯_{𝑖}$ for some scalars, and by (4.5.1), $𝐰 =$ $𝑇 (𝑇^{- 1} (𝐰)) =$ $\sum_{𝑖} 𝑐_{𝑖} 𝑇 (𝐯_{𝑖})$ . So $𝑇 (ℒ︀)$ spans $𝑊$ .

For the converses, recall that $𝑇^{- 1} \in L (𝑊, 𝑉)$ is an isomorphism (Proposition 2.5.17(a)) and that $𝑇^{- 1} (𝑇 (ℒ︀)) = ℒ︀$ . Applying what we just proved to $𝑇^{- 1}$ and the list $𝑇 (ℒ︀)$ shows that if $𝑇 (ℒ︀)$ is independent or spanning, then so is $ℒ︀$ . The statement about bases combines the two.

(b) By Proposition 4.2.3(c), $𝑇 ↾_{𝑈} \in L (𝑈, 𝑊)$ , so its image is a subspace of $𝑊$ (Proposition 2.6.8). It is injective by Proposition 4.2.3(a), and it maps $𝑈$ onto its own image. So it is a linear bijection $𝑈 \to im (𝑇 ↾_{𝑈})$ , hence an isomorphism (Proposition 2.7.7).

(c) If $ℒ︀$ is a finite list spanning $𝑉$ , then $𝑇 (ℒ︀)$ is a finite list spanning $𝑊$ by (a); applying this to $𝑇^{- 1}$ gives the converse. If $𝑉$ is finite-dimensional with basis $ℬ︀$ , then $𝑇 (ℬ︀)$ is a basis of $𝑊$ of the same length by (a), so $dim 𝑊 = dim 𝑉$ . Finally, if $𝑈$ is a finite-dimensional subspace of $𝑉$ , apply this to the isomorphism $𝑇 ↾_{𝑈} : 𝑈 \to$ $im (𝑇 ↾_{𝑈})$ of (b). □

Now the main theorem. By (c), isomorphic spaces have equal dimensions. The converse is the surprise: for finite-dimensional spaces, dimension is the only thing that can tell two vector spaces apart.

Theorem 4.5.6 (Classification by dimension). Let

𝑉

and

𝑊

be finite-dimensional vector spaces over

𝐹

. Then

𝑉 ≅ 𝑊

if and only if

dim 𝑉 = dim 𝑊

. In particular, for

𝑛 \geq 1

, a vector space is isomorphic to

𝐹^{𝑛}

if and only if it is finite-dimensional of dimension

𝑛

We give two proofs. The first works with bases directly; the second assembles the isomorphism from coordinate maps and gets the dimension count from rank–nullity.

Proof (with bases).routine

Uses. Proposition 4.5.5 (Isomorphisms preserve linear structure), Theorem 2.2.1 (Linear maps are determined by a basis), Proposition 2.6.22 (Testing injectivity and surjectivity on a basis), Proposition 2.7.7 (Bijective linear maps are invertible), Definition 4.1.5 (Dimension)

Techniques. Construct a map by choosing it on a basis (Technique 2.2.4)

Hint 1. For “only if”, what does an isomorphism do to a basis?

Hint 2. For “if”, choose bases

𝐯_{1}, \dots, 𝐯_{𝑛}

𝑉

and

𝐰_{1}, \dots, 𝐰_{𝑛}

𝑊

. Which linear map is the obvious candidate?

Hint 3. The map with

𝑇 (𝐯_{𝑖}) = 𝐰_{𝑖}

sends a basis to a basis. Use Proposition 2.6.22(c).

Sketch. An isomorphism carries a basis to a basis of the same length (Proposition 4.5.5). Conversely, if both spaces have bases of length

𝑛

, the linear map sending the

𝑖

-th vector of one to the

𝑖

-th vector of the other carries a basis to a basis, so it is bijective. The last claim follows since

dim 𝐹^{𝑛} = 𝑛

If $𝑉 ≅ 𝑊$ , then $dim 𝑉 = dim 𝑊$ by Proposition 4.5.5(c).

Conversely, suppose $dim 𝑉 = dim 𝑊 = 𝑛$ . If $𝑛 = 0$ , then $𝑉 = {𝟎}$ and $𝑊 = {𝟎}$ , and the zero map is a bijection between them. If $𝑛 \geq 1$ , choose bases $𝐯_{1}, \dots, 𝐯_{𝑛}$ of $𝑉$ and $𝐰_{1}, \dots, 𝐰_{𝑛}$ of $𝑊$ . By Theorem 2.2.1 there is a linear map $𝑇 : 𝑉 \to 𝑊$ with $𝑇 (𝐯_{𝑖}) = 𝐰_{𝑖}$ for each $𝑖$ . It carries a basis of $𝑉$ to a basis of $𝑊$ , so it is bijective by Proposition 2.6.22(c), and an isomorphism by Proposition 2.7.7.

For the last statement: $dim 𝐹^{𝑛} = 𝑛$ , so a finite-dimensional space of dimension $𝑛$ is isomorphic to $𝐹^{𝑛}$ by what we just proved. Conversely, if $𝑉 ≅ 𝐹^{𝑛}$ , then $𝑉$ is finite-dimensional with $dim 𝑉 = 𝑛$ by Proposition 4.5.5(c). □

Proof (with coordinate maps and rank–nullity).routine

Uses. Example 2.5.14 (Invertible maps and matrices), Proposition 4.5.4 (Isomorphism is an equivalence relation), Theorem 4.2.11 (Rank–nullity), Theorem 2.6.11 (Injectivity is decided at $𝟎$ ), Proposition 4.5.5 (Isomorphisms preserve linear structure)

Techniques. Dimension counting (Technique 4.2.24)

Hint 1. For “if”, both

𝑉

and

𝑊

are isomorphic to the same space. Which one?

Hint 2. The coordinate maps give

𝑉 ≅ 𝐹^{𝑛}

and

𝑊 ≅ 𝐹^{𝑛}

; use Proposition 4.5.4. For “only if”, what are

dim ker 𝑇

and

dim im 𝑇

for an isomorphism

𝑇

Sketch. If

dim 𝑉 = dim 𝑊 = 𝑛 \geq 1

, the coordinate maps give

𝑉 ≅ 𝐹^{𝑛} ≅ 𝑊

, and the composite

𝐶_{𝒞︀}^{- 1} 𝐶_{ℬ︀}

is an isomorphism. If

𝑇 : 𝑉 \to 𝑊

is an isomorphism, rank–nullity gives

dim 𝑉 =

dim ker 𝑇 +

dim im 𝑇 =

0 +

dim 𝑊

Suppose $dim 𝑉 = dim 𝑊 = 𝑛$ . For $𝑛 = 0$ both spaces are ${𝟎}$ , as in the first proof. For $𝑛 \geq 1$ , choose bases $ℬ︀$ of $𝑉$ and $𝒞︀$ of $𝑊$ . The coordinate maps $𝐶_{ℬ︀} : 𝑉 \to 𝐹^{𝑛}$ and $𝐶_{𝒞︀} : 𝑊 \to 𝐹^{𝑛}$ are isomorphisms (Example 2.5.14(c)), so by Proposition 4.5.4, $𝐶_{𝒞︀}^{- 1} 𝐶_{ℬ︀} : 𝑉 \to 𝑊$ is an isomorphism. Explicitly, it sends $\sum_{𝑖} 𝑎_{𝑖} 𝐯_{𝑖}$ to $\sum_{𝑖} 𝑎_{𝑖} 𝐰_{𝑖}$ : it is the map of the first proof.

Conversely, let $𝑇 : 𝑉 \to 𝑊$ be an isomorphism. Then $ker 𝑇 = {𝟎}$ (Theorem 2.6.11) and $im 𝑇 = 𝑊$ , so by Theorem 4.2.11(b),

dim 𝑉 = dim ker 𝑇 + dim im 𝑇 = 0 + dim 𝑊 .

The last statement follows as in the first proof. □

Intuition (Choosing an isomorphism is choosing a basis). The theorem says that up to renaming, the only

𝑛

-dimensional vector space is

𝐹^{𝑛}

. It does not say that there is a preferred renaming. In fact the isomorphisms

𝑉 \to 𝐹^{𝑛}

are exactly the coordinate maps of the bases of

𝑉

(Exercise 1.6.37): an isomorphism

𝜑

is the coordinate map of the basis

𝜑^{- 1} (𝐞_{1}),

\dots,

𝜑^{- 1} (𝐞_{𝑛})

. So there are as many ways to identify

𝑉

with

𝐹^{𝑛}

as there are bases, and in a space like

{𝐹 [𝑥]}_{\leq 2}

, or a plane in

ℝ^{3}

, none of them is singled out. The rest of this section is about how the identifications compare.

Technique 4.5.7 (Transport along an isomorphism). An isomorphism $𝑇 : 𝑉 \to 𝑊$ carries every statement about $𝑉$ that is phrased in terms of the vector space operations (independence, span, bases, subspaces and their dimensions, kernels and images of linear maps) to the corresponding statement about $𝑊$ , and back. So to answer such a question about an unfamiliar space, transport it along an isomorphism to a space where it is easier, answer it there, and transport the answer back.

Reach for it when the space is made of polynomials, matrices, functions, or linear maps, and the question only involves linear structure. The usual target is $𝐹^{𝑛}$ , through coordinates: Proposition 1.6.20 was an early instance, and Theorem 4.5.6 says that every $𝑛$ -dimensional space can be transported to $𝐹^{𝑛}$ . The canonical example is the computation of $dim L (𝑉, 𝑊)$ below, which transports the question to matrices. Be careful to transport only statements made of the linear operations: a property such as “every entry is positive” or “the polynomial has a root” is not preserved by isomorphisms.

Used in 4 places and 2 practice exercises

Theorem 4.5.8, Exercise 4.5.41, Exercise 4.5.42, Exercise 4.8.12; 2 practice exercises.

The dimension of $L (𝑉, 𝑊)$

In Section 2.4 we saw that $𝑇 \mapsto {[𝑇]}_{𝒞︀ ℬ︀}$ is a linear bijection $L (𝑉, 𝑊) \to 𝐹^{𝑚 \times 𝑛}$ (Proposition 2.4.12). In the language of this section, it is an isomorphism, and Technique 4.5.7 gives the dimension of $L (𝑉, 𝑊)$ .

Theorem 4.5.8 (Dimension of $L (𝑉, 𝑊)$ ). Let $𝑉$ and $𝑊$ be finite-dimensional with $dim 𝑉 = 𝑛$ and $dim 𝑊 = 𝑚$ . Then $L (𝑉, 𝑊)$ is finite-dimensional, and

dim L (𝑉, 𝑊) = 𝑚 𝑛 .

If $𝑛, 𝑚 \geq 1$ and $ℬ︀$ , $𝒞︀$ are bases of $𝑉$ , $𝑊$ , then $𝑇 \mapsto {[𝑇]}_{𝒞︀ ℬ︀}$ is an isomorphism $L (𝑉, 𝑊) \to 𝐹^{𝑚 \times 𝑛}$ .

Proof (by transport to matrices).routine

Uses. Proposition 2.4.12 (Matrices of sums and scalar multiples), Theorem 2.3.20 (Linear maps correspond to matrices), Proposition 4.5.5 (Isomorphisms preserve linear structure), Example 4.1.7 (Dimensions of the standard spaces), Proposition 2.1.4 (Linear maps preserve $𝟎$ and negatives)

Techniques. Transport along an isomorphism (Technique 4.5.7)

Hint 1. Dispose of

𝑛 = 0

𝑚 = 0

first: what are the linear maps out of

{𝟎}

, or into

{𝟎}

Hint 2. For

𝑛, 𝑚 \geq 1

, Proposition 2.4.12 gives a linear bijection to a space whose dimension you know.

Sketch. If

𝑉

𝑊

is zero, the only linear map

𝑉 \to 𝑊

0

, so

dim L (𝑉, 𝑊) =

0 =

𝑚 𝑛

. Otherwise

𝑇 \mapsto {[𝑇]}_{𝒞︀ ℬ︀}

is a linear bijection onto

𝐹^{𝑚 \times 𝑛}

, which has dimension

𝑚 𝑛

, and isomorphisms preserve dimension.

If $𝑛 = 0$ , then $𝑉 = {𝟎}$ , and every linear map $𝑇 : 𝑉 \to 𝑊$ is zero since $𝑇 (𝟎) = 𝟎$ (Proposition 2.1.4). If $𝑚 = 0$ , then every map into $𝑊 = {𝟎}$ is zero. In both cases $L (𝑉, 𝑊) = {0}$ has dimension $0 = 𝑚 𝑛$ .

Now let $𝑛, 𝑚 \geq 1$ . By Proposition 2.4.12 and Theorem 2.3.20, $𝑇 \mapsto {[𝑇]}_{𝒞︀ ℬ︀}$ is a linear bijection from $L (𝑉, 𝑊)$ onto $𝐹^{𝑚 \times 𝑛}$ , that is, an isomorphism. Since $dim 𝐹^{𝑚 \times 𝑛} = 𝑚 𝑛$ (Example 4.1.7(a)), Proposition 4.5.5(c) shows that $L (𝑉, 𝑊)$ is finite-dimensional of dimension $𝑚 𝑛$ . □

The second proof writes down a basis of $L (𝑉, 𝑊)$ . It is longer, but it shows what the isomorphism does: it matches a natural basis of $L (𝑉, 𝑊)$ with the standard basis of $𝐹^{𝑚 \times 𝑛}$ .

Proof (with an explicit basis).standard

Uses. Theorem 2.2.1 (Linear maps are determined by a basis), Definition 2.3.1 (Matrix of a linear map), Definition 1.6.1 (Basis), Definition 1.5.1 (Linear independence), Definition 2.1.21 (Pointwise operations)

Techniques. Construct a map by choosing it on a basis (Technique 2.2.4), Check it on a basis (Technique 2.2.7)

Hint 1. For bases

𝐯_{1}, \dots, 𝐯_{𝑛}

𝑉

and

𝐰_{1}, \dots, 𝐰_{𝑚}

𝑊

, define

𝑚 𝑛

maps by saying what each does on the basis of

𝑉

. Which maps correspond to the matrices

𝑈_{𝑖 𝑗}

Hint 2. Let

𝜑_{𝑖 𝑗}

send

𝐯_{𝑗}

𝐰_{𝑖}

and every other

𝐯_{𝑘}

𝟎

. To prove that they span, compare

𝑇

with

\sum 𝑎_{𝑖 𝑗} 𝜑_{𝑖 𝑗}

on the basis. To prove independence, evaluate a relation at

𝐯_{𝑘}

Sketch. Let

𝜑_{𝑖 𝑗}

be the linear map with

𝜑_{𝑖 𝑗} (𝐯_{𝑘}) = 𝐰_{𝑖}

𝑘 = 𝑗

and

𝟎

otherwise. If

𝑇 (𝐯_{𝑘}) =

\sum_{𝑖} 𝑎_{𝑖 𝑘} 𝐰_{𝑖}

, then

\sum_{𝑖, 𝑗} 𝑎_{𝑖 𝑗} 𝜑_{𝑖 𝑗}

agrees with

𝑇

on each

𝐯_{𝑘}

, so they are equal. If

\sum 𝑐_{𝑖 𝑗} 𝜑_{𝑖 𝑗} = 0

, evaluating at

𝐯_{𝑘}

gives

\sum_{𝑖} 𝑐_{𝑖 𝑘} 𝐰_{𝑖} = 𝟎

, so all

𝑐_{𝑖 𝑘} = 0

. So the

𝑚 𝑛

maps form a basis.

The zero cases are as in the first proof, so let $𝑛, 𝑚 \geq 1$ , and let $𝐯_{1}, \dots, 𝐯_{𝑛}$ and $𝐰_{1}, \dots, 𝐰_{𝑚}$ be bases of $𝑉$ and $𝑊$ . For $1 \leq 𝑖 \leq 𝑚$ and $1 \leq 𝑗 \leq 𝑛$ , let $𝜑_{𝑖 𝑗} \in L (𝑉, 𝑊)$ be the linear map with

𝜑_{𝑖 𝑗} (𝐯_{𝑘}) = {\begin{cases} 𝐰_{𝑖} & if 𝑘 = 𝑗 \\ 𝟎 & if 𝑘 \neq 𝑗 \end{cases}

(Theorem 2.2.1). We show that the $𝑚 𝑛$ maps $𝜑_{𝑖 𝑗}$ , listed in any order, form a basis of $L (𝑉, 𝑊)$ .

They span. Let $𝑇 \in L (𝑉, 𝑊)$ , and write $𝑇 (𝐯_{𝑘}) =$ $\sum_{𝑖 = 1}^{𝑚} 𝑎_{𝑖 𝑘} 𝐰_{𝑖}$ for each $𝑘$ . Let $𝑆 =$ $\sum_{𝑖, 𝑗} 𝑎_{𝑖 𝑗} 𝜑_{𝑖 𝑗}$ . Since the operations on $L (𝑉, 𝑊)$ are pointwise,

𝑆 (𝐯_{𝑘}) = \sum_{𝑖 = 1}^{𝑚} \sum_{𝑗 = 1}^{𝑛} 𝑎_{𝑖 𝑗} 𝜑_{𝑖 𝑗} (𝐯_{𝑘}) = \sum_{𝑖 = 1}^{𝑚} 𝑎_{𝑖 𝑘} 𝐰_{𝑖} = 𝑇 (𝐯_{𝑘})

for each $𝑘$ , because only the terms with $𝑗 = 𝑘$ survive. Two linear maps that agree on a basis are equal (Theorem 2.2.1), so $𝑇 = 𝑆$ .

They are independent. Suppose $\sum_{𝑖, 𝑗} 𝑐_{𝑖 𝑗} 𝜑_{𝑖 𝑗} =$ $0$ . Evaluating at $𝐯_{𝑘}$ as above gives $\sum_{𝑖 = 1}^{𝑚} 𝑐_{𝑖 𝑘} 𝐰_{𝑖} =$ $𝟎$ , so $𝑐_{1 𝑘} = \dots = 𝑐_{𝑚 𝑘} = 0$ because $𝐰_{1}, \dots, 𝐰_{𝑚}$ is independent. This holds for every $𝑘$ .

So $L (𝑉, 𝑊)$ has a basis of $𝑚 𝑛$ maps, and $dim L (𝑉, 𝑊) = 𝑚 𝑛$ . □

The coefficients $𝑎_{𝑖 𝑗}$ in the second proof are the entries of ${[𝑇]}_{𝒞︀ ℬ︀}$ , and ${[𝜑_{𝑖 𝑗}]}_{𝒞︀ ℬ︀} =$ $𝑈_{𝑖 𝑗}$ , since $𝜑_{𝑖 𝑗}$ sends the $𝑗$ -th basis vector to $𝐰_{𝑖}$ and the others to $𝟎$ . So the isomorphism of the first proof carries the basis $𝜑_{𝑖 𝑗}$ to the standard basis $𝑈_{𝑖 𝑗}$ of $𝐹^{𝑚 \times 𝑛}$ .

Two special cases come up constantly. For an operator space, $dim L (𝑉) = 𝑛^{2}$ ; so for every $𝑇 \in L (𝑉)$ , the $𝑛^{2} + 1$ operators ${id}_{𝑉}, 𝑇, \dots, 𝑇^{𝑛^{2}}$ are dependent, which is Proposition 4.1.31(b) again. For $𝑊 = 𝐹$ , $dim L (𝑉, 𝐹) = 𝑛$ . The linear maps $𝑉 \to 𝐹$ are the linear functionals, and $L (𝑉, 𝐹)$ is the dual space of $𝑉$ , the subject of Section 13.1.

Exercise 4.5.9.routine Sort the following vector spaces over $𝐹$ into isomorphism classes.

(a) $𝐹^{2 \times 3}$ .
(b) ${𝐹 [𝑥]}_{\leq 5}$ .
(c) $L (𝐹^{2}, 𝐹^{3})$ .
(d)The upper triangular matrices in $𝐹^{3 \times 3}$ .
(e) ${𝑝 \in {𝐹 [𝑥]}_{\leq 6} | 𝑝 (0) = 0}$ .
(f) $L (𝐹^{3}, 𝐹)$ .
(g) ${𝐴 \in 𝐹^{2 \times 2} | tr 𝐴 = 0}$ .

Solution.

Uses. Theorem 4.5.6 (Classification by dimension), Theorem 4.5.8 (Dimension of $L (𝑉, 𝑊)$ ), Example 4.1.7 (Dimensions of the standard spaces), Definition 2.4.26 (Trace)

Hint 1. By Theorem 4.5.6, it is enough to compute dimensions.

Hint 2. For (d), (e), and (g), write down a basis: which matrix units or monomials are allowed?

Sketch. The dimensions are

6, 6, 6, 6, 6, 3, 3

. So (a)–(e) form one class and (f), (g) another.

By Theorem 4.5.6, two finite-dimensional spaces are isomorphic exactly when their dimensions agree. We compute the dimensions.

(a) $2 \cdot 3 = 6$ (Example 4.1.7). (b) $6$ , with basis $1, 𝑥, \dots, 𝑥^{5}$ . (c) $3 \cdot 2 = 6$ by Theorem 4.5.8. (d) An upper triangular matrix is $\sum_{𝑖 \leq 𝑗} 𝑎_{𝑖 𝑗} 𝑈_{𝑖 𝑗}$ with unique coefficients, so the six matrix units $𝑈_{𝑖 𝑗}$ with $𝑖 \leq 𝑗$ form a basis: dimension $6$ . (e) $𝑝 (0) = 0$ means that the constant coefficient vanishes, so $𝑥, 𝑥^{2}, \dots, 𝑥^{6}$ is a basis: dimension $6$ . (f) $1 \cdot 3 = 3$ by Theorem 4.5.8. (g) $(\begin{matrix} 𝑎 & 𝑏 \\ 𝑐 & 𝑑 \end{matrix})$ has trace $0$ when $𝑑 = - 𝑎$ , and then it equals $𝑎 (𝑈_{11} - 𝑈_{22}) +$ $𝑏 𝑈_{12} +$ $𝑐 𝑈_{21}$ , with unique coefficients. So $𝑈_{11} -$ $𝑈_{22},$ $𝑈_{12},$ $𝑈_{21}$ is a basis: dimension $3$ .

So the classes are ${(𝑎), (𝑏), (𝑐), (𝑑), (𝑒)}$ and ${(𝑓), (𝑔)}$ .

4.5.2Change of coordinates

Intuition (One vector, two grids). In

ℝ^{2}

, let

𝐛_{1} = (2, 1)

and

𝐛_{2} = (- 1, 1)

. The vector

𝐯 = (0, 3)

has standard coordinates

(0, 3)

. It is also

𝐛_{1} + 2 𝐛_{2}

, so in the basis

ℬ︀ = 𝐛_{1}, 𝐛_{2}

its coordinates are

(1, 2)

. Draw the grid of integer combinations of

𝐛_{1}

and

𝐛_{2}

on top of the standard grid: the arrow

𝐯

is the same arrow, and each grid reads off its own pair of numbers. Changing the basis does not move any vector. It changes the grid on which we read the vectors, and the question is how the two readings are related.

The vector

𝐯 = (0, 3)

on the standard grid and on the grid of

ℬ︀ = (2, 1), (- 1, 1)

. Read on the first grid,

{[𝐯]}_{ℰ︀} = (0, 3)

; read on the second,

𝐯 = 𝐛_{1} + 2 𝐛_{2}

, so

{[𝐯]}_{ℬ︀} = (1, 2)

. In the interactive version, drag

𝐛_{1}

𝐛_{2}

, or

𝐯

(or focus one and use the arrow keys); the panel shows

{[id]}_{ℰ︀ ℬ︀}

and the conversion

{[𝐯]}_{ℬ︀} =

{[id]}_{ℬ︀ ℰ︀} {[𝐯]}_{ℰ︀}

. When

𝐛_{1}

and

𝐛_{2}

become parallel, the second grid collapses and the coordinates are no longer defined.

The conversion between the two readings is the matrix of the identity map with different bases on the two sides, which we met in Section 2.3.5. We give it a name.

Definition 4.5.10 (Change-of-coordinates matrix). Let $𝑉$ be a vector space with $dim 𝑉 = 𝑛 \geq 1$ , and let $ℬ︀ = 𝐛_{1}, \dots, 𝐛_{𝑛}$ and $\overset{̊}{ℬ︀} = {\overset{̊}{𝐛}}_{1}, \dots, {\overset{̊}{𝐛}}_{𝑛}$ be bases of $𝑉$ . The change-of-coordinates matrix from $ℬ︀$ to $\overset{̊}{ℬ︀}$ is

{[{id}_{𝑉}]}_{\overset{̊}{ℬ︀} ℬ︀} = (\begin{matrix} {[𝐛_{1}]}_{\overset{̊}{ℬ︀}} & {[𝐛_{2}]}_{\overset{̊}{ℬ︀}} & \dots & {[𝐛_{𝑛}]}_{\overset{̊}{ℬ︀}} \end{matrix}) \in 𝐹^{𝑛 \times 𝑛},

whose $𝑗$ -th column holds the $\overset{̊}{ℬ︀}$ -coordinates of the $𝑗$ -th vector of $ℬ︀$ . By Theorem 2.3.9,

{[𝐯]}_{\overset{̊}{ℬ︀}} = {[{id}_{𝑉}]}_{\overset{̊}{ℬ︀} ℬ︀} {[𝐯]}_{ℬ︀} for every 𝐯 \in 𝑉 .

As with every matrix of a map, the subscripts are target first: ${[{id}_{𝑉}]}_{\overset{̊}{ℬ︀} ℬ︀}$ eats $ℬ︀$ -coordinates and returns $\overset{̊}{ℬ︀}$ -coordinates. Books disagree on which of the two matrices to call “the change-of-basis matrix from $ℬ︀$ to $\overset{̊}{ℬ︀}$ ”, so it is safest to trust the subscripts and not the words. We drop the subscript $𝑉$ from ${id}_{𝑉}$ when the space is clear.

Example 4.5.11 (Two grids in the plane). Let $𝐹$ be a field in which $3 \neq 0$ , let $ℰ︀$ be the standard basis of $𝐹^{2}$ , and let $ℬ︀ = (2, 1), (- 1, 1)$ . The columns of ${[id]}_{ℰ︀ ℬ︀}$ are the standard coordinates of the vectors of $ℬ︀$ , which are those vectors themselves. The columns of ${[id]}_{ℬ︀ ℰ︀}$ are the $ℬ︀$ -coordinates of $𝐞_{1}$ and $𝐞_{2}$ , which we find by solving: $𝐞_{1} =$ $\frac{1}{3} (2, 1) -$ $\frac{1}{3} (- 1, 1)$ and $𝐞_{2} =$ $\frac{1}{3} (2, 1) +$ $\frac{2}{3} (- 1, 1)$ . So

{[id]}_{ℰ︀ ℬ︀} = (\begin{matrix} 2 & - 1 \\ 1 & 1 \end{matrix}), {[id]}_{ℬ︀ ℰ︀} = \frac{1}{3} (\begin{matrix} 1 & 1 \\ - 1 & 2 \end{matrix}) .

For $𝐯 = (0, 3)$ , we get ${[𝐯]}_{ℬ︀} =$ $\frac{1}{3} (0 + 3, 0 + 6) =$ $(1, 2)$ , as in the picture. (If $3 = 0$ in $𝐹$ , then $(- 1, 1) = (2, 1)$ and $ℬ︀$ is not a basis.)

The general rules are consequences of the composition theorem.

Proposition 4.5.12 (Calculus of change of coordinates). Let $ℬ︀$ , ${ℬ︀}^{'}$ , and ${ℬ︀}^{″}$ be bases of $𝑉$ , where $dim 𝑉 = 𝑛 \geq 1$ .

(a) ${[id]}_{{ℬ︀}^{″} {ℬ︀}^{'}} {[id]}_{{ℬ︀}^{'} ℬ︀} =$ ${[id]}_{{ℬ︀}^{″} ℬ︀}$ , and ${[id]}_{ℬ︀ ℬ︀} = 𝐼_{𝑛}$ .
(b) ${[id]}_{{ℬ︀}^{'} ℬ︀}$ is invertible, with inverse ${[id]}_{ℬ︀ {ℬ︀}^{'}}$ .
(c) ${[id]}_{{ℬ︀}^{'} ℬ︀}$ is the only matrix $𝑃 \in 𝐹^{𝑛 \times 𝑛}$ with $𝑃 {[𝐯]}_{ℬ︀} = {[𝐯]}_{{ℬ︀}^{'}}$ for every $𝐯 \in 𝑉$ .

Proof.routine

Uses. Theorem 2.4.2 (The matrix of a composite), Example 2.3.6 (The zero map and the identity map), Definition 2.5.12 (Invertible), Theorem 2.3.9 (The matrix of a map computes the map), Definition 2.3.1 (Matrix of a linear map)

Techniques. Translate between maps and matrices (Technique 2.3.27)

Hint 1. For (a),

id \circ id = id

. Which theorem turns a composite into a product of matrices?

Hint 2. (b) follows from (a) with

{ℬ︀}^{″} = ℬ︀

. For (c), apply

𝑃

to the coordinates of a basis vector

𝐛_{𝑗}

Sketch. (a) is Theorem 2.4.2 for

id \circ id = id

, with the middle basis

{ℬ︀}^{'}

. (b) Taking

{ℬ︀}^{″} = ℬ︀

in (a), and then exchanging the roles of

ℬ︀

and

{ℬ︀}^{'}

, both products are

𝐼_{𝑛}

. (c) Existence is Theorem 2.3.9; if

𝑃

converts coordinates, then

𝑃 𝐞_{𝑗} =

𝑃 {[𝐛_{𝑗}]}_{ℬ︀} =

{[𝐛_{𝑗}]}_{{ℬ︀}^{'}}

is the

𝑗

-th column of

{[id]}_{{ℬ︀}^{'} ℬ︀}

(a) By Theorem 2.4.2 applied to ${id}_{𝑉} \circ {id}_{𝑉} = {id}_{𝑉}$ , with $ℬ︀$ on the domain, ${ℬ︀}^{'}$ in the middle, and ${ℬ︀}^{″}$ on the target, ${[id]}_{{ℬ︀}^{″} {ℬ︀}^{'}} {[id]}_{{ℬ︀}^{'} ℬ︀} =$ ${[id]}_{{ℬ︀}^{″} ℬ︀}$ . And ${[id]}_{ℬ︀ ℬ︀} = 𝐼_{𝑛}$ by Example 2.3.6.

(b) By (a) with ${ℬ︀}^{″} = ℬ︀$ , ${[id]}_{ℬ︀ {ℬ︀}^{'}} {[id]}_{{ℬ︀}^{'} ℬ︀} =$ ${[id]}_{ℬ︀ ℬ︀} =$ $𝐼_{𝑛}$ . Exchanging the roles of $ℬ︀$ and ${ℬ︀}^{'}$ gives ${[id]}_{{ℬ︀}^{'} ℬ︀} {[id]}_{ℬ︀ {ℬ︀}^{'}} =$ $𝐼_{𝑛}$ . So ${[id]}_{ℬ︀ {ℬ︀}^{'}}$ is a two-sided inverse of ${[id]}_{{ℬ︀}^{'} ℬ︀}$ (Definition 2.5.12). (This is also Example 2.5.21.)

(c) $𝑃 = {[id]}_{{ℬ︀}^{'} ℬ︀}$ converts coordinates by Theorem 2.3.9. Conversely, suppose $𝑃 {[𝐯]}_{ℬ︀} = {[𝐯]}_{{ℬ︀}^{'}}$ for every $𝐯$ . Taking $𝐯 = 𝐛_{𝑗}$ , the $𝑗$ -th vector of $ℬ︀$ , whose $ℬ︀$ -coordinates are $𝐞_{𝑗}$ , gives $𝑃 𝐞_{𝑗} = {[𝐛_{𝑗}]}_{{ℬ︀}^{'}}$ . The left side is the $𝑗$ -th column of $𝑃$ , and the right side is the $𝑗$ -th column of ${[id]}_{{ℬ︀}^{'} ℬ︀}$ (Definition 2.3.1). □

In $𝐹^{𝑛}$ there is a standard way to compute. For a basis $ℬ︀ = 𝐛_{1}, \dots, 𝐛_{𝑛}$ of $𝐹^{𝑛}$ , ${[id]}_{ℰ︀ ℬ︀}$ is the matrix with columns $𝐛_{1}, \dots, 𝐛_{𝑛}$ , since ${[𝐛_{𝑗}]}_{ℰ︀} = 𝐛_{𝑗}$ . For two bases $ℬ︀$ and $\overset{̊}{ℬ︀}$ of $𝐹^{𝑛}$ , route through the standard basis: by Proposition 4.5.12,

{[id]}_{\overset{̊}{ℬ︀} ℬ︀} = {[id]}_{\overset{̊}{ℬ︀} ℰ︀} {[id]}_{ℰ︀ ℬ︀} = {({[id]}_{ℰ︀ \overset{̊}{ℬ︀}})}^{- 1} {[id]}_{ℰ︀ ℬ︀} .

To compute $𝑀^{- 1} 𝑁$ without computing $𝑀^{- 1}$ first, row reduce $(𝑀 | 𝑁)$ : if $𝐺 𝑀 = 𝐼_{𝑛}$ , then $𝐺 = 𝑀^{- 1}$ , and the reduction ends at $(𝐼_{𝑛} | 𝑀^{- 1} 𝑁)$ (Theorem 3.3.10).

Exercise 4.5.13.routine Over $𝔽_{5}$ , let $ℬ︀ = (1, 2), (3, 4)$ and $\overset{̊}{ℬ︀} = (1, 1), (1, 4)$ .

(a)Show that both lists are bases of $𝔽_{5}^{2}$ , and compute ${[id]}_{\overset{̊}{ℬ︀} ℬ︀}$ .
(b)The vector $𝐯$ has ${[𝐯]}_{ℬ︀} = (1, 1)$ . Find $𝐯$ and ${[𝐯]}_{\overset{̊}{ℬ︀}}$ , and check your answer.

Solution.

Uses. Proposition 4.5.12 (Calculus of change of coordinates), Definition 4.5.10 (Change-of-coordinates matrix), Proposition 2.5.28 (Inverse of a $2 \times 2$ matrix), Proposition 1.1.24 ( $ℤ / 𝑛 ℤ$ is a field if and only if $𝑛$ is prime)

Hint 1.

{[id]}_{\overset{̊}{ℬ︀} ℬ︀} = 𝑀^{- 1} 𝑁

, where

𝑁

and

𝑀

have the vectors of

ℬ︀

and

\overset{̊}{ℬ︀}

as columns. For a

2 \times 2

matrix with

𝑎 𝑑 - 𝑏 𝑐 \neq 0

, the inverse is

{(𝑎 𝑑 - 𝑏 𝑐)}^{- 1} (\begin{matrix} 𝑑 & - 𝑏 \\ - 𝑐 & 𝑎 \end{matrix})

(Proposition 2.5.28).

Hint 2. In

𝔽_{5}

3^{- 1} = 2

. Alternatively, find the columns

{[𝐛_{𝑗}]}_{\overset{̊}{ℬ︀}}

by solving

𝑥 (1, 1) +

𝑦 (1, 4) =

𝐛_{𝑗}

directly.

Sketch.

𝑀 = (\begin{matrix} 1 & 1 \\ 1 & 4 \end{matrix})

has

𝑎 𝑑 - 𝑏 𝑐 = 3

, so

𝑀^{- 1} =

2 (\begin{matrix} 4 & - 1 \\ - 1 & 1 \end{matrix}) =

(\begin{matrix} 3 & 3 \\ 3 & 2 \end{matrix})

, and

{[id]}_{\overset{̊}{ℬ︀} ℬ︀} =

𝑀^{- 1} 𝑁 =

(\begin{matrix} 4 & 1 \\ 2 & 2 \end{matrix})

. Then

𝐯 = (4, 6) = (4, 1)

and

{[𝐯]}_{\overset{̊}{ℬ︀}} =

(5, 4) =

(0, 4)

; indeed

4 (1, 4) =

(4, 16) =

(4, 1)

(a) Let $𝑁 = (\begin{matrix} 1 & 3 \\ 2 & 4 \end{matrix})$ and $𝑀 = (\begin{matrix} 1 & 1 \\ 1 & 4 \end{matrix})$ . Then $1 \cdot 4 - 3 \cdot 2 = - 2 = 3$ and $1 \cdot 4 - 1 \cdot 1 = 3$ are nonzero in $𝔽_{5}$ , so both matrices are invertible (Proposition 2.5.28), their columns are independent (Corollary 3.5.25), and two independent vectors in $𝔽_{5}^{2}$ form a basis (Theorem 4.1.10(b)). Since $3 \cdot 2 = 6 = 1$ , we have $3^{- 1} = 2$ and

𝑀^{- 1} = 2 (\begin{matrix} 4 & - 1 \\ - 1 & 1 \end{matrix}) = (\begin{matrix} 8 & - 2 \\ - 2 & 2 \end{matrix}) = (\begin{matrix} 3 & 3 \\ 3 & 2 \end{matrix}) .

{[id]}_{\overset{̊}{ℬ︀} ℬ︀} = 𝑀^{- 1} 𝑁 = (\begin{matrix} 3 & 3 \\ 3 & 2 \end{matrix}) (\begin{matrix} 1 & 3 \\ 2 & 4 \end{matrix}) = (\begin{matrix} 9 & 21 \\ 7 & 17 \end{matrix}) = (\begin{matrix} 4 & 1 \\ 2 & 2 \end{matrix}) .

Check: $4 (1, 1) +$ $2 (1, 4) =$ $(6, 12) =$ $(1, 2)$ and $1 (1, 1) +$ $2 (1, 4) =$ $(3, 9) =$ $(3, 4)$ .

(b) $𝐯 =$ $(1, 2) +$ $(3, 4) =$ $(4, 6) =$ $(4, 1)$ , and

{[𝐯]}_{\overset{̊}{ℬ︀}} = (\begin{matrix} 4 & 1 \\ 2 & 2 \end{matrix}) (\begin{matrix} 1 \\ 1 \end{matrix}) = (\begin{matrix} 5 \\ 4 \end{matrix}) = (\begin{matrix} 0 \\ 4 \end{matrix}) .

Check: $0 (1, 1) +$ $4 (1, 4) =$ $(4, 16) =$ $(4, 1) =$ $𝐯$ .

The next example changes coordinates in a polynomial space, where the new coordinates have a meaning of their own.

Example 4.5.14 (Expanding around a point). Let $𝑉 = {𝐹 [𝑥]}_{\leq 2}$ , with $ℬ︀ = 1, 𝑥, 𝑥^{2}$ and $\overset{̊}{ℬ︀} =$ $1,$ $𝑥 -$ $1,$ ${(𝑥 - 1)}^{2}$ . Since

𝑥 = 1 + (𝑥 - 1), 𝑥^{2} = 1 + 2 (𝑥 - 1) + {(𝑥 - 1)}^{2},

the three vectors of $\overset{̊}{ℬ︀}$ span $𝑉$ (their span contains $1$ , $𝑥$ , and $𝑥^{2}$ ), so they form a basis (Corollary 4.1.11(b)). The displayed identities give the columns of the change-of-coordinates matrix, and expanding $(𝑥 - 1)$ and ${(𝑥 - 1)}^{2}$ in powers of $𝑥$ gives the columns of its inverse:

{[id]}_{\overset{̊}{ℬ︀} ℬ︀} = (\begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{matrix}), {[id]}_{ℬ︀ \overset{̊}{ℬ︀}} = (\begin{matrix} 1 & - 1 & 1 \\ 0 & 1 & - 2 \\ 0 & 0 & 1 \end{matrix}) .

(Multiply them to check that the product is $𝐼_{3}$ .) So for $𝑝 = 𝑎 + 𝑏 𝑥 + 𝑐 𝑥^{2}$ ,

{[𝑝]}_{\overset{̊}{ℬ︀}} = (𝑎 + 𝑏 + 𝑐, 𝑏 + 2 𝑐, 𝑐), that is, 𝑝 = (𝑎 + 𝑏 + 𝑐) + (𝑏 + 2 𝑐) (𝑥 - 1) + 𝑐 {(𝑥 - 1)}^{2} .

Over $ℝ$ , the new coordinates are $𝑝 (1)$ , $𝑝^{'} (1)$ , and $𝑝^{″} (1) / 2$ : changing to the basis $\overset{̊}{ℬ︀}$ is Taylor expansion at $1$ .

The matrix ${[id]}_{\overset{̊}{ℬ︀} ℬ︀}$ also appeared in Example 2.4.13, as the matrix in the basis $ℬ︀$ of the shift $𝑝 (𝑥) \mapsto 𝑝 (𝑥 + 1)$ . The next remark explains why.

Remark 4.5.15 (Moving the grid or moving the vectors). A change-of-coordinates matrix is also the matrix of an operator in a single basis. Given bases $ℬ︀$ and $\overset{̊}{ℬ︀}$ of $𝑉$ , let $𝑅 \in L (𝑉)$ be the operator with $𝑅 (𝐛_{𝑗}) = {\overset{̊}{𝐛}}_{𝑗}$ for each $𝑗$ (Theorem 2.2.1). The $𝑗$ -th column of ${[𝑅]}_{ℬ︀ ℬ︀}$ is ${[{\overset{̊}{𝐛}}_{𝑗}]}_{ℬ︀}$ , which is also the $𝑗$ -th column of ${[id]}_{ℬ︀ \overset{̊}{ℬ︀}}$ . So

{[id]}_{ℬ︀ \overset{̊}{ℬ︀}} = {[𝑅]}_{ℬ︀ ℬ︀}, and hence {[id]}_{\overset{̊}{ℬ︀} ℬ︀} = {({[𝑅]}_{ℬ︀ ℬ︀})}^{- 1} = {[𝑅^{- 1}]}_{ℬ︀ ℬ︀}

by Proposition 4.5.12(b) and Proposition 2.5.20(b) ( $𝑅$ is invertible because it carries a basis to a basis). In Example 4.5.14, $𝑅 (𝑥^{𝑗}) = {(𝑥 - 1)}^{𝑗}$ , so $𝑅$ is the substitution $𝑝 (𝑥) \mapsto 𝑝 (𝑥 - 1)$ and $𝑅^{- 1}$ is $𝑝 (𝑥) \mapsto 𝑝 (𝑥 + 1)$ .

So there are two ways to read the same matrix. In the passive reading, the vectors stay put and the grid changes from $ℬ︀$ to $\overset{̊}{ℬ︀}$ . In the active reading, the grid stays put and the vectors move by $𝑅^{- 1}$ . Moving the grid forward by $𝑅$ looks, in coordinates, exactly like moving the vectors back by $𝑅^{- 1}$ .

So every change-of-coordinates matrix is invertible. Conversely, every invertible matrix changes coordinates, from any given basis to a suitable new one. This is what makes the change-of-basis formula below reversible.

Proposition 4.5.16 (Every invertible matrix changes coordinates). Let $ℬ︀ = 𝐛_{1}, \dots, 𝐛_{𝑛}$ be a basis of $𝑉$ , and let $𝑆 =$ $(𝑠_{𝑖 𝑗}) \in$ $𝐹^{𝑛 \times 𝑛}$ be invertible.

(a)There is exactly one basis $\overset{̊}{ℬ︀}$ of $𝑉$ with ${[id]}_{ℬ︀ \overset{̊}{ℬ︀}} = 𝑆$ , namely
${\overset{̊}{𝐛}}_{𝑗} = 𝑠_{1 𝑗} 𝐛_{1} + 𝑠_{2 𝑗} 𝐛_{2} + \dots + 𝑠_{𝑛 𝑗} 𝐛_{𝑛} (1 \leq 𝑗 \leq 𝑛) .$
(b)There is exactly one basis ${ℬ︀}^{'}$ of $𝑉$ with ${[id]}_{{ℬ︀}^{'} ℬ︀} = 𝑆$ .

Proof (through coordinates).routine

Uses. Definition 4.5.10 (Change-of-coordinates matrix), Definition 1.6.14 (Coordinates), Corollary 3.5.25 (The invertible matrix theorem), Theorem 4.1.10 (Independent lists are short), Proposition 1.6.20 (Coordinates translate questions about $𝑉$ into $𝐹^{𝑛}$ ), Proposition 4.5.12 (Calculus of change of coordinates), Proposition 2.5.17 (Algebra of inverses)

Techniques. Uniqueness via two candidates (Technique 1.1.8)

Hint 1. The

𝑗

-th column of

{[id]}_{ℬ︀ \overset{̊}{ℬ︀}}

{[{\overset{̊}{𝐛}}_{𝑗}]}_{ℬ︀}

. So if

\overset{̊}{ℬ︀}

exists, its vectors are forced. Which vectors?

Hint 2. With

{\overset{̊}{𝐛}}_{𝑗}

defined as in the statement,

{[{\overset{̊}{𝐛}}_{𝑗}]}_{ℬ︀}

is the

𝑗

-th column of

𝑆

. Why do the columns of an invertible matrix form a basis of

𝐹^{𝑛}

, and how does that transfer to

𝑉

Hint 3. For (b), apply (a) to

𝑆^{- 1}

and use Proposition 4.5.12(b).

Sketch. (a) The equation

{[id]}_{ℬ︀ \overset{̊}{ℬ︀}} = 𝑆

says

{[{\overset{̊}{𝐛}}_{𝑗}]}_{ℬ︀} = 𝐬_{𝑗}

(the

𝑗

-th column of

𝑆

) for each

𝑗

, which determines

{\overset{̊}{𝐛}}_{𝑗}

. These vectors form a basis because their coordinate vectors, the columns of the invertible

𝑆

, form a basis of

𝐹^{𝑛}

(Proposition 1.6.20(c)). (b)

{[id]}_{{ℬ︀}^{'} ℬ︀} = 𝑆

if and only if

{[id]}_{ℬ︀ {ℬ︀}^{'}} = 𝑆^{- 1}

; apply (a) to

𝑆^{- 1}

(a) Let $𝐬_{𝑗}$ be the $𝑗$ -th column of $𝑆$ . For any basis $\overset{̊}{ℬ︀}$ , the $𝑗$ -th column of ${[id]}_{ℬ︀ \overset{̊}{ℬ︀}}$ is ${[{\overset{̊}{𝐛}}_{𝑗}]}_{ℬ︀}$ (Definition 4.5.10). So ${[id]}_{ℬ︀ \overset{̊}{ℬ︀}} = 𝑆$ holds exactly when ${[{\overset{̊}{𝐛}}_{𝑗}]}_{ℬ︀} = 𝐬_{𝑗}$ for every $𝑗$ , that is, when ${\overset{̊}{𝐛}}_{𝑗} =$ $𝑠_{1 𝑗} 𝐛_{1} +$ $\dots +$ $𝑠_{𝑛 𝑗} 𝐛_{𝑛}$ (Definition 1.6.14). This proves uniqueness, and it remains to show that these vectors form a basis. Since $𝑆$ is invertible, its columns are independent (Corollary 3.5.25), and $𝑛$ independent vectors in $𝐹^{𝑛}$ form a basis (Theorem 4.1.10(b)). The coordinate vectors of ${\overset{̊}{𝐛}}_{1}, \dots, {\overset{̊}{𝐛}}_{𝑛}$ are $𝐬_{1}, \dots, 𝐬_{𝑛}$ , so $\overset{̊}{ℬ︀}$ is a basis of $𝑉$ by Proposition 1.6.20(c).

(b) By Proposition 4.5.12(b), for a basis ${ℬ︀}^{'}$ we have ${[id]}_{{ℬ︀}^{'} ℬ︀} = 𝑆$ if and only if ${[id]}_{ℬ︀ {ℬ︀}^{'}} = 𝑆^{- 1}$ . The matrix $𝑆^{- 1}$ is invertible (Proposition 2.5.17(a)), so by (a) there is exactly one such basis. □

Proof (by moving the basis).routine

Uses. Theorem 2.3.20 (Linear maps correspond to matrices), Proposition 2.5.20 (Inverses and matrices), Proposition 2.6.22 (Testing injectivity and surjectivity on a basis), Remark 4.5.15 (Moving the grid or moving the vectors)

Techniques. Translate between maps and matrices (Technique 2.3.27)

Hint 1. Use the active reading of Remark 4.5.15: look for an operator

𝑅

with

𝑅 (𝐛_{𝑗}) = {\overset{̊}{𝐛}}_{𝑗}

Hint 2. Let

𝑅 \in L (𝑉)

have

{[𝑅]}_{ℬ︀ ℬ︀} = 𝑆

. Why is

𝑅

invertible, and what is

𝑅 (ℬ︀)

Sketch. Let

𝑅 \in L (𝑉)

be the operator with

{[𝑅]}_{ℬ︀ ℬ︀} = 𝑆

. It is invertible because

𝑆

is, so

\overset{̊}{ℬ︀} = 𝑅 (ℬ︀)

is a basis, and

{[id]}_{ℬ︀ \overset{̊}{ℬ︀}} =

{[𝑅]}_{ℬ︀ ℬ︀} =

𝑆

by Remark 4.5.15. Uniqueness and (b) are as in the first proof.

(a) By Theorem 2.3.20(b) there is $𝑅 \in L (𝑉)$ with ${[𝑅]}_{ℬ︀ ℬ︀} = 𝑆$ , and $𝑅$ is invertible because $𝑆$ is (Proposition 2.5.20(b)). So $\overset{̊}{ℬ︀} =$ $𝑅 (𝐛_{1}),$ $\dots,$ $𝑅 (𝐛_{𝑛})$ is a basis of $𝑉$ (Proposition 2.6.22(c)), and by Remark 4.5.15, ${[id]}_{ℬ︀ \overset{̊}{ℬ︀}} =$ ${[𝑅]}_{ℬ︀ ℬ︀} =$ $𝑆$ . The $𝑗$ -th vector of $\overset{̊}{ℬ︀}$ is $𝑅 (𝐛_{𝑗}) =$ $\sum_{𝑖} 𝑠_{𝑖 𝑗} 𝐛_{𝑖}$ , because the $𝑗$ -th column of ${[𝑅]}_{ℬ︀ ℬ︀}$ is $𝐬_{𝑗}$ . Uniqueness is proved as in the first proof: the columns of ${[id]}_{ℬ︀ \overset{̊}{ℬ︀}}$ determine the vectors of $\overset{̊}{ℬ︀}$ .

(b) As in the first proof, apply (a) to $𝑆^{- 1}$ . □

In particular, for a fixed basis $ℬ︀$ , the map $\overset{̊}{ℬ︀} \mapsto {[id]}_{ℬ︀ \overset{̊}{ℬ︀}}$ is a bijection from the set of bases of $𝑉$ onto the set of invertible $𝑛 \times 𝑛$ matrices. Over a field with $𝑞$ elements, the number of invertible $𝑛 \times 𝑛$ matrices is therefore the number of bases of $𝐹^{𝑛}$ , which Corollary 4.1.35 computed: $(𝑞^{𝑛} - 1) (𝑞^{𝑛} - 𝑞) \dots (𝑞^{𝑛} - 𝑞^{𝑛 - 1})$ .

4.5.3The change-of-basis formula

Now let $𝑇 \in L (𝑉, 𝑊)$ , where $dim 𝑉 = 𝑛 \geq 1$ and $dim 𝑊 = 𝑚 \geq 1$ . If we replace the basis $ℬ︀$ of $𝑉$ by $\overset{̊}{ℬ︀}$ and the basis $𝒞︀$ of $𝑊$ by $\overset{̊}{𝒞︀}$ , how is the new matrix ${[𝑇]}_{\overset{̊}{𝒞︀} \overset{̊}{ℬ︀}}$ related to the old one?

Theorem 4.5.17 (Change of basis). Let $ℬ︀$ and $\overset{̊}{ℬ︀}$ be bases of $𝑉$ , let $𝒞︀$ and $\overset{̊}{𝒞︀}$ be bases of $𝑊$ , and let $𝑇 \in L (𝑉, 𝑊)$ . Then

{[𝑇]}_{\overset{̊}{𝒞︀} \overset{̊}{ℬ︀}} = {[{id}_{𝑊}]}_{\overset{̊}{𝒞︀} 𝒞︀} {[𝑇]}_{𝒞︀ ℬ︀} {[{id}_{𝑉}]}_{ℬ︀ \overset{̊}{ℬ︀}} .

In particular, if $𝑇 \in L (𝑉)$ and $𝑃 = {[{id}_{𝑉}]}_{ℬ︀ \overset{̊}{ℬ︀}}$ , then

{[𝑇]}_{\overset{̊}{ℬ︀} \overset{̊}{ℬ︀}} = 𝑃^{- 1} {[𝑇]}_{ℬ︀ ℬ︀} 𝑃 .

Intuition (Reading the formula). Read the right side from right to left, as the composite it is. Start with the $\overset{̊}{ℬ︀}$ -coordinates of a vector $𝐯$ . The factor ${[{id}_{𝑉}]}_{ℬ︀ \overset{̊}{ℬ︀}}$ converts them into $ℬ︀$ -coordinates, in which we know how to apply $𝑇$ ; the factor ${[𝑇]}_{𝒞︀ ℬ︀}$ does so and produces the $𝒞︀$ -coordinates of $𝑇 (𝐯)$ ; and ${[{id}_{𝑊}]}_{\overset{̊}{𝒞︀} 𝒞︀}$ converts those into $\overset{̊}{𝒞︀}$ -coordinates. The result is what ${[𝑇]}_{\overset{̊}{𝒞︀} \overset{̊}{ℬ︀}}$ does in one step.

There is also a bookkeeping rule. In a product of matrices of maps, adjacent subscripts must match, like dominoes: $\overset{̊}{𝒞︀} 𝒞︀ \cdot$ $𝒞︀ ℬ︀ \cdot$ $ℬ︀ \overset{̊}{ℬ︀}$ . The two outer subscripts, $\overset{̊}{𝒞︀}$ and $\overset{̊}{ℬ︀}$ , are the bases of the result. A formula whose subscripts do not chain this way is wrong, whatever its other merits.

The first proof is the lecture’s: the formula is the composition theorem applied to $𝑇 = {id}_{𝑊} \circ 𝑇 \circ {id}_{𝑉}$ , with each copy of a space carrying a different basis.

Proof (by composition).routine

Uses. Theorem 2.4.2 (The matrix of a composite), Proposition 2.1.24 (Algebra of composition), Proposition 4.5.12 (Calculus of change of coordinates)

Techniques. Translate between maps and matrices (Technique 2.3.27)

Hint 1. Write

𝑇

as a composite of three maps in which identity maps change the basis.

Hint 2.

𝑇 = {id}_{𝑊} \circ 𝑇 \circ {id}_{𝑉}

. Apply Theorem 2.4.2 twice, choosing the bases

\overset{̊}{ℬ︀}

ℬ︀

𝒞︀

\overset{̊}{𝒞︀}

on the four copies of

𝑉

and

𝑊

Sketch.

𝑇 = {id}_{𝑊} \circ 𝑇 \circ {id}_{𝑉}

(Proposition 2.1.24(b)). By Theorem 2.4.2, with

\overset{̊}{ℬ︀}

on the first copy of

𝑉

ℬ︀

on the second,

𝒞︀

on the first copy of

𝑊

, and

\overset{̊}{𝒞︀}

on the second, the matrix of the composite is the product of the three matrices. For operators,

{[id]}_{\overset{̊}{ℬ︀} ℬ︀} = 𝑃^{- 1}

by Proposition 4.5.12(b).

By Proposition 2.1.24(b), $𝑇 = {id}_{𝑊} \circ (𝑇 \circ {id}_{𝑉})$ . Apply Theorem 2.4.2 to $𝑇 \circ {id}_{𝑉}$ , with $\overset{̊}{ℬ︀}$ on the domain, $ℬ︀$ on the middle copy of $𝑉$ , and $𝒞︀$ on $𝑊$ :

{[𝑇]}_{𝒞︀ \overset{̊}{ℬ︀}} = {[𝑇 \circ {id}_{𝑉}]}_{𝒞︀ \overset{̊}{ℬ︀}} = {[𝑇]}_{𝒞︀ ℬ︀} {[{id}_{𝑉}]}_{ℬ︀ \overset{̊}{ℬ︀}} .

Apply it again to ${id}_{𝑊} \circ 𝑇$ , with $\overset{̊}{ℬ︀}$ on $𝑉$ , $𝒞︀$ on the middle copy of $𝑊$ , and $\overset{̊}{𝒞︀}$ on the target:

{[𝑇]}_{\overset{̊}{𝒞︀} \overset{̊}{ℬ︀}} = {[{id}_{𝑊}]}_{\overset{̊}{𝒞︀} 𝒞︀} {[𝑇]}_{𝒞︀ \overset{̊}{ℬ︀}} = {[{id}_{𝑊}]}_{\overset{̊}{𝒞︀} 𝒞︀} {[𝑇]}_{𝒞︀ ℬ︀} {[{id}_{𝑉}]}_{ℬ︀ \overset{̊}{ℬ︀}} .

For $𝑇 \in L (𝑉)$ , take $𝒞︀ = ℬ︀$ and $\overset{̊}{𝒞︀} = \overset{̊}{ℬ︀}$ . Then ${[{id}_{𝑉}]}_{\overset{̊}{ℬ︀} ℬ︀} = 𝑃^{- 1}$ by Proposition 4.5.12(b). □

The second proof chases a single vector through the diagram of Figure 4.5.1. It is the lecture’s proof of ${[𝐯]}_{\overset{̊}{ℬ︀}} =$ ${[id]}_{\overset{̊}{ℬ︀} ℬ︀} {[𝐯]}_{ℬ︀}$ , extended by two squares.

Figure 4.5.1. Three commuting squares. The top row composes to

𝑇

, and the bottom row multiplies three matrices. Each square says that two routes from its top left corner to its bottom right corner agree (Theorem 2.3.9), so a vector can be pushed along the bottom row instead of the top. The outer rectangle is the change-of-basis formula.

Proof (by chasing a vector).routine

Uses. Theorem 2.3.9 (The matrix of a map computes the map), Proposition 1.6.16 (The coordinate map is a bijection), Definition 2.3.8 (Matrix–vector product)

Techniques. Translate between maps and matrices (Technique 2.3.27), Fix an arbitrary element and chase it (Technique 1.3.13)

Hint 1. Two

𝑚 \times 𝑛

matrices are equal if they give the same result on every

𝐱 \in 𝐹^{𝑛}

: compare them on

𝐞_{𝑗}

Hint 2. Every

𝐱 \in 𝐹^{𝑛}

{[𝐯]}_{\overset{̊}{ℬ︀}}

for some

𝐯 \in 𝑉

. Multiply

{[𝐯]}_{\overset{̊}{ℬ︀}}

by the right side one factor at a time, using Theorem 2.3.9 at each step.

Sketch. For

𝐯 \in 𝑉

, multiplying

{[𝐯]}_{\overset{̊}{ℬ︀}}

by the three factors in turn gives

{[𝐯]}_{ℬ︀}

, then

{[𝑇 (𝐯)]}_{𝒞︀}

, then

{[𝑇 (𝐯)]}_{\overset{̊}{𝒞︀}}

, which is also

{[𝑇]}_{\overset{̊}{𝒞︀} \overset{̊}{ℬ︀}} {[𝐯]}_{\overset{̊}{ℬ︀}}

. Every column in

𝐹^{𝑛}

is a

\overset{̊}{ℬ︀}

-coordinate vector, in particular each

𝐞_{𝑗}

, so the two matrices have the same columns.

Let $𝑀$ be the right side of the formula, and let $𝐯 \in 𝑉$ . Applying Theorem 2.3.9 three times (to ${id}_{𝑉}$ , to $𝑇$ , and to ${id}_{𝑊}$ ),

\begin{aligned} 𝑀 {[𝐯]}_{\overset{̊}{ℬ︀}} & = {[{id}_{𝑊}]}_{\overset{̊}{𝒞︀} 𝒞︀} {[𝑇]}_{𝒞︀ ℬ︀} ({[{id}_{𝑉}]}_{ℬ︀ \overset{̊}{ℬ︀}} {[𝐯]}_{\overset{̊}{ℬ︀}}) \\ = {[{id}_{𝑊}]}_{\overset{̊}{𝒞︀} 𝒞︀} ({[𝑇]}_{𝒞︀ ℬ︀} {[𝐯]}_{ℬ︀}) \\ = {[{id}_{𝑊}]}_{\overset{̊}{𝒞︀} 𝒞︀} {[𝑇 (𝐯)]}_{𝒞︀} = {[𝑇 (𝐯)]}_{\overset{̊}{𝒞︀}}, \end{aligned}

and ${[𝑇]}_{\overset{̊}{𝒞︀} \overset{̊}{ℬ︀}} {[𝐯]}_{\overset{̊}{ℬ︀}} =$ ${[𝑇 (𝐯)]}_{\overset{̊}{𝒞︀}}$ as well. The coordinate map of $\overset{̊}{ℬ︀}$ is onto $𝐹^{𝑛}$ (Proposition 1.6.16), so for each $𝑗$ we may choose $𝐯$ with ${[𝐯]}_{\overset{̊}{ℬ︀}} = 𝐞_{𝑗}$ . Then the $𝑗$ -th columns $𝑀 𝐞_{𝑗}$ and ${[𝑇]}_{\overset{̊}{𝒞︀} \overset{̊}{ℬ︀}} 𝐞_{𝑗}$ agree (Definition 2.3.8), and the two matrices are equal. The statement for operators follows as in the first proof. □

The first proof is shorter, because Theorem 2.4.2 already did the vector chase once and for all. The second is the one to remember when the subscripts get confusing: follow one vector around the diagram, and the order of the factors takes care of itself.

Example 4.5.18 (An oblique projection). Let $ℓ_{1} = span ((1, 1))$ and $ℓ_{2} = span ((1, 2))$ in $𝐹^{2}$ . The vectors $𝐮_{1} = (1, 1)$ and $𝐮_{2} = (1, 2)$ form a basis $𝒰︀$ of $𝐹^{2}$ over every field, since $(\begin{matrix} 1 & 1 \\ 1 & 2 \end{matrix})$ has $1 \cdot 2 - 1 \cdot 1 = 1 \neq 0$ . Let $𝑃 \in L (𝐹^{2})$ be the projection onto $ℓ_{1}$ along $ℓ_{2}$ : $𝑃 (𝐮_{1}) = 𝐮_{1}$ and $𝑃 (𝐮_{2}) = 𝟎$ . Its matrix in $𝒰︀$ is obvious, and the theorem converts it into the standard matrix. With $𝑄 =$ ${[id]}_{ℰ︀ 𝒰︀} =$ $(\begin{matrix} 1 & 1 \\ 1 & 2 \end{matrix})$ , whose inverse is $(\begin{matrix} 2 & - 1 \\ - 1 & 1 \end{matrix})$ ,

{[𝑃]}_{ℰ︀ ℰ︀} = 𝑄 {[𝑃]}_{𝒰︀ 𝒰︀} 𝑄^{- 1} = (\begin{matrix} 1 & 1 \\ 1 & 2 \end{matrix}) (\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}) (\begin{matrix} 2 & - 1 \\ - 1 & 1 \end{matrix}) = (\begin{matrix} 2 & - 1 \\ 2 & - 1 \end{matrix}) .

Check: $(\begin{matrix} 2 & - 1 \\ 2 & - 1 \end{matrix}) (\begin{matrix} 1 \\ 1 \end{matrix}) = (\begin{matrix} 1 \\ 1 \end{matrix})$ and $(\begin{matrix} 2 & - 1 \\ 2 & - 1 \end{matrix}) (\begin{matrix} 1 \\ 2 \end{matrix}) = (\begin{matrix} 0 \\ 0 \end{matrix})$ . Here we used the theorem with $ℰ︀$ as the new basis: ${[𝑃]}_{ℰ︀ ℰ︀} =$ ${[id]}_{ℰ︀ 𝒰︀} {[𝑃]}_{𝒰︀ 𝒰︀} {[id]}_{𝒰︀ ℰ︀}$ .

Example 4.5.19 (Differentiation, revisited). In Example 2.3.22(c), with $2 \neq 0$ in $𝐹$ , we found the matrix of differentiation $𝐷$ on ${𝐹 [𝑥]}_{\leq 2}$ in the basis $\overset{̊}{ℬ︀} = 1, 𝑥, \frac{1}{2} 𝑥^{2}$ by applying $𝐷$ to each basis vector. The theorem gives it from ${[𝐷]}_{ℬ︀ ℬ︀}$ , where $ℬ︀ = 1, 𝑥, 𝑥^{2}$ . Here $𝑃 =$ ${[id]}_{ℬ︀ \overset{̊}{ℬ︀}} =$ $diag (1, 1, \frac{1}{2})$ , whose inverse is $diag (1, 1, 2)$ , and

𝑃^{- 1} {[𝐷]}_{ℬ︀ ℬ︀} 𝑃 = (\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{matrix}) (\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 2 \\ 0 & 0 & 0 \end{matrix}) (\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & \frac{1}{2} \end{matrix}) = (\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{matrix}),

as found there. Scaling the third basis vector by $\frac{1}{2}$ divides the third column by $2$ and multiplies the third row by $2$ .

Exercise 4.5.20.routine Let

𝐴 = (\begin{matrix} 4 & - 2 \\ 1 & 1 \end{matrix})

and

\overset{̊}{ℬ︀} = (1, 1), (2, 1)

, a basis of

𝐹^{2}

. Compute

{[𝑇_{𝐴}]}_{\overset{̊}{ℬ︀} \overset{̊}{ℬ︀}}

with Theorem 4.5.17, and check the result by computing

𝐴 (1, 1)

and

𝐴 (2, 1)

directly.

Solution.

Uses. Theorem 4.5.17 (Change of basis), Proposition 2.3.16 (The matrix of $𝑇_{𝐴}$ ), Definition 2.3.1 (Matrix of a linear map)

Hint 1. With

𝑃 =

{[id]}_{ℰ︀ \overset{̊}{ℬ︀}} =

(\begin{matrix} 1 & 2 \\ 1 & 1 \end{matrix})

, the formula is

{[𝑇_{𝐴}]}_{\overset{̊}{ℬ︀} \overset{̊}{ℬ︀}} =

𝑃^{- 1} 𝐴 𝑃

Hint 2.

𝑃^{- 1} = (\begin{matrix} - 1 & 2 \\ 1 & - 1 \end{matrix})

. For the check, write

𝐴 (1, 1)

and

𝐴 (2, 1)

as combinations of

(1, 1)

and

(2, 1)

Sketch.

𝐴 𝑃 = (\begin{matrix} 2 & 6 \\ 2 & 3 \end{matrix})

and

𝑃^{- 1} 𝐴 𝑃 = (\begin{matrix} 2 & 0 \\ 0 & 3 \end{matrix})

. Indeed

𝐴 (1, 1) =

(2, 2) =

2 (1, 1)

and

𝐴 (2, 1) =

(6, 3) =

3 (2, 1)

By Proposition 2.3.16, ${[𝑇_{𝐴}]}_{ℰ︀ ℰ︀} = 𝐴$ , and $𝑃 =$ ${[id]}_{ℰ︀ \overset{̊}{ℬ︀}} =$ $(\begin{matrix} 1 & 2 \\ 1 & 1 \end{matrix})$ has the vectors of $\overset{̊}{ℬ︀}$ as columns. It is invertible over every field, with $𝑃^{- 1} = (\begin{matrix} - 1 & 2 \\ 1 & - 1 \end{matrix})$ (multiply to check). By Theorem 4.5.17,

{[𝑇_{𝐴}]}_{\overset{̊}{ℬ︀} \overset{̊}{ℬ︀}} = 𝑃^{- 1} 𝐴 𝑃 = (\begin{matrix} - 1 & 2 \\ 1 & - 1 \end{matrix}) (\begin{matrix} 2 & 6 \\ 2 & 3 \end{matrix}) = (\begin{matrix} 2 & 0 \\ 0 & 3 \end{matrix}) .

Directly: $𝐴 (1, 1) =$ $(2, 2) =$ $2 (1, 1) +$ $0 (2, 1)$ and $𝐴 (2, 1) =$ $(6, 3) =$ $0 (1, 1) +$ $3 (2, 1)$ , so the columns of ${[𝑇_{𝐴}]}_{\overset{̊}{ℬ︀} \overset{̊}{ℬ︀}}$ are $(2, 0)$ and $(0, 3)$ . In the basis $\overset{̊}{ℬ︀}$ , $𝑇_{𝐴}$ just scales each basis vector. Finding such bases is the subject of Section 7.4.

Matrices of one map

Every pair of bases gives a matrix of $𝑇$ , and by the theorem they are all of the form $𝐺 𝐴 𝐻$ with $𝐺$ and $𝐻$ invertible. The converse also holds, because every invertible matrix is a change-of-coordinates matrix.

Corollary 4.5.21 (Matrices of one linear map). Let $𝑇 \in L (𝑉, 𝑊)$ with $dim 𝑉 = 𝑛 \geq 1$ and $dim 𝑊 = 𝑚 \geq 1$ , let $ℬ︀$ and $𝒞︀$ be bases of $𝑉$ and $𝑊$ , and let $𝐴 = {[𝑇]}_{𝒞︀ ℬ︀}$ . For $𝐴^{'} \in 𝐹^{𝑚 \times 𝑛}$ , the following are equivalent.

(a) $𝐴^{'} = {[𝑇]}_{\overset{̊}{𝒞︀} \overset{̊}{ℬ︀}}$ for some bases $\overset{̊}{ℬ︀}$ of $𝑉$ and $\overset{̊}{𝒞︀}$ of $𝑊$ .
(b) $𝐴^{'} = 𝐺 𝐴 𝐻$ for some invertible $𝐺 \in 𝐹^{𝑚 \times 𝑚}$ and $𝐻 \in 𝐹^{𝑛 \times 𝑛}$ .
(c) $rank 𝐴^{'} = rank 𝐴$ .

Proof.routine

Uses. Theorem 4.5.17 (Change of basis), Proposition 4.5.12 (Calculus of change of coordinates), Proposition 4.5.16 (Every invertible matrix changes coordinates), Proposition 4.3.1 (Invertible factors preserve rank), Theorem 3.5.27 (Rank normal form), Proposition 2.5.17 (Algebra of inverses)

Techniques. Proving TFAE (Technique 2.7.1)

Hint 1. (a)

\Rightarrow

(b) is the theorem. For (b)

\Rightarrow

(a), realize

𝐻

and

𝐺

as change-of-coordinates matrices with Proposition 4.5.16.

Hint 2. For (c)

\Rightarrow

(b), bring both

𝐴

and

𝐴^{'}

to the same rank normal form.

Sketch. (a)

\Rightarrow

(b): Theorem 4.5.17 with

𝐺 = {[{id}_{𝑊}]}_{\overset{̊}{𝒞︀} 𝒞︀}

and

𝐻 = {[{id}_{𝑉}]}_{ℬ︀ \overset{̊}{ℬ︀}}

. (b)

\Rightarrow

(a): choose

\overset{̊}{ℬ︀}

with

{[{id}_{𝑉}]}_{ℬ︀ \overset{̊}{ℬ︀}} = 𝐻

and

\overset{̊}{𝒞︀}

with

{[{id}_{𝑊}]}_{\overset{̊}{𝒞︀} 𝒞︀} = 𝐺

, and apply the theorem. (b)

\Rightarrow

\Rightarrow

(b):

𝐺_{1} 𝐴 𝐻_{1} =

𝑁_{𝑟} =

𝐺_{2} 𝐴^{'} 𝐻_{2}

, so

𝐴^{'} =

𝐺_{2}^{- 1} 𝐺_{1} 𝐴 𝐻_{1} 𝐻_{2}^{- 1}

We prove (a) $\Leftrightarrow$ (b) and (b) $\Leftrightarrow$ (c).

(a) $\Rightarrow$ (b). By Theorem 4.5.17, $𝐴^{'} =$ ${[{id}_{𝑊}]}_{\overset{̊}{𝒞︀} 𝒞︀} 𝐴 {[{id}_{𝑉}]}_{ℬ︀ \overset{̊}{ℬ︀}}$ , and both outer factors are invertible (Proposition 4.5.12(b)).

(b) $\Rightarrow$ (a). By Proposition 4.5.16(a) there is a basis $\overset{̊}{ℬ︀}$ of $𝑉$ with ${[{id}_{𝑉}]}_{ℬ︀ \overset{̊}{ℬ︀}} = 𝐻$ , and by part (b) there is a basis $\overset{̊}{𝒞︀}$ of $𝑊$ with ${[{id}_{𝑊}]}_{\overset{̊}{𝒞︀} 𝒞︀} = 𝐺$ . By Theorem 4.5.17, ${[𝑇]}_{\overset{̊}{𝒞︀} \overset{̊}{ℬ︀}} =$ $𝐺 𝐴 𝐻 =$ $𝐴^{'}$ .

(b) $\Rightarrow$ (c). This is Proposition 4.3.1.

(c) $\Rightarrow$ (b). Let $𝑟 = rank 𝐴 = rank 𝐴^{'}$ . By Theorem 3.5.27 there are invertible $𝐺_{1}, 𝐺_{2} \in 𝐹^{𝑚 \times 𝑚}$ and $𝐻_{1}, 𝐻_{2} \in 𝐹^{𝑛 \times 𝑛}$ with $𝐺_{1} 𝐴 𝐻_{1} =$ $𝑁_{𝑟} =$ $𝐺_{2} 𝐴^{'} 𝐻_{2}$ . Then $𝐴^{'} =$ $(𝐺_{2}^{- 1} 𝐺_{1}) 𝐴 (𝐻_{1} 𝐻_{2}^{- 1})$ , and both factors are invertible by Proposition 2.5.17. □

So when the domain and the target may carry unrelated bases, the matrix of $𝑇$ can be any matrix of rank $rank 𝑇$ , and nothing else about $𝑇$ is visible. Choosing the bases well, the matrix becomes as simple as possible.

Corollary 4.5.22 (Rank normal form for linear maps). Let $𝑇 \in L (𝑉, 𝑊)$ , where $𝑉$ and $𝑊$ are nonzero and finite-dimensional, and let $𝑟 = rank 𝑇$ . There are bases $ℬ︀$ of $𝑉$ and $𝒞︀$ of $𝑊$ with

{[𝑇]}_{𝒞︀ ℬ︀} = 𝑁_{𝑟} = (\begin{matrix} 𝐼_{𝑟} & 0 \\ 0 & 0 \end{matrix}) .

Proof (from the matrix version).routine

Uses. Corollary 4.5.21 (Matrices of one linear map), Proposition 4.2.15 (Rank through any matrix)

Hint 1. Start from any bases. What is the rank of the resulting matrix?

Hint 2.

𝑁_{𝑟}

has rank

𝑟

; use (c)

\Rightarrow

(a) of Corollary 4.5.21.

Sketch. For any bases,

rank {[𝑇]}_{{𝒞︀}_{0} {ℬ︀}_{0}} =

rank 𝑇 =

𝑟 =

rank 𝑁_{𝑟}

(Proposition 4.2.15), so by Corollary 4.5.21,

𝑁_{𝑟}

is a matrix of

𝑇

Choose any bases

{ℬ︀}_{0}

and

{𝒞︀}_{0}

. By Proposition 4.2.15,

rank {[𝑇]}_{{𝒞︀}_{0} {ℬ︀}_{0}} =

rank 𝑇 =

𝑟

, and

rank 𝑁_{𝑟} = 𝑟

. By Corollary 4.5.21, (c)

\Rightarrow

(a), there are bases

ℬ︀

and

𝒞︀

with

{[𝑇]}_{𝒞︀ ℬ︀} = 𝑁_{𝑟}

. □

Proof (by building the bases).standard

Uses. Theorem 4.1.16 (Extending to a basis), Theorem 4.2.11 (Rank–nullity), Proposition 2.6.22 (Testing injectivity and surjectivity on a basis), Definition 2.3.1 (Matrix of a linear map)

Techniques. Extend a basis of a subspace (Technique 4.1.18)

Hint 1. The last

𝑛 - 𝑟

vectors of

ℬ︀

must go to

𝟎

, and the first

𝑟

must go to the first

𝑟

vectors of

𝒞︀

. Where should the last

𝑛 - 𝑟

vectors come from?

Hint 2. Extend a basis of

ker 𝑇

to a basis of

𝑉

, putting the new vectors first. Their images form a basis of

im 𝑇

; extend that to a basis of

𝑊

Sketch. Extend a basis

𝐮_{1}, \dots, 𝐮_{𝑘}

ker 𝑇

to a basis

𝐯_{1},

\dots,

𝐯_{𝑟},

𝐮_{1},

\dots,

𝐮_{𝑘}

𝑉

. The images

𝑇 (𝐯_{1}),

\dots,

𝑇 (𝐯_{𝑟})

span

im 𝑇

, and

𝑟 = dim im 𝑇

by rank–nullity, so they form a basis of

im 𝑇

; extend them to a basis of

𝑊

. Then

𝑇 (𝐯_{𝑗}) = 𝐜_{𝑗}

and

𝑇 (𝐮_{𝑖}) = 𝟎

, which is

𝑁_{𝑟}

Let

𝐮_{1}, \dots, 𝐮_{𝑘}

be a basis of

ker 𝑇

, and extend it to a basis of

𝑉

(Theorem 4.1.16); list the new vectors first, so that

ℬ︀ =

𝐯_{1},

\dots,

𝐯_{𝑠},

𝐮_{1},

\dots,

𝐮_{𝑘}

. Since

𝑇 (𝐮_{𝑖}) = 𝟎

, the list

𝑇 (ℬ︀)

, which spans

im 𝑇

(Proposition 2.6.22(d)), still spans

im 𝑇

when the zeros are deleted. So

𝑇 (𝐯_{1}),

\dots,

𝑇 (𝐯_{𝑠})

spans

im 𝑇

. By Theorem 4.2.11(b),

𝑠 =

dim 𝑉 -

𝑘 =

dim im 𝑇 =

𝑟

, so these

𝑟

vectors span the

𝑟

-dimensional space

im 𝑇

, and they form a basis of it (Corollary 4.1.11(b)). Write

𝐜_{𝑗} = 𝑇 (𝐯_{𝑗})

and extend

𝐜_{1}, \dots, 𝐜_{𝑟}

to a basis

𝒞︀

𝑊

. Then the

𝑗

-th column of

{[𝑇]}_{𝒞︀ ℬ︀}

{[𝐜_{𝑗}]}_{𝒞︀} = 𝐞_{𝑗}

for

𝑗 \leq 𝑟

, and the remaining columns are

{[𝟎]}_{𝒞︀} = 𝟎

. So

{[𝑇]}_{𝒞︀ ℬ︀} = 𝑁_{𝑟}

. □

The second proof is the one sketched after the second proof of rank–nullity (Section 4.2), and the first is Corollary 3.5.29 transported to abstract spaces. For operators, where the same basis must be used on both sides, much more survives. That is the subject of the next subsection.

4.5.4Similarity

For $𝑇 \in L (𝑉)$ we compare $𝑇 (𝐯)$ with $𝐯$ , so we use one basis for both. By Theorem 4.5.17, the matrices of $𝑇$ in two bases are related by ${[𝑇]}_{\overset{̊}{ℬ︀} \overset{̊}{ℬ︀}} =$ $𝑃^{- 1} {[𝑇]}_{ℬ︀ ℬ︀} 𝑃$ , with the same invertible $𝑃$ on both sides.

Definition 4.5.23 (Similar matrices). Matrices $𝐴, 𝐵 \in 𝐹^{𝑛 \times 𝑛}$ are similar if

𝐵 = 𝑆^{- 1} 𝐴 𝑆

for some invertible $𝑆 \in 𝐹^{𝑛 \times 𝑛}$ . We then write $𝐴 \sim 𝐵$ .

The lecture writes the same condition as $𝐵 = 𝑄 𝐴 𝑄^{- 1}$ ; take $𝑄 = 𝑆^{- 1}$ . To check a proposed $𝑆$ , it is easier to verify $𝐴 𝑆 = 𝑆 𝐵$ (and that $𝑆$ is invertible) than to compute $𝑆^{- 1}$ .

Proposition 4.5.24 (Similarity is an equivalence relation). For all

𝐴, 𝐵, 𝐶 \in 𝐹^{𝑛 \times 𝑛}

𝐴 \sim 𝐴

; if

𝐴 \sim 𝐵

then

𝐵 \sim 𝐴

; and if

𝐴 \sim 𝐵

and

𝐵 \sim 𝐶

then

𝐴 \sim 𝐶

Proof.routine

Uses. Definition 4.5.23 (Similar matrices), Proposition 2.5.17 (Algebra of inverses), Proposition 2.4.14 (Algebra of matrices)

Hint 1. Each claim asks for an invertible matrix. Try

𝐼_{𝑛}

, then

𝑆^{- 1}

, then a product.

Hint 2. If

𝐵 = 𝑆^{- 1} 𝐴 𝑆

and

𝐶 = 𝑅^{- 1} 𝐵 𝑅

, substitute and regroup.

Sketch.

𝐴 = 𝐼^{- 1} 𝐴 𝐼

. If

𝐵 = 𝑆^{- 1} 𝐴 𝑆

, then

𝐴 =

𝑆 𝐵 𝑆^{- 1} =

{(𝑆^{- 1})}^{- 1} 𝐵 𝑆^{- 1}

. If also

𝐶 = 𝑅^{- 1} 𝐵 𝑅

, then

𝐶 =

𝑅^{- 1} 𝑆^{- 1} 𝐴 𝑆 𝑅 =

{(𝑆 𝑅)}^{- 1} 𝐴 (𝑆 𝑅)

First, $𝐴 = 𝐼_{𝑛}^{- 1} 𝐴 𝐼_{𝑛}$ . If $𝐵 = 𝑆^{- 1} 𝐴 𝑆$ , multiplying by $𝑆$ on the left and $𝑆^{- 1}$ on the right gives $𝐴 =$ $𝑆 𝐵 𝑆^{- 1} =$ ${(𝑆^{- 1})}^{- 1} 𝐵 𝑆^{- 1}$ , and $𝑆^{- 1}$ is invertible (Proposition 2.5.17(a)). If also $𝐶 = 𝑅^{- 1} 𝐵 𝑅$ , then by associativity (Proposition 2.4.14) and Proposition 2.5.17(b),

𝐶 = 𝑅^{- 1} 𝑆^{- 1} 𝐴 𝑆 𝑅 = {(𝑆 𝑅)}^{- 1} 𝐴 (𝑆 𝑅),

and $𝑆 𝑅$ is invertible. □

The definition is a statement about matrices, but its meaning is a statement about operators.

Theorem 4.5.25 (Similar matrices are one operator in two bases). Let $𝐴, 𝐵 \in 𝐹^{𝑛 \times 𝑛}$ with $𝑛 \geq 1$ . The following are equivalent.

(a) $𝐴 \sim 𝐵$ .
(b)Whenever $dim 𝑉 = 𝑛$ , $𝑇 \in L (𝑉)$ , and $ℬ︀$ is a basis of $𝑉$ with ${[𝑇]}_{ℬ︀ ℬ︀} = 𝐴$ , there is a basis $\overset{̊}{ℬ︀}$ of $𝑉$ with ${[𝑇]}_{\overset{̊}{ℬ︀} \overset{̊}{ℬ︀}} = 𝐵$ .
(c)There are a vector space $𝑉$ with $dim 𝑉 = 𝑛$ , an operator $𝑇 \in L (𝑉)$ , and bases $ℬ︀$ , $\overset{̊}{ℬ︀}$ of $𝑉$ with ${[𝑇]}_{ℬ︀ ℬ︀} = 𝐴$ and ${[𝑇]}_{\overset{̊}{ℬ︀} \overset{̊}{ℬ︀}} = 𝐵$ .

Moreover, if $𝐵 = 𝑆^{- 1} 𝐴 𝑆$ , then $𝐵 = {[𝑇_{𝐴}]}_{\overset{̊}{ℬ︀} \overset{̊}{ℬ︀}}$ , where $\overset{̊}{ℬ︀}$ is the list of columns of $𝑆$ .

Proof.routine

Uses. Theorem 4.5.17 (Change of basis), Proposition 4.5.16 (Every invertible matrix changes coordinates), Proposition 4.5.12 (Calculus of change of coordinates), Proposition 2.3.16 (The matrix of $𝑇_{𝐴}$ )

Techniques. Proving TFAE (Technique 2.7.1), Translate between maps and matrices (Technique 2.3.27)

Hint 1. Prove (a)

\Rightarrow

(b)

\Rightarrow

(c)

\Rightarrow

(a). Two of the three steps are Theorem 4.5.17 read in the two directions.

Hint 2. For (a)

\Rightarrow

(b), given

𝐵 = 𝑆^{- 1} 𝐴 𝑆

, use Proposition 4.5.16 to find

\overset{̊}{ℬ︀}

with

{[id]}_{ℬ︀ \overset{̊}{ℬ︀}} = 𝑆

. For (b)

\Rightarrow

(c), you need one example of

𝑉

𝑇

ℬ︀

: take the obvious one.

Sketch. (a)

\Rightarrow

(b): if

𝐵 = 𝑆^{- 1} 𝐴 𝑆

, choose

\overset{̊}{ℬ︀}

with

{[id]}_{ℬ︀ \overset{̊}{ℬ︀}} = 𝑆

; then

{[𝑇]}_{\overset{̊}{ℬ︀} \overset{̊}{ℬ︀}} =

𝑆^{- 1} 𝐴 𝑆 =

𝐵

. (b)

\Rightarrow

(c): apply (b) to

𝑇_{𝐴}

𝐹^{𝑛}

with the standard basis. (c)

\Rightarrow

(a): Theorem 4.5.17 with

𝑃 = {[id]}_{ℬ︀ \overset{̊}{ℬ︀}}

. For the last claim, the basis of

𝐹^{𝑛}

with

{[id]}_{ℰ︀ \overset{̊}{ℬ︀}} = 𝑆

is the list of columns of

𝑆

We prove (a) $\Rightarrow$ (b) $\Rightarrow$ (c) $\Rightarrow$ (a).

(a) $\Rightarrow$ (b). Let $𝐵 = 𝑆^{- 1} 𝐴 𝑆$ , and let $𝑉$ , $𝑇$ , and $ℬ︀$ be as in (b). By Proposition 4.5.16(a), there is a basis $\overset{̊}{ℬ︀}$ of $𝑉$ with ${[id]}_{ℬ︀ \overset{̊}{ℬ︀}} = 𝑆$ . By Theorem 4.5.17, ${[𝑇]}_{\overset{̊}{ℬ︀} \overset{̊}{ℬ︀}} =$ $𝑆^{- 1} {[𝑇]}_{ℬ︀ ℬ︀} 𝑆 =$ $𝑆^{- 1} 𝐴 𝑆 =$ $𝐵$ .

(b) $\Rightarrow$ (c). Take $𝑉 = 𝐹^{𝑛}$ , $𝑇 = 𝑇_{𝐴}$ , and the standard basis $ℰ︀$ ; then ${[𝑇_{𝐴}]}_{ℰ︀ ℰ︀} = 𝐴$ (Proposition 2.3.16), and (b) provides $\overset{̊}{ℬ︀}$ .

(c) $\Rightarrow$ (a). Let $𝑃 = {[id]}_{ℬ︀ \overset{̊}{ℬ︀}}$ , which is invertible (Proposition 4.5.12(b)). By Theorem 4.5.17, $𝐵 =$ ${[𝑇]}_{\overset{̊}{ℬ︀} \overset{̊}{ℬ︀}} =$ $𝑃^{- 1} 𝐴 𝑃$ .

For the last claim, apply the proof of (a) $\Rightarrow$ (b) to $𝑇_{𝐴}$ and $ℬ︀ = ℰ︀$ . By Proposition 4.5.16(a), the $𝑗$ -th vector of $\overset{̊}{ℬ︀}$ is $\sum_{𝑖} 𝑠_{𝑖 𝑗} 𝐞_{𝑖}$ , the $𝑗$ -th column of $𝑆$ . □

Intuition (Similarity is a change of viewpoint). Two similar matrices describe one operator, recorded in two coordinate systems, and

𝑆

is the dictionary between them: its columns are the new basis, written in the old coordinates. So a question about similarity is a question about whether two matrices can be the same operator. Any property of the operator itself, defined without reference to a basis, must be shared by all of its matrices. That is how we will tell matrices apart.

Example 4.5.26 (A similarity and a non-similarity).

(a)Let $𝐹$ be a field with $2 \neq 0$ , and let $𝐴 = (\begin{matrix} 1 & 1 \\ 1 & 1 \end{matrix})$ . Then $𝐴 (1, 1) =$ $(2, 2) =$ $2 (1, 1)$ and $𝐴 (1, - 1) = (0, 0)$ . So in the basis $\overset{̊}{ℬ︀} = (1, 1), (1, - 1)$ , the operator $𝑇_{𝐴}$ doubles the first basis vector and kills the second, and with $𝑆 = (\begin{matrix} 1 & 1 \\ 1 & - 1 \end{matrix})$ , the matrix whose columns are these vectors,

𝑆^{- 1} 𝐴 𝑆 = (\begin{matrix} 2 & 0 \\ 0 & 0 \end{matrix}) .

(Check: $𝐴 𝑆 =$ $(\begin{matrix} 2 & 0 \\ 2 & 0 \end{matrix}) =$ $𝑆 (\begin{matrix} 2 & 0 \\ 0 & 0 \end{matrix})$ , and $𝑆$ is invertible since $1 \cdot (- 1) - 1 \cdot 1 = - 2 \neq 0$ .) Similar matrices need not share a single entry.

(a)Over $𝔽_{2}$ , the same matrix $𝐴 = (\begin{matrix} 1 & 1 \\ 1 & 1 \end{matrix})$ is not similar to $(\begin{matrix} 2 & 0 \\ 0 & 0 \end{matrix})$ , which is now the zero matrix: $𝑆^{- 1} 0 𝑆 = 0$ for every $𝑆$ , so the only matrix similar to $0$ is $0$ . Whether two matrices are similar can depend on the field.

Similarity invariants

A similarity invariant is a quantity or property of square matrices that is the same for any two similar matrices. Here are the ones available so far.

Proposition 4.5.27 (Similarity invariants). Let $𝐴, 𝐵 \in 𝐹^{𝑛 \times 𝑛}$ with $𝐴 \sim 𝐵$ . Then:

(a) $rank 𝐴 = rank 𝐵$ ;
(b) $tr 𝐴 = tr 𝐵$ ;
(c) $𝐴^{𝑘} \sim 𝐵^{𝑘}$ for every $𝑘 \geq 0$ ; in particular $rank 𝐴^{𝑘} = rank 𝐵^{𝑘}$ ;
(d) $𝐴 -$ $𝑐 𝐼_{𝑛} \sim$ $𝐵 -$ $𝑐 𝐼_{𝑛}$ for every $𝑐 \in 𝐹$ ; in particular $rank (𝐴 - 𝑐 𝐼_{𝑛}) =$ $rank (𝐵 - 𝑐 𝐼_{𝑛})$ ;
(e) $𝐴$ is invertible if and only if $𝐵$ is, and then $𝐴^{- 1} \sim 𝐵^{- 1}$ ;
(f) $𝐴^{𝖳} \sim 𝐵^{𝖳}$ .

Proof (by matrix algebra).routine

Uses. Definition 4.5.23 (Similar matrices), Proposition 4.3.1 (Invertible factors preserve rank), Proposition 2.4.28 (Cyclicity of the trace), Proposition 2.4.14 (Algebra of matrices), Proposition 2.5.17 (Algebra of inverses), Proposition 2.4.23 (Transpose of a product)

Techniques. Cyclicity of trace (Technique 2.4.29)

Hint 1. Write

𝐵 = 𝑆^{- 1} 𝐴 𝑆

. For (a), invertible factors do not change the rank. For (b), move a factor around the trace.

Hint 2. For (c), the inner factors telescope:

(𝑆^{- 1} 𝐴 𝑆) (𝑆^{- 1} 𝐴 𝑆) =

𝑆^{- 1} 𝐴^{2} 𝑆

. For (d),

𝑆^{- 1} (𝑐 𝐼) 𝑆 = 𝑐 𝐼

. For (f), transpose

𝐵 = 𝑆^{- 1} 𝐴 𝑆

Sketch. With

𝐵 = 𝑆^{- 1} 𝐴 𝑆

: (a) Proposition 4.3.1. (b)

tr (𝑆^{- 1} (𝐴 𝑆)) =

tr ((𝐴 𝑆) 𝑆^{- 1}) =

tr 𝐴

. (c)

𝐵^{𝑘} = 𝑆^{- 1} 𝐴^{𝑘} 𝑆

by induction. (d)

𝑆^{- 1} (𝐴 - 𝑐 𝐼) 𝑆 =

𝐵 -

𝑐 𝐼

. (e)

𝐵^{- 1} = 𝑆^{- 1} 𝐴^{- 1} 𝑆

. (f)

𝐵^{𝖳} =

𝑆^{𝖳} 𝐴^{𝖳} {(𝑆^{𝖳})}^{- 1}

Write $𝐵 = 𝑆^{- 1} 𝐴 𝑆$ with $𝑆$ invertible.

(a) By Proposition 4.3.1, $rank (𝑆^{- 1} 𝐴 𝑆) =$ $rank 𝐴$ .

(b) By Proposition 2.4.28 with the factors $𝑆^{- 1}$ and $𝐴 𝑆$ ,

tr 𝐵 = tr (𝑆^{- 1} (𝐴 𝑆)) = tr ((𝐴 𝑆) 𝑆^{- 1}) = tr 𝐴 .

(c) We show $𝐵^{𝑘} = 𝑆^{- 1} 𝐴^{𝑘} 𝑆$ by induction on $𝑘$ . For $𝑘 = 0$ both sides are $𝐼_{𝑛}$ . If it holds for $𝑘$ , then

𝐵^{𝑘 + 1} = 𝐵^{𝑘} 𝐵 = (𝑆^{- 1} 𝐴^{𝑘} 𝑆) (𝑆^{- 1} 𝐴 𝑆) = 𝑆^{- 1} 𝐴^{𝑘} (𝑆 𝑆^{- 1}) 𝐴 𝑆 = 𝑆^{- 1} 𝐴^{𝑘 + 1} 𝑆 .

So $𝐴^{𝑘} \sim 𝐵^{𝑘}$ , and their ranks agree by (a).

(d) $𝑆^{- 1} (𝐴 - 𝑐 𝐼_{𝑛}) 𝑆 =$ $𝑆^{- 1} 𝐴 𝑆 -$ $𝑐 𝑆^{- 1} 𝑆 =$ $𝐵 -$ $𝑐 𝐼_{𝑛}$ (Proposition 2.4.14), and the ranks agree by (a).

(e) If $𝐴$ is invertible, then $𝐵$ is a product of invertible matrices, hence invertible, with $𝐵^{- 1} = 𝑆^{- 1} 𝐴^{- 1} 𝑆$ (Proposition 2.5.17(b)). So $𝐴^{- 1} \sim 𝐵^{- 1}$ . The converse follows by symmetry, since $𝐴 = 𝑆 𝐵 𝑆^{- 1}$ .

(f) By Proposition 2.4.23 and Proposition 2.5.17(d), $𝐵^{𝖳} =$ $𝑆^{𝖳} 𝐴^{𝖳} {(𝑆^{- 1})}^{𝖳} =$ $𝑆^{𝖳} 𝐴^{𝖳} {(𝑆^{𝖳})}^{- 1}$ . With $𝑅 = {(𝑆^{𝖳})}^{- 1}$ , this is $𝐵^{𝖳} = 𝑅^{- 1} 𝐴^{𝖳} 𝑅$ . □

The second proof uses Theorem 4.5.25: since $𝐴$ and $𝐵$ are matrices of one operator, anything defined from the operator alone is the same for both. It reaches (a), (c), (d), and (e), but not (b) or (f), because the trace and the transpose are defined through matrices.

Proof (through the operator).routine

Uses. Theorem 4.5.25 (Similar matrices are one operator in two bases), Proposition 4.2.15 (Rank through any matrix), Theorem 2.4.2 (The matrix of a composite), Proposition 2.4.12 (Matrices of sums and scalar multiples), Example 2.3.6 (The zero map and the identity map), Proposition 2.5.20 (Inverses and matrices)

Techniques. Translate between maps and matrices (Technique 2.3.27)

Hint 1. By Theorem 4.5.25,

𝐴 = {[𝑇]}_{ℬ︀ ℬ︀}

and

𝐵 = {[𝑇]}_{\overset{̊}{ℬ︀} \overset{̊}{ℬ︀}}

for a single operator

𝑇

. What is the rank of a matrix of

𝑇

Hint 2. Which operators have matrices

𝐴^{𝑘}

𝐴 - 𝑐 𝐼_{𝑛}

, and

𝐴^{- 1}

in the basis

ℬ︀

? Apply Theorem 4.5.25 to them in the other direction.

Sketch. Let

𝐴 = {[𝑇]}_{ℬ︀ ℬ︀}

and

𝐵 = {[𝑇]}_{\overset{̊}{ℬ︀} \overset{̊}{ℬ︀}}

. Then

rank 𝐴 =

rank 𝑇 =

rank 𝐵

. Also

𝐴^{𝑘}

and

𝐵^{𝑘}

are the matrices of

𝑇^{𝑘}

in the two bases,

𝐴 - 𝑐 𝐼_{𝑛}

and

𝐵 - 𝑐 𝐼_{𝑛}

those of

𝑇 - 𝑐 {id}_{𝑉}

, and

𝐴^{- 1}

𝐵^{- 1}

those of

𝑇^{- 1}

(which exists iff either matrix is invertible); so each pair is similar by the theorem.

By Theorem 4.5.25, (a) $\Rightarrow$ (c), there are $𝑉$ , $𝑇 \in L (𝑉)$ , and bases $ℬ︀$ , $\overset{̊}{ℬ︀}$ with $𝐴 = {[𝑇]}_{ℬ︀ ℬ︀}$ and $𝐵 = {[𝑇]}_{\overset{̊}{ℬ︀} \overset{̊}{ℬ︀}}$ .

(a) By Proposition 4.2.15, $rank 𝐴 =$ $rank 𝑇 =$ $rank 𝐵$ .

(c) By Theorem 2.4.2 and induction on $𝑘$ , ${[𝑇^{𝑘}]}_{ℬ︀ ℬ︀} = 𝐴^{𝑘}$ and ${[𝑇^{𝑘}]}_{\overset{̊}{ℬ︀} \overset{̊}{ℬ︀}} = 𝐵^{𝑘}$ (with $𝑇^{0} = {id}_{𝑉}$ and Example 2.3.6 for $𝑘 = 0$ ). So $𝐴^{𝑘}$ and $𝐵^{𝑘}$ are matrices of the single operator $𝑇^{𝑘}$ , and they are similar by Theorem 4.5.25, (c) $\Rightarrow$ (a).

(d) By Proposition 2.4.12 and Example 2.3.6, $𝐴 - 𝑐 𝐼_{𝑛}$ and $𝐵 - 𝑐 𝐼_{𝑛}$ are the matrices of $𝑇 - 𝑐 {id}_{𝑉}$ in the two bases.

(e) By Proposition 2.5.20(b), $𝐴$ is invertible if and only if $𝑇$ is, if and only if $𝐵$ is; and then $𝐴^{- 1}$ and $𝐵^{- 1}$ are the matrices of $𝑇^{- 1}$ in the two bases. □

Remark 4.5.28 (What the invariants cannot do). Equal invariants do not make matrices similar. The matrices

𝑁 = (\begin{matrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{matrix}), 𝐷 = (\begin{matrix} 1 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & 0 \end{matrix})

both have trace $0$ and rank $2$ . But $𝑁^{3} = 0$ , since $𝑁$ shifts $𝐞_{1} \mapsto 𝐞_{2} \mapsto 𝐞_{3} \mapsto 𝟎$ , while $𝐷^{3} = 𝐷$ has rank $2$ . By (c), $rank 𝑁^{3} \neq rank 𝐷^{3}$ rules out $𝑁 \sim 𝐷$ . Similarly, $𝐼_{2}$ and $𝐽 = (\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix})$ have the same rank and trace, and $𝐽^{𝑘}$ has rank $2$ like $𝐼_{2}^{𝑘}$ ; but $rank (𝐽 - 𝐼_{2}) =$ $1 \neq$ $0 =$ $rank (𝐼_{2} - 𝐼_{2})$ , so they are not similar. (More simply, the only matrix similar to $𝐼_{𝑛}$ is $𝐼_{𝑛}$ , since $𝑆^{- 1} 𝐼_{𝑛} 𝑆 = 𝐼_{𝑛}$ .)

The determinant is one more invariant (Proposition 5.4.17), and the characteristic polynomial, which contains both the trace and the determinant, is a stronger one (Section 7.3). None of these decides similarity on its own. Over $ℂ$ , the ranks of ${(𝐴 - 𝑐 𝐼_{𝑛})}^{𝑘}$ for all $𝑐$ and $𝑘$ do decide it; this is the content of the Jordan form (Section 10.3).

Technique 4.5.29 (Rule out similarity with an invariant). To show that $𝐴$ and $𝐵$ are not similar, find a similarity invariant on which they differ: trace, rank, rank of a power $𝐴^{𝑘}$ , rank of $𝐴 - 𝑐 𝐼_{𝑛}$ for a well-chosen $𝑐$ , and later the determinant and the characteristic polynomial. To show that they are similar, invariants are not enough: produce an $𝑆$ with $𝐴 𝑆 = 𝑆 𝐵$ , usually by finding a basis in which the operator $𝑇_{𝐴}$ has matrix $𝐵$ (Theorem 4.5.25).

Reach for the cheapest invariants first. Trace and rank take seconds. If they agree, look for a structural difference: is one matrix nilpotent, or a scalar multiple of $𝐼_{𝑛}$ , or does $𝐴 - 𝑐 𝐼_{𝑛}$ drop rank for some $𝑐$ where $𝐵 - 𝑐 𝐼_{𝑛}$ does not? The canonical example is Remark 4.5.28.

Used in 9 places and 4 practice exercises

Exercise 4.5.32, Exercise 4.5.44, Exercise 4.5.45, Exercise 4.5.46, Exercise 4.8.4, Problem 4.13, Exercise 7.3.38, Exercise 7.4.43, Exercise 7.6.43; 4 practice exercises.

Invariants of an operator

Because the trace is a similarity invariant, it can be assigned to an operator, just as the rank was in Definition 4.2.14.

Definition 4.5.30 (Trace of an operator). Let $𝑉$ be finite-dimensional with $dim 𝑉 \geq 1$ , and let $𝑇 \in L (𝑉)$ . The trace of $𝑇$ is

tr 𝑇 = tr {[𝑇]}_{ℬ︀ ℬ︀},

where $ℬ︀$ is any basis of $𝑉$ .

The definition does not depend on $ℬ︀$ : by Theorem 4.5.17, the matrices of $𝑇$ in two bases are similar, and similar matrices have the same trace (Proposition 4.5.27(b)). In the same way, $rank {[𝑇]}_{ℬ︀ ℬ︀} =$ $rank 𝑇$ for every basis (Proposition 4.2.15), and in Definition 5.4.18 the determinant joins them. For $𝑉 = 𝐹^{𝑛}$ and the standard basis, $tr 𝑇_{𝐴} = tr 𝐴$ .

Example 4.5.31 (The trace of transposition). Let $𝑇 \in L (𝐹^{𝑛 \times 𝑛})$ be $𝑇 (𝑋) = 𝑋^{𝖳}$ . In the basis of matrix units $𝑈_{𝑖 𝑗}$ (Proposition 1.6.7(b)), $𝑇 (𝑈_{𝑖 𝑗}) = 𝑈_{𝑗 𝑖}$ . The diagonal entry of $[𝑇]$ in the position belonging to $𝑈_{𝑖 𝑗}$ is the $𝑈_{𝑖 𝑗}$ -coordinate of $𝑇 (𝑈_{𝑖 𝑗}) = 𝑈_{𝑗 𝑖}$ , which is $1$ if $𝑖 = 𝑗$ and $0$ otherwise. So

tr 𝑇 = 𝑛 (that is, 𝑛 \cdot 1 \in 𝐹) .

Over $ℝ$ there is a second basis to check this against. $𝑇$ fixes the symmetric matrices, a subspace of dimension $𝑛 (𝑛 + 1) / 2$ , and negates the skew-symmetric ones, of dimension $𝑛 (𝑛 - 1) / 2$ . In a basis made of a basis of each, $[𝑇]$ is diagonal with $𝑛 (𝑛 + 1) / 2$ entries $1$ and $𝑛 (𝑛 - 1) / 2$ entries $- 1$ , and its trace is again $𝑛 (𝑛 + 1) / 2 -$ $𝑛 (𝑛 - 1) / 2 =$ $𝑛$ .

Exercise 4.5.32.standard Let $𝐽 = (\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}) \in ℚ^{2 \times 2}$ . Decide which of the following matrices are similar to $𝐽$ . For each one that is, find $𝑆$ with $𝑆^{- 1} 𝑀 𝑆 = 𝐽$ .

(a) $𝑀 = 𝐼_{2}$ .
(b) $𝑀 = (\begin{matrix} 2 & 1 \\ - 1 & 0 \end{matrix})$ .
(c) $𝑀 = (\begin{matrix} 1 & 1 \\ 0 & 2 \end{matrix})$ .

Solution.

Uses. Proposition 4.5.27 (Similarity invariants), Theorem 4.5.25 (Similar matrices are one operator in two bases), Definition 4.5.23 (Similar matrices)

Techniques. Rule out similarity with an invariant (Technique 4.5.29)

Hint 1. Compare traces first, and remember that only

𝐼_{2}

is similar to

𝐼_{2}

Hint 2. For (b),

𝑁 = 𝑀 - 𝐼_{2}

satisfies

𝑁^{2} = 0

and

𝑁 \neq 0

. Find

𝐰

with

𝑁 𝐰 \neq 𝟎

; what is the matrix of

𝑇_{𝑀}

in the basis

𝑁 𝐰, 𝐰

Sketch. (a) No:

𝑆^{- 1} 𝐼_{2} 𝑆 = 𝐼_{2} \neq 𝐽

. (c) No:

tr = 3 \neq 2

. (b) Yes: with

𝐰 = 𝐞_{1}

𝑁 𝐰 = (1, - 1)

; in the basis

(1, - 1), (1, 0)

𝑇_{𝑀}

fixes the first vector and sends the second to itself plus the first, so

𝑆 = (\begin{matrix} 1 & 1 \\ - 1 & 0 \end{matrix})

works. Check

𝑀 𝑆 = 𝑆 𝐽

(a) Not similar: $𝑆^{- 1} 𝐼_{2} 𝑆 = 𝐼_{2} \neq 𝐽$ for every invertible $𝑆$ .

(c) Not similar: $tr 𝑀 = 3 \neq 2 = tr 𝐽$ (Proposition 4.5.27(b)).

(b) Similar. Both matrices have trace $2$ and rank $2$ , which only says that similarity is possible. Let $𝑁 =$ $𝑀 -$ $𝐼_{2} =$ $(\begin{matrix} 1 & 1 \\ - 1 & - 1 \end{matrix})$ . Then $𝑁^{2} = (\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix})$ , while $𝑁 𝐞_{1} = (1, - 1) \neq 𝟎$ . Let $\overset{̊}{ℬ︀} = (1, - 1), (1, 0)$ , a basis since the vectors are not parallel. Then $𝑀 (1, - 1) =$ $(1, - 1) +$ $𝑁 (1, - 1) =$ $(1, - 1)$ , because $𝑁 (1, - 1) =$ $𝑁^{2} 𝐞_{1} =$ $𝟎$ , and $𝑀 (1, 0) =$ $(1, 0) +$ $𝑁 𝐞_{1} =$ $(1, 0) +$ $(1, - 1)$ . So the columns of ${[𝑇_{𝑀}]}_{\overset{̊}{ℬ︀} \overset{̊}{ℬ︀}}$ are $(1, 0)$ and $(1, 1)$ , that is, ${[𝑇_{𝑀}]}_{\overset{̊}{ℬ︀} \overset{̊}{ℬ︀}} = 𝐽$ . By Theorem 4.5.25 and Theorem 4.5.17, $𝑆 = (\begin{matrix} 1 & 1 \\ - 1 & 0 \end{matrix})$ , whose columns are the vectors of $\overset{̊}{ℬ︀}$ , satisfies $𝑆^{- 1} 𝑀 𝑆 = 𝐽$ . Check:

𝑀 𝑆 = (\begin{matrix} 2 & 1 \\ - 1 & 0 \end{matrix}) (\begin{matrix} 1 & 1 \\ - 1 & 0 \end{matrix}) = (\begin{matrix} 1 & 2 \\ - 1 & - 1 \end{matrix}), 𝑆 𝐽 = (\begin{matrix} 1 & 1 \\ - 1 & 0 \end{matrix}) (\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}) = (\begin{matrix} 1 & 2 \\ - 1 & - 1 \end{matrix}) .

4.5.5Choosing a convenient basis

The change-of-basis formula also gives a proof technique: if a statement does not depend on the basis, we may prove it in whichever basis makes it easiest.

Technique 4.5.33 (Choose a convenient basis). Many statements about a linear map $𝑇$ (or a matrix $𝐴$ ) are invariant under change of basis: they hold for one matrix of $𝑇$ if and only if they hold for all of them, or for $𝐴$ if and only if they hold for $𝐺 𝐴 𝐻$ (or for $𝑆^{- 1} 𝐴 𝑆$ ). To prove such a statement, first check this invariance, then choose bases in which the matrix is as simple as possible, prove the statement there, and transport the conclusion back.

For maps between two spaces, the best possible matrix is the rank normal form $𝑁_{𝑟}$ (Corollary 4.5.22), and in matrix language this says: to prove something about $𝐴$ that survives $𝐴 \mapsto 𝐺 𝐴 𝐻$ , assume $𝐴 = 𝑁_{𝑟}$ . For operators, the bases must agree on both sides, and the convenient bases come from the operator itself: bases adapted to a subspace that $𝑇$ preserves (below), bases of the form $𝐯, 𝑇 𝐯, 𝑇^{2} 𝐯, \dots$ (Exercise 4.5.50, Exercise 4.5.43), and later bases of eigenvectors (Section 7.4) and triangularizing bases (Section 7.5).

Used in 17 places and 6 practice exercises

Exercise 4.5.41, Exercise 4.5.42, Exercise 4.5.43, Exercise 4.5.49, Exercise 4.5.50, Exercise 4.5.51, Exercise 4.5.52, Exercise 4.5.53, Exercise 4.5.55, Exercise 4.8.12, Theorem 7.5.22, Problem 7.6, Problem 7.10, Problem 7.11, Proposition 8.2.2, Corollary 8.2.28, Exercise 8.3.43; 6 practice exercises.

Example 4.5.34 (A matrix with $𝐴 𝑋 𝐴 = 𝐴$ ). We show that for every $𝐴 \in 𝐹^{𝑚 \times 𝑛}$ there is $𝑋 \in 𝐹^{𝑛 \times 𝑚}$ with $𝐴 𝑋 𝐴 = 𝐴$ . (Such an $𝑋$ is a generalized inverse of $𝐴$ ; if $𝐴$ is invertible, $𝑋 = 𝐴^{- 1}$ is the only choice.)

For $𝐴 = 𝑁_{𝑟}$ the claim is easy: $𝑋 = 𝑁_{𝑟}^{𝖳} \in 𝐹^{𝑛 \times 𝑚}$ works. By block multiplication (Theorem 2.4.34), $𝑁_{𝑟} 𝑁_{𝑟}^{𝖳} =$ $(\begin{matrix} 𝐼_{𝑟} & 0 \\ 0 & 0 \end{matrix}) \in$ $𝐹^{𝑚 \times 𝑚}$ , and then $𝑁_{𝑟} 𝑁_{𝑟}^{𝖳} 𝑁_{𝑟} =$ $𝑁_{𝑟}$ .

The statement survives $𝐴 \mapsto 𝐺 𝐴 𝐻$ : if $𝑁_{𝑟} 𝑋 𝑁_{𝑟} = 𝑁_{𝑟}$ and $𝐴 = 𝐺 𝑁_{𝑟} 𝐻$ with $𝐺$ and $𝐻$ invertible, then $𝑋^{'} = 𝐻^{- 1} 𝑋 𝐺^{- 1}$ satisfies

𝐴 𝑋^{'} 𝐴 = 𝐺 𝑁_{𝑟} 𝐻 𝐻^{- 1} 𝑋 𝐺^{- 1} 𝐺 𝑁_{𝑟} 𝐻 = 𝐺 (𝑁_{𝑟} 𝑋 𝑁_{𝑟}) 𝐻 = 𝐺 𝑁_{𝑟} 𝐻 = 𝐴 .

By Theorem 3.5.27, every $𝐴$ is $𝐺 𝑁_{𝑟} 𝐻$ for some invertible $𝐺$ , $𝐻$ (namely, invert the factors in $𝐺^{'} 𝐴 𝐻^{'} = 𝑁_{𝑟}$ ). So every $𝐴$ has a generalized inverse, $𝑋^{'} =$ $𝐻^{- 1} 𝑁_{𝑟}^{𝖳} 𝐺^{- 1}$ .

In terms of maps: if ${[𝑇]}_{𝒞︀ ℬ︀} = 𝑁_{𝑟}$ with $ℬ︀ = 𝐛_{1}, \dots, 𝐛_{𝑛}$ and $𝒞︀ = 𝐜_{1}, \dots, 𝐜_{𝑚}$ , let $𝑆 \in L (𝑊, 𝑉)$ send $𝐜_{𝑗} \mapsto 𝐛_{𝑗}$ for $𝑗 \leq 𝑟$ and $𝐜_{𝑗} \mapsto 𝟎$ for $𝑗 > 𝑟$ . Then $𝑇 𝑆 𝑇 = 𝑇$ , as one checks on the basis $ℬ︀$ . The matrix computation is this argument in coordinates.

Bases adapted to a subspace

For an operator, the convenient basis must work on both sides at once. The first source of such bases is a subspace that $𝑇$ maps into itself.

Proposition 4.5.35 (Preserved subspaces give block triangular matrices). Let $dim 𝑉 = 𝑛 \geq 1$ , $𝑇 \in L (𝑉)$ , and $ℬ︀ = 𝐛_{1}, \dots, 𝐛_{𝑛}$ a basis of $𝑉$ . Let $1 \leq 𝑘 \leq 𝑛$ and $𝑈 =$ $span (𝐛_{1}, \dots, 𝐛_{𝑘})$ . Then $im (𝑇 ↾_{𝑈}) \subseteq 𝑈$ if and only if

{[𝑇]}_{ℬ︀ ℬ︀} = (\begin{matrix} 𝐴 & 𝐵 \\ 0 & 𝐷 \end{matrix})

for some $𝐴 \in 𝐹^{𝑘 \times 𝑘}$ , $𝐵 \in 𝐹^{𝑘 \times (𝑛 - 𝑘)}$ , and $𝐷 \in$ $𝐹^{(𝑛 - 𝑘) \times (𝑛 - 𝑘)}$ , where $0$ is the zero block of size $(𝑛 - 𝑘) \times 𝑘$ . In that case, $𝐴 =$ ${[𝑇 ↾_{𝑈}]}_{{ℬ︀}_{𝑈} {ℬ︀}_{𝑈}}$ , where ${ℬ︀}_{𝑈} = 𝐛_{1}, \dots, 𝐛_{𝑘}$ and $𝑇 ↾_{𝑈}$ is regarded as an operator on $𝑈$ .

Proof.routine

Uses. Definition 2.3.1 (Matrix of a linear map), Definition 1.6.14 (Coordinates), Proposition 2.6.22 (Testing injectivity and surjectivity on a basis), Definition 4.2.1 (Restriction)

Techniques. Check it on a basis (Technique 2.2.7)

Hint 1. The

𝑗

-th column of

{[𝑇]}_{ℬ︀ ℬ︀}

holds the coefficients of

𝑇 (𝐛_{𝑗})

. When does a vector lie in

𝑈

, in terms of its

ℬ︀

-coordinates?

Hint 2.

im (𝑇 ↾_{𝑈})

is spanned by

𝑇 (𝐛_{1}),

\dots,

𝑇 (𝐛_{𝑘})

, so it lies in

𝑈

exactly when each

𝑇 (𝐛_{𝑗})

with

𝑗 \leq 𝑘

does.

Sketch. A vector lies in

𝑈

exactly when its last

𝑛 - 𝑘

coordinates are

0

. The image of

𝑇 ↾_{𝑈}

is spanned by

𝑇 (𝐛_{1}),

\dots,

𝑇 (𝐛_{𝑘})

, so it lies in

𝑈

exactly when the first

𝑘

columns of

{[𝑇]}_{ℬ︀ ℬ︀}

vanish below row

𝑘

. Their top parts are then the coordinates of

𝑇 (𝐛_{𝑗})

{ℬ︀}_{𝑈}

, the columns of

[𝑇 ↾_{𝑈}]

A vector $𝐯$ lies in $𝑈$ if and only if its $ℬ︀$ -coordinates $(𝑐_{1}, \dots, 𝑐_{𝑛})$ satisfy $𝑐_{𝑘 + 1} = \dots = 𝑐_{𝑛} = 0$ : if they do, $𝐯 =$ $\sum_{𝑖 \leq 𝑘} 𝑐_{𝑖} 𝐛_{𝑖} \in$ $𝑈$ ; conversely, a vector of $𝑈$ is $\sum_{𝑖 \leq 𝑘} 𝑐_{𝑖} 𝐛_{𝑖}$ , and by uniqueness of coordinates (Definition 1.6.14) these are its coordinates, with zeros after position $𝑘$ .

The list $𝐛_{1}, \dots, 𝐛_{𝑘}$ is a basis of $𝑈$ , so $𝑇 (𝐛_{1}),$ $\dots,$ $𝑇 (𝐛_{𝑘})$ spans $im (𝑇 ↾_{𝑈})$ (Proposition 2.6.22(d), applied to $𝑇 ↾_{𝑈}$ ). Since $𝑈$ is a subspace, $im (𝑇 ↾_{𝑈}) \subseteq 𝑈$ if and only if $𝑇 (𝐛_{𝑗}) \in 𝑈$ for each $𝑗 \leq 𝑘$ , that is, if and only if the $𝑗$ -th column ${[𝑇 (𝐛_{𝑗})]}_{ℬ︀}$ of ${[𝑇]}_{ℬ︀ ℬ︀}$ has zeros in rows $𝑘 + 1, \dots, 𝑛$ for each $𝑗 \leq 𝑘$ . This is the block form.

In that case $𝑇 (𝐛_{𝑗}) =$ $\sum_{𝑖 \leq 𝑘} 𝑎_{𝑖 𝑗} 𝐛_{𝑖}$ for $𝑗 \leq 𝑘$ , where $(𝑎_{𝑖 𝑗})$ is the top left block. So the ${ℬ︀}_{𝑈}$ -coordinates of $𝑇 ↾_{𝑈} (𝐛_{𝑗})$ are the $𝑗$ -th column of $𝐴$ , and $𝐴 =$ ${[𝑇 ↾_{𝑈}]}_{{ℬ︀}_{𝑈} {ℬ︀}_{𝑈}}$ (Definition 2.3.1). □

If also the span of the last $𝑛 - 𝑘$ basis vectors is mapped into itself, the same argument shows that the block $𝐵$ is zero, and ${[𝑇]}_{ℬ︀ ℬ︀}$ is block diagonal. Subspaces with $im (𝑇 ↾_{𝑈}) \subseteq 𝑈$ are called invariant subspaces of $𝑇$ ; they are studied in Section 7.2.

Technique 4.5.36 (Choose a basis adapted to a preserved subspace). If $𝑇$ maps a subspace $𝑈$ into itself, take a basis of $𝑈$ and extend it to a basis of $𝑉$ (Theorem 4.1.16). In this basis the matrix of $𝑇$ is block upper triangular (Proposition 4.5.35), the top left block is the matrix of $𝑇$ on $𝑈$ , and a question about $𝑇$ splits into a question about $𝑇$ on $𝑈$ and one about the bottom right block. If $𝑉$ is assembled from several preserved subspaces, list bases of each in turn, and the matrix becomes block diagonal.

Reach for it when you can see a subspace that $𝑇$ preserves: a kernel or image of something commuting with $𝑇$ , the span of a vector fixed or scaled by $𝑇$ , or the span of $𝐯, 𝑇 𝐯, \dots, 𝑇^{𝑘} 𝐯$ . It is the first step of triangularization (Section 7.5) and of the Jordan form (Section 10.3).

Used in 8 places and 2 practice exercises

Exercise 4.5.42, Problem 4.4, Problem 4.12, Problem 4.13, Corollary 7.3.16, Exercise 7.3.41, Theorem 7.4.15, Lemma 7.5.11; 2 practice exercises.

Example 4.5.37 (The cyclic permutation of coordinates). Let $𝑍 \in 𝐹^{3 \times 3}$ be the matrix with $𝑍 (𝑥, 𝑦, 𝑧) =$ $(𝑧, 𝑥, 𝑦)$ , so that $𝑍^{3} = 𝐼_{3}$ . Two subspaces are visibly preserved by $𝑇_{𝑍}$ : the line $𝑈 = span (𝐮_{1})$ with $𝐮_{1} = (1, 1, 1)$ , which $𝑍$ fixes, and the plane $𝑈^{'} =$ ${(𝑥, 𝑦, 𝑧) | 𝑥 + 𝑦 + 𝑧 = 0}$ , since permuting the coordinates does not change their sum. Let $𝐮_{2} = (1, - 1, 0)$ and $𝐮_{3} = (0, 1, - 1)$ , both in $𝑈^{'}$ .

Suppose $3 \neq 0$ in $𝐹$ . Then $𝒰︀ = 𝐮_{1}, 𝐮_{2}, 𝐮_{3}$ is a basis of $𝐹^{3}$ : if $𝑎 𝐮_{1} +$ $𝑏 𝐮_{2} +$ $𝑐 𝐮_{3} =$ $(𝑎 + 𝑏, 𝑎 - 𝑏 + 𝑐, 𝑎 - 𝑐) =$ $𝟎$ , then $𝑏 = - 𝑎$ and $𝑐 = 𝑎$ , so the middle entry is $3 𝑎 = 0$ , and $𝑎 = 𝑏 = 𝑐 = 0$ . Now

𝑍 𝐮_{1} = 𝐮_{1}, 𝑍 𝐮_{2} = (0, 1, - 1) = 𝐮_{3}, 𝑍 𝐮_{3} = (- 1, 0, 1) = - 𝐮_{2} - 𝐮_{3},

{[𝑇_{𝑍}]}_{𝒰︀ 𝒰︀} = (\begin{matrix} 1 & 0 & 0 \\ 0 & 0 & - 1 \\ 0 & 1 & - 1 \end{matrix}) = (\begin{matrix} 1 & 0 \\ 0 & 𝑅 \end{matrix}), 𝑅 = (\begin{matrix} 0 & - 1 \\ 1 & - 1 \end{matrix}) .

The matrix is block diagonal because both $𝑈$ and $span (𝐮_{2}, 𝐮_{3}) = 𝑈^{'}$ are preserved. The trace is $0 = 1 + tr 𝑅$ , as it must be, and $𝑍^{3} = 𝐼_{3}$ becomes $𝑅^{3} = 𝐼_{2}$ , which is easy to check by hand. Over $ℝ$ , $𝑅$ turns out to be similar to the rotation by $2 𝜋 / 3$ (Exercise 4.5.43), which matches the picture of $𝑍$ as a rotation of $ℝ^{3}$ about the line $𝑈$ .

If $3 = 0$ in $𝐹$ , then $𝐮_{1} \in 𝑈^{'}$ , so $𝑈$ and $𝑈^{'}$ no longer fill $𝐹^{3}$ ; the list $𝐮_{1}, 𝐮_{2}, 𝐞_{1}$ still gives a block upper triangular matrix, but not a block diagonal one.

The normal forms ahead are all answers to one question: in the best basis, how simple can the matrix of an operator be? For maps between two spaces, Corollary 4.5.22 gave the complete answer. For operators the answer is diagonal when there is a basis of eigenvectors (Section 7.4), upper triangular over $ℂ$ (Section 7.5), and the Jordan form in general over $ℂ$ (Section 10.3).

Exercises

Exercise 4.5.38.routine Over $𝔽_{3}$ , let $ℬ︀ =$ $(1, 1, 0),$ $(0, 1, 1),$ $(1, 0, 1)$ and $\overset{̊}{ℬ︀} =$ $(1, 0, 0),$ $(1, 1, 0),$ $(1, 1, 1)$ .

(a)Show that $ℬ︀$ is a basis of $𝔽_{3}^{3}$ , and compute ${[id]}_{ℬ︀ ℰ︀}$ .
(b)Find ${[(1, 2, 0)]}_{ℬ︀}$ .
(c)Compute ${[id]}_{\overset{̊}{ℬ︀} ℬ︀}$ .
(d)Is the same list $ℬ︀$ a basis of $𝔽_{2}^{3}$ ?

Solution.

Uses. Proposition 4.5.12 (Calculus of change of coordinates), Corollary 3.5.25 (The invertible matrix theorem), Theorem 4.1.10 (Independent lists are short), Theorem 3.3.10 ( $𝐺 𝐴 = rref (𝐴)$ )

Hint 1.

{[id]}_{ℬ︀ ℰ︀}

is the inverse of the matrix

𝑀

whose columns are the vectors of

ℬ︀

. Row reduce

(𝑀 | 𝐼_{3})

over

𝔽_{3}

, where

2^{- 1} = 2

Hint 2. For (c),

{[id]}_{\overset{̊}{ℬ︀} ℬ︀} =

{({[id]}_{ℰ︀ \overset{̊}{ℬ︀}})}^{- 1} {[id]}_{ℰ︀ ℬ︀}

. For (d), add the three vectors.

Sketch.

𝑀^{- 1} = (\begin{matrix} 2 & 2 & 1 \\ 1 & 2 & 2 \\ 2 & 1 & 2 \end{matrix})

over

𝔽_{3}

(over

ℚ

it is

\frac{1}{2} (\begin{matrix} 1 & 1 & - 1 \\ - 1 & 1 & 1 \\ 1 & - 1 & 1 \end{matrix})

, and

1 / 2 = 2

𝔽_{3}

). Then

{[(1, 2, 0)]}_{ℬ︀} =

(0, 2, 1)

, and

{[id]}_{\overset{̊}{ℬ︀} ℬ︀} =

(\begin{matrix} 0 & 2 & 1 \\ 1 & 0 & 2 \\ 0 & 1 & 1 \end{matrix})

. Over

𝔽_{2}

the three vectors add to

𝟎

(a) Let $𝑀 =$ ${[id]}_{ℰ︀ ℬ︀} =$ $(\begin{matrix} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{matrix})$ . Row reducing $(𝑀 | 𝐼_{3})$ over $𝔽_{3}$ (or inverting over $ℚ$ and reading $1 / 2$ as $2$ , since $2 \cdot 2 = 4 = 1$ ) gives

𝑀^{- 1} = (\begin{matrix} 2 & 2 & 1 \\ 1 & 2 & 2 \\ 2 & 1 & 2 \end{matrix}),

and one checks $𝑀 𝑀^{- 1} = 𝐼_{3}$ over $𝔽_{3}$ ; for instance, the first row of the product is $(1 \cdot 2 + 1 \cdot 2, 1 \cdot 2 + 1 \cdot 1, 1 \cdot 1 + 1 \cdot 2) =$ $(4, 3, 3) =$ $(1, 0, 0)$ . So $𝑀$ is invertible, its columns form a basis of $𝔽_{3}^{3}$ (Corollary 3.5.25, Theorem 4.1.10(b)), and ${[id]}_{ℬ︀ ℰ︀} = 𝑀^{- 1}$ by Proposition 4.5.12(b).

(b) ${[(1, 2, 0)]}_{ℬ︀} =$ $𝑀^{- 1} (1, 2, 0) =$ $(2 + 4, 1 + 4, 2 + 2) =$ $(6, 5, 4) =$ $(0, 2, 1)$ . Check: $2 (0, 1, 1) +$ $(1, 0, 1) =$ $(1, 2, 3) =$ $(1, 2, 0)$ .

(c) ${[id]}_{ℰ︀ \overset{̊}{ℬ︀}} =$ $(\begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{matrix})$ has inverse $(\begin{matrix} 1 & - 1 & 0 \\ 0 & 1 & - 1 \\ 0 & 0 & 1 \end{matrix})$ , so

{[id]}_{\overset{̊}{ℬ︀} ℬ︀} = (\begin{matrix} 1 & - 1 & 0 \\ 0 & 1 & - 1 \\ 0 & 0 & 1 \end{matrix}) (\begin{matrix} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{matrix}) = (\begin{matrix} 0 & - 1 & 1 \\ 1 & 0 & - 1 \\ 0 & 1 & 1 \end{matrix}) = (\begin{matrix} 0 & 2 & 1 \\ 1 & 0 & 2 \\ 0 & 1 & 1 \end{matrix}) .

Check the first column: $0 (1, 0, 0) +$ $1 (1, 1, 0) +$ $0 (1, 1, 1) =$ $(1, 1, 0)$ , the first vector of $ℬ︀$ .

(d) No: over $𝔽_{2}$ , $(1, 1, 0) +$ $(0, 1, 1) +$ $(1, 0, 1) =$ $(2, 2, 2) =$ $𝟎$ , a nontrivial dependence.

Exercise 4.5.39.routine Prove or disprove.

(a)Every change-of-coordinates matrix is invertible.
(b)If $𝐴 \sim 𝐵$ , then $𝐴$ and $𝐵$ have the same diagonal entries, possibly in a different order.
(c)If $𝑇 \in L (𝑉, 𝑊)$ is injective and $dim 𝑉 = dim 𝑊 < \infty$ , then $𝑇$ is an isomorphism.
(d)Two matrices in $𝐹^{𝑛 \times 𝑛}$ with the same rank and the same trace are similar.
(e)If $dim 𝑉 = 2$ and $dim L (𝑉, 𝑊) = 6$ , then $𝑊 ≅ 𝐹^{3}$ .
(f)The only matrix similar to $𝑐 𝐼_{𝑛}$ is $𝑐 𝐼_{𝑛}$ .

Solution.

Uses. Proposition 4.5.12 (Calculus of change of coordinates), Example 4.5.26 (A similarity and a non-similarity), Theorem 4.2.11 (Rank–nullity), Proposition 4.1.21 (Dimensions of subspaces), Remark 4.5.28 (What the invariants cannot do), Theorem 4.5.8 (Dimension of $L (𝑉, 𝑊)$ ), Theorem 4.5.6 (Classification by dimension)

Techniques. Finding counterexamples and minimal examples (Technique 1.4.18)

Hint 1. Two of the six are false, and the counterexamples are in this section.

Hint 2. For (c), compute

dim im 𝑇

. For (f), compute

𝑆^{- 1} (𝑐 𝐼_{𝑛}) 𝑆

Sketch. (a) True: the inverse of

{[id]}_{{ℬ︀}^{'} ℬ︀}

{[id]}_{ℬ︀ {ℬ︀}^{'}}

. (b) False:

(\begin{matrix} 1 & 1 \\ 1 & 1 \end{matrix}) \sim (\begin{matrix} 2 & 0 \\ 0 & 0 \end{matrix})

over

ℚ

. (c) True by rank–nullity. (d) False:

𝐼_{2}

and

(\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix})

. (e) True:

2 dim 𝑊 = 6

. (f) True:

𝑆^{- 1} (𝑐 𝐼_{𝑛}) 𝑆 =

𝑐 𝐼_{𝑛}

(a) True, by Proposition 4.5.12(b).

(b) False. Over $ℚ$ , $(\begin{matrix} 1 & 1 \\ 1 & 1 \end{matrix}) \sim (\begin{matrix} 2 & 0 \\ 0 & 0 \end{matrix})$ (Example 4.5.26), and the diagonals $(1, 1)$ and $(2, 0)$ differ.

(c) True. $ker 𝑇 = {𝟎}$ , so by Theorem 4.2.11, $dim im 𝑇 =$ $dim 𝑉 =$ $dim 𝑊$ . Then $im 𝑇 = 𝑊$ by Proposition 4.1.21(b), so $𝑇$ is bijective, hence an isomorphism.

(d) False. $𝐼_{2}$ and $(\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix})$ both have rank $2$ and trace $2$ , and they are not similar (Remark 4.5.28).

(e) True. $𝑊$ is finite-dimensional: if $𝐯_{1}, 𝐯_{2}$ is a basis of $𝑉$ , then $𝑇 \mapsto 𝑇 (𝐯_{1})$ is a surjective linear map $L (𝑉, 𝑊) \to 𝑊$ (every $𝐰$ is $𝑇 (𝐯_{1})$ for the $𝑇$ with $𝐯_{1} \mapsto 𝐰$ and $𝐯_{2} \mapsto 𝟎$ ), so $dim 𝑊 \leq 6$ by Theorem 4.2.11. Then Theorem 4.5.8 gives $2 dim 𝑊 = 6$ , so $dim 𝑊 = 3$ and $𝑊 ≅ 𝐹^{3}$ (Theorem 4.5.6).

(f) True. $𝑆^{- 1} (𝑐 𝐼_{𝑛}) 𝑆 =$ $𝑐 𝑆^{- 1} 𝑆 =$ $𝑐 𝐼_{𝑛}$ for every invertible $𝑆$ .

Exercise 4.5.40.standard Let $𝑎 \in 𝐹$ and $𝑛 \geq 1$ . In ${𝐹 [𝑥]}_{\leq 𝑛}$ , let $ℬ︀ = 1, 𝑥, \dots, 𝑥^{𝑛}$ and ${ℬ︀}_{𝑎} =$ $1,$ $𝑥 -$ $𝑎,$ ${(𝑥 - 𝑎)}^{2},$ $\dots,$ ${(𝑥 - 𝑎)}^{𝑛}$ . Index rows and columns from $0$ to $𝑛$ .

(a)Show that ${ℬ︀}_{𝑎}$ is a basis, and that the $(𝑖, 𝑗)$ entry of ${[id]}_{{ℬ︀}_{𝑎} ℬ︀}$ is $(\binom{𝑗}{𝑖}) 𝑎^{𝑗 - 𝑖}$ for $𝑖 \leq 𝑗$ and $0$ for $𝑖 > 𝑗$ .
(b)Find the entries of ${[id]}_{ℬ︀ {ℬ︀}_{𝑎}}$ in the same way.
(c)Taking $𝐹 = ℚ$ and $𝑎 = 1$ , deduce that $\sum_{𝑘 = 𝑖}^{𝑗} {(- 1)}^{𝑘 - 𝑖} (\binom{𝑗}{𝑘}) (\binom{𝑘}{𝑖})$ is $1$ if $𝑖 = 𝑗$ and $0$ if $𝑖 < 𝑗$ .

Solution.

Uses. Corollary 4.1.11 (Spanning lists are long), Definition 4.5.10 (Change-of-coordinates matrix), Proposition 4.5.12 (Calculus of change of coordinates)

Hint 1. Write

𝑥 = (𝑥 - 𝑎) + 𝑎

and expand

𝑥^{𝑗}

by the binomial theorem, which holds in

𝐹 [𝑥]

because multiplication of polynomials is commutative.

Hint 2. For (b), expand

{(𝑥 - 𝑎)}^{𝑗}

in powers of

𝑥

. For (c), the two matrices of (a) and (b) are inverse to each other; compute the

(𝑖, 𝑗)

entry of their product.

Sketch.

𝑥^{𝑗} =

\sum_{𝑖} (\binom{𝑗}{𝑖}) 𝑎^{𝑗 - 𝑖} {(𝑥 - 𝑎)}^{𝑖}

shows that

{ℬ︀}_{𝑎}

spans, hence is a basis, and gives the columns of

{[id]}_{{ℬ︀}_{𝑎} ℬ︀}

. Likewise

{(𝑥 - 𝑎)}^{𝑗} =

\sum_{𝑖} (\binom{𝑗}{𝑖}) {(- 𝑎)}^{𝑗 - 𝑖} 𝑥^{𝑖}

. The product

{[id]}_{ℬ︀ {ℬ︀}_{𝑎}} {[id]}_{{ℬ︀}_{𝑎} ℬ︀} =

𝐼

has

(𝑖, 𝑗)

entry

𝑎^{𝑗 - 𝑖} \sum_{𝑘} {(- 1)}^{𝑘 - 𝑖} (\binom{𝑘}{𝑖}) (\binom{𝑗}{𝑘})

Here an integer such as $(\binom{𝑗}{𝑖})$ stands for the corresponding sum of $1$ ′s in $𝐹$ .

(a) By the binomial theorem in $𝐹 [𝑥]$ ,

𝑥^{𝑗} = {((𝑥 - 𝑎) + 𝑎)}^{𝑗} = \sum_{𝑖 = 0}^{𝑗} (\binom{𝑗}{𝑖}) 𝑎^{𝑗 - 𝑖} {(𝑥 - 𝑎)}^{𝑖} .

So $1, 𝑥, \dots, 𝑥^{𝑛}$ lie in $span ({ℬ︀}_{𝑎})$ , which is therefore all of ${𝐹 [𝑥]}_{\leq 𝑛}$ . A spanning list of length $𝑛 + 1 = dim {𝐹 [𝑥]}_{\leq 𝑛}$ is a basis (Corollary 4.1.11(b)). The display gives ${[𝑥^{𝑗}]}_{{ℬ︀}_{𝑎}}$ , the $𝑗$ -th column of ${[id]}_{{ℬ︀}_{𝑎} ℬ︀}$ : its $𝑖$ -th entry is $(\binom{𝑗}{𝑖}) 𝑎^{𝑗 - 𝑖}$ for $𝑖 \leq 𝑗$ and $0$ for $𝑖 > 𝑗$ .

(b) Similarly ${(𝑥 - 𝑎)}^{𝑗} =$ $\sum_{𝑖 = 0}^{𝑗} (\binom{𝑗}{𝑖}) {(- 𝑎)}^{𝑗 - 𝑖} 𝑥^{𝑖}$ , so the $(𝑖, 𝑗)$ entry of ${[id]}_{ℬ︀ {ℬ︀}_{𝑎}}$ is $(\binom{𝑗}{𝑖}) {(- 𝑎)}^{𝑗 - 𝑖}$ for $𝑖 \leq 𝑗$ and $0$ otherwise.

(c) By Proposition 4.5.12(b), the two matrices are inverse to each other. The $(𝑖, 𝑗)$ entry of ${[id]}_{ℬ︀ {ℬ︀}_{𝑎}} {[id]}_{{ℬ︀}_{𝑎} ℬ︀}$ is, for $𝑖 \leq 𝑗$ ,

\sum_{𝑘 = 𝑖}^{𝑗} (\binom{𝑘}{𝑖}) {(- 𝑎)}^{𝑘 - 𝑖} (\binom{𝑗}{𝑘}) 𝑎^{𝑗 - 𝑘} = 𝑎^{𝑗 - 𝑖} \sum_{𝑘 = 𝑖}^{𝑗} {(- 1)}^{𝑘 - 𝑖} (\binom{𝑗}{𝑘}) (\binom{𝑘}{𝑖}),

since only $𝑖 \leq 𝑘 \leq 𝑗$ contribute. With $𝐹 = ℚ$ and $𝑎 = 1$ , this entry of $𝐼_{𝑛 + 1}$ is $1$ if $𝑖 = 𝑗$ and $0$ if $𝑖 < 𝑗$ , which is the identity. (The case $𝑗 > 𝑛$ is covered by taking $𝑛$ larger.)

Exercise 4.5.41.standard Let $𝑉$ and $𝑊$ be finite-dimensional with $dim 𝑉 = 𝑛 \geq 1$ and $dim 𝑊 = 𝑚$ , and let $𝐯_{0} \in 𝑉$ be nonzero. Show that

𝐾 = {𝑇 \in L (𝑉, 𝑊) | 𝑇 (𝐯_{0}) = 𝟎}

is a subspace of $L (𝑉, 𝑊)$ of dimension $(𝑛 - 1) 𝑚$ . Give two proofs: one by choosing a basis of $𝑉$ that contains $𝐯_{0}$ , and one by rank–nullity.

Solution (with a convenient basis).

Uses. Theorem 4.1.16 (Extending to a basis), Theorem 4.5.8 (Dimension of $L (𝑉, 𝑊)$ ), Proposition 4.5.5 (Isomorphisms preserve linear structure), Proposition 1.6.7 (Standard bases)

Techniques. Choose a convenient basis (Technique 4.5.33), Transport along an isomorphism (Technique 4.5.7), Dimension counting (Technique 4.2.24)

Hint 1. Extend

𝐯_{0}

to a basis

ℬ︀

𝑉

with

𝐯_{0}

first. What does

𝑇 (𝐯_{0}) = 𝟎

say about

{[𝑇]}_{𝒞︀ ℬ︀}

Hint 2. The condition is that the first column is zero. Transport

𝐾

to matrices with Theorem 4.5.8.

Sketch. With

𝐯_{0}

the first vector of

ℬ︀

𝑇 \in 𝐾

iff the first column of

{[𝑇]}_{𝒞︀ ℬ︀}

𝟎

. The isomorphism

𝑇 \mapsto {[𝑇]}_{𝒞︀ ℬ︀}

carries

𝐾

onto the matrices with zero first column, a subspace with basis

𝑈_{𝑖 𝑗}

𝑗 \geq 2

, of dimension

(𝑛 - 1) 𝑚

If $𝑚 = 0$ , then $L (𝑉, 𝑊) = {0} = 𝐾$ , of dimension $0$ . Let $𝑚 \geq 1$ . Extend $𝐯_{0}$ to a basis $ℬ︀ = 𝐯_{0}, 𝐯_{2}, \dots, 𝐯_{𝑛}$ of $𝑉$ (Theorem 4.1.16), and choose a basis $𝒞︀$ of $𝑊$ . The first column of ${[𝑇]}_{𝒞︀ ℬ︀}$ is ${[𝑇 (𝐯_{0})]}_{𝒞︀}$ , which is $𝟎$ exactly when $𝑇 (𝐯_{0}) = 𝟎$ . So the isomorphism $Φ : 𝑇 \mapsto {[𝑇]}_{𝒞︀ ℬ︀}$ of Theorem 4.5.8 carries $𝐾$ onto

𝑍 = {𝐴 \in 𝐹^{𝑚 \times 𝑛} | the first column of 𝐴 is 𝟎} .

$𝑍$ consists of the combinations of the matrix units $𝑈_{𝑖 𝑗}$ with $𝑗 \geq 2$ , which are independent (Proposition 1.6.7(b)); so $𝑍$ is a subspace with $dim 𝑍 = 𝑚 (𝑛 - 1)$ . Then $𝐾 = im (Φ^{- 1} ↾_{𝑍})$ is a subspace of $L (𝑉, 𝑊)$ with $dim 𝐾 = dim 𝑍$ , by Proposition 4.5.5(b) and (c) applied to the isomorphism $Φ^{- 1}$ .

Solution (by rank–nullity).

Uses. Definition 2.1.21 (Pointwise operations), Theorem 4.1.16 (Extending to a basis), Theorem 2.2.1 (Linear maps are determined by a basis), Theorem 4.2.11 (Rank–nullity), Theorem 4.5.8 (Dimension of $L (𝑉, 𝑊)$ ), Proposition 2.6.8 (Kernels and images are subspaces)

Techniques. Dimension counting (Technique 4.2.24), Choose a convenient basis (Technique 4.5.33)

Hint 1.

𝐾

is the kernel of a linear map

L (𝑉, 𝑊) \to 𝑊

. Which one?

Hint 2. Evaluation

𝜀 (𝑇) = 𝑇 (𝐯_{0})

is linear and surjective. Apply rank–nullity in

L (𝑉, 𝑊)

, whose dimension you know.

Sketch.

𝜀 : 𝑇 \mapsto 𝑇 (𝐯_{0})

is linear with kernel

𝐾

, and surjective because a map can send

𝐯_{0}

anywhere. So

dim 𝐾 = 𝑛 𝑚 - 𝑚

Let $𝜀 : L (𝑉, 𝑊) \to 𝑊$ , $𝜀 (𝑇) = 𝑇 (𝐯_{0})$ . It is linear because the operations on $L (𝑉, 𝑊)$ are pointwise: $(𝑆 + 𝑐 𝑇) (𝐯_{0}) =$ $𝑆 (𝐯_{0}) +$ $𝑐 𝑇 (𝐯_{0})$ . Its kernel is $𝐾$ , which is therefore a subspace (Proposition 2.6.8). It is surjective: given $𝐰 \in 𝑊$ , extend $𝐯_{0}$ to a basis $𝐯_{0}, 𝐯_{2}, \dots, 𝐯_{𝑛}$ of $𝑉$ , and let $𝑇$ be the linear map with $𝑇 (𝐯_{0}) = 𝐰$ and $𝑇 (𝐯_{𝑗}) = 𝟎$ for $𝑗 \geq 2$ (Theorem 2.2.1). By Theorem 4.2.11(b) and Theorem 4.5.8,

dim 𝐾 = dim L (𝑉, 𝑊) - dim 𝑊 = 𝑛 𝑚 - 𝑚 = (𝑛 - 1) 𝑚 .

Exercise 4.5.42.hard Let $𝑉$ and $𝑊$ be finite-dimensional with $dim 𝑉 = 𝑛 \geq 1$ and $dim 𝑊 = 𝑚 \geq 1$ , and let $𝑈 \subseteq 𝑉$ and $𝑈^{'} \subseteq 𝑊$ be subspaces with $dim 𝑈 = 𝑘$ and $dim 𝑈^{'} = 𝑘^{'}$ . Show that

{𝑇 \in L (𝑉, 𝑊) | im (𝑇 ↾_{𝑈}) \subseteq 𝑈^{'}}

is a subspace of $L (𝑉, 𝑊)$ of dimension $𝑚 𝑛 - 𝑘 (𝑚 - 𝑘^{'})$ . In particular, the operators $𝑇 \in L (𝑉)$ with $im (𝑇 ↾_{𝑈}) \subseteq 𝑈$ form a subspace of $L (𝑉)$ of dimension $𝑛^{2} - 𝑘 (𝑛 - 𝑘)$ .

Solution.

Uses. Theorem 4.1.16 (Extending to a basis), Theorem 4.5.8 (Dimension of $L (𝑉, 𝑊)$ ), Proposition 4.5.5 (Isomorphisms preserve linear structure), Proposition 4.5.35 (Preserved subspaces give block triangular matrices), Proposition 1.6.7 (Standard bases), Proposition 2.6.22 (Testing injectivity and surjectivity on a basis)

Techniques. Choose a convenient basis (Technique 4.5.33), Choose a basis adapted to a preserved subspace (Technique 4.5.36), Transport along an isomorphism (Technique 4.5.7)

Hint 1. Choose a basis of

𝑉

that starts with a basis of

𝑈

, and a basis of

𝑊

that starts with a basis of

𝑈^{'}

Hint 2. In these bases, the condition says that certain entries of

{[𝑇]}_{𝒞︀ ℬ︀}

vanish. Which ones, and how many?

Hint 3. Compare the proof of Proposition 4.5.35. Then transport along

𝑇 \mapsto {[𝑇]}_{𝒞︀ ℬ︀}

Sketch. With adapted bases,

im (𝑇 ↾_{𝑈}) \subseteq 𝑈^{'}

iff

𝑇 (𝐮_{𝑗}) \in 𝑈^{'}

for

𝑗 \leq 𝑘

, iff the first

𝑘

columns of

{[𝑇]}_{𝒞︀ ℬ︀}

vanish in rows

𝑘^{'} + 1, \dots, 𝑚

. Those

𝑘 (𝑚 - 𝑘^{'})

entries are forced to be

0

and the other

𝑚 𝑛 - 𝑘 (𝑚 - 𝑘^{'})

are free, so the set corresponds to the span of the remaining matrix units.

Let $𝐮_{1}, \dots, 𝐮_{𝑘}$ be a basis of $𝑈$ , extended to a basis $ℬ︀$ of $𝑉$ , and $𝐮_{1}^{'}, \dots, 𝐮_{𝑘^{'}}^{'}$ a basis of $𝑈^{'}$ , extended to a basis $𝒞︀$ of $𝑊$ (Theorem 4.1.16). A vector of $𝑊$ lies in $𝑈^{'}$ if and only if its $𝒞︀$ -coordinates vanish after position $𝑘^{'}$ , as in the proof of Proposition 4.5.35. Since $𝑇 (𝐮_{1}),$ $\dots,$ $𝑇 (𝐮_{𝑘})$ spans $im (𝑇 ↾_{𝑈})$ (Proposition 2.6.22(d)) and $𝑈^{'}$ is a subspace, $im (𝑇 ↾_{𝑈}) \subseteq 𝑈^{'}$ holds if and only if $𝑇 (𝐮_{𝑗}) \in 𝑈^{'}$ for $𝑗 = 1, \dots, 𝑘$ , that is, if and only if the entries of ${[𝑇]}_{𝒞︀ ℬ︀}$ in rows $𝑘^{'} + 1, \dots, 𝑚$ and columns $1, \dots, 𝑘$ are all $0$ .

So the isomorphism $𝑇 \mapsto {[𝑇]}_{𝒞︀ ℬ︀}$ (Theorem 4.5.8) carries our set onto the set $𝑍$ of matrices whose entries in those $𝑘 (𝑚 - 𝑘^{'})$ positions vanish. $𝑍$ is the span of the matrix units $𝑈_{𝑖 𝑗}$ in the other $𝑚 𝑛 - 𝑘 (𝑚 - 𝑘^{'})$ positions, which are independent (Proposition 1.6.7(b)); so $𝑍$ is a subspace of that dimension. By Proposition 4.5.5(b), (c), applied to the inverse isomorphism, our set is a subspace of $L (𝑉, 𝑊)$ of the same dimension. For $𝑉 = 𝑊$ and $𝑈 = 𝑈^{'}$ , take $𝑚 = 𝑛$ and $𝑘^{'} = 𝑘$ .

Exercise 4.5.43.hard For $𝐴 =$ $(\begin{matrix} 𝑎 & 𝑏 \\ 𝑐 & 𝑑 \end{matrix}) \in$ $𝐹^{2 \times 2}$ , let $𝛿 (𝐴) = 𝑎 𝑑 - 𝑏 𝑐$ .

(a)Show that if $𝐴 𝐯 \in span (𝐯)$ for every $𝐯 \in 𝐹^{2}$ , then $𝐴 = 𝑐 𝐼_{2}$ for some $𝑐 \in 𝐹$ .
(b)Suppose $𝐴$ is not a scalar multiple of $𝐼_{2}$ . Show that there is $𝐯 \in 𝐹^{2}$ such that $𝐯, 𝐴 𝐯$ is a basis of $𝐹^{2}$ , and that in this basis $𝑇_{𝐴}$ has the matrix
$𝐶_{𝐴} = (\begin{matrix} 0 & - 𝛿 (𝐴) \\ 1 & tr 𝐴 \end{matrix}) .$
(c)Deduce that two matrices in $𝐹^{2 \times 2}$ , neither a scalar multiple of $𝐼_{2}$ , with the same trace and the same value of $𝛿$ , are similar.
(d)Show that $𝐴 \sim 𝐴^{𝖳}$ for every $𝐴 \in 𝐹^{2 \times 2}$ .
(e)Show that the matrix $𝑅 = (\begin{matrix} 0 & - 1 \\ 1 & - 1 \end{matrix})$ of Example 4.5.37 is similar over $ℝ$ to the rotation matrix $(\begin{matrix} cos (2 𝜋 / 3) & - sin (2 𝜋 / 3) \\ sin (2 𝜋 / 3) & cos (2 𝜋 / 3) \end{matrix})$ .

Solution.

Uses. Exercise 2.5.46, Theorem 4.1.10 (Independent lists are short), Theorem 4.5.25 (Similar matrices are one operator in two bases), Proposition 4.5.24 (Similarity is an equivalence relation), Proposition 1.5.14 (Appending a vector)

Techniques. Choose a convenient basis (Technique 4.5.33)

Hint 1. (a) Apply the hypothesis to

𝐞_{1}

𝐞_{2}

, and

𝐞_{1} + 𝐞_{2}

Hint 2. (b) By (a) there is

𝐯

with

𝐴 𝐯 \notin span (𝐯)

. For the matrix, you need

𝐴 (𝐴 𝐯)

in terms of

𝐯

and

𝐴 𝐯

: use

𝐴^{2} -

(tr 𝐴) 𝐴 +

𝛿 (𝐴) 𝐼_{2} =

0

(Exercise 2.5.46(a)).

Hint 3. (c) Both matrices are similar to the same

𝐶

. (d) What are the trace and

𝛿

𝐴^{𝖳}

, and when is

𝐴^{𝖳}

scalar?

Sketch. (a)

𝐴 𝐞_{1} = 𝜆 𝐞_{1}

𝐴 𝐞_{2} = 𝜇 𝐞_{2}

, and

𝐴 (𝐞_{1} + 𝐞_{2}) =

𝜈 (𝐞_{1} + 𝐞_{2})

force

𝜆 = 𝜈 = 𝜇

. (b) Pick

𝐯

with

𝐴 𝐯 \notin span (𝐯)

; then

𝐯, 𝐴 𝐯

is independent, and

𝐴^{2} 𝐯 =

(tr 𝐴) 𝐴 𝐯 -

𝛿 (𝐴) 𝐯

gives the second column. (c) Both are similar to the same

𝐶_{𝐴}

. (d)

𝐴^{𝖳}

has the same trace and

𝛿

, and is scalar iff

𝐴

is. (e) Both matrices have trace

- 1

and

𝛿 = 1

and are not scalar.

(a) By hypothesis $𝐴 𝐞_{1} = 𝜆 𝐞_{1}$ , $𝐴 𝐞_{2} = 𝜇 𝐞_{2}$ , and $𝐴 (𝐞_{1} + 𝐞_{2}) =$ $𝜈 (𝐞_{1} + 𝐞_{2})$ for some scalars. The last equation reads $𝜆 𝐞_{1} +$ $𝜇 𝐞_{2} =$ $𝜈 𝐞_{1} +$ $𝜈 𝐞_{2}$ , so $𝜆 = 𝜈 = 𝜇$ . The columns of $𝐴$ are then $𝜆 𝐞_{1}$ and $𝜆 𝐞_{2}$ , that is, $𝐴 = 𝜆 𝐼_{2}$ .

(b) By (a) there is $𝐯$ with $𝐴 𝐯 \notin span (𝐯)$ . Then $𝐯 \neq 𝟎$ (as $𝐴 𝟎 = 𝟎$ lies in every span), so the list $𝐯$ is independent, and $𝐯, 𝐴 𝐯$ is independent by Proposition 1.5.14. Two independent vectors in $𝐹^{2}$ form a basis (Theorem 4.1.10(b)). In this basis, $𝑇_{𝐴}$ sends the first vector $𝐯$ to the second vector $𝐴 𝐯$ , with coordinates $(0, 1)$ . By Exercise 2.5.46(a), $𝐴^{2} =$ $(tr 𝐴) 𝐴 -$ $𝛿 (𝐴) 𝐼_{2}$ , so

𝐴 (𝐴 𝐯) = - 𝛿 (𝐴) 𝐯 + (tr 𝐴) 𝐴 𝐯,

with coordinates $(- 𝛿 (𝐴), tr 𝐴)$ . So the matrix is $𝐶_{𝐴}$ .

(c) Let $𝐴$ and $𝐵$ be non-scalar with $tr 𝐴 = tr 𝐵$ and $𝛿 (𝐴) = 𝛿 (𝐵)$ . Then $𝐶_{𝐴} = 𝐶_{𝐵}$ . By (b) and Theorem 4.5.25, $𝐴 \sim 𝐶_{𝐴}$ and $𝐵 \sim 𝐶_{𝐵} = 𝐶_{𝐴}$ , so $𝐴 \sim 𝐵$ (Proposition 4.5.24).

(d) If $𝐴 = 𝑐 𝐼_{2}$ , then $𝐴^{𝖳} = 𝐴$ . Otherwise $𝐴^{𝖳}$ is not scalar either, since ${(𝑐 𝐼_{2})}^{𝖳} = 𝑐 𝐼_{2}$ and ${(𝐴^{𝖳})}^{𝖳} = 𝐴$ . The diagonal of $𝐴^{𝖳}$ is that of $𝐴$ , and $𝛿 (𝐴^{𝖳}) =$ $𝑎 𝑑 -$ $𝑐 𝑏 =$ $𝛿 (𝐴)$ . So $𝐴 \sim 𝐴^{𝖳}$ by (c).

(e) $𝑅$ is not scalar, $tr 𝑅 = - 1$ , and $𝛿 (𝑅) =$ $0 \cdot$ $(- 1) -$ $(- 1) \cdot$ $1 =$ $1$ . The rotation matrix is not scalar, its trace is $2 cos (2 𝜋 / 3) = - 1$ , and its $𝛿$ is ${cos}^{2} + {sin}^{2} = 1$ . By (c), the two are similar over $ℝ$ .

Exercise 4.5.44.hard Keep the notation $𝛿 (𝐴) = 𝑎 𝑑 - 𝑏 𝑐$ of Exercise 4.5.43.

(a)Show that $𝛿 (𝑋 𝑌) =$ $𝛿 (𝑋) 𝛿 (𝑌)$ for all $𝑋, 𝑌 \in 𝐹^{2 \times 2}$ , and deduce that $𝛿$ is a similarity invariant.
(b)Show that every $𝐴 \in 𝐹^{2 \times 2}$ is similar to exactly one matrix in the list consisting of $𝑐 𝐼_{2}$ for $𝑐 \in 𝐹$ and $(\begin{matrix} 0 & - 𝑑 \\ 1 & 𝑡 \end{matrix})$ for $𝑡, 𝑑 \in 𝐹$ .
(c)Deduce that over $𝔽_{𝑞}$ there are exactly $𝑞^{2} + 𝑞$ similarity classes of $2 \times 2$ matrices.

Solution.

Uses. Exercise 4.5.43, Proposition 4.5.27 (Similarity invariants), Definition 4.5.23 (Similar matrices)

Techniques. Rule out similarity with an invariant (Technique 4.5.29)

Hint 1. (a) is a direct expansion; then

𝛿 (𝑆^{- 1}) 𝛿 (𝑆) =

𝛿 (𝐼_{2}) =

1

Hint 2. (b) Existence is Exercise 4.5.43(b). For uniqueness, which invariants separate the matrices in the list? What are the trace and

𝛿

(\begin{matrix} 0 & - 𝑑 \\ 1 & 𝑡 \end{matrix})

Sketch. (a) Expanding,

𝛿 (𝑋 𝑌) =

(𝑎 𝑑 - 𝑏 𝑐) (𝑒 ℎ - 𝑓 𝑔)

for

𝑋 = (\begin{matrix} 𝑎 & 𝑏 \\ 𝑐 & 𝑑 \end{matrix})

𝑌 = (\begin{matrix} 𝑒 & 𝑓 \\ 𝑔 & ℎ \end{matrix})

; so

𝛿 (𝑆^{- 1} 𝐴 𝑆) =

𝛿 (𝑆^{- 1}) 𝛿 (𝐴) 𝛿 (𝑆) =

𝛿 (𝐴)

. (b) A scalar matrix is similar only to itself; a non-scalar

𝐴

is similar to

(\begin{matrix} 0 & - 𝛿 (𝐴) \\ 1 & tr 𝐴 \end{matrix})

, which is not scalar; and

(\begin{matrix} 0 & - 𝑑 \\ 1 & 𝑡 \end{matrix})

has trace

𝑡

and

𝛿 = 𝑑

, so different

(𝑡, 𝑑)

give non-similar matrices. (c)

𝑞

scalars plus

𝑞^{2}

pairs

(𝑡, 𝑑)

(a) For $𝑋 = (\begin{matrix} 𝑎 & 𝑏 \\ 𝑐 & 𝑑 \end{matrix})$ and $𝑌 = (\begin{matrix} 𝑒 & 𝑓 \\ 𝑔 & ℎ \end{matrix})$ , $𝑋 𝑌 =$ $(\begin{matrix} 𝑎 𝑒 + 𝑏 𝑔 & 𝑎 𝑓 + 𝑏 ℎ \\ 𝑐 𝑒 + 𝑑 𝑔 & 𝑐 𝑓 + 𝑑 ℎ \end{matrix})$ , and expanding,

\begin{aligned} 𝛿 (𝑋 𝑌) & = (𝑎 𝑒 + 𝑏 𝑔) (𝑐 𝑓 + 𝑑 ℎ) - (𝑎 𝑓 + 𝑏 ℎ) (𝑐 𝑒 + 𝑑 𝑔) \\ = 𝑎 𝑑 𝑒 ℎ + 𝑏 𝑐 𝑓 𝑔 - 𝑎 𝑑 𝑓 𝑔 - 𝑏 𝑐 𝑒 ℎ = (𝑎 𝑑 - 𝑏 𝑐) (𝑒 ℎ - 𝑓 𝑔), \end{aligned}

since the terms $𝑎 𝑐 𝑒 𝑓$ and $𝑏 𝑑 𝑔 ℎ$ cancel. If $𝐵 = 𝑆^{- 1} 𝐴 𝑆$ , then $𝛿 (𝑆^{- 1}) 𝛿 (𝑆) =$ $𝛿 (𝑆^{- 1} 𝑆) =$ $𝛿 (𝐼_{2}) =$ $1$ , and since scalars commute,

𝛿 (𝐵) = 𝛿 (𝑆^{- 1}) 𝛿 (𝐴) 𝛿 (𝑆) = 𝛿 (𝐴) .

(b) Existence. If $𝐴$ is scalar, it is in the list. Otherwise $𝐴 \sim$ $(\begin{matrix} 0 & - 𝛿 (𝐴) \\ 1 & tr 𝐴 \end{matrix})$ by Exercise 4.5.43(b) and Theorem 4.5.25.

Uniqueness. A scalar matrix $𝑐 𝐼_{2}$ is similar only to itself (Exercise 4.5.39(f)), so it is not similar to any other matrix of the list; the matrices $(\begin{matrix} 0 & - 𝑑 \\ 1 & 𝑡 \end{matrix})$ are not scalar, because of the entry $1$ . Finally, $(\begin{matrix} 0 & - 𝑑 \\ 1 & 𝑡 \end{matrix})$ has trace $𝑡$ and $𝛿 =$ $0 \cdot$ $𝑡 -$ $(- 𝑑) \cdot$ $1 =$ $𝑑$ . By (a) and Proposition 4.5.27(b), two matrices of this form with different pairs $(𝑡, 𝑑)$ are not similar.

(c) Over $𝔽_{𝑞}$ the list has $𝑞$ scalar matrices and $𝑞^{2}$ matrices $(\begin{matrix} 0 & - 𝑑 \\ 1 & 𝑡 \end{matrix})$ , and by (b) each similarity class contains exactly one of them. So there are $𝑞^{2} + 𝑞$ classes.

Exercise 4.5.45.routine Find the error in the following argument, and show that its conclusion is false.

Claim. Any two matrices $𝐴, 𝐵 \in 𝐹^{𝑛 \times 𝑛}$ of the same rank $𝑟$ are similar.

Proof. By Theorem 3.5.27 there are invertible $𝐺, 𝐻, 𝐺^{'}, 𝐻^{'}$ with $𝐺 𝐴 𝐻 =$ $𝑁_{𝑟} =$ $𝐺^{'} 𝐵 𝐻^{'}$ . Then $𝐵 =$ $𝐺^{' - 1} 𝐺 𝐴 𝐻 𝐻^{' - 1}$ . Let $𝑆 = 𝐻 𝐻^{' - 1}$ , which is invertible. Then $𝐵 = (𝐺^{' - 1} 𝐺) 𝐴 𝑆$ , a product of $𝐴$ with invertible matrices on both sides, so $𝐵$ is similar to $𝐴$ .

Solution.

Uses. Definition 4.5.23 (Similar matrices), Corollary 4.5.21 (Matrices of one linear map), Proposition 4.5.27 (Similarity invariants)

Techniques. Rule out similarity with an invariant (Technique 4.5.29)

Hint 1. Compare the last sentence with Definition 4.5.23. Which relation between the left and right factors does similarity require?

Hint 2. For a counterexample, look for two

2 \times 2

matrices of rank

1

that differ in an invariant.

Sketch. Similarity needs

𝐵 = 𝑆^{- 1} 𝐴 𝑆

, with the left factor the inverse of the right one; nothing makes

𝐺^{' - 1} 𝐺 = 𝑆^{- 1}

. The argument proves only

𝐵 = 𝐺 𝐴 𝐻

, which is Corollary 4.5.21. Counterexample:

(\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix})

and

(\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix})

have rank

1

but traces

0

and

1

The error is in the last sentence. Similarity requires $𝐵 = 𝑆^{- 1} 𝐴 𝑆$ : the left factor must be the inverse of the right factor. The argument produces $𝐵 = 𝐺^{″} 𝐴 𝑆$ with $𝐺^{″} = 𝐺^{' - 1} 𝐺$ and $𝑆 = 𝐻 𝐻^{' - 1}$ both invertible, but nothing forces $𝐺^{″} = 𝑆^{- 1}$ . What it proves is that $𝐴$ and $𝐵$ are matrices of one linear map with possibly different bases on the domain and the target (Corollary 4.5.21).

The claim is false: $(\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix})$ and $(\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix})$ both have rank $1$ , but their traces are $0$ and $1$ , so they are not similar (Proposition 4.5.27(b)).

Exercise 4.5.46.standard Let $𝐴,$ $𝐵,$ $𝐶,$ $𝐷 \in$ $𝐹^{𝑛 \times 𝑛}$ , $𝑛 \geq 2$ . Prove or disprove.

(a)If $𝐴 \sim 𝐵$ and $𝐶 \sim 𝐷$ , then $𝐴 + 𝐶 \sim 𝐵 + 𝐷$ .
(b)If $𝐴 \sim 𝐵$ , then $𝐴 𝐶 \sim 𝐵 𝐶$ .
(c)If $𝐴 \sim 𝐵$ , then $𝐴 +$ $𝑐 𝐼_{𝑛} \sim$ $𝐵 +$ $𝑐 𝐼_{𝑛}$ for every $𝑐 \in 𝐹$ .
(d)If $𝐴^{2} \sim 𝐵^{2}$ , then $𝐴 \sim 𝐵$ .

Solution.

Uses. Definition 4.5.23 (Similar matrices), Proposition 4.5.27 (Similarity invariants), Lemma 2.4.24 (Matrix units)

Techniques. Finding counterexamples and minimal examples (Technique 1.4.18), Rule out similarity with an invariant (Technique 4.5.29)

Hint 1. Only one statement is true. For counterexamples, use matrix units, and remember that only

0

is similar to

0

and only

𝐼_{𝑛}

𝐼_{𝑛}

Hint 2.

𝑈_{11} \sim 𝑈_{22}

(swap the basis vectors). Try

𝐶 = 𝑈_{11}

in (a) and (b).

Sketch. (a) False:

𝑈_{11} \sim 𝑈_{22}

and

𝑈_{11} \sim 𝑈_{11}

, but

2 𝑈_{11}

(rank

1

0

) is not similar to

𝑈_{11} + 𝑈_{22}

(rank

2

). (b) False:

𝑈_{11} 𝑈_{11} = 𝑈_{11}

but

𝑈_{22} 𝑈_{11} = 0

. (c) True by Proposition 4.5.27(d). (d) False:

𝑈_{12}^{2} = 0 = 0^{2}

We write $𝑈_{𝑖 𝑗}$ for the matrix units (here in $𝐹^{𝑛 \times 𝑛}$ ). Swapping $𝐞_{1}$ and $𝐞_{2}$ in the standard basis shows $𝑈_{11} \sim 𝑈_{22}$ : with $𝑆$ the matrix of the swap, $𝑆^{- 1} 𝑈_{11} 𝑆 = 𝑈_{22}$ .

(a) False. Take $𝐴 = 𝐶 = 𝐷 = 𝑈_{11}$ and $𝐵 = 𝑈_{22}$ . Then $𝐴 + 𝐶 = 2 𝑈_{11}$ , while $𝐵 + 𝐷 = 𝑈_{11} + 𝑈_{22}$ has rank $2$ . If $2 \neq 0$ in $𝐹$ , then $2 𝑈_{11}$ has rank $1$ ; if $2 = 0$ , it is the zero matrix, of rank $0$ . Either way the ranks differ, so the sums are not similar.

(b) False. With $𝐴 = 𝑈_{11}$ , $𝐵 = 𝑈_{22}$ , and $𝐶 = 𝑈_{11}$ , we get $𝐴 𝐶 = 𝑈_{11}$ and $𝐵 𝐶 = 𝑈_{22} 𝑈_{11} = 0$ (Lemma 2.4.24(c)). Only $0$ is similar to $0$ .

(c) True: apply Proposition 4.5.27(d) with $- 𝑐$ in place of $𝑐$ .

(d) False. $𝐴 = 𝑈_{12}$ and $𝐵 = 0$ satisfy $𝐴^{2} = 0 = 𝐵^{2}$ , but $𝐴 \neq 0$ is not similar to $0$ .

Exercise 4.5.47.hard Let

𝑓 : 𝐹^{𝑛 \times 𝑛} \to 𝐹

be a linear map such that

𝑓 (𝑆^{- 1} 𝐴 𝑆) =

𝑓 (𝐴)

for every

𝐴 \in 𝐹^{𝑛 \times 𝑛}

and every invertible

𝑆 \in 𝐹^{𝑛 \times 𝑛}

. Show that

𝑓 = 𝑐 tr

for some

𝑐 \in 𝐹

. That is, up to a scalar factor, the trace is the only linear similarity invariant.

Solution.

Uses. Lemma 2.4.24 (Matrix units), Proposition 2.4.27 (The trace is linear), Definition 2.4.26 (Trace), Proposition 1.6.7 (Standard bases)

Techniques. Cyclicity of trace (Technique 2.4.29), Check it on a basis (Technique 2.2.7)

Hint 1. Apply the hypothesis to

𝐴 = 𝑆 𝑋

: it says

𝑓 (𝑋 𝑆) = 𝑓 (𝑆 𝑋)

for every

𝑋

and every invertible

𝑆

Hint 2. For

𝑘 \neq 𝑙

, the matrix

𝐼_{𝑛} + 𝑈_{𝑘 𝑙}

is invertible, with inverse

𝐼_{𝑛} - 𝑈_{𝑘 𝑙}

. Deduce

𝑓 (𝑋 𝑈_{𝑘 𝑙}) =

𝑓 (𝑈_{𝑘 𝑙} 𝑋)

Hint 3. Take

𝑋 = 𝑈_{𝑖 𝑘}

. Then

𝑋 𝑈_{𝑘 𝑙} = 𝑈_{𝑖 𝑙}

and

𝑈_{𝑘 𝑙} 𝑋 =

𝛿_{𝑙 𝑖} 𝑈_{𝑘 𝑘}

. What does this say about

𝑓 (𝑈_{𝑖 𝑙})

for

𝑖 \neq 𝑙

, and about

𝑓 (𝑈_{𝑙 𝑙})

Sketch. Invariance with

𝐴 = 𝑆 𝑋

gives

𝑓 (𝑋 𝑆) = 𝑓 (𝑆 𝑋)

. With

𝑆 = 𝐼 + 𝑈_{𝑘 𝑙}

(

𝑘 \neq 𝑙

) and linearity,

𝑓 (𝑋 𝑈_{𝑘 𝑙}) =

𝑓 (𝑈_{𝑘 𝑙} 𝑋)

. For

𝑋 = 𝑈_{𝑖 𝑘}

this reads

𝑓 (𝑈_{𝑖 𝑙}) =

𝛿_{𝑙 𝑖} 𝑓 (𝑈_{𝑘 𝑘})

. So

𝑓 (𝑈_{𝑖 𝑙}) = 0

for

𝑖 \neq 𝑙

and

𝑓 (𝑈_{𝑙 𝑙}) =

𝑓 (𝑈_{𝑘 𝑘})

for all

𝑘 \neq 𝑙

. With

𝑐

the common value,

𝑓 (𝐴) =

\sum 𝑎_{𝑖 𝑗} 𝑓 (𝑈_{𝑖 𝑗}) =

𝑐 tr 𝐴

If $𝑛 = 1$ , then $𝑓 (𝑎) =$ $𝑎 𝑓 (1) =$ $𝑓 (1) tr (𝑎)$ by linearity. Let $𝑛 \geq 2$ .

Step 1. For $𝑋 \in 𝐹^{𝑛 \times 𝑛}$ and invertible $𝑆$ , the hypothesis with $𝐴 = 𝑆 𝑋$ gives $𝑓 (𝑆^{- 1} 𝑆 𝑋 𝑆) =$ $𝑓 (𝑆 𝑋)$ , that is,

𝑓 (𝑋 𝑆) = 𝑓 (𝑆 𝑋) .

Step 2. Let $𝑘 \neq 𝑙$ . By Lemma 2.4.24(c), $𝑈_{𝑘 𝑙} 𝑈_{𝑘 𝑙} =$ $𝛿_{𝑙 𝑘} 𝑈_{𝑘 𝑙} =$ $0$ , so $(𝐼_{𝑛} + 𝑈_{𝑘 𝑙}) (𝐼_{𝑛} - 𝑈_{𝑘 𝑙}) =$ $𝐼_{𝑛} -$ $𝑈_{𝑘 𝑙}^{2} =$ $𝐼_{𝑛}$ , and likewise in the other order. So $𝑆 = 𝐼_{𝑛} + 𝑈_{𝑘 𝑙}$ is invertible, and Step 1 gives $𝑓 (𝑋 + 𝑋 𝑈_{𝑘 𝑙}) =$ $𝑓 (𝑋 + 𝑈_{𝑘 𝑙} 𝑋)$ . By linearity,

𝑓 (𝑋 𝑈_{𝑘 𝑙}) = 𝑓 (𝑈_{𝑘 𝑙} 𝑋) for all 𝑋 and all 𝑘 \neq 𝑙 .

Step 3. Take $𝑋 = 𝑈_{𝑖 𝑘}$ . By Lemma 2.4.24(c), $𝑋 𝑈_{𝑘 𝑙} = 𝑈_{𝑖 𝑙}$ and $𝑈_{𝑘 𝑙} 𝑋 =$ $𝛿_{𝑙 𝑖} 𝑈_{𝑘 𝑘}$ , so

𝑓 (𝑈_{𝑖 𝑙}) = 𝛿_{𝑙 𝑖} 𝑓 (𝑈_{𝑘 𝑘}) for all 𝑖 and all 𝑘 \neq 𝑙 .

If $𝑖 \neq 𝑙$ , choose any $𝑘 \neq 𝑙$ (possible since $𝑛 \geq 2$ ); then $𝑓 (𝑈_{𝑖 𝑙}) = 0$ . If $𝑖 = 𝑙$ , we get $𝑓 (𝑈_{𝑙 𝑙}) =$ $𝑓 (𝑈_{𝑘 𝑘})$ for every $𝑘 \neq 𝑙$ . So all $𝑓 (𝑈_{𝑘 𝑘})$ are equal to a single $𝑐 \in 𝐹$ .

Conclusion. For $𝐴 =$ $\sum_{𝑖, 𝑗} 𝑎_{𝑖 𝑗} 𝑈_{𝑖 𝑗}$ (Proposition 1.6.7(b)), linearity gives

𝑓 (𝐴) = \sum_{𝑖, 𝑗} 𝑎_{𝑖 𝑗} 𝑓 (𝑈_{𝑖 𝑗}) = 𝑐 \sum_{𝑖} 𝑎_{𝑖 𝑖} = 𝑐 tr 𝐴 .

Conversely, every $𝑐 tr$ is linear and similarity-invariant (Proposition 2.4.27, Proposition 4.5.27(b)).

Exercise 4.5.48.standard For

𝐴, 𝐵 \in 𝐹^{𝑛 \times 𝑛}

, let

𝑇 \in L (𝐹^{𝑛 \times 𝑛})

𝑇 (𝑋) = 𝐴 𝑋 𝐵

. Show that

tr 𝑇 = (tr 𝐴) (tr 𝐵)

. Deduce the traces of the operators

𝑋 \mapsto 𝐴 𝑋

𝑋 \mapsto 𝑋 𝐵

, and

𝑋 \mapsto 𝐴 𝑋 - 𝑋 𝐴

Solution.

Uses. Definition 4.5.30 (Trace of an operator), Lemma 2.4.24 (Matrix units), Proposition 1.6.7 (Standard bases), Proposition 2.4.12 (Matrices of sums and scalar multiples), Proposition 2.4.27 (The trace is linear)

Techniques. Study a linear map on a space of matrices (Technique 4.2.29)

Hint 1. Use the basis of matrix units. The diagonal entry of

[𝑇]

belonging to

𝑈_{𝑖 𝑗}

is the

𝑈_{𝑖 𝑗}

-coordinate of

𝑇 (𝑈_{𝑖 𝑗})

, which is the

(𝑖, 𝑗)

entry of

𝐴 𝑈_{𝑖 𝑗} 𝐵

Hint 2. Write

𝑈_{𝑖 𝑗} = 𝐞_{𝑖} 𝐞_{𝑗}^{𝖳}

, so the

(𝑖, 𝑗)

entry of

𝐴 𝑈_{𝑖 𝑗} 𝐵

(𝐞_{𝑖}^{𝖳} 𝐴 𝐞_{𝑖}) (𝐞_{𝑗}^{𝖳} 𝐵 𝐞_{𝑗})

Sketch. The

(𝑖, 𝑗)

entry of

𝐴 𝑈_{𝑖 𝑗} 𝐵

𝑎_{𝑖 𝑖} 𝑏_{𝑗 𝑗}

, so

tr 𝑇 =

\sum_{𝑖, 𝑗} 𝑎_{𝑖 𝑖} 𝑏_{𝑗 𝑗} =

(tr 𝐴) (tr 𝐵)

. With

𝐵 = 𝐼_{𝑛}

𝐴 = 𝐼_{𝑛}

the traces are

𝑛 tr 𝐴

and

𝑛 tr 𝐵

, and

𝑋 \mapsto 𝐴 𝑋 - 𝑋 𝐴

has trace

𝑛 tr 𝐴 - 𝑛 tr 𝐴 = 0

In the basis of matrix units $𝑈_{𝑖 𝑗} = 𝐞_{𝑖} 𝐞_{𝑗}^{𝖳}$ (Proposition 1.6.7(b), Lemma 2.4.24), the $𝑈_{𝑖 𝑗}$ -coordinate of a matrix is its $(𝑖, 𝑗)$ entry. So the diagonal entry of $[𝑇]$ in the position of $𝑈_{𝑖 𝑗}$ is the $(𝑖, 𝑗)$ entry of $𝑇 (𝑈_{𝑖 𝑗}) =$ $𝐴 𝐞_{𝑖} 𝐞_{𝑗}^{𝖳} 𝐵$ , which by Lemma 2.4.24(b) is

𝐞_{𝑖}^{𝖳} 𝐴 𝐞_{𝑖} 𝐞_{𝑗}^{𝖳} 𝐵 𝐞_{𝑗} = 𝑎_{𝑖 𝑖} 𝑏_{𝑗 𝑗} .

Summing over all $𝑖$ and $𝑗$ (Definition 4.5.30),

tr 𝑇 = \sum_{𝑖 = 1}^{𝑛} \sum_{𝑗 = 1}^{𝑛} 𝑎_{𝑖 𝑖} 𝑏_{𝑗 𝑗} = (tr 𝐴) (tr 𝐵) .

With $𝐵 = 𝐼_{𝑛}$ , the operator $𝑋 \mapsto 𝐴 𝑋$ has trace $(tr 𝐴) (tr 𝐼_{𝑛}) =$ $𝑛 tr 𝐴$ ; with $𝐴 = 𝐼_{𝑛}$ , $𝑋 \mapsto 𝑋 𝐵$ has trace $𝑛 tr 𝐵$ . The trace of an operator is linear in the operator, by Proposition 2.4.12 and Proposition 2.4.27, so $𝑋 \mapsto 𝐴 𝑋 - 𝑋 𝐴$ has trace $𝑛 tr 𝐴 - 𝑛 tr 𝐴 = 0$ .

Exercise 4.5.49.standard Let $dim 𝑉 = 𝑛 \geq 1$ , and let $𝑃 \in L (𝑉)$ satisfy $𝑃^{2} = 𝑃$ . Let $𝑟 = rank 𝑃$ .

(a)Show that $𝑃$ fixes every vector of $im 𝑃$ , that $ker 𝑃 \cap im 𝑃 = {𝟎}$ , and that a basis of $im 𝑃$ followed by a basis of $ker 𝑃$ is a basis of $𝑉$ .
(b)Deduce that in this basis $[𝑃] =$ $𝑁_{𝑟} =$ $diag (1, \dots, 1, 0, \dots, 0)$ , with $𝑟$ ones, and that $tr 𝑃 = 𝑟 \cdot 1$ .
(c)Show that two matrices $𝐴, 𝐵 \in 𝐹^{𝑛 \times 𝑛}$ with $𝐴^{2} = 𝐴$ and $𝐵^{2} = 𝐵$ are similar if and only if $rank 𝐴 = rank 𝐵$ .

Solution.

Uses. Theorem 4.2.11 (Rank–nullity), Proposition 1.5.14 (Appending a vector), Theorem 4.1.10 (Independent lists are short), Definition 4.5.30 (Trace of an operator), Theorem 4.5.25 (Similar matrices are one operator in two bases), Proposition 4.5.27 (Similarity invariants), Proposition 4.5.24 (Similarity is an equivalence relation)

Techniques. Choose a convenient basis (Technique 4.5.33)

Hint 1. If

𝐯 = 𝑃 𝐰

, compute

𝑃 𝐯

Hint 2. For independence of the combined list, split a relation into a part in

im 𝑃

and a part in

ker 𝑃

. Count with rank–nullity.

Hint 3. For (c), both matrices are similar to the same

𝑁_{𝑟}

Sketch. If

𝐯 = 𝑃 𝐰

then

𝑃 𝐯 = 𝑃^{2} 𝐰 = 𝐯

, so a vector in

ker 𝑃 \cap im 𝑃

equals

𝑃 𝐯 = 𝟎

. A relation among the combined list gives a vector in both subspaces, hence zero, so the list is independent; it has

𝑟 + (𝑛 - 𝑟) = 𝑛

vectors.

𝑃

fixes the first

𝑟

and kills the rest, so its matrix is

𝑁_{𝑟}

and

tr 𝑃 = 𝑟

. In (c),

𝐴 \sim 𝑁_{𝑟} \sim 𝐵

(a) If $𝐯 = 𝑃 𝐰 \in im 𝑃$ , then $𝑃 𝐯 =$ $𝑃^{2} 𝐰 =$ $𝑃 𝐰 =$ $𝐯$ . If also $𝐯 \in ker 𝑃$ , then $𝐯 = 𝑃 𝐯 = 𝟎$ . Let $𝐩_{1}, \dots, 𝐩_{𝑟}$ be a basis of $im 𝑃$ and $𝐤_{1}, \dots, 𝐤_{𝑠}$ a basis of $ker 𝑃$ , where $𝑠 = 𝑛 - 𝑟$ by Theorem 4.2.11. Suppose $\sum_{𝑖} 𝑎_{𝑖} 𝐩_{𝑖} +$ $\sum_{𝑗} 𝑏_{𝑗} 𝐤_{𝑗} =$ $𝟎$ . The vector $𝐱 =$ $\sum_{𝑖} 𝑎_{𝑖} 𝐩_{𝑖} =$ $- \sum_{𝑗} 𝑏_{𝑗} 𝐤_{𝑗}$ lies in $im 𝑃 \cap ker 𝑃 = {𝟎}$ , so all $𝑎_{𝑖}$ and all $𝑏_{𝑗}$ are $0$ by independence of the two bases. So the combined list of $𝑛$ vectors is independent, hence a basis of $𝑉$ (Theorem 4.1.10(b)).

(b) $𝑃 𝐩_{𝑖} = 𝐩_{𝑖}$ and $𝑃 𝐤_{𝑗} = 𝟎$ , so the matrix of $𝑃$ in this basis has columns $𝐞_{1},$ $\dots,$ $𝐞_{𝑟},$ $𝟎,$ $\dots,$ $𝟎$ ; it is $𝑁_{𝑟}$ . Hence $tr 𝑃 = tr 𝑁_{𝑟} = 𝑟 \cdot 1$ (Definition 4.5.30). Over a field of characteristic $𝑝$ , this determines $𝑟$ only modulo $𝑝$ .

(c) If $𝐴 \sim 𝐵$ , then $rank 𝐴 = rank 𝐵$ (Proposition 4.5.27(a)). Conversely, if both have rank $𝑟$ , apply (b) to $𝑇_{𝐴}$ and $𝑇_{𝐵}$ on $𝐹^{𝑛}$ : each has matrix $𝑁_{𝑟}$ in a suitable basis, so $𝐴 \sim 𝑁_{𝑟}$ and $𝐵 \sim 𝑁_{𝑟}$ by Theorem 4.5.25, and $𝐴 \sim 𝐵$ by Proposition 4.5.24.

Exercise 4.5.50.standard Let $dim 𝑉 = 𝑛 \geq 1$ , and let $𝑁 \in L (𝑉)$ satisfy $𝑁^{𝑛} = 0$ and $𝑁^{𝑛 - 1} \neq 0$ . Let $𝐽_{𝑛} \in 𝐹^{𝑛 \times 𝑛}$ be the matrix with $1$ in positions $(2, 1),$ $(3, 2),$ $\dots,$ $(𝑛, 𝑛 - 1)$ and $0$ elsewhere.

(a)Show that ${[𝑁]}_{ℬ︀ ℬ︀} = 𝐽_{𝑛}$ for some basis $ℬ︀$ of $𝑉$ .
(b)Deduce that any two matrices $𝐴, 𝐵 \in 𝐹^{𝑛 \times 𝑛}$ with $𝐴^{𝑛} = 𝐵^{𝑛} = 0$ , $𝐴^{𝑛 - 1} \neq 0$ , and $𝐵^{𝑛 - 1} \neq 0$ are similar.
(c)Show that $𝐽_{𝑛} \sim 𝐽_{𝑛}^{𝖳}$ .

Solution.

Uses. Proposition 4.1.26 (Nilpotent chains are independent), Theorem 4.1.10 (Independent lists are short), Theorem 4.5.25 (Similar matrices are one operator in two bases), Proposition 4.5.24 (Similarity is an equivalence relation), Proposition 2.4.23 (Transpose of a product)

Techniques. Choose a convenient basis (Technique 4.5.33), Apply an operator to a relation to kill terms (Technique 4.1.25)

Hint 1. Choose

𝐯

with

𝑁^{𝑛 - 1} 𝐯 \neq 𝟎

, and look at

𝐯,

𝑁 𝐯,

\dots,

𝑁^{𝑛 - 1} 𝐯

Hint 2. These vectors are independent by Proposition 4.1.26. Where does

𝑁

send each of them?

Hint 3. For (c), check that

𝐽_{𝑛}^{𝖳}

satisfies the hypotheses of (b). The transpose of a power is the power of the transpose.

Sketch. Pick

𝐯

with

𝑁^{𝑛 - 1} 𝐯 \neq 𝟎

; since

𝑁^{𝑛} 𝐯 = 𝟎

, Proposition 4.1.26 makes

𝐯,

𝑁 𝐯,

\dots,

𝑁^{𝑛 - 1} 𝐯

independent, hence a basis.

𝑁

sends each vector to the next and the last to

𝟎

, which is

𝐽_{𝑛}

. (b)

𝑇_{𝐴}

and

𝑇_{𝐵}

both have matrix

𝐽_{𝑛}

. (c)

{(𝐽_{𝑛}^{𝖳})}^{𝑘} =

{(𝐽_{𝑛}^{𝑘})}^{𝖳}

(a) Since $𝑁^{𝑛 - 1} \neq 0$ , there is $𝐯$ with $𝑁^{𝑛 - 1} 𝐯 \neq 𝟎$ , and $𝑁^{𝑛} 𝐯 = 𝟎$ . By Proposition 4.1.26 with $𝑘 = 𝑛 - 1$ , the list $ℬ︀ =$ $𝐯,$ $𝑁 𝐯,$ $\dots,$ $𝑁^{𝑛 - 1} 𝐯$ is independent, and it has $𝑛$ vectors, so it is a basis (Theorem 4.1.10(b)). $𝑁$ sends the $𝑗$ -th vector $𝑁^{𝑗 - 1} 𝐯$ to $𝑁^{𝑗} 𝐯$ , the $(𝑗 + 1)$ -th vector, for $𝑗 < 𝑛$ , and the last vector to $𝑁^{𝑛} 𝐯 = 𝟎$ . So the columns of ${[𝑁]}_{ℬ︀ ℬ︀}$ are $𝐞_{2}, \dots, 𝐞_{𝑛}, 𝟎$ , and ${[𝑁]}_{ℬ︀ ℬ︀} = 𝐽_{𝑛}$ .

(b) $𝑇_{𝐴}$ and $𝑇_{𝐵}$ satisfy the hypotheses of (a) on $𝐹^{𝑛}$ (since ${(𝑇_{𝐴})}^{𝑘} = 𝑇_{𝐴^{𝑘}}$ ). So $𝐴 \sim 𝐽_{𝑛}$ and $𝐵 \sim 𝐽_{𝑛}$ by Theorem 4.5.25, and $𝐴 \sim 𝐵$ (Proposition 4.5.24).

(c) $𝐽_{𝑛} 𝐞_{𝑗} = 𝐞_{𝑗 + 1}$ for $𝑗 < 𝑛$ and $𝐽_{𝑛} 𝐞_{𝑛} = 𝟎$ , so $𝐽_{𝑛}^{𝑛 - 1} 𝐞_{1} =$ $𝐞_{𝑛} \neq$ $𝟎$ and $𝐽_{𝑛}^{𝑛} = 0$ . By Proposition 2.4.23, ${(𝐽_{𝑛}^{𝖳})}^{𝑘} =$ ${(𝐽_{𝑛}^{𝑘})}^{𝖳}$ , so $𝐽_{𝑛}^{𝖳}$ satisfies the same two conditions, and $𝐽_{𝑛} \sim 𝐽_{𝑛}^{𝖳}$ by (b). Concretely, reversing the order of the basis $ℬ︀$ turns $𝐽_{𝑛}$ into $𝐽_{𝑛}^{𝖳}$ .

Exercise 4.5.51.hard Let

𝑛 \geq 2

, let

𝐴 \in 𝐹^{𝑛 \times 𝑛}

have rank

1

, and let

𝑡 = tr 𝐴

. Show that

𝐴 \sim 𝑡 𝑈_{11}

𝑡 \neq 0

, and

𝐴 \sim 𝑈_{21}

𝑡 = 0

. Conclude that two matrices of rank

1

are similar if and only if they have the same trace.

Solution.

Uses. Proposition 4.3.16 (Rank-one matrices), Proposition 2.4.28 (Cyclicity of the trace), Theorem 4.2.11 (Rank–nullity), Theorem 4.1.16 (Extending to a basis), Proposition 1.5.14 (Appending a vector), Theorem 4.1.10 (Independent lists are short), Theorem 4.5.25 (Similar matrices are one operator in two bases), Proposition 4.5.27 (Similarity invariants)

Techniques. Write a rank-one matrix as $𝐮 𝐯^{𝖳}$ (Technique 4.3.17), Choose a convenient basis (Technique 4.5.33)

Hint 1. Write

𝐴 = 𝐮 𝐯^{𝖳}

with

𝐮, 𝐯 \neq 𝟎

. Then

𝐴 𝐱 = (𝐯^{𝖳} 𝐱) 𝐮

, so

ker 𝐴 =

{𝐱 | 𝐯^{𝖳} 𝐱 = 0}

, and

𝑡 = 𝐯^{𝖳} 𝐮

Hint 2. If

𝑡 \neq 0

, then

𝐮 \notin ker 𝐴

; use

𝐮

followed by a basis of

ker 𝐴

. If

𝑡 = 0

, then

𝐮 \in ker 𝐴

; pick

𝐰

with

𝐴 𝐰 = 𝐮

and use

𝐰, 𝐮

, followed by more vectors of

ker 𝐴

Sketch.

𝐴 = 𝐮 𝐯^{𝖳}

𝑡 = 𝐯^{𝖳} 𝐮

ker 𝐴 =

{𝐱 | 𝐯^{𝖳} 𝐱 = 0}

has dimension

𝑛 - 1

. If

𝑡 \neq 0

: in the basis

𝐮

, then a basis of

ker 𝐴

𝐴

has matrix

𝑡 𝑈_{11}

. If

𝑡 = 0

: choose

𝐰

with

𝐯^{𝖳} 𝐰 = 1

, so

𝐴 𝐰 = 𝐮

; extend

𝐮

to a basis of

ker 𝐴

; in the basis

𝐰, 𝐮, \dots

, the matrix is

𝑈_{21}

By Proposition 4.3.16, $𝐴 = 𝐮 𝐯^{𝖳}$ with $𝐮, 𝐯 \neq 𝟎$ . Then $𝐴 𝐱 =$ $𝐮 (𝐯^{𝖳} 𝐱) =$ $(𝐯^{𝖳} 𝐱) 𝐮$ , so $ker 𝐴 =$ ${𝐱 | 𝐯^{𝖳} 𝐱 = 0}$ , which has dimension $𝑛 - 1$ by Theorem 4.2.11. By Proposition 2.4.28, $𝑡 =$ $tr (𝐮 𝐯^{𝖳}) =$ $tr (𝐯^{𝖳} 𝐮) =$ $𝐯^{𝖳} 𝐮$ .

Case $𝑡 \neq 0$ . Then $𝐴 𝐮 = 𝑡 𝐮 \neq 𝟎$ , so $𝐮 \notin ker 𝐴$ . Let $𝐤_{2}, \dots, 𝐤_{𝑛}$ be a basis of $ker 𝐴$ ; by Proposition 1.5.14, $𝐤_{2}, \dots, 𝐤_{𝑛}, 𝐮$ is independent, so (reordering) $𝐮, 𝐤_{2}, \dots, 𝐤_{𝑛}$ is a basis of $𝐹^{𝑛}$ (Theorem 4.1.10(b)). $𝑇_{𝐴}$ sends $𝐮$ to $𝑡 𝐮$ and each $𝐤_{𝑗}$ to $𝟎$ , so its matrix in this basis is $𝑡 𝑈_{11}$ .

Case $𝑡 = 0$ . Then $𝐮 \in ker 𝐴$ . Some entry $𝑣_{𝑖}$ of $𝐯$ is nonzero; let $𝐰 = 𝑣_{𝑖}^{- 1} 𝐞_{𝑖}$ , so $𝐯^{𝖳} 𝐰 = 1$ and $𝐴 𝐰 = 𝐮$ . Extend $𝐮$ to a basis $𝐮, 𝐤_{3}, \dots, 𝐤_{𝑛}$ of $ker 𝐴$ (Theorem 4.1.16). Since $𝐴 𝐰 \neq 𝟎$ , $𝐰 \notin ker 𝐴$ , so as before $𝐰, 𝐮, 𝐤_{3}, \dots, 𝐤_{𝑛}$ is a basis of $𝐹^{𝑛}$ . $𝑇_{𝐴}$ sends $𝐰$ to $𝐮$ , the second basis vector, and the other basis vectors to $𝟎$ . So its matrix is $𝑈_{21}$ .

In both cases Theorem 4.5.25 gives the similarity. Two matrices of rank $1$ with the same trace $𝑡$ are similar to the same matrix, hence to each other; conversely, similar matrices have the same trace (Proposition 4.5.27(b)).

Exercise 4.5.52.standard Let

𝐹

be a field with at least three elements. Show that every

𝐴 \in 𝐹^{𝑛 \times 𝑛}

is the sum of two invertible matrices. Show that this fails for the matrix

(1) \in 𝔽_{2}^{1 \times 1}

Solution.

Uses. Theorem 3.5.27 (Rank normal form), Proposition 2.5.17 (Algebra of inverses), Example 2.5.14 (Invertible maps and matrices)

Techniques. Choose a convenient basis (Technique 4.5.33)

Hint 1. The property “is a sum of two invertible matrices” survives

𝐴 \mapsto 𝐺 𝐴 𝐻

with

𝐺

𝐻

invertible. So assume

𝐴 = 𝑁_{𝑟}

Hint 2. Write

𝑁_{𝑟} =

diag (1, \dots, 1, 0, \dots, 0)

as the sum of two diagonal matrices with nonzero diagonal entries. You need a scalar

𝑐

with

𝑐 \neq 0

and

1 - 𝑐 \neq 0

Sketch. If

𝐴 = 𝐺 𝑁_{𝑟} 𝐻

and

𝑁_{𝑟} = 𝑋 + 𝑌

, then

𝐴 = 𝐺 𝑋 𝐻 + 𝐺 𝑌 𝐻

. Choose

𝑐 \in 𝐹

with

𝑐 \neq 0, 1

, and split

𝑁_{𝑟} =

diag (𝑐, \dots, 𝑐, 1, \dots, 1) +

diag (1 - 𝑐, \dots, 1 - 𝑐, - 1, \dots, - 1)

. Over

𝔽_{2}

, the only invertible

1 \times 1

matrix is

(1)

, and

(1) + (1) = (0)

By Theorem 3.5.27, $𝐺^{'} 𝐴 𝐻^{'} = 𝑁_{𝑟}$ for some invertible $𝐺^{'}$ , $𝐻^{'}$ , where $𝑟 = rank 𝐴$ ; so $𝐴 = 𝐺 𝑁_{𝑟} 𝐻$ with $𝐺 = 𝐺^{' - 1}$ and $𝐻 = 𝐻^{' - 1}$ invertible. If $𝑁_{𝑟} = 𝑋 + 𝑌$ with $𝑋$ and $𝑌$ invertible, then $𝐴 = 𝐺 𝑋 𝐻 + 𝐺 𝑌 𝐻$ , and $𝐺 𝑋 𝐻$ and $𝐺 𝑌 𝐻$ are invertible (Proposition 2.5.17(b)). So it suffices to treat $𝑁_{𝑟}$ .

Choose $𝑐 \in 𝐹$ with $𝑐 \neq 0$ and $𝑐 \neq 1$ , which is possible because $𝐹$ has at least three elements. Let

𝑋 = diag (𝑐, \dots, 𝑐, 1, \dots, 1), 𝑌 = diag (1 - 𝑐, \dots, 1 - 𝑐, - 1, \dots, - 1),

with $𝑟$ entries $𝑐$ (respectively $1 - 𝑐$ ) followed by $𝑛 - 𝑟$ entries $1$ (respectively $- 1$ ). Then $𝑋 + 𝑌 = 𝑁_{𝑟}$ , and $𝑋$ and $𝑌$ are invertible because their diagonal entries are nonzero (Example 2.5.14(d)).

Over $𝔽_{2}$ , the only invertible $1 \times 1$ matrix is $(1)$ , and $(1) + (1) = (0) \neq (1)$ . So $(1)$ is not a sum of two invertible matrices.

Exercise 4.5.53.challenge Show that for

𝑛 \geq 2

, every

𝐴 \in 𝔽_{2}^{𝑛 \times 𝑛}

is the sum of two invertible matrices.

Solution.

Uses. Exercise 4.5.52, Theorem 3.5.27 (Rank normal form), Theorem 2.4.34 (Block multiplication), Proposition 2.5.17 (Algebra of inverses)

Techniques. Choose a convenient basis (Technique 4.5.33), Block decomposition (Technique 2.4.35)

Hint 1. As in Exercise 4.5.52, it suffices to treat

𝑁_{𝑟}

. Over

𝔽_{2}

a diagonal matrix with a

0

on the diagonal is not invertible, so the splitting must use non-diagonal blocks.

Hint 2. Block diagonal matrices with invertible blocks are invertible. Split

𝑁_{𝑟}

into diagonal blocks and decompose each block separately. Which small blocks can you decompose? Try

(0)

𝐼_{2}

diag (1, 0)

, and

𝐼_{3}

Hint 3.

(0) = (1) + (1)

𝐼_{2} = (\begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix}) + (\begin{matrix} 0 & 1 \\ 1 & 1 \end{matrix})

diag (1, 0) =

(\begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix}) +

(\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix})

. For

𝐼_{3}

, try

𝑋

with

𝑋 𝐞_{1} = 𝐞_{3}

𝑋 𝐞_{2} = 𝐞_{1} + 𝐞_{3}

𝑋 𝐞_{3} = 𝐞_{2}

, and

𝑌 = 𝐼_{3} + 𝑋

Sketch. Reduce to

𝑁_{𝑟}

. Decompose the blocks

(0) = (1) + (1)

𝐼_{2}

diag (1, 0)

, and

𝐼_{3}

explicitly over

𝔽_{2}

. Then cut

𝑁_{𝑟}

into diagonal blocks of these types: if

𝑟 = 0

, use

(0)

‘s; if

𝑟 = 1

, use

diag (1, 0)

and

(0)

’s (here

𝑛 \geq 2

is needed); if

𝑟 \geq 2

, write

𝐼_{𝑟}

with

𝐼_{2}

′s and at most one

𝐼_{3}

, and the rest with

(0)

‘s. Block diagonal sums of invertible blocks finish the proof.

As in the solution of Exercise 4.5.52, $𝐴 = 𝐺 𝑁_{𝑟} 𝐻$ with $𝐺$ , $𝐻$ invertible, and it suffices to write $𝑁_{𝑟} = 𝑋 + 𝑌$ with $𝑋$ , $𝑌$ invertible. All arithmetic is in $𝔽_{2}$ , where $1 + 1 = 0$ .

Block diagonal matrices. If $𝐷 =$ $diag (𝐷_{1}, \dots, 𝐷_{𝑠})$ is block diagonal and $𝐷_{𝑖} = 𝑋_{𝑖} + 𝑌_{𝑖}$ with $𝑋_{𝑖}$ , $𝑌_{𝑖}$ invertible, then $𝐷 =$ $diag (𝑋_{1}, \dots, 𝑋_{𝑠}) +$ $diag (𝑌_{1}, \dots, 𝑌_{𝑠})$ , and $diag (𝑋_{1}, \dots, 𝑋_{𝑠})$ is invertible with inverse $diag (𝑋_{1}^{- 1}, \dots, 𝑋_{𝑠}^{- 1})$ by Theorem 2.4.34; likewise for the $𝑌_{𝑖}$ .

Small blocks. Over $𝔽_{2}$ :

$(0) = (1) + (1)$ .
$𝐼_{2} = (\begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix}) + (\begin{matrix} 0 & 1 \\ 1 & 1 \end{matrix})$ , and both summands are invertible: $(\begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix})$ has inverse $(\begin{matrix} 0 & 1 \\ 1 & 1 \end{matrix})$ (multiply to check), and vice versa.
$diag (1, 0) =$ $(\begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix}) +$ $(\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix})$ , and $(\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix})$ is its own inverse.
$𝐼_{3} = 𝑋_{3} + 𝑌_{3}$ with
$𝑋_{3} = (\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 1 & 0 \end{matrix}), 𝑌_{3} = 𝐼_{3} + 𝑋_{3} = (\begin{matrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 1 & 1 \end{matrix}) .$
$𝑋_{3}$ is invertible: its columns $𝐞_{3}$ , $𝐞_{1} + 𝐞_{3}$ , $𝐞_{2}$ span $𝔽_{2}^{3}$ , since $𝐞_{1} = (𝐞_{1} + 𝐞_{3}) + 𝐞_{3}$ . $𝑌_{3}$ is invertible: if $𝑌_{3} (𝑥, 𝑦, 𝑧) =$ $(𝑥 + 𝑦, 𝑦 + 𝑧, 𝑥 + 𝑦 + 𝑧) =$ $𝟎$ , then $𝑥 = 𝑦 = 𝑧$ (from the first two entries) and the third entry is $3 𝑥 = 𝑥 = 0$ .

Cutting $𝑁_{𝑟}$ into blocks. If $𝑟 = 0$ , then $𝑁_{𝑟}$ is a block diagonal matrix of $𝑛$ blocks $(0)$ . If $𝑟 = 1$ , then since $𝑛 \geq 2$ , $𝑁_{1} =$ $diag (diag (1, 0), (0), \dots, (0))$ . If $𝑟 \geq 2$ , write $𝑟 = 2 𝑎$ or $𝑟 = 2 𝑎 + 3$ with $𝑎 \geq 0$ ; then $𝐼_{𝑟}$ is block diagonal with $𝑎$ blocks $𝐼_{2}$ and, if $𝑟$ is odd, one block $𝐼_{3}$ , and $𝑁_{𝑟} =$ $diag (𝐼_{𝑟}, (0), \dots, (0))$ . In every case each block has a decomposition from the list above, and the block diagonal step gives $𝑁_{𝑟} = 𝑋 + 𝑌$ with $𝑋$ and $𝑌$ invertible.

Exercise 4.5.54.standard This exercise shows what goes wrong in infinite dimensions.

(a)Let $𝑥 𝐹 [𝑥]$ be the subspace of polynomials in $𝐹 [𝑥]$ whose constant coefficient is $0$ . Show that $𝑆 : 𝐹 [𝑥] \to$ $𝑥 𝐹 [𝑥]$ , $𝑆 (𝑝) = 𝑥 𝑝$ , is an isomorphism. So $𝐹 [𝑥]$ is isomorphic to a proper subspace of itself, and $𝑆$ , regarded as an operator on $𝐹 [𝑥]$ , is injective but not surjective.
(b)Explain why a finite-dimensional space is never isomorphic to a proper subspace of itself.
(c)Show that differentiation $𝐷 \in L (ℝ [𝑥])$ is surjective but not injective.
(d)Show that if $𝑉 ≅ 𝑊$ and $𝑉$ is infinite-dimensional, then so is $𝑊$ .

Solution.

Uses. Definition 1.2.13 (Polynomials), Proposition 2.7.7 (Bijective linear maps are invertible), Proposition 4.5.5 (Isomorphisms preserve linear structure), Proposition 4.1.21 (Dimensions of subspaces), Example 2.1.9 (Maps of polynomials)

Hint 1. (a) In terms of coefficient sequences,

𝑆

shifts every coefficient up by one place.

Hint 2. (b) Compare dimensions. (c) Every polynomial has an antiderivative; which polynomials have derivative

0

? (d) Contrapositive of Proposition 4.5.5(c).

Sketch. (a)

𝑆 (𝑎_{0} + 𝑎_{1} 𝑥 + \dots + 𝑎_{𝑑} 𝑥^{𝑑}) =

𝑎_{0} 𝑥 +

\dots +

𝑎_{𝑑} 𝑥^{𝑑 + 1}

is linear, injective, and onto

𝑥 𝐹 [𝑥]

;

1 \notin 𝑥 𝐹 [𝑥]

. (b) A subspace isomorphic to

𝑉

has dimension

dim 𝑉

, so it is

𝑉

. (c)

𝐷 (\sum 𝑎_{𝑘} 𝑥^{𝑘 + 1} / (𝑘 + 1)) =

\sum 𝑎_{𝑘} 𝑥^{𝑘}

, and

𝐷 (1) = 0

. (d) If

𝑊

were finite-dimensional, so would

𝑉

be.

(a) For $𝑝 =$ $𝑎_{0} +$ $𝑎_{1} 𝑥 +$ $\dots +$ $𝑎_{𝑑} 𝑥^{𝑑}$ , $𝑆 (𝑝) =$ $𝑎_{0} 𝑥 +$ $𝑎_{1} 𝑥^{2} +$ $\dots +$ $𝑎_{𝑑} 𝑥^{𝑑 + 1}$ , whose coefficient sequence is that of $𝑝$ shifted by one place. So $𝑆$ is linear, and $𝑆 (𝑝) = 𝟎$ forces all $𝑎_{𝑘} = 0$ , so $𝑆$ is injective. Every $𝑞 =$ $𝑏_{1} 𝑥 +$ $\dots +$ $𝑏_{𝑒} 𝑥^{𝑒} \in$ $𝑥 𝐹 [𝑥]$ is $𝑆 (𝑏_{1} + 𝑏_{2} 𝑥 + \dots + 𝑏_{𝑒} 𝑥^{𝑒 - 1})$ . So $𝑆$ is a linear bijection onto $𝑥 𝐹 [𝑥]$ , an isomorphism (Proposition 2.7.7). The subspace $𝑥 𝐹 [𝑥]$ is proper, since $1 \notin 𝑥 𝐹 [𝑥]$ ; so $𝑆$ is not surjective as a map $𝐹 [𝑥] \to 𝐹 [𝑥]$ .

(b) If $𝑈$ is a subspace of a finite-dimensional $𝑉$ and $𝑈 ≅ 𝑉$ , then $dim 𝑈 = dim 𝑉$ (Proposition 4.5.5(c)), so $𝑈 = 𝑉$ by Proposition 4.1.21(b).

(c) $𝐷 (𝑎_{0} + 𝑎_{1} 𝑥 + \dots + 𝑎_{𝑑} 𝑥^{𝑑}) =$ $𝑎_{1} +$ $2 𝑎_{2} 𝑥 +$ $\dots +$ $𝑑 𝑎_{𝑑} 𝑥^{𝑑 - 1}$ (Example 2.1.9). Given $𝑞 =$ $\sum_{𝑘 = 0}^{𝑒} 𝑏_{𝑘} 𝑥^{𝑘} \in$ $ℝ [𝑥]$ , the polynomial $𝑝 =$ $\sum_{𝑘 = 0}^{𝑒} \frac{𝑏_{𝑘}}{𝑘 + 1} 𝑥^{𝑘 + 1}$ satisfies $𝐷 (𝑝) = 𝑞$ , so $𝐷$ is surjective. But $𝐷 (1) = 0$ with $1 \neq 0$ , so $𝐷$ is not injective. Rank–nullity forbids both phenomena for operators on finite-dimensional spaces (Corollary 4.2.18).

(d) If $𝑊$ were finite-dimensional, then $𝑉$ , being isomorphic to $𝑊$ , would be finite-dimensional by Proposition 4.5.5(c), applied to an isomorphism $𝑊 \to 𝑉$ .

Exercise 4.5.55.standard Let $𝑅 = (\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix})$ , the rotation by $𝜋 / 2$ .

(a)Show that $𝑅$ is similar over $ℂ$ to $diag (𝑖, - 𝑖)$ , by finding $𝑆 \in ℂ^{2 \times 2}$ .
(b)Show that $𝑅$ is not similar over $ℝ$ to any diagonal matrix in $ℝ^{2 \times 2}$ .
(c)Show that $𝑅$ is similar over $ℝ$ to $- 𝑅$ , and interpret the similarity geometrically.

Solution.

Uses. Theorem 4.5.25 (Similar matrices are one operator in two bases), Definition 4.5.23 (Similar matrices), Proposition 2.5.28 (Inverse of a $2 \times 2$ matrix)

Techniques. Choose a convenient basis (Technique 4.5.33)

Hint 1. (a) Look for

𝐯 \in ℂ^{2}

with

𝑅 𝐯 = 𝑖 𝐯

and

𝐰

with

𝑅 𝐰 = - 𝑖 𝐰

. Then use them as the columns of

𝑆

Hint 2. (b) If

𝑆^{- 1} 𝑅 𝑆 =

diag (𝑝, 𝑞)

with

𝑆

real, then the first column

𝐬

𝑆

satisfies

𝑅 𝐬 = 𝑝 𝐬

. Write this out.

Hint 3. (c) Try

𝑆 = diag (1, - 1)

, a reflection.

Sketch. (a)

𝑅 (1, - 𝑖) =

(𝑖, 1) =

𝑖 (1, - 𝑖)

and

𝑅 (1, 𝑖) =

- 𝑖 (1, 𝑖)

, so

𝑆 = (\begin{matrix} 1 & 1 \\ - 𝑖 & 𝑖 \end{matrix})

works. (b)

𝑅 (𝑥, 𝑦) =

𝑝 (𝑥, 𝑦)

means

- 𝑦 = 𝑝 𝑥

and

𝑥 = 𝑝 𝑦

, so

(1 + 𝑝^{2}) 𝑥 = 0

, giving

𝑥 = 𝑦 = 0

; but columns of an invertible

𝑆

are nonzero. (c)

diag (1, - 1) 𝑅 diag (1, - 1) =

- 𝑅

: conjugating a rotation by a reflection reverses its direction.

(a) $𝑅 (𝑥, 𝑦) =$ $(- 𝑦, 𝑥)$ . So $𝑅 (1, - 𝑖) =$ $(𝑖, 1) =$ $𝑖 (1, - 𝑖)$ and $𝑅 (1, 𝑖) =$ $(- 𝑖, 1) =$ $- 𝑖 (1, 𝑖)$ . Let $𝑆 = (\begin{matrix} 1 & 1 \\ - 𝑖 & 𝑖 \end{matrix})$ ; it is invertible, since $1 \cdot$ $𝑖 -$ $1 \cdot$ $(- 𝑖) =$ $2 𝑖 \neq$ $0$ (Proposition 2.5.28). In the basis of its columns, $𝑇_{𝑅}$ has matrix $diag (𝑖, - 𝑖)$ , so $𝑆^{- 1} 𝑅 𝑆 =$ $diag (𝑖, - 𝑖)$ (Theorem 4.5.25). Equivalently, $𝑅 𝑆 =$ $(\begin{matrix} 𝑖 & - 𝑖 \\ 1 & 1 \end{matrix}) =$ $𝑆 diag (𝑖, - 𝑖)$ .

(b) Suppose $𝑆^{- 1} 𝑅 𝑆 =$ $diag (𝑝, 𝑞)$ with $𝑆 \in ℝ^{2 \times 2}$ invertible and $𝑝, 𝑞 \in ℝ$ . Then $𝑅 𝑆 =$ $𝑆 diag (𝑝, 𝑞)$ , and comparing first columns, $𝑅 𝐬 = 𝑝 𝐬$ for the first column $𝐬 = (𝑥, 𝑦)$ of $𝑆$ . That is, $- 𝑦 = 𝑝 𝑥$ and $𝑥 = 𝑝 𝑦$ , so $𝑥 = 𝑝 𝑦 = - 𝑝^{2} 𝑥$ and $(1 + 𝑝^{2}) 𝑥 = 0$ . Since $1 + 𝑝^{2} > 0$ , $𝑥 = 0$ , and then $𝑦 = - 𝑝 𝑥 = 0$ . But $𝐬 \neq 𝟎$ , because $𝑆$ is invertible. So no such $𝑆$ exists, and $𝑅$ is not similar over $ℝ$ to a real diagonal matrix.

(c) With $𝑆 =$ $diag (1, - 1) =$ $𝑆^{- 1}$ ,

𝑆^{- 1} 𝑅 𝑆 = (\begin{matrix} 1 & 0 \\ 0 & - 1 \end{matrix}) (\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix}) (\begin{matrix} 1 & 0 \\ 0 & - 1 \end{matrix}) = (\begin{matrix} 0 & 1 \\ - 1 & 0 \end{matrix}) = - 𝑅 .

Geometrically, $- 𝑅$ is the rotation by $- 𝜋 / 2$ , and the similarity is a change of basis from $𝐞_{1}, 𝐞_{2}$ to $𝐞_{1}, - 𝐞_{2}$ , which reverses the orientation of the plane. A counterclockwise quarter turn, viewed in a mirror, is a clockwise one. So over $ℝ$ , a rotation and its inverse are similar, although they are different operators.

Exercise 4.5.56.routine Let

𝑇 \in L (𝐹^{2})

satisfy

𝑇 (1, 2) = (3, 6)

and

𝑇 (1, 1) = (- 1, - 1)

. Use Theorem 4.5.17 to find

{[𝑇]}_{ℰ︀ ℰ︀}

. Check your answer, and compute

tr 𝑇

in two ways.

Solution.

Uses. Theorem 4.5.17 (Change of basis), Definition 4.5.30 (Trace of an operator), Proposition 2.5.28 (Inverse of a $2 \times 2$ matrix)

Hint 1. In the basis

𝒰︀ = (1, 2), (1, 1)

, the matrix of

𝑇

is diagonal. Which diagonal matrix?

Hint 2.

{[𝑇]}_{ℰ︀ ℰ︀} =

𝑄 {[𝑇]}_{𝒰︀ 𝒰︀} 𝑄^{- 1}

with

𝑄 =

{[id]}_{ℰ︀ 𝒰︀} =

(\begin{matrix} 1 & 1 \\ 2 & 1 \end{matrix})

Sketch.

{[𝑇]}_{𝒰︀ 𝒰︀} =

diag (3, - 1)

𝑄^{- 1} = (\begin{matrix} - 1 & 1 \\ 2 & - 1 \end{matrix})

, and

{[𝑇]}_{ℰ︀ ℰ︀} = (\begin{matrix} - 5 & 4 \\ - 8 & 7 \end{matrix})

. The trace is

2

either way.

$𝑇 (1, 2) = 3 (1, 2)$ and $𝑇 (1, 1) = - (1, 1)$ , so in the basis $𝒰︀ = (1, 2), (1, 1)$ (a basis over every field, since $1 \cdot 1 - 1 \cdot 2 = - 1 \neq 0$ ), ${[𝑇]}_{𝒰︀ 𝒰︀} =$ $diag (3, - 1)$ . With $𝑄 =$ ${[id]}_{ℰ︀ 𝒰︀} =$ $(\begin{matrix} 1 & 1 \\ 2 & 1 \end{matrix})$ and $𝑄^{- 1} =$ ${[id]}_{𝒰︀ ℰ︀} =$ $(\begin{matrix} - 1 & 1 \\ 2 & - 1 \end{matrix})$ (Proposition 2.5.28), Theorem 4.5.17 gives

{[𝑇]}_{ℰ︀ ℰ︀} = 𝑄 (\begin{matrix} 3 & 0 \\ 0 & - 1 \end{matrix}) 𝑄^{- 1} = (\begin{matrix} 3 & - 1 \\ 6 & - 1 \end{matrix}) (\begin{matrix} - 1 & 1 \\ 2 & - 1 \end{matrix}) = (\begin{matrix} - 5 & 4 \\ - 8 & 7 \end{matrix}) .

Check: $(\begin{matrix} - 5 & 4 \\ - 8 & 7 \end{matrix}) (\begin{matrix} 1 \\ 2 \end{matrix}) = (\begin{matrix} 3 \\ 6 \end{matrix})$ and $(\begin{matrix} - 5 & 4 \\ - 8 & 7 \end{matrix}) (\begin{matrix} 1 \\ 1 \end{matrix}) =$ $(\begin{matrix} - 1 \\ - 1 \end{matrix})$ . The trace is $- 5 + 7 = 2$ from the standard matrix and $3 + (- 1) = 2$ from the diagonal one, as Definition 4.5.30 requires.

More practice

The practice bank has 29 exercises on this section’s topics.