7.2 Eigenvalues, eigenvectors, and invariant subspaces

A diagonal matrix $𝐷 =$ $diag (𝑑_{1}, \dots, 𝑑_{𝑛})$ is the simplest kind of operator: it multiplies the $𝑖$ -th coordinate by $𝑑_{𝑖}$ , so it sends each standard basis vector $𝐞_{𝑖}$ to the multiple $𝑑_{𝑖} 𝐞_{𝑖}$ of itself. A general operator mixes directions, but it may still have some directions that it only stretches, that is, nonzero vectors $𝐯$ with $𝑇 𝐯 = 𝜆 𝐯$ . If $𝑉$ has a basis of such vectors, the matrix of $𝑇$ in that basis is diagonal, and questions about $𝑇$ (its powers, its kernel, its long-run behavior) become questions about the scalars $𝜆$ . This section introduces these vectors, the eigenvectors, together with the more general subspaces that an operator maps into themselves. Computing eigenvalues is the subject of Section 7.3, and the question of when a basis of eigenvectors exists is the subject of Section 7.4.

Throughout, $𝑉$ is a vector space over a field $𝐹$ , possibly infinite-dimensional unless we say otherwise, and $𝑇 \in L (𝑉)$ is an operator on $𝑉$ (Notation 2.1.2). We write $𝑇 𝐯$ for $𝑇 (𝐯)$ when no confusion can arise.

7.2.1Invariant subspaces

To understand an operator, we would like to cut $𝑉$ into pieces that $𝑇$ does not mix. The minimal requirement on a piece $𝑈$ is that $𝑇$ does not move vectors of $𝑈$ out of $𝑈$ . Then $𝑇$ restricts to an operator on $𝑈$ , which we can study on its own.

Definition 7.2.1 (Invariant subspace). Let

𝑇 \in L (𝑉)

. A subspace

𝑈

𝑉

is invariant under $𝑇$ (or $𝑇$ -invariant) if

𝑇 𝐮 \in 𝑈

for every

𝐮 \in 𝑈

, that is, if

im (𝑇 ↾_{𝑈}) \subseteq 𝑈

. In that case we regard

𝑇 ↾_{𝑈}

as an operator on

𝑈

𝑇 ↾_{𝑈} \in L (𝑈)

Example 7.2.2 (Invariant subspaces).

(a)The subspaces ${𝟎}$ and $𝑉$ are invariant under every operator on $𝑉$ .
(b) $ker 𝑇$ and $im 𝑇$ are $𝑇$ -invariant. If $𝐮 \in ker 𝑇$ , then $𝑇 𝐮 = 𝟎 \in ker 𝑇$ ; and $𝑇 𝐮 \in im 𝑇$ for every $𝐮 \in 𝑉$ , in particular for every $𝐮 \in im 𝑇$ .
(c)Every subspace is invariant under a scaling map $𝜆 {id}_{𝑉}$ , since $𝜆 𝐮 \in 𝑈$ whenever $𝐮 \in 𝑈$ .
(d)Differentiation $𝐷$ on $𝐹 [𝑥]$ (Example 2.1.9) lowers degrees, so for each $𝑛 \geq 0$ the subspace ${𝐹 [𝑥]}_{\leq 𝑛}$ is $𝐷$ -invariant. Multiplication by $𝑥$ , $𝑀 (𝑝) = 𝑥 𝑝$ , raises degrees, and ${𝐹 [𝑥]}_{\leq 𝑛}$ is not $𝑀$ -invariant, as $𝑀 (𝑥^{𝑛}) = 𝑥^{𝑛 + 1}$ . The subspace ${𝑝 | 𝑝 (0) = 0}$ is $𝑀$ -invariant, since $(𝑥 𝑝) (0) = 0$ for every $𝑝$ .
(e)The cyclic permutation $𝑍 (𝑥, 𝑦, 𝑧) =$ $(𝑧, 𝑥, 𝑦)$ of Example 4.5.37 leaves the line spanned by $(1, 1, 1)$ and the plane $𝑥 + 𝑦 + 𝑧 = 0$ invariant.

Non-example 7.2.3 (A line that moves). Let

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

, the shear

(𝑥, 𝑦) \mapsto

(𝑥 + 𝑦, 𝑦)

𝐹^{2}

. Then

𝐴 𝐞_{1} = 𝐞_{1}

, so the line

span (𝐞_{1})

is invariant. But

𝐴 𝐞_{2} =

(1, 1) \notin

span (𝐞_{2})

, so the line

span (𝐞_{2})

is not, although it is a complement of an invariant line. Invariance is a property of a subspace relative to

𝑇

, and complements of invariant subspaces need not inherit it.

In the last two examples we checked invariance only on spanning vectors. That is always enough.

Proposition 7.2.4 (Invariance is checked on a spanning list). Let

𝑇 \in L (𝑉)

and

𝑈 =

span (𝐮_{1}, \dots, 𝐮_{𝑘})

. Then

𝑈

𝑇

-invariant if and only if

𝑇 𝐮_{𝑖} \in 𝑈

for

𝑖 = 1, \dots, 𝑘

Proof.routine

Uses. Definition 7.2.1 (Invariant subspace), Definition 1.3.10 (Span), Definition 2.1.1 (Linear map), Proposition 1.4.14 (The span is the smallest subspace containing a list)

Techniques. Check it on a basis (Technique 2.2.7)

Hint 1. One direction is immediate. For the other, write an arbitrary

𝐮 \in 𝑈

as a combination of the

𝐮_{𝑖}

Hint 2. Apply

𝑇

to the combination and use that

𝑈

is closed under linear combinations.

Sketch. If

𝑈

is invariant, then in particular

𝑇 𝐮_{𝑖} \in 𝑈

. Conversely,

𝑇 (\sum 𝑎_{𝑖} 𝐮_{𝑖}) =

\sum 𝑎_{𝑖} 𝑇 𝐮_{𝑖}

is a linear combination of vectors of the subspace

𝑈

, so it lies in

𝑈

If $𝑈$ is $𝑇$ -invariant, then $𝑇 𝐮_{𝑖} \in 𝑈$ because $𝐮_{𝑖} \in 𝑈$ . Conversely, suppose $𝑇 𝐮_{𝑖} \in 𝑈$ for each $𝑖$ , and let $𝐮 \in 𝑈$ . By Definition 1.3.10, $𝐮 =$ $𝑎_{1} 𝐮_{1} +$ $\dots +$ $𝑎_{𝑘} 𝐮_{𝑘}$ for some $𝑎_{𝑖} \in 𝐹$ , and by linearity

𝑇 𝐮 = 𝑎_{1} 𝑇 𝐮_{1} + \dots + 𝑎_{𝑘} 𝑇 𝐮_{𝑘} .

This is a linear combination of vectors of the subspace $𝑈$ , so it lies in $𝑈$ (Proposition 1.4.14). □

Remark 7.2.5 (Invariant subspaces and block matrices). Suppose $𝑉$ is finite-dimensional. By Proposition 4.5.35, a subspace $𝑈$ is $𝑇$ -invariant exactly when, in a basis of $𝑉$ that starts with a basis of $𝑈$ , the matrix of $𝑇$ has the block form

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

and then $𝐴$ is the matrix of $𝑇 ↾_{𝑈}$ . If $𝑉 = 𝑈 \oplus 𝑊$ with both $𝑈$ and $𝑊$ invariant, then $𝐵 = 0$ as well, and $𝑇$ splits into the two operators $𝑇 ↾_{𝑈}$ and $𝑇 ↾_{𝑊}$ (Technique 4.5.36). So invariant subspaces are what make matrices block triangular, and invariant complements are what make them block diagonal. The smallest nonzero blocks have size $1 \times 1$ , and they come from invariant lines.

7.2.2Eigenvalues and eigenvectors

Intuition (Invariant lines). A line

span (𝐯)

, with

𝐯 \neq 𝟎

, is

𝑇

-invariant exactly when

𝑇 𝐯 \in span (𝐯)

(Proposition 7.2.4), that is, when

𝑇 𝐯 = 𝜆 𝐯

for some scalar

𝜆

. On such a line

𝑇

acts as multiplication by

𝜆

𝑇 (𝑐 𝐯) =

𝑐 𝜆 𝐯 =

𝜆 (𝑐 𝐯)

. So the one-dimensional invariant subspaces are the directions that

𝑇

stretches (by

𝜆

, possibly

0

or negative) without turning.

Definition 7.2.6 (Eigenvalue, eigenvector). Let $𝑇 \in L (𝑉)$ . A scalar $𝜆 \in 𝐹$ is an eigenvalue of $𝑇$ if there is a vector $𝐯 \in 𝑉$ with

𝐯 \neq 𝟎 and 𝑇 𝐯 = 𝜆 𝐯 .

Such a vector $𝐯$ is an eigenvector of $𝑇$ with eigenvalue $𝜆$ (or an eigenvector for $𝜆$ ). For $𝐴 \in 𝐹^{𝑛 \times 𝑛}$ , the eigenvalues and eigenvectors of $𝐴$ are those of $𝑇_{𝐴}$ : nonzero $𝐯 \in 𝐹^{𝑛}$ with $𝐴 𝐯 = 𝜆 𝐯$ .

Three remarks on the definition. First, eigenvectors are nonzero by definition: $𝑇 𝟎 = 𝜆 𝟎$ holds for every $𝜆$ , so allowing $𝟎$ would make every scalar an eigenvalue. The eigenvalue $0$ is allowed, and it occurs exactly when $ker 𝑇 \neq {𝟎}$ . Second, an eigenvector determines its eigenvalue: if $𝜆 𝐯 = 𝜇 𝐯$ with $𝐯 \neq 𝟎$ , then $(𝜆 - 𝜇) 𝐯 = 𝟎$ , so $𝜆 = 𝜇$ by Proposition 1.2.35. Third, eigenvalues are elements of $𝐹$ . The same matrix can have eigenvalues over $ℂ$ and none over $ℝ$ (Example 7.2.18), so the field is part of the question.

Proposition 7.2.7 (Invariant lines are spanned by eigenvectors). Let

𝑇 \in L (𝑉)

and

𝐯 \in 𝑉

with

𝐯 \neq 𝟎

. Then

span (𝐯)

𝑇

-invariant if and only if

𝐯

is an eigenvector of

𝑇

. In that case every nonzero vector of

span (𝐯)

is an eigenvector of

𝑇

with the same eigenvalue.

Proof.routine

Uses. Proposition 7.2.4 (Invariance is checked on a spanning list), Definition 7.2.6 (Eigenvalue, eigenvector), Definition 1.3.10 (Span)

Hint 1. Apply Proposition 7.2.4 to the list consisting of

𝐯

alone.

Hint 2. For the last claim, compute

𝑇 (𝑐 𝐯)

Sketch. By Proposition 7.2.4,

span (𝐯)

is invariant iff

𝑇 𝐯 \in

span (𝐯) =

{𝑐 𝐯 | 𝑐 \in 𝐹}

, which says

𝑇 𝐯 = 𝜆 𝐯

for some

𝜆

. Then

𝑇 (𝑐 𝐯) = 𝜆 (𝑐 𝐯)

By Proposition 7.2.4,

span (𝐯)

𝑇

-invariant if and only if

𝑇 𝐯 \in

span (𝐯) =

{𝑐 𝐯 | 𝑐 \in 𝐹}

, that is, if and only if

𝑇 𝐯 = 𝜆 𝐯

for some

𝜆 \in 𝐹

. Since

𝐯 \neq 𝟎

, this says that

𝐯

is an eigenvector. If

𝑇 𝐯 = 𝜆 𝐯

and

𝑐 \neq 0

, then

𝑐 𝐯 \neq 𝟎

and

𝑇 (𝑐 𝐯) =

𝑐 𝜆 𝐯 =

𝜆 (𝑐 𝐯)

. □

Example 7.2.8 (Eigenvectors of a symmetric $2 \times 2$ matrix). Let $𝐴 = (\begin{matrix} 2 & 1 \\ 1 & 2 \end{matrix}) \in ℝ^{2 \times 2}$ . Then

𝐴 (\begin{matrix} 1 \\ 1 \end{matrix}) = (\begin{matrix} 3 \\ 3 \end{matrix}) = 3 (\begin{matrix} 1 \\ 1 \end{matrix}), 𝐴 (\begin{matrix} 1 \\ - 1 \end{matrix}) = (\begin{matrix} 1 \\ - 1 \end{matrix}), 𝐴 (\begin{matrix} 1 \\ 0 \end{matrix}) = (\begin{matrix} 2 \\ 1 \end{matrix}) .

So $(1, 1)$ is an eigenvector with eigenvalue $3$ , $(1, - 1)$ is an eigenvector with eigenvalue $1$ , and $𝐞_{1}$ is not an eigenvector, since $(2, 1)$ is not a multiple of $(1, 0)$ . The lines $𝑦 = 𝑥$ and $𝑦 = - 𝑥$ are invariant. In the basis $(1, 1), (1, - 1)$ the matrix of $𝑇_{𝐴}$ is $diag (3, 1)$ : $𝐴$ stretches the first diagonal by $3$ and fixes the second pointwise. We will see in Corollary 7.2.22 that $𝐴$ has no other eigenvalues.

The map of

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

. The dashed lines

𝑦 = 𝑥

and

𝑦 = - 𝑥

are invariant:

𝐴 (1, 1) = (3, 3)

, while

(1, - 1)

is fixed. The vector

𝐞_{1}

is turned toward the diagonal. In the interactive version, drag

𝐴 𝐞_{1}

and

𝐴 𝐞_{2}

and look for lines through

𝟎

that are mapped to themselves. The far corner of the image of the unit square is

𝐴 (1, 1) =

𝐴 𝐞_{1} +

𝐴 𝐞_{2}

, so the diagonal

𝑦 = 𝑥

is invariant exactly when that corner stays on it, that is, when the two rows of

𝐴

have the same sum. The

𝑥

-axis is invariant exactly when

𝐴 𝐞_{1}

lies on it. Try a rotation by an angle that is not a multiple of

𝜋

, which leaves no line in place, and a shear, which keeps exactly one.

To find all eigenvalues, rewrite $𝑇 𝐯 = 𝜆 𝐯$ as $(𝑇 - 𝜆 id) 𝐯 = 𝟎$ , which turns the search for eigenvalues into a question about kernels.

Theorem 7.2.9 (Eigenvalues and invertibility). Let $𝑇 \in L (𝑉)$ and $𝜆 \in 𝐹$ , and consider the conditions

(I) $𝜆$ is an eigenvalue of $𝑇$ ;
(II) $ker (𝑇 - 𝜆 id) \neq {𝟎}$ , that is, $𝑇 - 𝜆 id$ is not injective;
(III) $𝑇 - 𝜆 id$ is not surjective;
(IV) $𝑇 - 𝜆 id$ is not invertible.

Then (I) and (II) are equivalent. If $𝑉$ is finite-dimensional, all four are equivalent. In particular, $𝜆$ is an eigenvalue of $𝐴 \in 𝐹^{𝑛 \times 𝑛}$ if and only if $𝐴 - 𝜆 𝐼_{𝑛}$ is not invertible, if and only if $rank (𝐴 - 𝜆 𝐼_{𝑛}) <$ $𝑛$ , if and only if $det (𝐴 - 𝜆 𝐼_{𝑛}) = 0$ .

Proof.routine

Uses. Definition 7.2.6 (Eigenvalue, eigenvector), Definition 2.1.21 (Pointwise operations), Theorem 2.6.11 (Injectivity is decided at $𝟎$ ), Corollary 4.2.18 (Injective if and only if surjective), Proposition 2.7.7 (Bijective linear maps are invertible), Corollary 3.5.25 (The invertible matrix theorem), Theorem 3.5.12 (First Fundamental Theorem, completed), Theorem 5.4.7 (Invertibility and the determinant)

Hint 1. Rewrite

𝑇 𝐯 = 𝜆 𝐯

using the operator

𝑇 - 𝜆 id

Hint 2. For (I)

\Leftrightarrow

(II), use Theorem 2.6.11. In finite dimensions, Corollary 4.2.18 compares (II), (III), and (IV).

Hint 3. For matrices,

𝑇_{𝐴 - 𝜆 𝐼_{𝑛}} =

𝑇_{𝐴} -

𝜆 id

. Translate “not injective” with Corollary 3.5.25 and Theorem 5.4.7.

Sketch.

(𝑇 - 𝜆 id) 𝐯 =

𝑇 𝐯 -

𝜆 𝐯

, so the eigenvectors for

𝜆

are exactly the nonzero vectors of

ker (𝑇 - 𝜆 id)

, and such vectors exist iff

𝑇 - 𝜆 id

is not injective. If

𝑉

is finite-dimensional, an operator on

𝑉

is injective iff surjective iff bijective iff invertible. For

𝐴

, apply this to

𝑇_{𝐴} -

𝜆 id =

𝑇_{𝐴 - 𝜆 𝐼_{𝑛}}

and use the invertible matrix theorem and

det

By Definition 2.1.21, $(𝑇 - 𝜆 id) 𝐯 =$ $𝑇 𝐯 -$ $𝜆 𝐯$ , so $𝑇 𝐯 = 𝜆 𝐯$ if and only if $𝐯 \in ker (𝑇 - 𝜆 id)$ . Hence $𝜆$ is an eigenvalue if and only if $ker (𝑇 - 𝜆 id)$ contains a nonzero vector, which proves (I) $\Leftrightarrow$ (II). The two forms of (II) are equivalent by Theorem 2.6.11, applied to the linear map $𝑇 - 𝜆 id$ .

Now let $𝑉$ be finite-dimensional. By Corollary 4.2.18, the operator $𝑇 - 𝜆 id$ on $𝑉$ is injective if and only if it is surjective, if and only if it is bijective, and by Proposition 2.7.7 a bijective linear map is invertible (conversely, an invertible map is bijective). Negating, (II), (III), and (IV) are equivalent.

For $𝐴 \in 𝐹^{𝑛 \times 𝑛}$ , $(𝐴 - 𝜆 𝐼_{𝑛}) 𝐱 =$ $𝐴 𝐱 -$ $𝜆 𝐱$ for every $𝐱$ , so $𝑇_{𝐴 - 𝜆 𝐼_{𝑛}} =$ $𝑇_{𝐴} -$ $𝜆 id$ . Thus $𝜆$ is an eigenvalue of $𝐴$ if and only if $𝑇_{𝐴 - 𝜆 𝐼_{𝑛}}$ is not injective. By Theorem 3.5.12 and Corollary 3.5.25, this happens if and only if $rank (𝐴 - 𝜆 𝐼_{𝑛}) <$ $𝑛$ , if and only if $𝐴 - 𝜆 𝐼_{𝑛}$ is not invertible, and by Theorem 5.4.7 if and only if $det (𝐴 - 𝜆 𝐼_{𝑛}) = 0$ . □

The determinant condition is the practical one: $det (𝑥 𝐼_{𝑛} - 𝐴)$ is a polynomial in $𝑥$ whose roots are the eigenvalues (Section 7.3). Until then we find eigenvalues by solving $𝐴 𝐯 = 𝜆 𝐯$ directly, as in the examples below.

Remark 7.2.10 (Infinite dimensions). Without finite dimension, (III) and (IV) are not equivalent to (I). Multiplication by

𝑥

𝐹 [𝑥]

𝑀 (𝑝) = 𝑥 𝑝

, is injective but not surjective (the constant

1

is not of the form

𝑥 𝑝

). So

𝑀 - 0 id = 𝑀

is not invertible, yet

0

is not an eigenvalue of

𝑀

, since

𝑥 𝑝 = 𝟎

forces

𝑝 = 𝟎

. In fact

𝑀

has no eigenvalues at all (Example 7.2.20).

Exercise 7.2.11.routine Let $𝐴 =$ $(\begin{matrix} 1 & 2 & 0 \\ 2 & 1 & 0 \\ 0 & 0 & - 1 \end{matrix}) \in$ $ℝ^{3 \times 3}$ . Decide which of the vectors

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

are eigenvectors of $𝐴$ , and give the eigenvalue of each one that is.

Solution.

Uses. Definition 7.2.6 (Eigenvalue, eigenvector)

Hint 1. Multiply each vector by

𝐴

and compare with the vector itself.

Hint 2.

𝐴 𝐯

must be a scalar multiple of

𝐯

. Look at a nonzero coordinate of

𝐯

to find the only possible scalar, then check the other coordinates.

Sketch.

𝐴 (1, 1, 0) =

3 (1, 1, 0)

;

𝐴 (1, - 1, 0) =

- (1, - 1, 0)

;

𝐴 (0, 0, 1) =

- (0, 0, 1)

;

𝐴 (1, 0, 0) = (1, 2, 0)

, not an eigenvector;

𝐴 (1, - 1, 1) =

- (1, - 1, 1)

;

𝐴 (1, 1, 1) =

(3, 3, - 1)

, not an eigenvector.

We compute $𝐴 𝐯$ in each case:

\begin{aligned} 𝐴 (1, 1, 0) & = (3, 3, 0) = 3 (1, 1, 0), & 𝐴 (1, - 1, 0) & = (- 1, 1, 0) = - (1, - 1, 0), \\ 𝐴 (0, 0, 1) & = (0, 0, - 1) = - (0, 0, 1), & 𝐴 (1, 0, 0) & = (1, 2, 0), \\ 𝐴 (1, - 1, 1) & = (- 1, 1, - 1) = - (1, - 1, 1), & 𝐴 (1, 1, 1) & = (3, 3, - 1) . \end{aligned}

So $(1, 1, 0)$ is an eigenvector with eigenvalue $3$ , and $(1, - 1, 0)$ , $(0, 0, 1)$ , $(1, - 1, 1)$ are eigenvectors with eigenvalue $- 1$ . The vector $(1, 0, 0)$ is not an eigenvector: a multiple of it has second coordinate $0$ . Neither is $(1, 1, 1)$ : a multiple $𝑐 (1, 1, 1)$ has three equal coordinates.

The sum of the two eigenvectors $(1, - 1, 0)$ and $(0, 0, 1)$ for the eigenvalue $- 1$ is again an eigenvector for $- 1$ , but the sum $(1, 1, 1)$ of the eigenvectors $(1, 1, 0)$ and $(0, 0, 1)$ , which have different eigenvalues, is not. The definition below and Theorem 7.2.21 explain both observations.

Exercise 7.2.12.routine Let

𝑇 \in L (𝑉)

be invertible. Show that

0

is not an eigenvalue of

𝑇

, and that for

𝜆 \neq 0

, a vector

𝐯

is an eigenvector of

𝑇

with eigenvalue

𝜆

if and only if it is an eigenvector of

𝑇^{- 1}

with eigenvalue

𝜆^{- 1}

Solution.

Uses. Theorem 7.2.9 (Eigenvalues and invertibility), Definition 2.5.12 (Invertible), Definition 7.2.6 (Eigenvalue, eigenvector)

Hint 1. An invertible map is injective. What does Theorem 7.2.9 say about

𝜆 = 0

Hint 2. Apply

𝑇^{- 1}

to both sides of

𝑇 𝐯 = 𝜆 𝐯

, and divide by

𝜆

Sketch.

𝑇 - 0 id = 𝑇

is injective, so

0

is not an eigenvalue. If

𝑇 𝐯 = 𝜆 𝐯

, applying

𝑇^{- 1}

gives

𝐯 = 𝜆 𝑇^{- 1} 𝐯

, so

𝑇^{- 1} 𝐯 = 𝜆^{- 1} 𝐯

. The converse is the same argument for

𝑇^{- 1}

, whose inverse is

𝑇

Since $𝑇$ has a left inverse, it is injective: $𝑇 𝐯 = 𝟎$ gives $𝐯 = 𝑇^{- 1} 𝑇 𝐯 = 𝟎$ . So $𝑇 - 0 id = 𝑇$ is injective, and $0$ is not an eigenvalue by Theorem 7.2.9.

Let $𝜆 \neq 0$ and suppose $𝐯 \neq 𝟎$ and $𝑇 𝐯 = 𝜆 𝐯$ . Applying the linear map $𝑇^{- 1}$ gives $𝐯 =$ $𝑇^{- 1} (𝜆 𝐯) =$ $𝜆 𝑇^{- 1} 𝐯$ , and multiplying by $𝜆^{- 1}$ gives $𝑇^{- 1} 𝐯 = 𝜆^{- 1} 𝐯$ . Conversely, $𝑇^{- 1}$ is invertible with inverse $𝑇$ , so the same argument with the roles of $𝑇$ and $𝑇^{- 1}$ exchanged shows that $𝑇^{- 1} 𝐯 = 𝜆^{- 1} 𝐯$ implies $𝑇 𝐯 =$ ${(𝜆^{- 1})}^{- 1} 𝐯 =$ $𝜆 𝐯$ .

The eigenvectors for a fixed $𝜆$ , together with $𝟎$ , form the kernel in Theorem 7.2.9(II). That kernel gets a name.

Definition 7.2.13 (Eigenspace). Let $𝑇 \in L (𝑉)$ and $𝜆 \in 𝐹$ . The eigenspace of $𝑇$ for $𝜆$ is

𝐸_{𝜆} = 𝐸_{𝜆} (𝑇) = ker (𝑇 - 𝜆 id) = {𝐯 \in 𝑉 | 𝑇 𝐯 = 𝜆 𝐯} .

For $𝐴 \in 𝐹^{𝑛 \times 𝑛}$ , $𝐸_{𝜆} (𝐴) =$ $ker (𝐴 - 𝜆 𝐼_{𝑛})$ is the null space of $𝐴 - 𝜆 𝐼_{𝑛}$ .

The eigenspace is defined for every $𝜆 \in 𝐹$ , and it is a subspace, being a kernel (Proposition 2.6.8). By Theorem 7.2.9, $𝐸_{𝜆} \neq {𝟎}$ exactly when $𝜆$ is an eigenvalue, and then $𝐸_{𝜆}$ consists of $𝟎$ and the eigenvectors for $𝜆$ . Each eigenspace is $𝑇$ -invariant, since $𝑇$ acts on it as the scaling $𝜆 id$ . The eigenvectors of $𝑇$ for all eigenvalues together, with $𝟎$ added, do not form a subspace in general, as Exercise 7.2.11 showed. Its dimension, $dim 𝐸_{𝜆}$ , is the geometric multiplicity of $𝜆$ (Section 7.4).

Example 7.2.14 (Computing eigenspaces). Let $𝐽 \in ℝ^{3 \times 3}$ be the matrix whose entries are all $1$ , and let

𝐴 = 2 𝐼_{3} + 𝐽 = (\begin{matrix} 3 & 1 & 1 \\ 1 & 3 & 1 \\ 1 & 1 & 3 \end{matrix}) .

Since $𝐽 (𝑥, 𝑦, 𝑧) =$ $(𝑠, 𝑠, 𝑠)$ with $𝑠 = 𝑥 + 𝑦 + 𝑧$ , we have $𝐴 𝐯 =$ $2 𝐯 +$ $𝑠 (1, 1, 1)$ . So

𝐸_{2} = ker 𝐽 = {(𝑥, 𝑦, 𝑧) | 𝑥 + 𝑦 + 𝑧 = 0} = span ((1, - 1, 0), (0, 1, - 1)),

a plane, and $𝐴 (1, 1, 1) = (5, 5, 5)$ , so $5$ is an eigenvalue. To find all of $𝐸_{5}$ , row reduce $𝐴 - 5 𝐼_{3}$ :

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

so $𝐸_{5} =$ ${(𝑥, 𝑦, 𝑧) | 𝑥 = 𝑧, 𝑦 = 𝑧} =$ $span ((1, 1, 1))$ . We will see right after Theorem 7.2.21 that $2$ and $5$ are the only eigenvalues of $𝐴$ .

7.2.3A gallery of examples

The following operators will serve as examples and counterexamples for the rest of the chapter. In each case we solve $𝑇 𝐯 = 𝜆 𝐯$ by hand.

Example 7.2.15 (Diagonal matrices). Let

𝐷 =

diag (𝑑_{1}, \dots, 𝑑_{𝑛})

. Then

𝐷 𝐞_{𝑖} = 𝑑_{𝑖} 𝐞_{𝑖}

, so each

𝑑_{𝑖}

is an eigenvalue, with eigenvector

𝐞_{𝑖}

. There are no others: if

𝜆 \neq 𝑑_{𝑖}

for every

𝑖

, then

𝐷 -

𝜆 𝐼_{𝑛} =

diag (𝑑_{1} - 𝜆, \dots, 𝑑_{𝑛} - 𝜆)

has inverse

diag ({(𝑑_{1} - 𝜆)}^{- 1}, \dots, {(𝑑_{𝑛} - 𝜆)}^{- 1})

, so

𝜆

is not an eigenvalue by Theorem 7.2.9. The eigenspace

𝐸_{𝜆}

is spanned by the

𝐞_{𝑖}

with

𝑑_{𝑖} = 𝜆

, because

(𝐷 - 𝜆 𝐼_{𝑛}) 𝐱

has coordinates

(𝑑_{𝑖} - 𝜆) 𝑥_{𝑖}

, which all vanish exactly when

𝑥_{𝑖} = 0

for every

𝑖

with

𝑑_{𝑖} \neq 𝜆

Example 7.2.16 (Projections). Let $𝑃 \in L (𝑉)$ satisfy $𝑃^{2} = 𝑃$ , for instance the projection ${Pr}_{ℓ}$ of $ℝ^{2}$ onto a line $ℓ$ (Example 2.1.11). If $𝑃 𝐯 = 𝜆 𝐯$ with $𝐯 \neq 𝟎$ , then

𝜆 𝐯 = 𝑃 𝐯 = 𝑃 (𝑃 𝐯) = 𝑃 (𝜆 𝐯) = 𝜆^{2} 𝐯,

so $(𝜆^{2} - 𝜆) 𝐯 = 𝟎$ and $𝜆 (𝜆 - 1) = 0$ . So the only possible eigenvalues are $0$ and $1$ . Their eigenspaces are

𝐸_{1} = im 𝑃, 𝐸_{0} = ker 𝑃 .

The second is the definition. For the first, if $𝐯 = 𝑃 𝐰$ , then $𝑃 𝐯 =$ $𝑃^{2} 𝐰 =$ $𝑃 𝐰 =$ $𝐯$ , and conversely $𝑃 𝐯 = 𝐯$ exhibits $𝐯$ as a value of $𝑃$ . For ${Pr}_{ℓ}$ , $𝐸_{1} = ℓ$ and $𝐸_{0}$ is the line through $𝟎$ perpendicular to $ℓ$ . Every vector is the sum of its components in these two eigenspaces, $𝐯 = 𝑃 𝐯 + (𝐯 - 𝑃 𝐯)$ .

Example 7.2.17 (Reflections, and the field $𝔽_{2}$ ). Let $𝑆 \in L (𝑉)$ satisfy $𝑆^{2} = id$ , for instance the reflection $𝑆_{ℓ}$ of $ℝ^{2}$ across a line $ℓ$ . If $𝑆 𝐯 = 𝜆 𝐯$ with $𝐯 \neq 𝟎$ , then $𝐯 = 𝑆^{2} 𝐯 = 𝜆^{2} 𝐯$ , so $𝜆^{2} = 1$ , that is, $(𝜆 - 1) (𝜆 + 1) = 0$ , and $𝜆 = \pm 1$ . For $𝑆_{ℓ}$ , $𝐸_{1} = ℓ$ and $𝐸_{- 1}$ is the perpendicular line.

Now take the swap $𝜎 (𝑥, 𝑦) = (𝑦, 𝑥)$ on $𝐹^{2}$ , the reflection across the diagonal. Then $𝜎 (1, 1) = (1, 1)$ and $𝜎 (1, - 1) =$ $(- 1, 1) =$ $- (1, - 1)$ . If $1 + 1 \neq 0$ in $𝐹$ , then $1 \neq - 1$ , and $(1, 1)$ , $(1, - 1)$ are eigenvectors for the two different eigenvalues $1$ and $- 1$ ; they form a basis, in which $𝜎$ has matrix $diag (1, - 1)$ . Over $𝔽_{2}$ , however, $- 1 = 1$ , and $1$ is the only possible eigenvalue. Its eigenspace is $ker (𝜎 - id)$ , and $(𝜎 - id) (𝑥, 𝑦) =$ $(𝑦 - 𝑥, 𝑥 - 𝑦)$ , so $𝐸_{1} = span ((1, 1))$ is a line. Over $𝔽_{2}$ the swap has no basis of eigenvectors, although it still satisfies $𝜎^{2} = id$ .

Example 7.2.18 (Rotations). Let $𝐴 = (\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix})$ , so that $𝐴 (𝑥, 𝑦) =$ $(- 𝑦, 𝑥)$ ; over $ℝ$ this is the rotation $𝑅_{𝜋 / 2}$ by a right angle. Suppose $𝐴 (𝑥, 𝑦) =$ $𝜆 (𝑥, 𝑦)$ , that is, $- 𝑦 = 𝜆 𝑥$ and $𝑥 = 𝜆 𝑦$ . Substituting each equation into the other gives

𝑥 = 𝜆 𝑦 = - 𝜆^{2} 𝑥, 𝑦 = - 𝜆 𝑥 = - 𝜆^{2} 𝑦, so (1 + 𝜆^{2}) 𝑥 = (1 + 𝜆^{2}) 𝑦 = 0 .

So if $𝜆^{2} \neq - 1$ , then $(𝑥, 𝑦) = 𝟎$ , and $𝜆$ is not an eigenvalue. If $𝜆^{2} = - 1$ , then $(1, - 𝜆)$ is an eigenvector: $𝐴 (1, - 𝜆) =$ $(𝜆, 1) =$ $𝜆 (1, - 𝜆)$ , since $- 𝜆^{2} = 1$ . So the eigenvalues of $𝐴$ are exactly the square roots of $- 1$ in $𝐹$ :

Over $ℝ$ : none, since $𝜆^{2} \geq 0 > - 1$ . A rotation by a right angle turns every line, so its only invariant subspaces are ${𝟎}$ and $ℝ^{2}$ (by Proposition 7.2.7 and Proposition 4.1.21, an invariant subspace other than these would be an invariant line).
Over $ℂ$ : $𝜆 = 𝑖$ with eigenvector $(1, - 𝑖)$ , and $𝜆 = - 𝑖$ with eigenvector $(1, 𝑖)$ .
Over $𝔽_{5}$ : $2^{2} = 4 = - 1$ and $3^{2} = 9 = - 1$ , so the eigenvalues are $2$ and $3$ , with eigenvectors $(1, - 2) = (1, 3)$ and $(1, - 3) = (1, 2)$ .
Over $𝔽_{3}$ : the squares are $0, 1, 1$ , and $- 1 = 2$ is not among them, so there are no eigenvalues.
Over $𝔽_{2}$ : $- 1 = 1$ , so $𝜆 = 1$ , and $𝐴$ is the swap of Example 7.2.17, with the single eigenvector line $span ((1, 1))$ .

This is the polynomial $𝑥^{2} + 1$ of Section 7.1.5 in disguise: it has no real root, and accordingly the rotation has no eigenvalue.

Example 7.2.19 (Differentiation). Let $𝐷$ be differentiation on $𝐹 [𝑥]$ , over any field. We claim that $0$ is its only eigenvalue. Suppose $𝐷 𝑝 = 𝜆 𝑝$ with $𝑝 \neq 𝟎$ . If $𝑝$ is a nonzero constant, then $𝐷 𝑝 = 𝟎$ , so $𝜆 𝑝 = 𝟎$ and $𝜆 = 0$ . If $deg 𝑝 \geq 1$ , then $𝐷 𝑝$ is either $𝟎$ or of degree less than $deg 𝑝$ (Example 2.1.9), while $𝜆 𝑝$ has degree $deg 𝑝$ unless $𝜆 = 0$ ; so again $𝜆 = 0$ . And $0$ is an eigenvalue, as $𝐷 1 = 𝟎$ . Its eigenspace $𝐸_{0} = ker 𝐷$ is the constants over $ℚ$ , $ℝ$ , and $ℂ$ ; over $𝔽_{𝑝}$ it is larger, since $𝐷 (𝑥^{𝑝}) =$ $𝑝 𝑥^{𝑝 - 1} =$ $𝟎$ (Example 4.1.28).

Contrast this with differentiation on the real vector space of infinitely differentiable functions $ℝ \to ℝ$ . There, for every $𝜆 \in ℝ$ , the function $𝑓_{𝜆} (𝑡) = 𝑒^{𝜆 𝑡}$ satisfies $𝑓_{𝜆}' = 𝜆 𝑓_{𝜆}$ , so every real number is an eigenvalue. For $𝜆 \neq 0$ these eigenvectors are not polynomials.

Example 7.2.20 (Shifts). Write $𝑝 \in 𝐹 [𝑥]$ by its coefficient sequence $(𝑎_{0}, 𝑎_{1}, 𝑎_{2}, \dots)$ (Definition 1.2.13). Multiplication by $𝑥$ , $𝑀$ , shifts the sequence to the right, and the map $𝐿$ of Exercise 2.5.41,

𝐿 (𝑎_{0}, 𝑎_{1}, 𝑎_{2}, \dots) = (𝑎_{1}, 𝑎_{2}, 𝑎_{3}, \dots),

shifts it to the left. $𝑀$ has no eigenvalues: if $𝑥 𝑝 = 𝜆 𝑝$ with $𝑝 \neq 𝟎$ , the left side has degree $deg 𝑝 + 1$ and the right side has degree at most $deg 𝑝$ . $𝐿$ lowers degrees like $𝐷$ , and the argument of Example 7.2.19 shows that $0$ is its only eigenvalue, with $𝐸_{0} = ker 𝐿$ the constants.

On the space $𝐹^{𝑋}$ of all sequences $(𝑥_{0}, 𝑥_{1}, \dots)$ , where $𝑋 = {0, 1, 2, \dots}$ (Remark 1.2.11), the left shift $𝐿$ given by the same formula behaves differently. $𝐿 𝐱 = 𝜆 𝐱$ says $𝑥_{𝑛 + 1} = 𝜆 𝑥_{𝑛}$ for all $𝑛$ , that is, $𝑥_{𝑛} = 𝜆^{𝑛} 𝑥_{0}$ . So every $𝜆 \in 𝐹$ is an eigenvalue, with

𝐸_{𝜆} = span ((1, 𝜆, 𝜆^{2}, 𝜆^{3}, \dots)),

a line of geometric sequences. Only for $𝜆 = 0$ is the eigenvector a polynomial.

Finally, on $𝐹^{𝑛}$ the shift $𝑁 (𝑥_{1}, \dots, 𝑥_{𝑛}) =$ $(𝑥_{2}, \dots, 𝑥_{𝑛}, 0)$ of Section 4.1.7 satisfies $𝑁^{𝑛} = 0$ . If $𝑁 𝐯 = 𝜆 𝐯$ , then applying $𝑁$ repeatedly gives $𝟎 = 𝑁^{𝑛} 𝐯 = 𝜆^{𝑛} 𝐯$ , so $𝜆 = 0$ ; and $𝐸_{0} =$ $ker 𝑁 =$ $span (𝐞_{1})$ .

So an operator may have no eigenvalues (a rotation of $ℝ^{2}$ , multiplication by $𝑥$ ), a few, or every scalar (the left shift on sequences). In finite dimensions there are at most $dim 𝑉$ eigenvalues, as we now show; over an infinite field, this rules out the last behavior.

7.2.4Eigenvectors for distinct eigenvalues

In every example so far, eigenvectors for different eigenvalues pointed in different directions. They are in fact always linearly independent. The reason is that $𝑇$ treats them differently, and that difference can be used to separate them.

Intuition (Separating eigenvectors). Let $𝑇 𝐯_{𝑙} = 𝜆_{𝑙} 𝐯_{𝑙}$ . The operator $𝑇 - 𝜇 id$ multiplies each $𝐯_{𝑙}$ by $𝜆_{𝑙} - 𝜇$ :

(𝑇 - 𝜇 id) 𝐯_{𝑙} = 𝑇 𝐯_{𝑙} - 𝜇 𝐯_{𝑙} = (𝜆_{𝑙} - 𝜇) 𝐯_{𝑙} .

Choosing $𝜇 = 𝜆_{𝑗}$ kills $𝐯_{𝑗}$ and only rescales the others, by nonzero factors when the eigenvalues are distinct. So, given a linear relation among eigenvectors, we can apply such operators to destroy chosen terms of the relation while keeping control of the rest. There are two ways to organize this. We can kill all terms but one at once, using a product of such operators, or we can kill one term at a time and argue that a shortest relation cannot exist.

Theorem 7.2.21 (Eigenvectors for distinct eigenvalues are independent). Let

𝑇 \in L (𝑉)

, and let

𝐯_{1}, \dots, 𝐯_{𝑚}

be eigenvectors of

𝑇

with eigenvalues

𝜆_{1}, \dots, 𝜆_{𝑚}

, which are pairwise distinct. Then

𝐯_{1}, \dots, 𝐯_{𝑚}

is linearly independent.

Proof (kill all terms but one).standard

Uses. Definition 7.2.6 (Eigenvalue, eigenvector), Definition 2.1.21 (Pointwise operations), Proposition 1.1.17 (No zero divisors), Proposition 1.2.35 (Zero products), Theorem 1.5.5 (Independence is decided at zero)

Techniques. Apply an operator to a relation to kill terms (Technique 4.1.25)

Hint 1. Take a relation

𝑎_{1} 𝐯_{1} +

\dots +

𝑎_{𝑚} 𝐯_{𝑚} =

𝟎

and fix an index

𝑖

. Find an operator that sends

𝐯_{𝑙}

𝟎

for every

𝑙 \neq 𝑖

, but not

𝐯_{𝑖}

Hint 2. Try the composite

𝑃_{𝑖}

of the operators

𝑇 - 𝜆_{𝑗} id

for all

𝑗 \neq 𝑖

. Each factor multiplies

𝐯_{𝑙}

𝜆_{𝑙} - 𝜆_{𝑗}

Hint 3. So

𝑃_{𝑖} 𝐯_{𝑙} =

\prod_{𝑗 \neq 𝑖} (𝜆_{𝑙} - 𝜆_{𝑗}) 𝐯_{𝑙}

. For

𝑙 \neq 𝑖

the product contains the factor

𝜆_{𝑙} - 𝜆_{𝑙} = 0

; for

𝑙 = 𝑖

every factor is nonzero.

Sketch. Let

𝑃_{𝑖}

be the composite of the

𝑇 - 𝜆_{𝑗} id

𝑗 \neq 𝑖

. Since

(𝑇 - 𝜆_{𝑗} id) 𝐯_{𝑙} =

(𝜆_{𝑙} - 𝜆_{𝑗}) 𝐯_{𝑙}

, applying the factors one at a time gives

𝑃_{𝑖} 𝐯_{𝑙} =

\prod_{𝑗 \neq 𝑖} (𝜆_{𝑙} - 𝜆_{𝑗}) 𝐯_{𝑙}

, which is

𝟎

for

𝑙 \neq 𝑖

and a nonzero multiple of

𝐯_{𝑖}

for

𝑙 = 𝑖

. Applying

𝑃_{𝑖}

to a relation

\sum 𝑎_{𝑙} 𝐯_{𝑙} = 𝟎

leaves

𝑎_{𝑖} 𝑐_{𝑖} 𝐯_{𝑖} = 𝟎

with

𝑐_{𝑖} \neq 0

, so

𝑎_{𝑖} = 0

First, for every $𝑙$ and every $𝜇 \in 𝐹$ ,

(𝑇 - 𝜇 id) 𝐯_{𝑙} = 𝑇 𝐯_{𝑙} - 𝜇 𝐯_{𝑙} = (𝜆_{𝑙} - 𝜇) 𝐯_{𝑙} .

(7.2.1)

Suppose $𝑎_{1} 𝐯_{1} +$ $\dots +$ $𝑎_{𝑚} 𝐯_{𝑚} =$ $𝟎$ , and fix $𝑖 \in {1, \dots, 𝑚}$ . Let $𝑃_{𝑖}$ be the composite, in increasing order of $𝑗$ , of the operators $𝑇 - 𝜆_{𝑗} id$ with $𝑗 \neq 𝑖$ . It is linear, as a composite of linear maps. Applying its factors to $𝐯_{𝑙}$ one at a time, (7.2.1) shows that each factor multiplies $𝐯_{𝑙}$ by a scalar, so

𝑃_{𝑖} 𝐯_{𝑙} = 𝑐_{𝑖 𝑙} 𝐯_{𝑙}, where 𝑐_{𝑖 𝑙} = \prod_{𝑗 \neq 𝑖} (𝜆_{𝑙} - 𝜆_{𝑗}) .

If $𝑙 \neq 𝑖$ , the product $𝑐_{𝑖 𝑙}$ contains the factor $𝜆_{𝑙} - 𝜆_{𝑙} = 0$ (the factor with $𝑗 = 𝑙$ ), so $𝑐_{𝑖 𝑙} = 0$ . If $𝑙 = 𝑖$ , every factor $𝜆_{𝑖} - 𝜆_{𝑗}$ is nonzero because the eigenvalues are distinct, so $𝑐_{𝑖 𝑖} \neq 0$ by Proposition 1.1.17. Applying $𝑃_{𝑖}$ to the relation therefore gives

𝟎 = 𝑃_{𝑖} (\sum_{𝑙 = 1}^{𝑚} 𝑎_{𝑙} 𝐯_{𝑙}) = \sum_{𝑙 = 1}^{𝑚} 𝑎_{𝑙} 𝑐_{𝑖 𝑙} 𝐯_{𝑙} = 𝑎_{𝑖} 𝑐_{𝑖 𝑖} 𝐯_{𝑖} .

Since $𝐯_{𝑖} \neq 𝟎$ , Proposition 1.2.35 gives $𝑎_{𝑖} 𝑐_{𝑖 𝑖} = 0$ , and since $𝑐_{𝑖 𝑖} \neq 0$ , $𝑎_{𝑖} = 0$ . This holds for every $𝑖$ , so the list is independent by Theorem 1.5.5. □

Proof (a shortest relation).standard

Uses. Definition 7.2.6 (Eigenvalue, eigenvector), Definition 1.5.3 (Linear relation), Proposition 1.2.35 (Zero products), Theorem 1.5.5 (Independence is decided at zero)

Techniques. Extremal choice and minimal counterexamples (Technique 1.5.9)

Hint 1. Argue by contradiction. Among all nontrivial relations

\sum 𝑎_{𝑙} 𝐯_{𝑙} = 𝟎

, choose one with as few nonzero coefficients as possible.

Hint 2. Pick

𝑘

with

𝑎_{𝑘} \neq 0

and apply

𝑇 - 𝜆_{𝑘} id

to the relation. Which coefficients of the new relation are nonzero?

Hint 3. The new relation has coefficients

𝑎_{𝑙} (𝜆_{𝑙} - 𝜆_{𝑘})

: zero at

𝑙 = 𝑘

, and nonzero exactly where

𝑎_{𝑙} \neq 0

for

𝑙 \neq 𝑘

. It is shorter, so it must be trivial.

Sketch. Take a nontrivial relation with the fewest nonzero coefficients, and

𝑘

with

𝑎_{𝑘} \neq 0

. Applying

𝑇 - 𝜆_{𝑘} id

gives the relation

\sum 𝑎_{𝑙} (𝜆_{𝑙} - 𝜆_{𝑘}) 𝐯_{𝑙} =

𝟎

, which has strictly fewer nonzero coefficients, hence is trivial. So

𝑎_{𝑙} = 0

for all

𝑙 \neq 𝑘

, and the original relation reads

𝑎_{𝑘} 𝐯_{𝑘} = 𝟎

with

𝑎_{𝑘} \neq 0

, contradicting

𝐯_{𝑘} \neq 𝟎

Suppose, for contradiction, that the list is linearly dependent. Then it has nontrivial linear relations (Theorem 1.5.5). Choose one, $𝑎_{1} 𝐯_{1} +$ $\dots +$ $𝑎_{𝑚} 𝐯_{𝑚} =$ $𝟎$ , in which the number of nonzero coefficients is as small as possible, and choose $𝑘$ with $𝑎_{𝑘} \neq 0$ . Applying $𝑇 - 𝜆_{𝑘} id$ and using (7.2.1),

\sum_{𝑙 = 1}^{𝑚} 𝑎_{𝑙} (𝜆_{𝑙} - 𝜆_{𝑘}) 𝐯_{𝑙} = 𝟎 .

In this relation the coefficient of $𝐯_{𝑘}$ is $0$ , and for $𝑙 \neq 𝑘$ the coefficient $𝑎_{𝑙} (𝜆_{𝑙} - 𝜆_{𝑘})$ is nonzero exactly when $𝑎_{𝑙} \neq 0$ , since $𝜆_{𝑙} \neq 𝜆_{𝑘}$ . So it has exactly one nonzero coefficient fewer than the chosen relation. By minimality it is not nontrivial: all its coefficients are $0$ , so $𝑎_{𝑙} = 0$ for every $𝑙 \neq 𝑘$ . The chosen relation then reads $𝑎_{𝑘} 𝐯_{𝑘} = 𝟎$ with $𝑎_{𝑘} \neq 0$ , so $𝐯_{𝑘} = 𝟎$ by Proposition 1.2.35. This contradicts that $𝐯_{𝑘}$ is an eigenvector. □

The first proof is constructive: the operator $𝑃_{𝑖}$ isolates the $𝑖$ -th term of any combination of the $𝐯_{𝑙}$ , and after dividing by $𝑐_{𝑖 𝑖}$ it extracts the coefficient $𝑎_{𝑖}$ . In the language of Section 7.2.5 below, $𝑐_{𝑖 𝑖}^{- 1} 𝑃_{𝑖}$ is $𝐿_{𝑖} (𝑇)$ , where $𝐿_{𝑖}$ is the Lagrange polynomial (Definition 3.4.26) for the nodes $𝜆_{1}, \dots, 𝜆_{𝑚}$ with $𝐿_{𝑖} (𝜆_{𝑙}) = 0$ for $𝑙 \neq 𝑖$ and $𝐿_{𝑖} (𝜆_{𝑖}) = 1$ . The second proof needs only one operator at a time, and it is the argument the lecture gives as an induction on $𝑚$ : assuming the result for $𝑚 - 1$ eigenvectors, apply $𝑇 - 𝜆_{𝑚} id$ . Choosing a shortest relation does that induction in one step.

Corollary 7.2.22 (At most

dim 𝑉

eigenvalues). If

𝑉

is finite-dimensional, an operator

𝑇 \in L (𝑉)

has at most

dim 𝑉

distinct eigenvalues. In particular,

𝐴 \in 𝐹^{𝑛 \times 𝑛}

has at most

𝑛

eigenvalues.

Proof.routine

Uses. Theorem 7.2.21 (Eigenvectors for distinct eigenvalues are independent), Theorem 4.1.10 (Independent lists are short)

Hint 1. Choose one eigenvector for each of

𝑚

distinct eigenvalues.

Hint 2. Compare the length of that list with

dim 𝑉

Sketch. Eigenvectors

𝐯_{1}, \dots, 𝐯_{𝑚}

for

𝑚

distinct eigenvalues form an independent list by Theorem 7.2.21, so

𝑚 \leq dim 𝑉

Let

𝜆_{1}, \dots, 𝜆_{𝑚}

be distinct eigenvalues of

𝑇

, and choose an eigenvector

𝐯_{𝑖}

for each

𝜆_{𝑖}

. By Theorem 7.2.21 the list

𝐯_{1}, \dots, 𝐯_{𝑚}

is independent, so

𝑚 \leq dim 𝑉

by Theorem 4.1.10(a). The statement for

𝐴

is the case

𝑇 = 𝑇_{𝐴}

𝑉 = 𝐹^{𝑛}

. □

So the symmetric matrix of Example 7.2.8 has no eigenvalues besides $3$ and $1$ . The theorem also says something about whole eigenspaces.

Corollary 7.2.23 (Eigenspaces are independent). Let $𝜆_{1}, \dots, 𝜆_{𝑚} \in 𝐹$ be distinct. Then the sum $𝐸_{𝜆_{1}} + \dots + 𝐸_{𝜆_{𝑚}}$ of the eigenspaces of $𝑇$ is direct. If $𝑉$ is finite-dimensional, then

dim 𝐸_{𝜆_{1}} + \dots + dim 𝐸_{𝜆_{𝑚}} \leq dim 𝑉 .

Proof.standard

Uses. Theorem 7.2.21 (Eigenvectors for distinct eigenvalues are independent), Theorem 4.4.7 (Criteria for direct sums), Definition 4.4.4 (Direct sum), Theorem 4.1.10 (Independent lists are short), Theorem 1.5.5 (Independence is decided at zero)

Hint 1. By Theorem 4.4.7, it suffices to show: if

𝐮_{1} + \dots + 𝐮_{𝑚} = 𝟎

with

𝐮_{𝑖} \in 𝐸_{𝜆_{𝑖}}

, then every

𝐮_{𝑖} = 𝟎

. Which of the

𝐮_{𝑖}

are eigenvectors?

Hint 2. For the dimension bound, concatenate bases of the eigenspaces and show that the long list is independent.

Sketch. The nonzero

𝐮_{𝑖}

𝐮_{1} + \dots + 𝐮_{𝑚} = 𝟎

are eigenvectors for distinct eigenvalues, and the equation is a relation among them with all coefficients

1

, so by Theorem 7.2.21 there are none. For the bound, a relation among the concatenated bases groups into

𝐮_{1} + \dots + 𝐮_{𝑚} = 𝟎

with

𝐮_{𝑖} \in 𝐸_{𝜆_{𝑖}}

, so each

𝐮_{𝑖} = 𝟎

, and independence of each basis kills the coefficients.

Suppose $𝐮_{1} + \dots + 𝐮_{𝑚} = 𝟎$ with $𝐮_{𝑖} \in 𝐸_{𝜆_{𝑖}}$ , and let $𝐼 = {𝑖 | 𝐮_{𝑖} \neq 𝟎}$ . For $𝑖 \in 𝐼$ , $𝐮_{𝑖}$ is an eigenvector with eigenvalue $𝜆_{𝑖}$ , and these eigenvalues are distinct. If $𝐼$ were nonempty, $\sum_{𝑖 \in 𝐼} 1 \cdot 𝐮_{𝑖} = 𝟎$ would be a nontrivial relation among them, contradicting Theorem 7.2.21. So $𝐼$ is empty, every $𝐮_{𝑖}$ is $𝟎$ , and the sum is direct by the implication (II) $\Rightarrow$ (I) of Theorem 4.4.7.

Now let $𝑉$ be finite-dimensional, let $𝑑_{𝑖} = dim 𝐸_{𝜆_{𝑖}}$ , and choose a basis $𝐛_{𝑖 1}, \dots, 𝐛_{𝑖 𝑑_{𝑖}}$ of each $𝐸_{𝜆_{𝑖}}$ . Suppose $\sum_{𝑖 = 1}^{𝑚} \sum_{𝑗 = 1}^{𝑑_{𝑖}} 𝑎_{𝑖 𝑗} 𝐛_{𝑖 𝑗} =$ $𝟎$ . With $𝐮_{𝑖} =$ $\sum_{𝑗} 𝑎_{𝑖 𝑗} 𝐛_{𝑖 𝑗} \in$ $𝐸_{𝜆_{𝑖}}$ this reads $𝐮_{1} + \dots + 𝐮_{𝑚} = 𝟎$ , so every $𝐮_{𝑖} = 𝟎$ by the first part, and then every $𝑎_{𝑖 𝑗} = 0$ because the $𝐛_{𝑖 𝑗}$ with fixed $𝑖$ are independent. So the concatenated list, of length $𝑑_{1} + \dots + 𝑑_{𝑚}$ , is independent, and $𝑑_{1} + \dots + 𝑑_{𝑚} \leq dim 𝑉$ by Theorem 4.1.10(a). □

For the matrix $𝐴 = 2 𝐼_{3} + 𝐽$ of Example 7.2.14, $dim 𝐸_{2} +$ $dim 𝐸_{5} =$ $2 +$ $1 =$ $3$ . If $𝐴$ had a third eigenvalue $𝜆$ , its eigenspace would add at least $1$ to this sum, which would then exceed $dim ℝ^{3} = 3$ . So $2$ and $5$ are the only eigenvalues. When the dimensions of the eigenspaces add up to $dim 𝑉$ , concatenating their bases gives a basis of eigenvectors; this is the starting point of Section 7.4.

Example 7.2.24 (Independence for free). Two independence results from earlier chapters are instances of the theorem.

(a)The geometric sequences $(1, 𝜆_{𝑖}, 𝜆_{𝑖}^{2}, \dots)$ for distinct $𝜆_{1}, \dots, 𝜆_{𝑘} \in 𝐹$ are eigenvectors of the left shift on $𝐹^{𝑋}$ with the distinct eigenvalues $𝜆_{𝑖}$ (Example 7.2.20). So they are independent, which is Exercise 1.5.44.
(b)The functions $𝑡 \mapsto 𝑒^{𝜆_{𝑖} 𝑡}$ for distinct real $𝜆_{𝑖}$ are eigenvectors of differentiation with distinct eigenvalues (Example 7.2.19), so they are independent, which is Example 1.5.27 and its generalization.

Exercise 7.2.25.routine Let

𝐯_{1}, \dots, 𝐯_{𝑚}

be eigenvectors of

𝑇

with distinct eigenvalues

𝜆_{1}, \dots, 𝜆_{𝑚}

, where

𝑚 \geq 2

. Show that

𝐯_{1} + \dots + 𝐯_{𝑚}

is not an eigenvector of

𝑇

Solution.

Uses. Theorem 7.2.21 (Eigenvectors for distinct eigenvalues are independent), Theorem 1.5.5 (Independence is decided at zero)

Hint 1. Suppose

𝑇 (\sum 𝐯_{𝑙}) = 𝜇 \sum 𝐯_{𝑙}

. Expand both sides in terms of the

𝐯_{𝑙}

Hint 2. You get a linear relation with coefficients

𝜆_{𝑙} - 𝜇

. What does independence say about them?

Sketch. If

𝑇 (\sum 𝐯_{𝑙}) = 𝜇 \sum 𝐯_{𝑙}

, then

\sum (𝜆_{𝑙} - 𝜇) 𝐯_{𝑙} = 𝟎

, so

𝜆_{𝑙} = 𝜇

for all

𝑙

by independence. This contradicts

𝜆_{1} \neq 𝜆_{2}

Let $𝐰 = 𝐯_{1} + \dots + 𝐯_{𝑚}$ , and suppose $𝑇 𝐰 = 𝜇 𝐰$ for some $𝜇 \in 𝐹$ . Then

𝟎 = 𝑇 𝐰 - 𝜇 𝐰 = \sum_{𝑙 = 1}^{𝑚} 𝜆_{𝑙} 𝐯_{𝑙} - \sum_{𝑙 = 1}^{𝑚} 𝜇 𝐯_{𝑙} = \sum_{𝑙 = 1}^{𝑚} (𝜆_{𝑙} - 𝜇) 𝐯_{𝑙} .

By Theorem 7.2.21 the list $𝐯_{1}, \dots, 𝐯_{𝑚}$ is independent, so $𝜆_{𝑙} - 𝜇 = 0$ for every $𝑙$ (Theorem 1.5.5). Then $𝜆_{1} = 𝜇 = 𝜆_{2}$ , contradicting distinctness. So $𝐰$ is not an eigenvector. (It is nonzero, again by independence, so the only way to fail is the one we ruled out.)

7.2.5Polynomials in an operator

The proofs above composed operators of the form $𝑇 - 𝜆 id$ . Such composites are polynomials in $𝑇$ , and polynomials in $𝑇$ are the natural source of operators that interact well with $𝑇$ .

Definition 7.2.26 (Polynomials in an operator). Let $𝑇 \in L (𝑉)$ and $𝑝 =$ $𝑎_{0} +$ $𝑎_{1} 𝑥 +$ $\dots +$ $𝑎_{𝑑} 𝑥^{𝑑} \in$ $𝐹 [𝑥]$ . Then

𝑝 (𝑇) = 𝑎_{0} {id}_{𝑉} + 𝑎_{1} 𝑇 + 𝑎_{2} 𝑇^{2} + \dots + 𝑎_{𝑑} 𝑇^{𝑑} \in L (𝑉),

where $𝑇^{𝑘}$ is the $𝑘$ -fold composite (Notation 2.1.18) and $𝑇^{0} = {id}_{𝑉}$ . For a matrix $𝐴 \in 𝐹^{𝑛 \times 𝑛}$ , $𝑝 (𝐴) =$ $𝑎_{0} 𝐼_{𝑛} +$ $𝑎_{1} 𝐴 +$ $\dots +$ $𝑎_{𝑑} 𝐴^{𝑑}$ , as in Remark 7.1.11.

For example, if $𝑝 =$ $𝑥^{2} -$ $3 𝑥 +$ $2 =$ $(𝑥 - 1) (𝑥 - 2)$ , then $𝑝 (𝑇) =$ $𝑇^{2} -$ $3 𝑇 +$ $2 id$ , and the next proposition shows that this equals $(𝑇 - id) (𝑇 - 2 id)$ , and also $(𝑇 - 2 id) (𝑇 - id)$ .

Proposition 7.2.27 (Evaluating polynomials at an operator). Let $𝑇 \in L (𝑉)$ , $𝑝, 𝑞 \in 𝐹 [𝑥]$ , and $𝑐 \in 𝐹$ .

(a) $(𝑝 + 𝑞) (𝑇) =$ $𝑝 (𝑇) +$ $𝑞 (𝑇)$ , $(𝑐 𝑝) (𝑇) =$ $𝑐 𝑝 (𝑇)$ , and $(𝑝 𝑞) (𝑇) =$ $𝑝 (𝑇) 𝑞 (𝑇)$ , where the product of operators is composition.
(b)In particular $𝑝 (𝑇) 𝑞 (𝑇) =$ $𝑞 (𝑇) 𝑝 (𝑇)$ , and $𝑝 (𝑇)$ commutes with $𝑇$ .
(c)If $𝑉$ is finite-dimensional and $ℬ︀$ is a basis of $𝑉$ , then ${[𝑝 (𝑇)]}_{ℬ︀ ℬ︀} =$ $𝑝 ({[𝑇]}_{ℬ︀ ℬ︀})$ .
(d)Every $𝑇$ -invariant subspace is $𝑝 (𝑇)$ -invariant.

Proof.routine

Uses. Proposition 7.1.10 (Evaluation respects sums and products), Remark 7.1.11 (Polynomials in a matrix), Proposition 2.1.24 (Algebra of composition), Notation 2.1.18, Theorem 2.4.2 (The matrix of a composite), Proposition 7.2.4 (Invariance is checked on a spanning list)

Techniques. Check it on a basis (Technique 2.2.7)

Hint 1. For (a), copy the proof of Proposition 7.1.10. Which facts about multiplication did it use, and do they hold for composition of operators?

Hint 2. It used

𝑡^{𝑖} 𝑡^{𝑗} = 𝑡^{𝑖 + 𝑗}

, distributivity, and that scalars commute with powers of

𝑡

. For operators these are Notation 2.1.18 and Proposition 2.1.24. Then (b) is (a) with

𝑝 𝑞 = 𝑞 𝑝

Hint 3. For (d), show by induction that

𝑇^{𝑘} 𝐮 \in 𝑈

for

𝐮 \in 𝑈

, then take a combination.

Sketch. The proof of Proposition 7.1.10 used only

𝑡^{𝑖} 𝑡^{𝑗} = 𝑡^{𝑖 + 𝑗}

, that multiplication distributes over sums, and that scalars commute with powers of

𝑡

; all three hold for composition of operators (Proposition 2.1.24), so (a) follows, as for matrices in Remark 7.1.11. (b) is

𝑝 (𝑇) 𝑞 (𝑇) =

(𝑝 𝑞) (𝑇) =

(𝑞 𝑝) (𝑇) =

𝑞 (𝑇) 𝑝 (𝑇)

with

𝑞 = 𝑥

. (c):

𝑇 \mapsto {[𝑇]}_{ℬ︀ ℬ︀}

respects sums, scalar multiples, and composites. (d):

𝑇^{𝑘}

maps

𝑈

into

𝑈

by induction, and

𝑝 (𝑇) 𝐮

is a combination of the

𝑇^{𝑘} 𝐮

(a) The proof of Proposition 7.1.10, with $𝑡 = 𝑇$ , used only three facts: $𝑡^{𝑖} 𝑡^{𝑗} = 𝑡^{𝑖 + 𝑗}$ ; multiplication distributes over sums on both sides; and scalars can be moved past powers of $𝑡$ , $(𝑎 𝑡^{𝑖}) (𝑏 𝑡^{𝑗}) =$ $(𝑎 𝑏) 𝑡^{𝑖 + 𝑗}$ . For composition of operators, the first is Notation 2.1.18 and the other two are Proposition 2.1.24. Commutativity of $𝑡$ with itself was never needed beyond $𝑡^{𝑖} 𝑡^{𝑗} = 𝑡^{𝑖 + 𝑗}$ . So the same computation proves (a), exactly as for matrices in Remark 7.1.11.

(b) Since $𝐹 [𝑥]$ is commutative, $𝑝 𝑞 = 𝑞 𝑝$ , so by (a), $𝑝 (𝑇) 𝑞 (𝑇) =$ $(𝑝 𝑞) (𝑇) =$ $(𝑞 𝑝) (𝑇) =$ $𝑞 (𝑇) 𝑝 (𝑇)$ . Taking $𝑞 = 𝑥$ , for which $𝑞 (𝑇) = 𝑇$ , shows that $𝑝 (𝑇) 𝑇 = 𝑇 𝑝 (𝑇)$ .

(c) Write $𝐴 = {[𝑇]}_{ℬ︀ ℬ︀}$ . By Theorem 2.4.2 and induction on $𝑘$ , ${[𝑇^{𝑘}]}_{ℬ︀ ℬ︀} = 𝐴^{𝑘}$ , and ${[{id}_{𝑉}]}_{ℬ︀ ℬ︀} = 𝐼_{𝑛}$ with $𝑛 = dim 𝑉$ . The map $𝑆 \mapsto {[𝑆]}_{ℬ︀ ℬ︀}$ respects sums and scalar multiples, so ${[\sum_{𝑘} 𝑎_{𝑘} 𝑇^{𝑘}]}_{ℬ︀ ℬ︀} =$ $\sum_{𝑘} 𝑎_{𝑘} 𝐴^{𝑘} =$ $𝑝 (𝐴)$ .

(d) Let $𝑈$ be $𝑇$ -invariant and $𝐮 \in 𝑈$ . Then $𝑇^{0} 𝐮 = 𝐮 \in 𝑈$ , and if $𝑇^{𝑘} 𝐮 \in 𝑈$ , then $𝑇^{𝑘 + 1} 𝐮 =$ $𝑇 (𝑇^{𝑘} 𝐮) \in$ $𝑈$ . So every $𝑇^{𝑘} 𝐮$ lies in $𝑈$ , and so does the linear combination $𝑝 (𝑇) 𝐮 =$ $\sum_{𝑘} 𝑎_{𝑘} 𝑇^{𝑘} 𝐮$ . □

Proposition 7.2.28 (Polynomials act on eigenvectors by their values). Let $𝑇 \in L (𝑉)$ , and let $𝐯$ be an eigenvector of $𝑇$ with eigenvalue $𝜆$ . Then for every $𝑝 \in 𝐹 [𝑥]$ ,

𝑝 (𝑇) 𝐯 = 𝑝 (𝜆) 𝐯 .

Consequently, if $𝑝 (𝑇) = 0$ , then every eigenvalue of $𝑇$ is a root of $𝑝$ .

Proof.routine

Uses. Definition 7.2.26 (Polynomials in an operator), Definition 1.2.17 (Evaluation), Proposition 1.2.35 (Zero products)

Hint 1. First show

𝑇^{𝑘} 𝐯 = 𝜆^{𝑘} 𝐯

by induction on

𝑘

Hint 2. For the consequence, apply

𝑝 (𝑇) = 0

to an eigenvector.

Sketch. By induction,

𝑇^{𝑘} 𝐯 = 𝜆^{𝑘} 𝐯

, so

𝑝 (𝑇) 𝐯 =

\sum 𝑎_{𝑘} 𝜆^{𝑘} 𝐯 =

𝑝 (𝜆) 𝐯

. If

𝑝 (𝑇) = 0

and

𝐯

is an eigenvector for

𝜆

, then

𝑝 (𝜆) 𝐯 = 𝟎

with

𝐯 \neq 𝟎

, so

𝑝 (𝜆) = 0

We have $𝑇^{0} 𝐯 = 𝐯 = 𝜆^{0} 𝐯$ , and if $𝑇^{𝑘} 𝐯 = 𝜆^{𝑘} 𝐯$ , then $𝑇^{𝑘 + 1} 𝐯 =$ $𝑇 (𝜆^{𝑘} 𝐯) =$ $𝜆^{𝑘} 𝑇 𝐯 =$ $𝜆^{𝑘 + 1} 𝐯$ . So for $𝑝 = \sum_{𝑘} 𝑎_{𝑘} 𝑥^{𝑘}$ ,

𝑝 (𝑇) 𝐯 = \sum_{𝑘} 𝑎_{𝑘} 𝑇^{𝑘} 𝐯 = \sum_{𝑘} 𝑎_{𝑘} 𝜆^{𝑘} 𝐯 = 𝑝 (𝜆) 𝐯 .

If $𝑝 (𝑇) = 0$ and $𝜆$ is an eigenvalue with eigenvector $𝐯$ , then $𝟎 =$ $𝑝 (𝑇) 𝐯 =$ $𝑝 (𝜆) 𝐯$ , so $𝑝 (𝜆) = 0$ by Proposition 1.2.35. □

This explains the computations in Example 7.2.16, Example 7.2.17, and Example 7.2.20 at once. A projection satisfies $𝑝 (𝑃) = 0$ for $𝑝 = 𝑥^{2} - 𝑥$ , so its eigenvalues lie among the roots $0, 1$ . An involution satisfies $𝑥^{2} - 1$ , with roots $\pm 1$ . A nilpotent operator satisfies $𝑥^{𝑚}$ , whose only root is $0$ . The cyclic permutation $𝑍$ of Example 7.2.2 satisfies $𝑍^{3} = id$ , so its eigenvalues are cube roots of $1$ : over $ℝ$ the only candidate is $1$ , and over $ℂ$ the candidates are $1, 𝜔, 𝜔^{2}$ , where $𝜔 = 𝑒^{2 𝜋 𝑖 / 3}$ . The converse fails: the identity satisfies $𝑥^{2} - 𝑥$ , but $0$ is not one of its eigenvalues. A polynomial with $𝑝 (𝑇) = 0$ only bounds the eigenvalues; Section 7.6 finds the polynomial that pins them down exactly.

Exercise 7.2.29.routine Let $𝑇 \in L (𝑉)$ satisfy $𝑇^{2} = 5 𝑇 - 6 id$ .

(a)Show that every eigenvalue of $𝑇$ is $2$ or $3$ .
(b)Assume $1 + 1 \neq 0$ in $𝐹$ . Show that $𝑇 - 4 id$ is invertible, and write its inverse as a polynomial in $𝑇$ .

Solution.

Uses. Proposition 7.2.28 (Polynomials act on eigenvectors by their values), Proposition 7.2.27 (Evaluating polynomials at an operator), Definition 2.5.12 (Invertible)

Hint 1.

𝑇

is a root of

𝑥^{2} - 5 𝑥 + 6

. Factor it and apply Proposition 7.2.28.

Hint 2. For (b), divide

𝑥^{2} - 5 𝑥 + 6

𝑥 - 4

with remainder, and evaluate the resulting identity at

𝑇

Sketch. (a)

𝑝 =

𝑥^{2} -

5 𝑥 +

6 =

(𝑥 - 2) (𝑥 - 3)

satisfies

𝑝 (𝑇) = 0

. (b)

𝑥^{2} -

5 𝑥 +

6 =

(𝑥 - 4) (𝑥 - 1) +

2

, so

0 =

(𝑇 - 4 id) (𝑇 - id) +

2 id

, and

- (1 / 2) (𝑇 - id)

is a two-sided inverse of

𝑇 - 4 id

(polynomials in

𝑇

commute). Over a field with

2 = 0

, (b) fails.

(a) Let $𝑝 =$ $𝑥^{2} -$ $5 𝑥 +$ $6 =$ $(𝑥 - 2) (𝑥 - 3)$ . By hypothesis $𝑝 (𝑇) =$ $𝑇^{2} -$ $5 𝑇 +$ $6 id =$ $0$ , so by Proposition 7.2.28 every eigenvalue $𝜆$ satisfies $(𝜆 - 2) (𝜆 - 3) = 0$ , that is, $𝜆 \in {2, 3}$ .

(b) In $𝐹 [𝑥]$ , $(𝑥 - 4) (𝑥 - 1) =$ $𝑥^{2} -$ $5 𝑥 +$ $4$ , so $𝑝 = (𝑥 - 4) (𝑥 - 1) + 2$ . Evaluating at $𝑇$ with Proposition 7.2.27(a),

0 = 𝑝 (𝑇) = (𝑇 - 4 id) (𝑇 - id) + 2 id, so (𝑇 - 4 id) (𝑇 - id) = - 2 id .

Since $2 \neq 0$ in $𝐹$ , we may let $𝑆 = - 2^{- 1} (𝑇 - id)$ . Then $(𝑇 - 4 id) 𝑆 = id$ , and $𝑆 (𝑇 - 4 id) = id$ as well, because polynomials in $𝑇$ commute (Proposition 7.2.27(b)). So $𝑇 - 4 id$ is invertible with inverse $- 2^{- 1} (𝑇 - id)$ .

The hypothesis $1 + 1 \neq 0$ is needed. Over $𝔽_{2}$ , $5 = 1$ , $6 = 0$ , and $4 = 0$ , so the relation reads $𝑇^{2} = 𝑇$ and $𝑇 - 4 id = 𝑇$ , which is not invertible when $𝑇$ is, say, the zero operator on $𝑉 \neq {𝟎}$ .

The commutativity in Proposition 7.2.27(b) has a geometric consequence, which is the main source of invariant subspaces in this book.

Proposition 7.2.30 (Kernels and images of commuting operators are invariant). Let $𝑆, 𝑇 \in L (𝑉)$ with $𝑆 𝑇 = 𝑇 𝑆$ . Then $ker 𝑆$ and $im 𝑆$ are $𝑇$ -invariant. In particular:

(a)for every $𝑝 \in 𝐹 [𝑥]$ , $ker 𝑝 (𝑇)$ and $im 𝑝 (𝑇)$ are $𝑇$ -invariant;
(b)every eigenspace $𝐸_{𝜆} (𝑆)$ of $𝑆$ is $𝑇$ -invariant.

Proof.routine

Uses. Definition 2.6.5 (Kernel), Definition 2.6.6 (Image), Proposition 7.2.27 (Evaluating polynomials at an operator), Definition 7.2.13 (Eigenspace), Proposition 2.1.24 (Algebra of composition)

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

Hint 1. Take

𝐯 \in ker 𝑆

and compute

𝑆 (𝑇 𝐯)

, moving

𝑇

to the left with

𝑆 𝑇 = 𝑇 𝑆

Hint 2. For

im 𝑆

, write the vector as

𝑆 𝐰

and rewrite

𝑇 𝑆 𝐰

as a value of

𝑆

Hint 3. For (a) and (b), find an operator that commutes with

𝑇

and has the given kernel or image.

Sketch. If

𝑆 𝐯 = 𝟎

, then

𝑆 𝑇 𝐯 = 𝑇 𝑆 𝐯 = 𝟎

. And

𝑇 (𝑆 𝐰) =

𝑆 (𝑇 𝐰) \in

im 𝑆

. Then (a) is the case

𝑆 = 𝑝 (𝑇)

, which commutes with

𝑇

by Proposition 7.2.27(b), and (b) is the case

𝑆 - 𝜆 id

, which commutes with

𝑇

because

𝑆

and

id

do.

Let $𝐯 \in ker 𝑆$ . Then $𝑆 (𝑇 𝐯) =$ $𝑇 (𝑆 𝐯) =$ $𝑇 𝟎 =$ $𝟎$ , so $𝑇 𝐯 \in ker 𝑆$ . Let $𝐮 \in im 𝑆$ , say $𝐮 = 𝑆 𝐰$ . Then $𝑇 𝐮 =$ $𝑇 𝑆 𝐰 =$ $𝑆 (𝑇 𝐰) \in$ $im 𝑆$ . (This is part (a) of Exercise 2.6.52.)

(a) $𝑝 (𝑇)$ commutes with $𝑇$ by Proposition 7.2.27(b), so the first part applies with $𝑆 = 𝑝 (𝑇)$ .

(b) By Proposition 2.1.24, $(𝑆 - 𝜆 id) 𝑇 =$ $𝑆 𝑇 -$ $𝜆 𝑇 =$ $𝑇 𝑆 -$ $𝜆 𝑇 =$ $𝑇 (𝑆 - 𝜆 id)$ . Apply the first part to $𝑆 - 𝜆 id$ , whose kernel is $𝐸_{𝜆} (𝑆)$ . □

Technique 7.2.31 (Look for invariant subspaces among kernels and images of commuting operators). To find subspaces that $𝑇$ maps into themselves, look for operators $𝑆$ with $𝑆 𝑇 = 𝑇 𝑆$ : the kernel and the image of each such $𝑆$ , and each eigenspace of $𝑆$ , are $𝑇$ -invariant. Polynomials in $𝑇$ always commute with $𝑇$ (Proposition 7.2.27), so ${ker (𝑇 - 𝜆 id)}^{𝑘}$ , ${im (𝑇 - 𝜆 id)}^{𝑘}$ , and $ker 𝑝 (𝑇)$ are always available. Conversely, if the operators at hand commute, an eigenspace of one is a space on which the others act, and the problem can often be reduced to that smaller space.

Reach for it when a problem involves two operators that commute, or when you need an invariant subspace of $𝑇$ and have a polynomial identity for $𝑇$ . The canonical example is Proposition 7.2.30. Generalized eigenspaces ${ker (𝑇 - 𝜆 id)}^{𝑘}$ (Section 10.1) and common eigenvectors of commuting operators (Section 7.6) rest on the same observation.

Used in 15 places and 3 practice exercises

Exercise 7.2.47, Exercise 7.2.48, Exercise 7.2.55, Exercise 7.4.36, Exercise 7.4.50, Exercise 7.4.51, Proposition 7.6.12, Theorem 7.6.28, Theorem 7.6.31, Theorem 7.6.34, Exercise 7.7.6, Exercise 7.7.12, Exercise 7.7.13, Problem 7.4, Exercise 8.2.36; 3 practice exercises.

Example 7.2.32 (What commutes with a diagonal matrix). Let

𝐷 =

diag (𝑑_{1}, \dots, 𝑑_{𝑛})

with distinct

𝑑_{𝑖}

, and suppose

𝑆 𝐷 = 𝐷 𝑆

for some

𝑆 \in 𝐹^{𝑛 \times 𝑛}

. Each eigenspace

𝐸_{𝑑_{𝑖}} (𝐷) =

span (𝐞_{𝑖})

(Example 7.2.15) is

𝑆

-invariant by Proposition 7.2.30(b), so

𝐞_{𝑖}

is an eigenvector of

𝑆

by Proposition 7.2.7:

𝑆 𝐞_{𝑖} = 𝑠_{𝑖} 𝐞_{𝑖}

. That is, the

𝑖

-th column of

𝑆

𝑠_{𝑖} 𝐞_{𝑖}

, and

𝑆

is diagonal. This recovers Exercise 2.4.42 without computing a single entry of

𝑆 𝐷 - 𝐷 𝑆

. If two of the

𝑑_{𝑖}

coincide, the eigenspace is a plane, and

𝑆

only has to preserve the plane, so non-diagonal

𝑆

appear.

7.2.6Krylov subspaces and chains

Given a single vector $𝐯$ , what is the smallest $𝑇$ -invariant subspace containing it? It must contain $𝑇 𝐯$ , then $𝑇^{2} 𝐯$ , and so on, and all their linear combinations. These combinations are exactly the vectors $𝑝 (𝑇) 𝐯$ .

Definition 7.2.33 (Krylov subspace). Let $𝑇 \in L (𝑉)$ and $𝐯 \in 𝑉$ . The Krylov subspace (or cyclic subspace) of $𝐯$ under $𝑇$ is

𝐾_{𝑇} (𝐯) = {𝑝 (𝑇) 𝐯 | 𝑝 \in 𝐹 [𝑥]},

the set of all finite linear combinations of $𝐯, 𝑇 𝐯, 𝑇^{2} 𝐯, \dots$ .

Proposition 7.2.34 (The smallest invariant subspace containing a vector). Let

𝑇 \in L (𝑉)

and

𝐯 \in 𝑉

. Then

𝐾_{𝑇} (𝐯)

is a

𝑇

-invariant subspace containing

𝐯

, and it is contained in every

𝑇

-invariant subspace that contains

𝐯

Proof.routine

Uses. Definition 7.2.33 (Krylov subspace), Proposition 7.2.27 (Evaluating polynomials at an operator), Definition 1.4.1 (Subspace), Definition 7.2.1 (Invariant subspace)

Hint 1. Use Proposition 7.2.27(a):

𝑝 (𝑇) 𝐯 +

𝑐 𝑞 (𝑇) 𝐯 =

(𝑝 + 𝑐 𝑞) (𝑇) 𝐯

and

𝑇 𝑝 (𝑇) 𝐯 =

(𝑥 𝑝) (𝑇) 𝐯

Hint 2. For minimality, take an invariant

𝑈

containing

𝐯

, and use Proposition 7.2.27(d).

Sketch.

𝐾_{𝑇} (𝐯)

contains

𝟎 = 𝟎 (𝑇) 𝐯

and

𝐯 = 1 (𝑇) 𝐯

, is closed under combinations by

(𝑝 + 𝑐 𝑞) (𝑇) =

𝑝 (𝑇) +

𝑐 𝑞 (𝑇)

, and is invariant because

𝑇 𝑝 (𝑇) 𝐯 =

(𝑥 𝑝) (𝑇) 𝐯

. If

𝑈

is invariant and

𝐯 \in 𝑈

, then

𝑝 (𝑇) 𝐯 \in 𝑈

by Proposition 7.2.27(d).

The constant polynomials $𝟎$ and $1$ give $𝟎 = 𝟎 (𝑇) 𝐯$ and $𝐯 = id 𝐯 = 1 (𝑇) 𝐯$ , so $𝟎, 𝐯 \in 𝐾_{𝑇} (𝐯)$ . For $𝑝, 𝑞 \in 𝐹 [𝑥]$ and $𝑐 \in 𝐹$ , Proposition 7.2.27(a) gives

𝑝 (𝑇) 𝐯 + 𝑐 𝑞 (𝑇) 𝐯 = (𝑝 + 𝑐 𝑞) (𝑇) 𝐯 \in 𝐾_{𝑇} (𝐯), 𝑇 (𝑝 (𝑇) 𝐯) = (𝑥 𝑝) (𝑇) 𝐯 \in 𝐾_{𝑇} (𝐯) .

So $𝐾_{𝑇} (𝐯)$ is a subspace (Definition 1.4.1), and it is $𝑇$ -invariant. If $𝑈$ is a $𝑇$ -invariant subspace with $𝐯 \in 𝑈$ , then $𝑈$ is $𝑝 (𝑇)$ -invariant for every $𝑝$ by Proposition 7.2.27(d), so $𝑝 (𝑇) 𝐯 \in 𝑈$ , and $𝐾_{𝑇} (𝐯) \subseteq 𝑈$ . □

In particular, $𝐾_{𝑇} (𝐯)$ is a line exactly when $𝐯$ is an eigenvector, and ${𝟎}$ exactly when $𝐯 = 𝟎$ . In finite dimensions the sequence $𝐯, 𝑇 𝐯, 𝑇^{2} 𝐯, \dots$ must eventually become dependent, and the first dependence tells us everything about $𝐾_{𝑇} (𝐯)$ .

Proposition 7.2.35 (A basis of a Krylov subspace). Let $𝑇 \in L (𝑉)$ and $𝐯 \in 𝑉$ with $𝐯 \neq 𝟎$ , and suppose $𝐾_{𝑇} (𝐯)$ is finite-dimensional (for instance, $𝑉$ is). Then there is a smallest integer $𝑑 \geq 1$ with

𝑇^{𝑑} 𝐯 \in span (𝐯, 𝑇 𝐯, \dots, 𝑇^{𝑑 - 1} 𝐯),

and for this $𝑑$ :

(a) $𝐯,$ $𝑇 𝐯,$ $\dots,$ $𝑇^{𝑑 - 1} 𝐯$ is a basis of $𝐾_{𝑇} (𝐯)$ , so $dim 𝐾_{𝑇} (𝐯) = 𝑑$ ;
(b)writing $𝑇^{𝑑} 𝐯 =$ $- (𝑐_{0} 𝐯 + 𝑐_{1} 𝑇 𝐯 + \dots + 𝑐_{𝑑 - 1} 𝑇^{𝑑 - 1} 𝐯)$ , the monic polynomial $𝑝 =$ $𝑥^{𝑑} +$ $𝑐_{𝑑 - 1} 𝑥^{𝑑 - 1} +$ $\dots +$ $𝑐_{0}$ satisfies $𝑝 (𝑇) 𝐯 = 𝟎$ , and no nonzero polynomial of degree less than $𝑑$ sends $𝐯$ to $𝟎$ .

Proof.standard

Uses. Lemma 1.5.10 (Linear dependence lemma), Theorem 4.1.10 (Independent lists are short), Proposition 7.2.4 (Invariance is checked on a spanning list), Proposition 7.2.34 (The smallest invariant subspace containing a vector), Theorem 1.5.5 (Independence is decided at zero)

Techniques. Too many vectors force a dependence (Technique 4.1.30), Extremal choice and minimal counterexamples (Technique 1.5.9)

Hint 1. For existence of

𝑑

: if

dim 𝐾_{𝑇} (𝐯) = 𝑚

, the

𝑚 + 1

vectors

𝐯, \dots, 𝑇^{𝑚} 𝐯

lie in

𝐾_{𝑇} (𝐯)

. Apply the linear dependence lemma.

Hint 2. Independence of

𝐯, \dots, 𝑇^{𝑑 - 1} 𝐯

: if they were dependent, the linear dependence lemma would produce a smaller

𝑑

Hint 3. Spanning: show that

𝑊 =

span (𝐯, \dots, 𝑇^{𝑑 - 1} 𝐯)

𝑇

-invariant by checking the spanning vectors, then use Proposition 7.2.34.

Sketch. The

𝑚 + 1

vectors

𝐯, \dots, 𝑇^{𝑚} 𝐯

in the

𝑚

-dimensional

𝐾_{𝑇} (𝐯)

are dependent, and the linear dependence lemma gives some

𝑇^{𝑗} 𝐯

in the span of the earlier ones, with

𝑗 \geq 1

𝐯 \neq 𝟎

; so

𝑑

exists. The same lemma shows that

𝐯, \dots, 𝑇^{𝑑 - 1} 𝐯

is independent, by minimality of

𝑑

. Their span

𝑊

is invariant, since

𝑇

sends each

𝑇^{𝑖} 𝐯

to the next one and

𝑇^{𝑑} 𝐯 \in 𝑊

; so

𝐾_{𝑇} (𝐯) \subseteq

𝑊 \subseteq

𝐾_{𝑇} (𝐯)

. In (b), a polynomial of degree less than

𝑑

killing

𝐯

is a relation among the basis vectors.

Let $𝑚 = dim 𝐾_{𝑇} (𝐯)$ . The vectors $𝐯, 𝑇 𝐯, \dots, 𝑇^{𝑚} 𝐯$ lie in $𝐾_{𝑇} (𝐯)$ , and there are $𝑚 + 1$ of them, so they are dependent (Theorem 4.1.10(a)). By Lemma 1.5.10 some $𝑇^{𝑗} 𝐯$ , $0 \leq 𝑗 \leq 𝑚$ , lies in the span of $𝐯, \dots, 𝑇^{𝑗 - 1} 𝐯$ . Here $𝑗 \neq 0$ , because the span of the empty list is ${𝟎}$ and $𝐯 \neq 𝟎$ . So the set of integers $𝑑 \geq 1$ with $𝑇^{𝑑} 𝐯 \in$ $span (𝐯, \dots, 𝑇^{𝑑 - 1} 𝐯)$ is nonempty, and it has a smallest element $𝑑$ .

(a) If $𝐯, \dots, 𝑇^{𝑑 - 1} 𝐯$ were dependent, Lemma 1.5.10 would give $𝑗 \leq 𝑑 - 1$ with $𝑇^{𝑗} 𝐯 \in$ $span (𝐯, \dots, 𝑇^{𝑗 - 1} 𝐯)$ , and as before $𝑗 \geq 1$ ; this contradicts the minimality of $𝑑$ . So the list is independent. Let $𝑊$ be its span. For $𝑖 < 𝑑 - 1$ , $𝑇 (𝑇^{𝑖} 𝐯) =$ $𝑇^{𝑖 + 1} 𝐯 \in$ $𝑊$ , and $𝑇 (𝑇^{𝑑 - 1} 𝐯) =$ $𝑇^{𝑑} 𝐯 \in$ $𝑊$ by the choice of $𝑑$ . So $𝑊$ is $𝑇$ -invariant by Proposition 7.2.4, and it contains $𝐯$ , so $𝐾_{𝑇} (𝐯) \subseteq 𝑊$ by Proposition 7.2.34. Conversely, each $𝑇^{𝑖} 𝐯$ lies in $𝐾_{𝑇} (𝐯)$ , so $𝑊 \subseteq 𝐾_{𝑇} (𝐯)$ . Hence $𝑊 = 𝐾_{𝑇} (𝐯)$ , and the list is a basis.

(b) By the choice of $𝑐_{0}, \dots, 𝑐_{𝑑 - 1}$ ,

𝑝 (𝑇) 𝐯 = 𝑇^{𝑑} 𝐯 + 𝑐_{𝑑 - 1} 𝑇^{𝑑 - 1} 𝐯 + \dots + 𝑐_{0} 𝐯 = 𝟎 .

If $𝑞 =$ $𝑏_{0} +$ $𝑏_{1} 𝑥 +$ $\dots +$ $𝑏_{𝑑 - 1} 𝑥^{𝑑 - 1}$ satisfies $𝑞 (𝑇) 𝐯 = 𝟎$ , then $𝑏_{0} 𝐯 +$ $\dots +$ $𝑏_{𝑑 - 1} 𝑇^{𝑑 - 1} 𝐯 =$ $𝟎$ is a linear relation among the basis vectors of (a), so every $𝑏_{𝑖} = 0$ and $𝑞 = 𝟎$ (Theorem 1.5.5). □

In the basis of (a), $𝑇$ shifts each basis vector to the next, except that the last one goes to $𝑇^{𝑑} 𝐯 =$ $- \sum 𝑐_{𝑖} 𝑇^{𝑖} 𝐯$ . So the matrix of $𝑇 ↾_{𝐾_{𝑇} (𝐯)}$ has $1$ ′s just below the diagonal, the numbers $- 𝑐_{0}, \dots, - 𝑐_{𝑑 - 1}$ in its last column, and $0$ ′s elsewhere. This is the companion matrix of $𝑝$ , which reappears in Section 7.3.

Example 7.2.36 (Krylov subspaces of the cyclic permutation). Let $𝑍 (𝑥, 𝑦, 𝑧) =$ $(𝑧, 𝑥, 𝑦)$ on $𝐹^{3}$ .

(a)For $𝐯 = (1, 1, 1)$ , $𝑍 𝐯 = 𝐯$ , so $𝑑 = 1$ , $𝑝 = 𝑥 - 1$ , and $𝐾_{𝑍} (𝐯)$ is the line $span (𝐯)$ .
(b)For $𝐯 = (1, - 1, 0)$ , $𝑍 𝐯 = (0, 1, - 1)$ , which is not a multiple of $𝐯$ , and $𝑍^{2} 𝐯 =$ $(- 1, 0, 1) =$ $- 𝐯 -$ $𝑍 𝐯$ . So $𝑑 = 2$ , $𝑝 = 𝑥^{2} + 𝑥 + 1$ , and $𝐾_{𝑍} (𝐯) =$ $span ((1, - 1, 0), (0, 1, - 1))$ is the plane $𝑥 + 𝑦 + 𝑧 = 0$ .
(c)For $𝐯 = 𝐞_{1}$ , the vectors $𝐞_{1},$ $𝑍 𝐞_{1} =$ $𝐞_{2},$ $𝑍^{2} 𝐞_{1} =$ $𝐞_{3}$ are independent and $𝑍^{3} 𝐞_{1} = 𝐞_{1}$ . So $𝑑 = 3$ , $𝑝 = 𝑥^{3} - 1$ , and $𝐾_{𝑍} (𝐞_{1}) = 𝐹^{3}$ .

The polynomials fit together: $𝑥^{3} -$ $1 =$ $(𝑥 - 1) (𝑥^{2} + 𝑥 + 1)$ , the line in (a) and the plane in (b) are the two invariant subspaces of Example 7.2.2, and $𝐞_{1} =$ $\frac{1}{3} (1, 1, 1) +$ $\frac{1}{3} (2, - 1, - 1)$ splits along them (when $3 \neq 0$ in $𝐹$ ).

Chains

The Krylov sequence of $𝐯$ can also end by reaching $𝟎$ . In Section 4.1.7 we proved that if $𝑇^{𝑘} 𝐯 \neq 𝟎$ and $𝑇^{𝑘 + 1} 𝐯 = 𝟎$ , then $𝐯, 𝑇 𝐯, \dots, 𝑇^{𝑘} 𝐯$ is independent (Proposition 4.1.26). The same holds with $𝑇$ replaced by $𝑇 - 𝜆 id$ , and in that form the chain carries information about eigenvalues: its last nonzero vector is an eigenvector.

Proposition 7.2.37 (Chains are independent). Let $𝑇 \in L (𝑉)$ , $𝜆 \in 𝐹$ , and $𝑁 = 𝑇 - 𝜆 id$ . Let $𝐯 \in 𝑉$ and $𝑘 \geq 0$ satisfy

𝑁^{𝑘} 𝐯 \neq 𝟎 and 𝑁^{𝑘 + 1} 𝐯 = 𝟎 .

Then $𝐯,$ $𝑁 𝐯,$ $𝑁^{2} 𝐯,$ $\dots,$ $𝑁^{𝑘} 𝐯$ is linearly independent.

For $𝜆 = 0$ this is Proposition 4.1.26. Every power $𝑁^{𝑗}$ with $𝑗 \geq 𝑘 + 1$ kills $𝐯$ , since $𝑁^{𝑗} 𝐯 =$ $𝑁^{𝑗 - 𝑘 - 1} (𝑁^{𝑘 + 1} 𝐯) =$ $𝟎$ . So applying a power of $𝑁$ to a combination of the chain pushes every term toward the end of the chain, and terms that are pushed past $𝑁^{𝑘} 𝐯$ disappear. The two proofs below use this in different ways.

Proof (peel off the coefficients).standard

Uses. Theorem 1.5.5 (Independence is decided at zero), Notation 2.1.18, Proposition 2.1.4 (Linear maps preserve $𝟎$ and negatives), Proposition 1.2.35 (Zero products)

Techniques. Apply an operator to a relation to kill terms (Technique 4.1.25)

Hint 1. Take a relation

𝑎_{0} 𝐯 +

𝑎_{1} 𝑁 𝐯 +

\dots +

𝑎_{𝑘} 𝑁^{𝑘} 𝐯 =

𝟎

and apply

𝑁^{𝑘}

. Which terms survive?

Hint 2. Only

𝑎_{0} 𝑁^{𝑘} 𝐯

survives, so

𝑎_{0} = 0

. Now apply

𝑁^{𝑘 - 1}

to what is left.

Hint 3. Prove

𝑎_{𝑖} = 0

by induction on

𝑖

: if

𝑎_{0} = \dots = 𝑎_{𝑖 - 1} = 0

, apply

𝑁^{𝑘 - 𝑖}

Sketch. Let

\sum_{𝑗 = 0}^{𝑘} 𝑎_{𝑗} 𝑁^{𝑗} 𝐯 =

𝟎

. Assume

𝑎_{𝑗} = 0

for

𝑗 < 𝑖

and apply

𝑁^{𝑘 - 𝑖}

: the term

𝑗 = 𝑖

becomes

𝑎_{𝑖} 𝑁^{𝑘} 𝐯

, and every term with

𝑗 > 𝑖

becomes

𝑁^{𝑘 + (𝑗 - 𝑖)} 𝐯 = 𝟎

. So

𝑎_{𝑖} 𝑁^{𝑘} 𝐯 = 𝟎

, and

𝑎_{𝑖} = 0

because

𝑁^{𝑘} 𝐯 \neq 𝟎

. By induction all

𝑎_{𝑖} = 0

Suppose that

𝑎_{0} 𝐯 + 𝑎_{1} 𝑁 𝐯 + \dots + 𝑎_{𝑘} 𝑁^{𝑘} 𝐯 = 𝟎 .

(7.2.2)

We show $𝑎_{𝑖} = 0$ by induction on $𝑖 \in {0, \dots, 𝑘}$ . Let $𝑖$ be given and assume $𝑎_{𝑗} = 0$ for all $𝑗 < 𝑖$ (no assumption when $𝑖 = 0$ ). Then (7.2.2) reads $\sum_{𝑗 = 𝑖}^{𝑘} 𝑎_{𝑗} 𝑁^{𝑗} 𝐯 =$ $𝟎$ . Applying the linear map $𝑁^{𝑘 - 𝑖}$ ,

𝟎 = \sum_{𝑗 = 𝑖}^{𝑘} 𝑎_{𝑗} 𝑁^{𝑘 - 𝑖 + 𝑗} 𝐯 = 𝑎_{𝑖} 𝑁^{𝑘} 𝐯 + \sum_{𝑗 = 𝑖 + 1}^{𝑘} 𝑎_{𝑗} 𝑁^{𝑘 + (𝑗 - 𝑖)} 𝐯 = 𝑎_{𝑖} 𝑁^{𝑘} 𝐯,

because each exponent $𝑘 + (𝑗 - 𝑖)$ in the last sum is at least $𝑘 + 1$ . As $𝑁^{𝑘} 𝐯 \neq 𝟎$ , Proposition 1.2.35 gives $𝑎_{𝑖} = 0$ . So all $𝑎_{𝑖} = 0$ , and the list is independent by Theorem 1.5.5. □

Proof (lowest nonzero coefficient).standard

Uses. Theorem 1.5.5 (Independence is decided at zero), Notation 2.1.18, Proposition 2.1.4 (Linear maps preserve $𝟎$ and negatives), Proposition 1.2.35 (Zero products)

Techniques. Extremal choice and minimal counterexamples (Technique 1.5.9), Apply an operator to a relation to kill terms (Technique 4.1.25)

Hint 1. Suppose there is a nontrivial relation, and let

𝑚

be the smallest index with

𝑎_{𝑚} \neq 0

Hint 2. Apply a single power of

𝑁

that moves

𝑁^{𝑚} 𝐯

exactly to the end of the chain. Where do the terms with

𝑗 > 𝑚

go?

Hint 3. Applying

𝑁^{𝑘 - 𝑚}

leaves only

𝑎_{𝑚} 𝑁^{𝑘} 𝐯

Sketch. If

\sum_{𝑗 = 0}^{𝑘} 𝑎_{𝑗} 𝑁^{𝑗} 𝐯 =

𝟎

is nontrivial, let

𝑚

be the least index with

𝑎_{𝑚} \neq 0

. Applying

𝑁^{𝑘 - 𝑚}

\sum_{𝑗 \geq 𝑚} 𝑎_{𝑗} 𝑁^{𝑗} 𝐯 =

𝟎

sends the term

𝑗 = 𝑚

𝑎_{𝑚} 𝑁^{𝑘} 𝐯

and every term with

𝑗 > 𝑚

past the end of the chain, to

𝟎

. So

𝑎_{𝑚} 𝑁^{𝑘} 𝐯 = 𝟎

, forcing

𝑎_{𝑚} = 0

, a contradiction.

Suppose, for contradiction, that (7.2.2) holds with not all $𝑎_{𝑗}$ zero, and let $𝑚$ be the smallest index with $𝑎_{𝑚} \neq 0$ . Then $\sum_{𝑗 = 𝑚}^{𝑘} 𝑎_{𝑗} 𝑁^{𝑗} 𝐯 =$ $𝟎$ . Applying $𝑁^{𝑘 - 𝑚}$ ,

𝟎 = 𝑎_{𝑚} 𝑁^{𝑘} 𝐯 + \sum_{𝑗 = 𝑚 + 1}^{𝑘} 𝑎_{𝑗} 𝑁^{𝑘 + (𝑗 - 𝑚)} 𝐯 = 𝑎_{𝑚} 𝑁^{𝑘} 𝐯,

since $𝑘 + (𝑗 - 𝑚) \geq 𝑘 + 1$ for $𝑗 > 𝑚$ . As $𝑁^{𝑘} 𝐯 \neq 𝟎$ , this forces $𝑎_{𝑚} = 0$ (Proposition 1.2.35), contradicting the choice of $𝑚$ . So the relation is trivial, and the list is independent by Theorem 1.5.5. □

Intuition (Why the lowest index). The first proof peels off the coefficients from the front of the chain, one application at a time, and needs an induction to organize the peeling. The second proof asks which single term of a relation is hardest to kill, and kills everything else. Applying $𝑁^{𝑠}$ shifts every term $𝑠$ places toward the end, and the term that reaches the end last is the one with the smallest index. So the lowest index with a nonzero coefficient is the right extremal choice, and one application of $𝑁^{𝑘 - 𝑚}$ isolates it. Choosing the highest index instead does not work: applying $𝑁^{𝑘 - 𝑚^{'}}$ for the highest index $𝑚^{'}$ sends every lower term somewhere strictly inside the chain, so nothing is isolated.

The pattern “choose the extremal term, then apply the operator that kills everything except that term” is exactly the shortest-relation proof of Theorem 7.2.21, with “smallest index” in place of “fewest terms”. It solves Exercise 7.2.51, where several chains are mixed together.

Corollary 7.2.38 (The span of a chain). Under the hypotheses of Proposition 7.2.37:

(a) $𝑁^{𝑘} 𝐯$ is an eigenvector of $𝑇$ with eigenvalue $𝜆$ ;
(b) $span (𝐯, 𝑁 𝐯, \dots, 𝑁^{𝑘} 𝐯) =$ $𝐾_{𝑇} (𝐯)$ , so $dim 𝐾_{𝑇} (𝐯) = 𝑘 + 1$ ;
(c)in the basis $𝑁^{𝑘} 𝐯,$ $𝑁^{𝑘 - 1} 𝐯,$ $\dots,$ $𝑁 𝐯,$ $𝐯$ of $𝐾_{𝑇} (𝐯)$ , the matrix of $𝑇 ↾_{𝐾_{𝑇} (𝐯)}$ has every diagonal entry equal to $𝜆$ , every entry just above the diagonal equal to $1$ , and all other entries $0$ .

Proof.routine

Uses. Proposition 7.2.37 (Chains are independent), Proposition 7.2.34 (The smallest invariant subspace containing a vector), Proposition 7.2.4 (Invariance is checked on a spanning list), Proposition 7.2.27 (Evaluating polynomials at an operator), Definition 2.3.1 (Matrix of a linear map)

Hint 1. For (a), compute

(𝑇 - 𝜆 id) (𝑁^{𝑘} 𝐯)

Hint 2. For (b), each

𝑁^{𝑗} 𝐯 =

{(𝑥 - 𝜆)}^{𝑗} (𝑇) 𝐯

lies in

𝐾_{𝑇} (𝐯)

. For the other inclusion, show the span of the chain is

𝑇

-invariant, using

𝑇 = 𝑁 + 𝜆 id

Hint 3. For (c), compute

𝑇

on each basis vector:

𝑇 (𝑁^{𝑗} 𝐯) =

𝑁^{𝑗 + 1} 𝐯 +

𝜆 𝑁^{𝑗} 𝐯

Sketch. (a)

𝑁 (𝑁^{𝑘} 𝐯) =

𝑁^{𝑘 + 1} 𝐯 =

𝟎

and

𝑁^{𝑘} 𝐯 \neq 𝟎

. (b)

𝑁^{𝑗} 𝐯 =

({(𝑥 - 𝜆)}^{𝑗}) (𝑇) 𝐯 \in

𝐾_{𝑇} (𝐯)

, and

𝑇 (𝑁^{𝑗} 𝐯) =

𝑁^{𝑗 + 1} 𝐯 +

𝜆 𝑁^{𝑗} 𝐯

lies in the span

𝑊

of the chain, so

𝑊

is invariant and contains

𝐾_{𝑇} (𝐯)

. The chain is independent, so

dim = 𝑘 + 1

. (c) Read off the coordinates of

𝑇 (𝑁^{𝑗} 𝐯)

(a) $(𝑇 - 𝜆 id) (𝑁^{𝑘} 𝐯) =$ $𝑁^{𝑘 + 1} 𝐯 =$ $𝟎$ , so $𝑇 (𝑁^{𝑘} 𝐯) =$ $𝜆 𝑁^{𝑘} 𝐯$ , and $𝑁^{𝑘} 𝐯 \neq 𝟎$ by hypothesis.

(b) Let $𝑊 =$ $span (𝐯, 𝑁 𝐯, \dots, 𝑁^{𝑘} 𝐯)$ . Each $𝑁^{𝑗} = {(𝑇 - 𝜆 id)}^{𝑗}$ equals $𝑞_{𝑗} (𝑇)$ for $𝑞_{𝑗} = {(𝑥 - 𝜆)}^{𝑗}$ (Proposition 7.2.27(a)), so $𝑁^{𝑗} 𝐯 \in 𝐾_{𝑇} (𝐯)$ and $𝑊 \subseteq 𝐾_{𝑇} (𝐯)$ . Since $𝑇 = 𝑁 + 𝜆 id$ ,

𝑇 (𝑁^{𝑗} 𝐯) = 𝑁^{𝑗 + 1} 𝐯 + 𝜆 𝑁^{𝑗} 𝐯,

(7.2.3)

which lies in $𝑊$ for $𝑗 < 𝑘$ , and equals $𝜆 𝑁^{𝑘} 𝐯 \in 𝑊$ for $𝑗 = 𝑘$ . So $𝑊$ is $𝑇$ -invariant (Proposition 7.2.4) and contains $𝐯$ , hence $𝐾_{𝑇} (𝐯) \subseteq 𝑊$ (Proposition 7.2.34). Thus $𝑊 = 𝐾_{𝑇} (𝐯)$ , and by Proposition 7.2.37 the chain is a basis of it, of length $𝑘 + 1$ .

(c) Number the basis as $𝐛_{1} =$ $𝑁^{𝑘} 𝐯,$ $𝐛_{2} =$ $𝑁^{𝑘 - 1} 𝐯,$ $\dots,$ $𝐛_{𝑘 + 1} =$ $𝐯$ , so $𝐛_{𝑖} = 𝑁^{𝑘 + 1 - 𝑖} 𝐯$ . By (7.2.3), $𝑇 𝐛_{1} = 𝜆 𝐛_{1}$ and $𝑇 𝐛_{𝑖} =$ $𝐛_{𝑖 - 1} +$ $𝜆 𝐛_{𝑖}$ for $𝑖 \geq 2$ . So the $𝑖$ -th column of the matrix has $𝜆$ in row $𝑖$ , $1$ in row $𝑖 - 1$ when $𝑖 \geq 2$ , and $0$ elsewhere (Definition 2.3.1). □

For $𝑘 = 2$ the matrix in (c) is

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

Matrices of this shape are Jordan blocks, and Section 10.3 shows that every operator on a finite-dimensional complex vector space is built out of chains like these. Here is the smallest example that is not diagonal in any basis.

Example 7.2.39 (A matrix with a single invariant line). Let $𝐴 = (\begin{matrix} 3 & 1 \\ - 1 & 1 \end{matrix}) \in ℝ^{2 \times 2}$ , $𝜆 = 2$ , and

𝑁 = 𝐴 - 2 𝐼_{2} = (\begin{matrix} 1 & 1 \\ - 1 & - 1 \end{matrix}), 𝑁^{2} = (\begin{matrix} 1 - 1 & 1 - 1 \\ - 1 + 1 & - 1 + 1 \end{matrix}) = 0 .

Take $𝐯 = 𝐞_{1}$ : $𝑁 𝐞_{1} = (1, - 1) \neq 𝟎$ and $𝑁^{2} 𝐞_{1} = 𝟎$ , so Proposition 7.2.37 applies with $𝑘 = 1$ . By Corollary 7.2.38, $(1, - 1)$ is an eigenvector with eigenvalue $2$ (indeed $𝐴 (1, - 1) = (2, - 2)$ ), and in the basis $(1, - 1), 𝐞_{1}$ the matrix of $𝑇_{𝐴}$ is $(\begin{matrix} 2 & 1 \\ 0 & 2 \end{matrix})$ .

The only eigenvalue is $2$ : if $𝐴 𝐰 = 𝜇 𝐰$ , then $𝑁 𝐰 = (𝜇 - 2) 𝐰$ and $𝟎 =$ $𝑁^{2} 𝐰 =$ ${(𝜇 - 2)}^{2} 𝐰$ , so $𝜇 = 2$ when $𝐰 \neq 𝟎$ . Its eigenspace $𝐸_{2} =$ $ker 𝑁 =$ $span ((1, - 1))$ is a line, since $𝑁 (𝑥, 𝑦) =$ $(𝑥 + 𝑦) (1, - 1)$ . So $𝐴$ has exactly one invariant line, and there is no basis of $ℝ^{2}$ consisting of eigenvectors of $𝐴$ .

The map of

𝐴 = (\begin{matrix} 3 & 1 \\ - 1 & 1 \end{matrix})

from Example 7.2.39. The dashed line

𝑦 = - 𝑥

is stretched by

2

and is the only invariant line. The vector

𝐞_{1}

goes to

𝐴 𝐞_{1} = 2 𝐞_{1} + 𝑁 𝐞_{1}

: twice itself plus a vector on the invariant line. In the interactive version, change the entries slightly and the single invariant line either splits into two or disappears. For

(\begin{matrix} 3 & 𝑏 \\ 𝑐 & 1 \end{matrix})

there are two invariant lines when

𝑏 𝑐 > - 1

, one when

𝑏 𝑐 = - 1

, and none when

𝑏 𝑐 < - 1

Exercise 7.2.40.routine Let $𝐴 =$ $(\begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{matrix}) \in$ $𝐹^{3 \times 3}$ and $𝑁 = 𝐴 - 𝐼_{3}$ .

(a)Compute $𝑁 𝐞_{3}$ , $𝑁^{2} 𝐞_{3}$ , and $𝑁^{3} 𝐞_{3}$ , and deduce that $𝐞_{3}, 𝑁 𝐞_{3}, 𝑁^{2} 𝐞_{3}$ is a basis of $𝐹^{3}$ .
(b)Show that $1$ is the only eigenvalue of $𝐴$ and that $𝐸_{1} (𝐴) = span (𝐞_{1})$ .

Solution.

Uses. Proposition 7.2.37 (Chains are independent), Corollary 7.2.38 (The span of a chain), Theorem 4.1.10 (Independent lists are short), Proposition 7.2.28 (Polynomials act on eigenvectors by their values)

Hint 1.

𝑁 𝐞_{3}

is the third column of

𝑁

Hint 2. For (b),

𝑁^{3}

kills a basis, so

𝑁^{3} = 0

. If

𝐴 𝐰 = 𝜇 𝐰

, what is

𝑁^{3} 𝐰

Sketch. (a)

𝑁 𝐞_{3} = (1, 1, 0)

𝑁^{2} 𝐞_{3} = (1, 0, 0)

𝑁^{3} 𝐞_{3} = 𝟎

; by Proposition 7.2.37 with

𝑘 = 2

the chain is independent, hence a basis of

𝐹^{3}

. (b)

𝑁^{3}

kills the basis, so

𝑁^{3} = 0

, and an eigenvalue

𝜇

satisfies

{(𝜇 - 1)}^{3} = 0

𝐸_{1} =

ker 𝑁 =

{𝑦 + 𝑧 = 0, 𝑧 = 0} =

span (𝐞_{1})

(a) $𝑁 = (\begin{matrix} 0 & 1 & 1 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{matrix})$ , so $𝑁 (𝑥, 𝑦, 𝑧) =$ $(𝑦 + 𝑧, 𝑧, 0)$ . Hence

𝑁 𝐞_{3} = (1, 1, 0), 𝑁^{2} 𝐞_{3} = 𝑁 (1, 1, 0) = (1, 0, 0), 𝑁^{3} 𝐞_{3} = 𝑁 (1, 0, 0) = 𝟎 .

By Proposition 7.2.37 with $𝜆 = 1$ and $𝑘 = 2$ , the list $𝐞_{3},$ $(1, 1, 0),$ $(1, 0, 0)$ is independent. It has length $3 = dim 𝐹^{3}$ , so it is a basis (Theorem 4.1.10(b)).

(b) $𝑁^{3}$ sends each of the three basis vectors $𝑁^{𝑗} 𝐞_{3}$ to $𝑁^{3 + 𝑗} 𝐞_{3} = 𝟎$ , so $𝑁^{3} = 0$ . That is, $𝑝 (𝐴) = 0$ for $𝑝 = {(𝑥 - 1)}^{3}$ , and by Proposition 7.2.28 every eigenvalue $𝜇$ of $𝐴$ satisfies ${(𝜇 - 1)}^{3} = 0$ , so $𝜇 = 1$ . And $1$ is an eigenvalue, with eigenvector $𝑁^{2} 𝐞_{3} = 𝐞_{1}$ (Corollary 7.2.38(a)). Finally $𝐸_{1} =$ $ker 𝑁 =$ ${(𝑥, 𝑦, 𝑧) | 𝑦 + 𝑧 = 0, 𝑧 = 0} =$ $span (𝐞_{1})$ .

Exercises

Exercise 7.2.41.routine Let

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

. Find all eigenvalues of

𝐵

, with an eigenvector for each, over

ℝ

𝔽_{2}

𝔽_{3}

𝔽_{5}

, and

𝔽_{7}

Solution.

Uses. Definition 7.2.6 (Eigenvalue, eigenvector), Proposition 1.2.35 (Zero products)

Hint 1. Write out

𝐵 (𝑥, 𝑦) =

𝜆 (𝑥, 𝑦)

as two equations and eliminate

𝑥

Hint 2. You should find

(𝜆^{2} - 2) 𝑦 = 0

and

𝑥 = 𝜆 𝑦

. So the eigenvalues are the square roots of

2

in the field. List the squares in each

𝔽_{𝑝}

Sketch.

𝐵 (𝑥, 𝑦) =

(2 𝑦, 𝑥)

, so

𝐵 𝐯 = 𝜆 𝐯

means

2 𝑦 = 𝜆 𝑥

and

𝑥 = 𝜆 𝑦

, hence

(𝜆^{2} - 2) 𝑦 = 0

, and

𝑦 = 0

forces

𝐯 = 𝟎

. So

𝜆

is an eigenvalue iff

𝜆^{2} = 2

, with eigenvector

(𝜆, 1)

ℝ

\pm \sqrt{2}

𝔽_{2}

𝜆 = 0

𝔽_{3}

𝔽_{5}

: none.

𝔽_{7}

3

and

4

$𝐵 (𝑥, 𝑦) =$ $(2 𝑦, 𝑥)$ , so $𝐵 (𝑥, 𝑦) =$ $𝜆 (𝑥, 𝑦)$ means $2 𝑦 = 𝜆 𝑥$ and $𝑥 = 𝜆 𝑦$ . Substituting the second into the first gives $(𝜆^{2} - 2) 𝑦 = 0$ . If $𝑦 = 0$ , then $𝑥 = 𝜆 𝑦 = 0$ . So a nonzero solution exists only if $𝜆^{2} = 2$ , and conversely, if $𝜆^{2} = 2$ , then $𝐵 (𝜆, 1) =$ $(2, 𝜆) =$ $(𝜆^{2}, 𝜆) =$ $𝜆 (𝜆, 1)$ . So the eigenvalues are the $𝜆 \in 𝐹$ with $𝜆^{2} = 2$ , with eigenvectors $(𝜆, 1)$ .

$ℝ$ : $𝜆 = \pm \sqrt{2}$ , eigenvectors $(\pm \sqrt{2}, 1)$ .
$𝔽_{2}$ : $2 = 0$ , so $𝜆^{2} = 0$ and $𝜆 = 0$ , with eigenvector $(0, 1)$ . (Here $𝐵 = (\begin{matrix} 0 & 0 \\ 1 & 0 \end{matrix})$ .)
$𝔽_{3}$ : the squares are $0, 1, 4 = 1$ , and $2$ is not one of them. No eigenvalues.
$𝔽_{5}$ : the squares are $0, 1, 4, 9 = 4, 16 = 1$ . No eigenvalues.
$𝔽_{7}$ : $3^{2} = 9 = 2$ and $4^{2} = 16 = 2$ . Eigenvalues $3$ and $4$ , with eigenvectors $(3, 1)$ and $(4, 1)$ .

Exercise 7.2.42.standard For $𝜃 \in ℝ$ , let $𝑅_{𝜃} =$ $(\begin{matrix} cos 𝜃 & - sin 𝜃 \\ sin 𝜃 & cos 𝜃 \end{matrix})$ , the rotation by $𝜃$ (Example 2.1.11).

(a)For which $𝜃$ does $𝑅_{𝜃}$ have a real eigenvalue? Find the eigenspaces in those cases.
(b)Regard $𝑅_{𝜃}$ as a matrix in $ℂ^{2 \times 2}$ . Show that $(1, - 𝑖)$ and $(1, 𝑖)$ are eigenvectors, with eigenvalues $cos 𝜃 + 𝑖 sin 𝜃$ and $cos 𝜃 - 𝑖 sin 𝜃$ .

Solution.

Uses. Theorem 7.2.9 (Eigenvalues and invertibility), Theorem 5.1.13 (The $2 \times 2$ determinant), Definition 7.2.6 (Eigenvalue, eigenvector)

Hint 1. By Theorem 7.2.9,

𝜆

is an eigenvalue iff

det (𝑅_{𝜃} - 𝜆 𝐼_{2}) = 0

. Compute this

2 \times 2

determinant.

Hint 2.

{(cos 𝜃 - 𝜆)}^{2} +

{sin}^{2} 𝜃

is a sum of two real squares. When is it

0

Sketch. (a)

det (𝑅_{𝜃} - 𝜆 𝐼_{2}) =

{(cos 𝜃 - 𝜆)}^{2} +

{sin}^{2} 𝜃

, which for real

𝜆

vanishes iff

sin 𝜃 = 0

and

𝜆 = cos 𝜃

. So

𝜃 \in 𝜋 ℤ

𝑅_{𝜃} = 𝐼_{2}

(eigenvalue

1

𝐸_{1} = ℝ^{2}

) or

𝑅_{𝜃} = - 𝐼_{2}

(eigenvalue

- 1

𝐸_{- 1} = ℝ^{2}

). (b) Direct multiplication:

𝑅_{𝜃} (1, - 𝑖) =

(cos 𝜃 + 𝑖 sin 𝜃, sin 𝜃 - 𝑖 cos 𝜃) =

(cos 𝜃 + 𝑖 sin 𝜃) (1, - 𝑖)

, and similarly for

(1, 𝑖)

(a) By Theorem 7.2.9 and Theorem 5.1.13, $𝜆 \in ℝ$ is an eigenvalue if and only if

det (𝑅_{𝜃} - 𝜆 𝐼_{2}) = {(cos 𝜃 - 𝜆)}^{2} + {sin}^{2} 𝜃 = 0 .

A sum of two squares of real numbers is $0$ only if both are $0$ , so this happens if and only if $sin 𝜃 = 0$ and $𝜆 = cos 𝜃$ . Thus $𝑅_{𝜃}$ has a real eigenvalue exactly when $𝜃$ is an integer multiple of $𝜋$ . If $𝜃 \in 2 𝜋 ℤ$ , then $𝑅_{𝜃} = 𝐼_{2}$ , the only eigenvalue is $1$ , and $𝐸_{1} = ℝ^{2}$ . If $𝜃 \in 𝜋 + 2 𝜋 ℤ$ , then $𝑅_{𝜃} = - 𝐼_{2}$ , the only eigenvalue is $- 1$ , and $𝐸_{- 1} = ℝ^{2}$ . Geometrically: a rotation that is not by a multiple of $𝜋$ moves every line through $𝟎$ .

(b) Write $𝑐 = cos 𝜃$ , $𝑠 = sin 𝜃$ . Then

𝑅_{𝜃} (\begin{matrix} 1 \\ - 𝑖 \end{matrix}) = (\begin{matrix} 𝑐 + 𝑖 𝑠 \\ 𝑠 - 𝑖 𝑐 \end{matrix}) = (𝑐 + 𝑖 𝑠) (\begin{matrix} 1 \\ - 𝑖 \end{matrix}),

because $(𝑐 + 𝑖 𝑠) (- 𝑖) =$ $- 𝑖 𝑐 -$ $𝑖^{2} 𝑠 =$ $𝑠 -$ $𝑖 𝑐$ . Similarly

𝑅_{𝜃} (\begin{matrix} 1 \\ 𝑖 \end{matrix}) = (\begin{matrix} 𝑐 - 𝑖 𝑠 \\ 𝑠 + 𝑖 𝑐 \end{matrix}) = (𝑐 - 𝑖 𝑠) (\begin{matrix} 1 \\ 𝑖 \end{matrix}),

because $(𝑐 - 𝑖 𝑠) 𝑖 =$ $𝑖 𝑐 +$ $𝑠$ . Both vectors are nonzero, so they are eigenvectors with eigenvalues $𝑐 + 𝑖 𝑠 = 𝑒^{𝑖 𝜃}$ and $𝑐 - 𝑖 𝑠 = 𝑒^{- 𝑖 𝜃}$ . These are distinct unless $𝑠 = 0$ , in agreement with (a).

Exercise 7.2.43.standard Let $𝑆, 𝑇 \in L (𝑉)$ . Prove or disprove:

(a)If $𝜆$ is an eigenvalue of $𝑆$ and $𝜇$ is an eigenvalue of $𝑇$ , then $𝜆 + 𝜇$ is an eigenvalue of $𝑆 + 𝑇$ .
(b)If $𝜆$ is an eigenvalue of $𝑆$ and $𝜇$ is an eigenvalue of $𝑇$ , then $𝜆 𝜇$ is an eigenvalue of $𝑆 𝑇$ .
(c)If $𝐯$ is an eigenvector of both $𝑆$ and $𝑇$ , then it is an eigenvector of $𝑆 + 𝑇$ and of $𝑆 𝑇$ .
(d)If $𝐯$ is an eigenvector of $𝑆 𝑇$ , then it is an eigenvector of $𝑇 𝑆$ .
(e)If $𝑆 𝑇 = 𝑇 𝑆$ , then every eigenvector of $𝑆$ is an eigenvector of $𝑇$ .

Solution.

Uses. Definition 7.2.6 (Eigenvalue, eigenvector), Example 7.2.15 (Diagonal matrices)

Techniques. Finding counterexamples and minimal examples (Technique 1.4.18)

Hint 1. For (a) and (b), try diagonal matrices whose eigenvectors are different standard basis vectors.

Hint 2. For (d), take

𝑆

a projection and

𝑇

with

𝑇 𝑆 \neq 𝑆 𝑇

, and look for

𝐯 \in ker (𝑆 𝑇)

. For (e), what commutes with everything?

Sketch. (a), (b) false:

𝑆 = diag (1, 0)

𝑇 = diag (0, 1)

have eigenvalue

1

, but

𝑆 + 𝑇 = 𝐼_{2}

has only

1 \neq 2

, and

𝑆 𝑇 = 0

has only

0 \neq 1

. (c) true:

(𝑆 + 𝑇) 𝐯 =

(𝜆 + 𝜇) 𝐯

𝑆 𝑇 𝐯 = 𝜇 𝜆 𝐯

. (d) false:

𝑆 = diag (1, 0)

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

𝐯 = (1, - 1)

𝑆 𝑇 𝐯 = 𝟎

but

𝑇 𝑆 𝐯 = (1, 0)

. (e) false:

𝑆 = id

(a) False. On $𝐹^{2}$ let $𝑆 = diag (1, 0)$ and $𝑇 = diag (0, 1)$ . Both have the eigenvalue $1$ (with eigenvectors $𝐞_{1}$ and $𝐞_{2}$ ). But $𝑆 + 𝑇 = 𝐼_{2}$ , whose only eigenvalue is $1 \neq 1 + 1$ when $1 + 1 \neq 1$ , which holds in every field.

(b) False, with the same $𝑆$ and $𝑇$ : $𝑆 𝑇 = 0$ , whose only eigenvalue is $0$ , while $1 \cdot 1 = 1$ .

(c) True. If $𝑆 𝐯 = 𝜆 𝐯$ and $𝑇 𝐯 = 𝜇 𝐯$ with $𝐯 \neq 𝟎$ , then $(𝑆 + 𝑇) 𝐯 =$ $(𝜆 + 𝜇) 𝐯$ and $𝑆 𝑇 𝐯 =$ $𝑆 (𝜇 𝐯) =$ $𝜇 𝜆 𝐯$ .

(d) False. Let $𝑆 = (\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix})$ and $𝑇 = (\begin{matrix} 1 & 1 \\ 0 & 0 \end{matrix})$ , so $𝑆 𝑇 = (\begin{matrix} 1 & 1 \\ 0 & 0 \end{matrix})$ and $𝑇 𝑆 = (\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix})$ . For $𝐯 = (1, - 1)$ , $𝑆 𝑇 𝐯 = 𝟎 = 0 𝐯$ , so $𝐯$ is an eigenvector of $𝑆 𝑇$ . But $𝑇 𝑆 𝐯 = (1, 0)$ , which is not a multiple of $(1, - 1)$ . (What is true: if $𝑆 𝑇 𝐯 = 𝜆 𝐯$ and $𝑇 𝐯 \neq 𝟎$ , then $𝑇 𝐯$ is an eigenvector of $𝑇 𝑆$ for $𝜆$ .)

(e) False. $𝑆 = {id}_{𝑉}$ commutes with every $𝑇$ , and every nonzero vector is an eigenvector of $𝑆$ , but the rotation $𝑇 = 𝑅_{𝜋 / 2}$ of $ℝ^{2}$ has no eigenvectors at all (Example 7.2.18). By Proposition 7.2.30, what is true is that $𝑇$ maps each eigenspace of $𝑆$ into itself.

Exercise 7.2.44.standard Let $𝐴 \in 𝐹^{𝑛 \times 𝑛}$ .

(a)Suppose every row of $𝐴$ has the same sum $𝑠$ . Show that $𝑠$ is an eigenvalue of $𝐴$ .
(b)Show that $𝐴$ and $𝐴^{𝖳}$ have the same eigenvalues.
(c)Suppose every column of $𝐴$ has the same sum $𝑠$ . Show that $𝑠$ is an eigenvalue of $𝐴$ . Must $(1, \dots, 1)$ be an eigenvector?

Solution.

Uses. Definition 7.2.6 (Eigenvalue, eigenvector), Theorem 7.2.9 (Eigenvalues and invertibility), Theorem 4.3.8 (Row rank equals column rank)

Hint 1. For (a), compute

𝐴 (1, 1, \dots, 1)

Hint 2. For (b), use the rank criterion in Theorem 7.2.9, and

{(𝐴 - 𝜆 𝐼_{𝑛})}^{𝖳} =

𝐴^{𝖳} -

𝜆 𝐼_{𝑛}

Sketch. (a)

𝐴 (1, \dots, 1)

is the vector of row sums,

𝑠 (1, \dots, 1)

. (b)

rank (𝐴^{𝖳} - 𝜆 𝐼_{𝑛}) =

rank ({(𝐴 - 𝜆 𝐼_{𝑛})}^{𝖳}) =

rank (𝐴 - 𝜆 𝐼_{𝑛})

, so one is less than

𝑛

iff the other is. (c) Apply (a) to

𝐴^{𝖳}

and use (b).

(1, \dots, 1)

need not be an eigenvector:

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

(a) The $𝑖$ -th coordinate of $𝐴 (1, \dots, 1)$ is $\sum_{𝑗} 𝑎_{𝑖 𝑗}$ , the $𝑖$ -th row sum, which is $𝑠$ . So $𝐴 (1, \dots, 1) =$ $𝑠 (1, \dots, 1)$ , and $(1, \dots, 1) \neq 𝟎$ is an eigenvector with eigenvalue $𝑠$ .

(b) For $𝜆 \in 𝐹$ , ${(𝐴 - 𝜆 𝐼_{𝑛})}^{𝖳} =$ $𝐴^{𝖳} -$ $𝜆 𝐼_{𝑛}$ , so by Theorem 4.3.8, $rank (𝐴^{𝖳} - 𝜆 𝐼_{𝑛}) =$ $rank (𝐴 - 𝜆 𝐼_{𝑛})$ . By Theorem 7.2.9, $𝜆$ is an eigenvalue of $𝐴$ if and only if this rank is less than $𝑛$ , if and only if $𝜆$ is an eigenvalue of $𝐴^{𝖳}$ .

(c) The rows of $𝐴^{𝖳}$ are the columns of $𝐴$ , so every row of $𝐴^{𝖳}$ sums to $𝑠$ , and $𝑠$ is an eigenvalue of $𝐴^{𝖳}$ by (a), hence of $𝐴$ by (b). The vector $(1, \dots, 1)$ need not be an eigenvector of $𝐴$ : for $𝐴 = (\begin{matrix} 1 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{matrix})$ , every column sums to $1$ , but $𝐴 (1, 1, 1) = (2, 1, 0)$ , which is not a multiple of $(1, 1, 1)$ because its last two coordinates differ. An eigenvector of $𝐴$ for the eigenvalue $1$ is $𝐞_{1}$ . (A $2 \times 2$ example such as $(\begin{matrix} 1 & 1 \\ 0 & 0 \end{matrix})$ works only when $1 + 1 \neq 0$ : over $𝔽_{2}$ it sends $(1, 1)$ to $𝟎$ .) (Matrices with nonnegative entries and column sums $1$ are the transition matrices of Markov chains, and this exercise shows they always have the eigenvalue $1$ ; see Section 11.1.)

Exercise 7.2.45.standard

(a)Show that an operator $𝑇 \in L (ℝ^{2})$ has $0$ , $1$ , $2$ , or infinitely many invariant lines, and give an example of each.
(b)Over $𝔽_{2}$ , show that every $𝑇 \in L (𝔽_{2}^{2})$ has $0$ , $1$ , $2$ , or $3$ invariant lines, and give an example of each.

Solution.

Uses. Proposition 7.2.7 (Invariant lines are spanned by eigenvectors), Corollary 7.2.22 (At most $dim 𝑉$ eigenvalues), Proposition 4.1.21 (Dimensions of subspaces), Example 7.2.18 (Rotations), Theorem 2.2.1 (Linear maps are determined by a basis)

Techniques. Dimension counting (Technique 4.2.24)

Hint 1. By Proposition 7.2.7, an invariant line is a line contained in some eigenspace. How many lines can an eigenspace contain?

Hint 2. If some

𝐸_{𝜆}

is all of

𝑉

, then

𝑇 = 𝜆 id

. Otherwise each

𝐸_{𝜆}

{𝟎}

or a line, and there are at most

2

eigenvalues.

Sketch. An invariant line is

span (𝐯)

for an eigenvector

𝐯

, so it lies in

𝐸_{𝜆}

for its eigenvalue. If

𝑇 = 𝑐 id

, every line is invariant: infinitely many over

ℝ

, all

3

lines of

𝔽_{2}^{2}

. Otherwise every eigenspace is a proper subspace of a

2

-dimensional space, hence

{𝟎}

or a line, and invariant lines correspond to eigenvalues, of which there are at most

2

. Examples: rotation, shear,

diag (1, 2)

id

over

ℝ

; over

𝔽_{2}

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

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

diag (0, 1)

id

Let $𝑉 = 𝐹^{2}$ with $𝐹 = ℝ$ or $𝔽_{2}$ , and $𝑇 \in L (𝑉)$ . By Proposition 7.2.7, a line $ℓ$ is invariant if and only if it is spanned by an eigenvector, and then $ℓ \subseteq 𝐸_{𝜆}$ for the eigenvalue $𝜆$ of that eigenvector. Conversely every line inside an eigenspace is invariant.

If $𝐸_{𝜆} = 𝑉$ for some $𝜆$ , then $𝑇 = 𝜆 id$ , and every line is invariant. Over $ℝ$ there are infinitely many lines. Over $𝔽_{2}$ there are exactly $3$ : each line ${𝟎, 𝐯}$ contains one nonzero vector, and $𝔽_{2}^{2}$ has $3$ nonzero vectors.

Otherwise each $𝐸_{𝜆}$ is a proper subspace of the $2$ -dimensional $𝑉$ , so it is ${𝟎}$ or a line (Proposition 4.1.21), and a line contains exactly one line. So the invariant lines are exactly the eigenspaces $𝐸_{𝜆}$ of the eigenvalues $𝜆$ , and their number is the number of eigenvalues, which is at most $2$ by Corollary 7.2.22. This proves the restrictions in (a) and (b). Examples:

(a) $𝑅_{𝜋 / 2}$ has no eigenvalues (Example 7.2.18): $0$ lines. The shear $𝐴 = (\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix})$ satisfies ${(𝐴 - 𝐼_{2})}^{2} = 0$ , so $1$ is its only possible eigenvalue (Proposition 7.2.28), and $𝐸_{1} =$ $ker (𝐴 - 𝐼_{2}) =$ $span (𝐞_{1})$ : $1$ line. $diag (1, 2)$ has $𝐸_{1} = span (𝐞_{1})$ and $𝐸_{2} = span (𝐞_{2})$ (Example 7.2.15): $2$ lines. The identity: infinitely many.

(b) Over $𝔽_{2}$ the only possible eigenvalues are $0$ and $1$ . For $𝐴 = (\begin{matrix} 0 & 1 \\ 1 & 1 \end{matrix})$ , $𝐴 (𝑥, 𝑦) =$ $(𝑦, 𝑥 + 𝑦)$ ; $𝐴 𝐯 = 𝟎$ forces $𝑦 = 0$ and then $𝑥 = 0$ , and $𝐴 𝐯 = 𝐯$ means $𝑦 = 𝑥$ and $𝑥 + 𝑦 = 𝑦$ , so $𝑥 = 0$ and $𝑦 = 0$ . So $𝐴$ has no eigenvalues: $0$ lines. The shear $(\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix})$ is invertible, so $0$ is not an eigenvalue, and $𝐸_{1} =$ $ker (\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}) =$ $span (𝐞_{1})$ : $1$ line. $diag (0, 1)$ : $2$ lines. The identity: $3$ lines.

Exercise 7.2.46.standard Let $𝐷$ be differentiation on $ℝ [𝑥]$ .

(a)Show that the $𝐷$ -invariant subspaces of $ℝ [𝑥]$ are exactly ${𝟎}$ , the spaces ${ℝ [𝑥]}_{\leq 𝑛}$ for $𝑛 \geq 0$ , and $ℝ [𝑥]$ itself.
(b)Over $𝔽_{𝑝}$ , find a $𝐷$ -invariant subspace of $𝔽_{𝑝} [𝑥]$ that is not of this form.

Solution.

Uses. Proposition 7.2.27 (Evaluating polynomials at an operator), Proposition 1.5.24 (Distinct degrees), Theorem 4.1.10 (Independent lists are short), Example 4.1.7 (Dimensions of the standard spaces), Example 2.1.9 (Maps of polynomials)

Techniques. Dimension counting (Technique 4.2.24)

Hint 1. Let

𝑈 \neq {𝟎}

be invariant and

𝑝 \in 𝑈

of degree

𝑛

. What are the degrees of

𝑝,

𝐷 𝑝,

𝐷^{2} 𝑝,

\dots,

𝐷^{𝑛} 𝑝

over

ℝ

Hint 2. These

𝑛 + 1

polynomials have distinct degrees

𝑛, 𝑛 - 1, \dots, 0

, so they are independent, and they lie in

𝑈 \cap {ℝ [𝑥]}_{\leq 𝑛}

. Count dimensions. Then split into the cases where the degrees of elements of

𝑈

are bounded or not.

Sketch. If

𝑝 \in 𝑈

has degree

𝑛

, then

𝐷^{𝑗} 𝑝

has degree

𝑛 - 𝑗

for

0 \leq 𝑗 \leq 𝑛

(leading coefficient

𝑛 (𝑛 - 1) \dots (𝑛 - 𝑗 + 1) 𝑎_{𝑛} \neq

0

ℝ

). These

𝑛 + 1

polynomials are independent and lie in

𝑈 \cap {ℝ [𝑥]}_{\leq 𝑛}

, so

{ℝ [𝑥]}_{\leq 𝑛} \subseteq 𝑈

. If degrees in

𝑈

are bounded with maximum

𝑛

, then

𝑈 = {ℝ [𝑥]}_{\leq 𝑛}

; if unbounded,

𝑈

contains every

{ℝ [𝑥]}_{\leq 𝑛}

, so

𝑈 = ℝ [𝑥]

. Over

𝔽_{𝑝}

span (𝑥^{𝑝})

is invariant since

𝐷 (𝑥^{𝑝}) = 𝟎

(a) Each listed subspace is invariant: $𝐷$ lowers degrees (Example 7.2.2). Conversely, let $𝑈$ be $𝐷$ -invariant with $𝑈 \neq {𝟎}$ .

Claim. If $𝑝 \in 𝑈$ has degree $𝑛$ , then ${ℝ [𝑥]}_{\leq 𝑛} \subseteq 𝑈$ . Write $𝑝 =$ $𝑎_{𝑛} 𝑥^{𝑛} +$ $(lower terms)$ with $𝑎_{𝑛} \neq 0$ . By the formula for $𝐷$ (Example 2.1.9), $𝐷^{𝑗} 𝑝 =$ $𝑛 (𝑛 - 1) \dots (𝑛 - 𝑗 + 1) 𝑎_{𝑛} 𝑥^{𝑛 - 𝑗} +$ $(lower terms)$ for $0 \leq 𝑗 \leq 𝑛$ , and the leading coefficient is nonzero in $ℝ$ . So $𝑝, 𝐷 𝑝, \dots, 𝐷^{𝑛} 𝑝$ have the distinct degrees $𝑛, 𝑛 - 1, \dots, 0$ , hence are independent (Proposition 1.5.24). They lie in $𝑈$ by Proposition 7.2.27(d), and in ${ℝ [𝑥]}_{\leq 𝑛}$ , which has dimension $𝑛 + 1$ (Example 4.1.7). So they form a basis of ${ℝ [𝑥]}_{\leq 𝑛}$ (Theorem 4.1.10(b)), and ${ℝ [𝑥]}_{\leq 𝑛} =$ $span (𝑝, \dots, 𝐷^{𝑛} 𝑝) \subseteq$ $𝑈$ .

If the degrees of the nonzero elements of $𝑈$ are bounded, let $𝑛$ be the largest. Then $𝑈 \subseteq {ℝ [𝑥]}_{\leq 𝑛}$ , and the claim gives equality. If they are unbounded, then for every $𝑚$ there is $𝑝 \in 𝑈$ with $deg 𝑝 \geq 𝑚$ , and the claim gives ${ℝ [𝑥]}_{\leq 𝑚} \subseteq$ ${ℝ [𝑥]}_{\leq deg 𝑝} \subseteq$ $𝑈$ . Every polynomial lies in some ${ℝ [𝑥]}_{\leq 𝑚}$ , so $𝑈 = ℝ [𝑥]$ .

(b) Over $𝔽_{𝑝}$ , $𝐷 (𝑥^{𝑝}) =$ $𝑝 𝑥^{𝑝 - 1} =$ $𝟎$ , so the line $span (𝑥^{𝑝})$ is $𝐷$ -invariant. It is not ${𝟎}$ , it is not $𝔽_{𝑝} [𝑥]$ , and it is not $𝔽_{𝑝} {[𝑥]}_{\leq 𝑛}$ for any $𝑛$ , since it does not contain the constant $1$ . The proof of (a) breaks down because the leading coefficient $𝑝$ of $𝐷 (𝑥^{𝑝})$ vanishes in $𝔽_{𝑝}$ .

Exercise 7.2.47.standard Let $dim 𝑉 = 𝑛$ , and let $𝑇 \in L (𝑉)$ have $𝑛$ distinct eigenvalues $𝜆_{1}, \dots, 𝜆_{𝑛}$ , with eigenvectors $𝐯_{1}, \dots, 𝐯_{𝑛}$ . Let $𝑆 \in L (𝑉)$ commute with $𝑇$ .

(a)Show that each $𝐯_{𝑖}$ is an eigenvector of $𝑆$ .
(b)Show that $𝑆 = 𝑝 (𝑇)$ for some $𝑝 \in 𝐹 [𝑥]$ of degree less than $𝑛$ .

Solution.

Uses. Corollary 7.2.23 (Eigenspaces are independent), Proposition 7.2.30 (Kernels and images of commuting operators are invariant), Proposition 7.2.7 (Invariant lines are spanned by eigenvectors), Theorem 7.2.21 (Eigenvectors for distinct eigenvalues are independent), Theorem 4.1.10 (Independent lists are short), Theorem 3.4.24 (Polynomial interpolation), Proposition 7.2.28 (Polynomials act on eigenvectors by their values), Corollary 2.2.6 (Agreeing on a spanning list)

Techniques. Look for invariant subspaces among kernels and images of commuting operators (Technique 7.2.31)

Hint 1. Each eigenspace

𝐸_{𝜆_{𝑖}} (𝑇)

𝑆

-invariant. What is its dimension?

Hint 2. The dimensions of the

𝑛

eigenspaces are each at least

1

and add up to at most

𝑛

Hint 3. For (b), if

𝑆 𝐯_{𝑖} = 𝜇_{𝑖} 𝐯_{𝑖}

, choose

𝑝

with

𝑝 (𝜆_{𝑖}) = 𝜇_{𝑖}

for all

𝑖

, and compare

𝑆

and

𝑝 (𝑇)

on the basis

𝐯_{1}, \dots, 𝐯_{𝑛}

Sketch. (a) By Corollary 7.2.23,

\sum dim 𝐸_{𝜆_{𝑖}} \leq 𝑛

with each term

\geq 1

, so each

𝐸_{𝜆_{𝑖}} = span (𝐯_{𝑖})

is a line. It is

𝑆

-invariant by Proposition 7.2.30, so

𝐯_{𝑖}

is an eigenvector of

𝑆

, say

𝑆 𝐯_{𝑖} = 𝜇_{𝑖} 𝐯_{𝑖}

. (b) Interpolate:

𝑝 (𝜆_{𝑖}) = 𝜇_{𝑖}

with

deg 𝑝 < 𝑛

. Then

𝑝 (𝑇) 𝐯_{𝑖} =

𝜇_{𝑖} 𝐯_{𝑖} =

𝑆 𝐯_{𝑖}

, and the

𝐯_{𝑖}

form a basis.

(a) Each $𝐸_{𝜆_{𝑖}} (𝑇)$ contains $𝐯_{𝑖} \neq 𝟎$ , so $dim 𝐸_{𝜆_{𝑖}} \geq 1$ , and by Corollary 7.2.23 the $𝑛$ dimensions add up to at most $𝑛$ . So each $dim 𝐸_{𝜆_{𝑖}} = 1$ , and $𝐸_{𝜆_{𝑖}} = span (𝐯_{𝑖})$ . By Proposition 7.2.30(b) (with the roles of $𝑆$ and $𝑇$ exchanged), $𝐸_{𝜆_{𝑖}} (𝑇)$ is $𝑆$ -invariant, so $𝐯_{𝑖}$ is an eigenvector of $𝑆$ by Proposition 7.2.7. Write $𝑆 𝐯_{𝑖} = 𝜇_{𝑖} 𝐯_{𝑖}$ .

(b) By Theorem 3.4.24 there is $𝑝 \in {𝐹 [𝑥]}_{\leq 𝑛 - 1}$ with $𝑝 (𝜆_{𝑖}) = 𝜇_{𝑖}$ for $𝑖 = 1, \dots, 𝑛$ , since the $𝜆_{𝑖}$ are distinct. By Proposition 7.2.28, $𝑝 (𝑇) 𝐯_{𝑖} =$ $𝑝 (𝜆_{𝑖}) 𝐯_{𝑖} =$ $𝜇_{𝑖} 𝐯_{𝑖} =$ $𝑆 𝐯_{𝑖}$ . The list $𝐯_{1}, \dots, 𝐯_{𝑛}$ is independent (Theorem 7.2.21) of length $𝑛$ , so it is a basis (Theorem 4.1.10(b)), and $𝑆 = 𝑝 (𝑇)$ by Corollary 2.2.6. Conversely every $𝑝 (𝑇)$ commutes with $𝑇$ , so the operators commuting with $𝑇$ are exactly the polynomials in $𝑇$ .

Exercise 7.2.48.hard Let

dim 𝑉 = 𝑛

, and let

𝑇 \in L (𝑉)

have

𝑛

distinct eigenvalues. Show that

𝑇

has exactly

2^{𝑛}

invariant subspaces.

Solution.

Uses. Theorem 7.2.21 (Eigenvectors for distinct eigenvalues are independent), Theorem 4.1.10 (Independent lists are short), Definition 3.4.26 (Lagrange polynomials), Proposition 7.2.28 (Polynomials act on eigenvectors by their values), Proposition 7.2.27 (Evaluating polynomials at an operator), Proposition 7.2.4 (Invariance is checked on a spanning list)

Techniques. Look for invariant subspaces among kernels and images of commuting operators (Technique 7.2.31)

Hint 1. Let

𝐯_{1}, \dots, 𝐯_{𝑛}

be eigenvectors for the distinct eigenvalues

𝜆_{1}, \dots, 𝜆_{𝑛}

; they form a basis. Guess the invariant subspaces.

Hint 2. For an invariant

𝑈

and

𝐮 = \sum 𝑐_{𝑖} 𝐯_{𝑖} \in 𝑈

, find a polynomial

𝑞

with

𝑞 (𝑇) 𝐮 = 𝑐_{𝑖} 𝐯_{𝑖}

. Lagrange polynomials do this.

Hint 3. Conclude that

𝑈 =

span (𝐯_{𝑖} | 𝐯_{𝑖} \in 𝑈)

, so invariant subspaces correspond to subsets of

{1, \dots, 𝑛}

Sketch. With an eigenbasis

𝐯_{𝑖}

, each

𝑈_{𝐼} =

span (𝐯_{𝑖} | 𝑖 \in 𝐼)

is invariant, and distinct

𝐼

give distinct

𝑈_{𝐼}

. Conversely, if

𝑈

is invariant and

𝐮 = \sum 𝑐_{𝑖} 𝐯_{𝑖} \in 𝑈

, then

𝐿_{𝑖} (𝑇) 𝐮 =

𝑐_{𝑖} 𝐯_{𝑖} \in

𝑈

, where

𝐿_{𝑖}

is the Lagrange polynomial with

𝐿_{𝑖} (𝜆_{𝑗}) =

[𝑖 = 𝑗]

. So

𝑈 = 𝑈_{𝐼}

for

𝐼 = {𝑖 | 𝐯_{𝑖} \in 𝑈}

. Hence

2^{𝑛}

Let $𝜆_{1}, \dots, 𝜆_{𝑛}$ be the eigenvalues and $𝐯_{𝑖}$ an eigenvector for $𝜆_{𝑖}$ . By Theorem 7.2.21 and Theorem 4.1.10(b), $𝐯_{1}, \dots, 𝐯_{𝑛}$ is a basis of $𝑉$ . For $𝐼 \subseteq {1, \dots, 𝑛}$ let $𝑈_{𝐼} =$ $span (𝐯_{𝑖} | 𝑖 \in 𝐼)$ (with $𝑈_{\emptyset} = {𝟎}$ ). Each $𝑈_{𝐼}$ is invariant by Proposition 7.2.4, since $𝑇 𝐯_{𝑖} =$ $𝜆_{𝑖} 𝐯_{𝑖} \in$ $𝑈_{𝐼}$ for $𝑖 \in 𝐼$ . Different subsets give different subspaces: if $𝑗 \in 𝐼$ and $𝑗 \notin 𝐼^{'}$ , then $𝐯_{𝑗} \in 𝑈_{𝐼}$ but $𝐯_{𝑗} \notin 𝑈_{𝐼^{'}}$ , because $𝐯_{𝑗}$ is not a combination of the other basis vectors.

Conversely, let $𝑈$ be $𝑇$ -invariant and $𝐼 = {𝑖 | 𝐯_{𝑖} \in 𝑈}$ . Clearly $𝑈_{𝐼} \subseteq 𝑈$ . Let $𝐮 \in 𝑈$ and write $𝐮 = \sum_{𝑖} 𝑐_{𝑖} 𝐯_{𝑖}$ . For each $𝑖$ , let $𝐿_{𝑖}$ be the Lagrange polynomial for the nodes $𝜆_{1}, \dots, 𝜆_{𝑛}$ with $𝐿_{𝑖} (𝜆_{𝑖}) = 1$ and $𝐿_{𝑖} (𝜆_{𝑗}) = 0$ for $𝑗 \neq 𝑖$ (Definition 3.4.26). By Proposition 7.2.28,

𝐿_{𝑖} (𝑇) 𝐮 = \sum_{𝑗} 𝑐_{𝑗} 𝐿_{𝑖} (𝜆_{𝑗}) 𝐯_{𝑗} = 𝑐_{𝑖} 𝐯_{𝑖},

and $𝐿_{𝑖} (𝑇) 𝐮 \in 𝑈$ by Proposition 7.2.27(d). So $𝑐_{𝑖} 𝐯_{𝑖} \in 𝑈$ for every $𝑖$ . If $𝑐_{𝑖} \neq 0$ , then $𝐯_{𝑖} =$ $𝑐_{𝑖}^{- 1} (𝑐_{𝑖} 𝐯_{𝑖}) \in$ $𝑈$ , so $𝑖 \in 𝐼$ . Hence $𝐮 =$ $\sum_{𝑖 \in 𝐼} 𝑐_{𝑖} 𝐯_{𝑖} \in$ $𝑈_{𝐼}$ , and $𝑈 = 𝑈_{𝐼}$ .

So the invariant subspaces are exactly the $𝑈_{𝐼}$ , one for each of the $2^{𝑛}$ subsets $𝐼$ . Without distinct eigenvalues the count can be infinite: every line in $ℝ^{2}$ is invariant under $id$ .

Exercise 7.2.49.standard Let

𝐯_{1}, \dots, 𝐯_{𝑛}

be eigenvectors of

𝑇 \in L (𝑉)

with distinct eigenvalues

𝜆_{1}, \dots, 𝜆_{𝑛}

, and let

𝐰 = 𝐯_{1} + \dots + 𝐯_{𝑛}

. Show that

𝐰,

𝑇 𝐰,

\dots,

𝑇^{𝑛 - 1} 𝐰

is linearly independent, and deduce that

𝐾_{𝑇} (𝐰) =

span (𝐯_{1}, \dots, 𝐯_{𝑛})

Solution.

Uses. Proposition 7.2.28 (Polynomials act on eigenvectors by their values), Theorem 7.2.21 (Eigenvectors for distinct eigenvalues are independent), Proposition 7.1.22 (Counting roots), Proposition 7.2.34 (The smallest invariant subspace containing a vector), Theorem 4.1.10 (Independent lists are short)

Hint 1. A relation

\sum_{𝑗 < 𝑛} 𝑐_{𝑗} 𝑇^{𝑗} 𝐰 =

𝟎

says

𝑝 (𝑇) 𝐰 = 𝟎

for

𝑝 = \sum 𝑐_{𝑗} 𝑥^{𝑗}

. Compute

𝑝 (𝑇) 𝐰

with Proposition 7.2.28.

Hint 2.

𝑝 (𝑇) 𝐰 =

\sum_{𝑖} 𝑝 (𝜆_{𝑖}) 𝐯_{𝑖}

. Independence of the

𝐯_{𝑖}

gives

𝑛

roots of

𝑝

, which has degree less than

𝑛

Sketch. If

𝑝 (𝑇) 𝐰 = 𝟎

with

deg 𝑝 < 𝑛

, then

\sum_{𝑖} 𝑝 (𝜆_{𝑖}) 𝐯_{𝑖} =

𝟎

, so

𝑝 (𝜆_{𝑖}) = 0

for all

𝑛

distinct

𝜆_{𝑖}

, and

𝑝 = 𝟎

(Proposition 7.1.22). So the

𝑛

vectors are independent. They lie in the invariant subspace

𝑊 = span (𝐯_{𝑖})

, which has dimension

𝑛

, so they span it; and

𝑊 \subseteq 𝐾_{𝑇} (𝐰) \subseteq 𝑊

Suppose $\sum_{𝑗 = 0}^{𝑛 - 1} 𝑐_{𝑗} 𝑇^{𝑗} 𝐰 =$ $𝟎$ , and let $𝑝 =$ $\sum_{𝑗 = 0}^{𝑛 - 1} 𝑐_{𝑗} 𝑥^{𝑗}$ . By Proposition 7.2.28,

𝟎 = 𝑝 (𝑇) 𝐰 = \sum_{𝑖 = 1}^{𝑛} 𝑝 (𝑇) 𝐯_{𝑖} = \sum_{𝑖 = 1}^{𝑛} 𝑝 (𝜆_{𝑖}) 𝐯_{𝑖} .

The $𝐯_{𝑖}$ are independent by Theorem 7.2.21, so $𝑝 (𝜆_{𝑖}) = 0$ for every $𝑖$ . Thus $𝑝$ has at least $𝑛$ distinct roots but degree less than $𝑛$ (if nonzero), so $𝑝 = 𝟎$ by Proposition 7.1.22, and all $𝑐_{𝑗} = 0$ .

Let $𝑊 =$ $span (𝐯_{1}, \dots, 𝐯_{𝑛})$ . It is $𝑇$ -invariant (Proposition 7.2.4) and contains $𝐰$ , so $𝐾_{𝑇} (𝐰) \subseteq 𝑊$ (Proposition 7.2.34). On the other hand, $𝐰, \dots, 𝑇^{𝑛 - 1} 𝐰$ are $𝑛$ independent vectors of $𝐾_{𝑇} (𝐰) \subseteq 𝑊$ , and $dim 𝑊 = 𝑛$ , so they span $𝑊$ (Theorem 4.1.10(b)) and $𝑊 \subseteq 𝐾_{𝑇} (𝐰)$ . Hence $𝐾_{𝑇} (𝐰) = 𝑊$ .

The independence can also be read off a determinant: in the basis $𝐯_{1}, \dots, 𝐯_{𝑛}$ of $𝑊$ , the coordinates of $𝑇^{𝑗} 𝐰$ are $(𝜆_{1}^{𝑗}, \dots, 𝜆_{𝑛}^{𝑗})$ , so the matrix with these columns is a Vandermonde matrix, which is invertible for distinct $𝜆_{𝑖}$ (Theorem 3.4.24).

Exercise 7.2.50.hard Let $dim 𝑉 = 𝑛 \geq 1$ , and let $𝑁 \in L (𝑉)$ satisfy $𝑁^{𝑛} = 0$ and $𝑁^{𝑛 - 1} \neq 0$ .

(a)Show that $dim ker 𝑁^{𝑗} = 𝑗$ for $𝑗 = 0, 1, \dots, 𝑛$ .
(b)Show that the $𝑁$ -invariant subspaces of $𝑉$ are exactly $ker 𝑁^{0},$ $ker 𝑁^{1},$ $\dots,$ $ker 𝑁^{𝑛}$ . In particular there are exactly $𝑛 + 1$ of them.

Solution.

Uses. Proposition 4.1.26 (Nilpotent chains are independent), Theorem 4.1.10 (Independent lists are short), Proposition 7.2.30 (Kernels and images of commuting operators are invariant), Proposition 4.1.21 (Dimensions of subspaces), Theorem 1.5.5 (Independence is decided at zero)

Techniques. Dimension counting (Technique 4.2.24)

Hint 1. Choose

𝐯

with

𝑁^{𝑛 - 1} 𝐯 \neq 𝟎

. The chain

𝐯,

𝑁 𝐯,

\dots,

𝑁^{𝑛 - 1} 𝐯

is a basis. Compute

𝑁^{𝑗}

on a general combination of it.

Hint 2. For (b), let

𝑈

be invariant and

𝑗

the least integer with

𝑁^{𝑗} (𝑈) = {𝟎}

. Find a chain of length

𝑗

inside

𝑈

Hint 3. Compare

dim 𝑈

with

dim ker 𝑁^{𝑗} = 𝑗

Sketch. (a) With

𝐯

as in the hint, the chain is independent (Proposition 4.1.26) of length

𝑛

, hence a basis, and

𝑁^{𝑗} (\sum_{𝑖} 𝑐_{𝑖} 𝑁^{𝑖} 𝐯) =

\sum_{𝑖 \leq 𝑛 - 1 - 𝑗} 𝑐_{𝑖} 𝑁^{𝑖 + 𝑗} 𝐯

vanishes iff

𝑐_{0} = \dots = 𝑐_{𝑛 - 1 - 𝑗} = 0

; so

ker 𝑁^{𝑗} =

span (𝑁^{𝑛 - 𝑗} 𝐯, \dots, 𝑁^{𝑛 - 1} 𝐯)

. (b) Kernels of powers are invariant. Conversely, if

𝑗

is least with

𝑈 \subseteq ker 𝑁^{𝑗}

and

𝑗 \geq 1

, some

𝐮 \in 𝑈

has

𝑁^{𝑗 - 1} 𝐮 \neq

𝟎 =

𝑁^{𝑗} 𝐮

, giving

𝑗

independent vectors in

𝑈

; so

𝑗 \leq

dim 𝑈 \leq

dim ker 𝑁^{𝑗} =

𝑗

, and

𝑈 = ker 𝑁^{𝑗}

(a) Choose $𝐯$ with $𝑁^{𝑛 - 1} 𝐯 \neq 𝟎$ ; then $𝑁^{𝑛} 𝐯 = 𝟎$ . By Proposition 4.1.26, $𝐯,$ $𝑁 𝐯,$ $\dots,$ $𝑁^{𝑛 - 1} 𝐯$ is independent, and it has length $𝑛 = dim 𝑉$ , so it is a basis (Theorem 4.1.10(b)). Let $0 \leq 𝑗 \leq 𝑛$ . For $𝐱 =$ $\sum_{𝑖 = 0}^{𝑛 - 1} 𝑐_{𝑖} 𝑁^{𝑖} 𝐯$ ,

𝑁^{𝑗} 𝐱 = \sum_{𝑖 = 0}^{𝑛 - 1} 𝑐_{𝑖} 𝑁^{𝑖 + 𝑗} 𝐯 = \sum_{𝑖 = 0}^{𝑛 - 1 - 𝑗} 𝑐_{𝑖} 𝑁^{𝑖 + 𝑗} 𝐯,

since $𝑁^{𝑖 + 𝑗} 𝐯 = 𝟎$ for $𝑖 + 𝑗 \geq 𝑛$ . The vectors $𝑁^{𝑖 + 𝑗} 𝐯$ , $0 \leq 𝑖 \leq 𝑛 - 1 - 𝑗$ , are distinct members of the basis, so $𝑁^{𝑗} 𝐱 = 𝟎$ if and only if $𝑐_{0} = \dots = 𝑐_{𝑛 - 1 - 𝑗} = 0$ . Hence $ker 𝑁^{𝑗} =$ $span (𝑁^{𝑛 - 𝑗} 𝐯, \dots, 𝑁^{𝑛 - 1} 𝐯)$ , which has dimension $𝑗$ .

(b) Each $ker 𝑁^{𝑗}$ is $𝑁$ -invariant by Proposition 7.2.30(a). Conversely, let $𝑈$ be $𝑁$ -invariant. Since $𝑁^{𝑛} = 0$ , there is a least $𝑗 \in {0, \dots, 𝑛}$ with $𝑁^{𝑗} 𝐮 = 𝟎$ for all $𝐮 \in 𝑈$ , that is, $𝑈 \subseteq ker 𝑁^{𝑗}$ . If $𝑗 = 0$ , then $𝑈 = {𝟎} = ker 𝑁^{0}$ . If $𝑗 \geq 1$ , minimality gives some $𝐮 \in 𝑈$ with $𝑁^{𝑗 - 1} 𝐮 \neq 𝟎$ , while $𝑁^{𝑗} 𝐮 = 𝟎$ . By Proposition 4.1.26, $𝐮,$ $𝑁 𝐮,$ $\dots,$ $𝑁^{𝑗 - 1} 𝐮$ is independent, and it lies in $𝑈$ because $𝑈$ is invariant. So

𝑗 \leq dim 𝑈 \leq dim ker 𝑁^{𝑗} = 𝑗,

and $𝑈 = ker 𝑁^{𝑗}$ by Proposition 4.1.21. By (a) the $𝑛 + 1$ subspaces $ker 𝑁^{0} ⊊ ker 𝑁^{1} ⊊ \dots ⊊ ker 𝑁^{𝑛} =$ $𝑉$ have different dimensions, so they are distinct.

Exercise 7.2.51.hard Let $𝑁 \in L (𝑉)$ , let $𝐮_{1}, \dots, 𝐮_{𝑟} \in 𝑉$ , and let $𝑘_{1}, \dots, 𝑘_{𝑟} \geq 0$ satisfy $𝑁^{𝑘_{𝑖} + 1} 𝐮_{𝑖} = 𝟎$ for each $𝑖$ . Suppose the “tops” $𝑁^{𝑘_{1}} 𝐮_{1},$ $\dots,$ $𝑁^{𝑘_{𝑟}} 𝐮_{𝑟}$ are linearly independent. Show that the list of all vectors

𝑁^{𝑗} 𝐮_{𝑖} (1 \leq 𝑖 \leq 𝑟, 0 \leq 𝑗 \leq 𝑘_{𝑖})

is linearly independent.

Solution.

Uses. Theorem 1.5.5 (Independence is decided at zero), Notation 2.1.18, Proposition 7.2.37 (Chains are independent)

Techniques. Extremal choice and minimal counterexamples (Technique 1.5.9), Apply an operator to a relation to kill terms (Technique 4.1.25)

Hint 1. For one chain this is Proposition 7.2.37. Imitate its proof by the lowest nonzero coefficient. A term

𝑎_{𝑖 𝑗} 𝑁^{𝑗} 𝐮_{𝑖}

sits

𝑘_{𝑖} - 𝑗

steps below the top of its chain.

Hint 2. Given a nontrivial relation, let

ℎ

be the largest value of

𝑘_{𝑖} - 𝑗

over the nonzero coefficients

𝑎_{𝑖 𝑗}

, and apply

𝑁^{ℎ}

Hint 3. Terms with

𝑘_{𝑖} - 𝑗 < ℎ

are pushed past the top and die; terms with

𝑘_{𝑖} - 𝑗 = ℎ

land on the tops. Use independence of the tops.

Sketch. Suppose

\sum_{𝑖, 𝑗} 𝑎_{𝑖 𝑗} 𝑁^{𝑗} 𝐮_{𝑖} =

𝟎

is nontrivial, and let

ℎ =

max {𝑘_{𝑖} - 𝑗 | 𝑎_{𝑖 𝑗} \neq 0}

. Applying

𝑁^{ℎ}

kills every term with

𝑘_{𝑖} - 𝑗 < ℎ

(as

𝑗 + ℎ \geq 𝑘_{𝑖} + 1

) and sends each term with

𝑘_{𝑖} - 𝑗 = ℎ

𝑎_{𝑖, 𝑘_{𝑖} - ℎ} 𝑁^{𝑘_{𝑖}} 𝐮_{𝑖}

; for each

𝑖

there is at most one such term. The result is a relation among the tops with at least one nonzero coefficient, a contradiction.

The hypothesis implies that each top $𝑁^{𝑘_{𝑖}} 𝐮_{𝑖}$ is nonzero. Suppose, for contradiction, that

\sum_{𝑖 = 1}^{𝑟} \sum_{𝑗 = 0}^{𝑘_{𝑖}} 𝑎_{𝑖 𝑗} 𝑁^{𝑗} 𝐮_{𝑖} = 𝟎

with not all $𝑎_{𝑖 𝑗}$ zero. Call $𝑘_{𝑖} - 𝑗$ the depth of the term $(𝑖, 𝑗)$ , and let $ℎ$ be the largest depth among the terms with $𝑎_{𝑖 𝑗} \neq 0$ . Then $0 \leq ℎ$ , and apply $𝑁^{ℎ}$ to the relation. For a term with $𝑎_{𝑖 𝑗} \neq 0$ :

if its depth is less than $ℎ$ , then $𝑗 + ℎ \geq 𝑘_{𝑖} + 1$ , so $𝑁^{ℎ} 𝑁^{𝑗} 𝐮_{𝑖} =$ $𝑁^{𝑗 + ℎ - 𝑘_{𝑖} - 1} 𝑁^{𝑘_{𝑖} + 1} 𝐮_{𝑖} =$ $𝟎$ ;
if its depth is $ℎ$ , then $𝑁^{ℎ} 𝑁^{𝑗} 𝐮_{𝑖} =$ $𝑁^{𝑘_{𝑖}} 𝐮_{𝑖}$ .

Terms with $𝑎_{𝑖 𝑗} = 0$ contribute nothing. For each $𝑖$ at most one $𝑗$ has depth $ℎ$ , namely $𝑗 = 𝑘_{𝑖} - ℎ$ . So, with $𝐼 =$ ${𝑖 | 𝑘_{𝑖} \geq ℎ and 𝑎_{𝑖, 𝑘_{𝑖} - ℎ} \neq 0}$ ,

\sum_{𝑖 \in 𝐼} 𝑎_{𝑖, 𝑘_{𝑖} - ℎ} 𝑁^{𝑘_{𝑖}} 𝐮_{𝑖} = 𝟎 .

By the choice of $ℎ$ , $𝐼$ is nonempty, so this is a nontrivial relation among the tops, contradicting their independence. Hence all $𝑎_{𝑖 𝑗} = 0$ , and the list is independent by Theorem 1.5.5.

For $𝑟 = 1$ the depth of the deepest nonzero term is $𝑘 - 𝑚$ , where $𝑚$ is the lowest index with $𝑎_{𝑚} \neq 0$ , and the argument is the second proof of Proposition 7.2.37. The statement is the independence half of the Jordan form (Section 10.3): to build a basis out of chains, it suffices to choose their tops independent.

Exercise 7.2.52.standard Let $𝑛 \geq 1$ and define $𝑇 : {𝐹 [𝑥]}_{\leq 𝑛} \to$ ${𝐹 [𝑥]}_{\leq 𝑛}$ by $𝑇 (𝑝) = 𝑥 𝑝^{'}$ , where $𝑝^{'}$ is the formal derivative.

(a)Over $ℝ$ , find the eigenvalues and eigenspaces of $𝑇$ .
(b)Over $𝔽_{3}$ with $𝑛 = 5$ , find the eigenvalues and eigenspaces of $𝑇$ .

Solution.

Uses. Example 2.1.9 (Maps of polynomials), Definition 7.2.13 (Eigenspace), Definition 1.2.13 (Polynomials)

Hint 1. Compute

𝑇 (𝑥^{𝑘})

Hint 2. For

𝑝 = \sum 𝑎_{𝑘} 𝑥^{𝑘}

, the equation

𝑇 (𝑝) = 𝜆 𝑝

says

(𝑘 - 𝜆) 𝑎_{𝑘} = 0

for every

𝑘

. In

𝔽_{3}

, which

𝑘 \in {0, \dots, 5}

are equal to a given

𝜆

Sketch.

𝑇 (𝑥^{𝑘}) = 𝑘 𝑥^{𝑘}

, so

𝑇 (\sum 𝑎_{𝑘} 𝑥^{𝑘}) =

𝜆 \sum 𝑎_{𝑘} 𝑥^{𝑘}

iff

(𝑘 - 𝜆) 𝑎_{𝑘} = 0

for all

𝑘

. Over

ℝ

: eigenvalues

0, 1, \dots, 𝑛

with

𝐸_{𝑘} = span (𝑥^{𝑘})

. Over

𝔽_{3}

𝑛 = 5

: eigenvalues

0, 1, 2

with

𝐸_{0} = span (1, 𝑥^{3})

𝐸_{1} = span (𝑥, 𝑥^{4})

𝐸_{2} = span (𝑥^{2}, 𝑥^{5})

For $𝑘 \geq 1$ , $𝑇 (𝑥^{𝑘}) =$ $𝑥 \cdot$ $𝑘 𝑥^{𝑘 - 1} =$ $𝑘 𝑥^{𝑘}$ , and $𝑇 (1) = 𝟎 = 0 \cdot 1$ . By linearity, for $𝑝 =$ $\sum_{𝑘 = 0}^{𝑛} 𝑎_{𝑘} 𝑥^{𝑘}$ ,

𝑇 (𝑝) - 𝜆 𝑝 = \sum_{𝑘 = 0}^{𝑛} (𝑘 - 𝜆) 𝑎_{𝑘} 𝑥^{𝑘},

which is $𝟎$ if and only if $(𝑘 - 𝜆) 𝑎_{𝑘} = 0$ for each $𝑘$ , where $𝑘$ means $1 + \dots + 1$ in $𝐹$ . So $𝐸_{𝜆} =$ $span (𝑥^{𝑘} | 0 \leq 𝑘 \leq 𝑛, 𝑘 = 𝜆 in 𝐹)$ , and $𝜆$ is an eigenvalue exactly when $𝜆$ equals some $𝑘 \in {0, \dots, 𝑛}$ in $𝐹$ .

(a) Over $ℝ$ the integers $0, 1, \dots, 𝑛$ are distinct, so the eigenvalues are $0, 1, \dots, 𝑛$ , with $𝐸_{𝑘} = span (𝑥^{𝑘})$ . These are $𝑛 + 1 = dim {ℝ [𝑥]}_{\leq 𝑛}$ distinct eigenvalues.

(b) Over $𝔽_{3}$ , $𝑘 = 𝜆$ in $𝔽_{3}$ when $𝑘 \equiv 𝜆$ (mod $3$ ). For $𝑘 \in {0, \dots, 5}$ the residues are $0, 1, 2, 0, 1, 2$ . So the eigenvalues are $0, 1, 2$ , with

𝐸_{0} = span (1, 𝑥^{3}), 𝐸_{1} = span (𝑥, 𝑥^{4}), 𝐸_{2} = span (𝑥^{2}, 𝑥^{5}) .

The dimensions add up to $6 = dim 𝔽_{3} {[𝑥]}_{\leq 5}$ , as Corollary 7.2.23 allows, and the monomials still form a basis of eigenvectors.

Exercise 7.2.53.standard Find the error in the following argument, and decide whether the claim is true.

Claim. Let $𝑇 \in L (𝑉)$ with $𝑉$ a complex vector space. Every eigenvector of $𝑇^{2}$ is an eigenvector of $𝑇$ .

“Proof.” Let $𝑇^{2} 𝐯 = 𝜇 𝐯$ with $𝐯 \neq 𝟎$ , and choose $𝜆 \in ℂ$ with $𝜆^{2} = 𝜇$ . Then

(𝑇 - 𝜆 id) (𝑇 + 𝜆 id) 𝐯 = (𝑇^{2} - 𝜆^{2} id) 𝐯 = 𝟎 .

So either $(𝑇 + 𝜆 id) 𝐯 = 𝟎$ or $(𝑇 - 𝜆 id) 𝐯 = 𝟎$ . In the first case $𝐯$ is an eigenvector of $𝑇$ for $- 𝜆$ , and in the second for $𝜆$ .

Solution.

Uses. Proposition 7.2.27 (Evaluating polynomials at an operator), Definition 7.2.6 (Eigenvalue, eigenvector)

Techniques. Finding counterexamples and minimal examples (Technique 1.4.18)

Hint 1. Let

𝐰 = (𝑇 + 𝜆 id) 𝐯

. The display says

(𝑇 - 𝜆 id) 𝐰 = 𝟎

. Does that say anything about

𝐯

when

𝐰 \neq 𝟎

Hint 2. Test the claim on

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

, for which

𝑇^{2} = 𝐼_{2}

Sketch. The display shows

(𝑇 - 𝜆 id) 𝐰 = 𝟎

for

𝐰 = (𝑇 + 𝜆 id) 𝐯

. If

𝐰 \neq 𝟎

, this makes

𝐰

, not

𝐯

, an eigenvector for

𝜆

; the “either/or” step confuses

𝐰

with

𝐯

. The claim is false:

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

has

𝑇^{2} = 𝐼_{2}

, so

𝐞_{1}

is an eigenvector of

𝑇^{2}

, but

𝑇 𝐞_{1} = 𝐞_{2}

The first display is correct, by Proposition 7.2.27(a). Write $𝐰 = (𝑇 + 𝜆 id) 𝐯$ ; the display says $(𝑇 - 𝜆 id) 𝐰 = 𝟎$ . The next sentence treats this as if it were a statement about $𝐯$ : from “ $𝐴 𝐵 𝐯 = 𝟎$ ” it concludes “ $𝐵 𝐯 = 𝟎$ or $𝐴 𝐯 = 𝟎$ ”. What actually follows is: either $𝐰 = 𝟎$ , in which case $𝐯$ is an eigenvector for $- 𝜆$ , or $𝐰 \neq 𝟎$ and $𝐰$ is an eigenvector of $𝑇$ for $𝜆$ . In the second case nothing has been shown about $𝐯$ .

The claim is false. Let $𝑇 = (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}) \in ℂ^{2 \times 2}$ , so $𝑇^{2} = 𝐼_{2}$ and every nonzero vector is an eigenvector of $𝑇^{2}$ . But $𝑇 𝐞_{1} = 𝐞_{2}$ is not a multiple of $𝐞_{1}$ . Following the corrected argument with $𝜇 = 1$ , $𝜆 = 1$ , $𝐯 = 𝐞_{1}$ : $𝐰 =$ $(𝑇 + id) 𝐞_{1} =$ $(1, 1)$ , which is indeed an eigenvector of $𝑇$ for $1$ . Over $ℝ$ the argument fails even earlier, since $𝜇$ may have no real square root: $𝑅_{𝜋 / 2}^{2} = - id$ (Example 7.2.18).

Exercise 7.2.54.standard Let $𝑇 \in L (𝑉)$ , and let $𝑈, 𝑊$ be subspaces of $𝑉$ . Prove or disprove:

(a)If $𝑈$ and $𝑊$ are $𝑇$ -invariant, then so are $𝑈 + 𝑊$ and $𝑈 \cap 𝑊$ .
(b)If $𝑈$ is $𝑇^{2}$ -invariant, then $𝑈$ is $𝑇$ -invariant.
(c)If $𝑉$ is finite-dimensional, $𝑇$ is invertible, and $𝑈$ is $𝑇$ -invariant, then $𝑈$ is $𝑇^{- 1}$ -invariant.
(d)Statement (c) holds without assuming that $𝑉$ is finite-dimensional.

Solution.

Uses. Definition 7.2.1 (Invariant subspace), Corollary 4.2.18 (Injective if and only if surjective), Example 7.2.18 (Rotations)

Techniques. Finding counterexamples and minimal examples (Technique 1.4.18)

Hint 1. (a) is a generic-element check. For (b), find a

𝑇

with

𝑇^{2} = - id

Hint 2. For (c),

𝑇 ↾_{𝑈}

is an injective operator on the finite-dimensional

𝑈

. For (d), look for an invertible shift on sequences indexed by all of

ℤ

Sketch. (a) True:

𝑇 (𝐮 + 𝐰) =

𝑇 𝐮 +

𝑇 𝐰 \in

𝑈 +

𝑊

, and

𝑇

maps

𝑈 \cap 𝑊

into both. (b) False:

𝑅_{𝜋 / 2}^{2} = - id

leaves every line invariant, but

𝑅_{𝜋 / 2}

leaves none. (c) True:

𝑇 ↾_{𝑈} \in L (𝑈)

is injective, hence surjective, so each

𝐮 \in 𝑈

𝑇 𝐮^{'}

with

𝐮^{'} \in 𝑈

, and

𝑇^{- 1} 𝐮 = 𝐮^{'} \in 𝑈

. (d) False: on

𝐹^{ℤ}

the right shift is invertible, and the sequences vanishing at negative indices form a subspace invariant under it but not under its inverse.

(a) True. If $𝐮 \in 𝑈$ and $𝐰 \in 𝑊$ , then $𝑇 (𝐮 + 𝐰) =$ $𝑇 𝐮 +$ $𝑇 𝐰 \in$ $𝑈 +$ $𝑊$ . If $𝐱 \in 𝑈 \cap 𝑊$ , then $𝑇 𝐱 \in 𝑈$ and $𝑇 𝐱 \in 𝑊$ .

(b) False. Let $𝑇 = 𝑅_{𝜋 / 2}$ on $ℝ^{2}$ . Then $𝑇^{2} = - id$ , under which every subspace is invariant, but no line is $𝑇$ -invariant (Example 7.2.18). Take $𝑈 = span (𝐞_{1})$ .

(c) True. Since $𝑇$ is injective, so is $𝑇 ↾_{𝑈} \in L (𝑈)$ , and $𝑈$ is finite-dimensional, so $𝑇 ↾_{𝑈}$ is surjective onto $𝑈$ (Corollary 4.2.18). Given $𝐮 \in 𝑈$ , choose $𝐮^{'} \in 𝑈$ with $𝑇 𝐮^{'} = 𝐮$ . Then $𝑇^{- 1} 𝐮 = 𝐮^{'} \in 𝑈$ .

(d) False. Let $𝑉 = 𝐹^{ℤ}$ be the space of two-sided sequences ${(𝑥_{𝑛})}_{𝑛 \in ℤ}$ , and let $𝑇$ be the right shift, ${(𝑇 𝐱)}_{𝑛} = 𝑥_{𝑛 - 1}$ . It is invertible, with inverse the left shift ${(𝑇^{- 1} 𝐱)}_{𝑛} = 𝑥_{𝑛 + 1}$ . The subspace $𝑈 =$ ${𝐱 | 𝑥_{𝑛} = 0 for all 𝑛 < 0}$ is $𝑇$ -invariant: if $𝑥_{𝑛} = 0$ for $𝑛 < 0$ , then ${(𝑇 𝐱)}_{𝑛} = 𝑥_{𝑛 - 1} = 0$ for $𝑛 < 0$ . But the sequence $𝐱$ with $𝑥_{0} = 1$ and all other terms $0$ lies in $𝑈$ , while ${(𝑇^{- 1} 𝐱)}_{- 1} = 𝑥_{0} = 1$ , so $𝑇^{- 1} 𝐱 \notin 𝑈$ . The proof of (c) fails because $𝑇 ↾_{𝑈}$ is injective but not surjective onto $𝑈$ .

Exercise 7.2.55.hardBeyond 21-242 Let $𝑇 \in L (𝑉)$ and let $𝑝, 𝑞 \in 𝐹 [𝑥]$ be coprime, with $𝑝 (𝑇) 𝑞 (𝑇) = 0$ . Show that

𝑉 = ker 𝑝 (𝑇) \oplus ker 𝑞 (𝑇),

and that both summands are $𝑇$ -invariant. Deduce that if $1 + 1 \neq 0$ in $𝐹$ and $𝑇^{2} = id$ , then $𝑉 = 𝐸_{1} \oplus 𝐸_{- 1}$ .

Solution.

Uses. Corollary 7.1.48 (Coprime polynomials), Proposition 7.2.27 (Evaluating polynomials at an operator), Proposition 7.2.30 (Kernels and images of commuting operators are invariant), Theorem 4.4.7 (Criteria for direct sums), Definition 4.4.4 (Direct sum)

Techniques. Look for invariant subspaces among kernels and images of commuting operators (Technique 7.2.31)

Hint 1. By Bézout (Corollary 7.1.48),

𝑢 𝑝 + 𝑣 𝑞 = 1

for some

𝑢, 𝑣 \in 𝐹 [𝑥]

. Evaluate at

𝑇

id =

𝑢 (𝑇) 𝑝 (𝑇) +

𝑣 (𝑇) 𝑞 (𝑇)

Hint 2. Apply this to

𝐯

. Which kernel does

𝑢 (𝑇) 𝑝 (𝑇) 𝐯

lie in, and which does

𝑣 (𝑇) 𝑞 (𝑇) 𝐯

lie in?

Hint 3. For the intersection, apply the same identity to a vector killed by both

𝑝 (𝑇)

and

𝑞 (𝑇)

Sketch. From

𝑢 𝑝 + 𝑣 𝑞 = 1

id =

𝑢 (𝑇) 𝑝 (𝑇) +

𝑣 (𝑇) 𝑞 (𝑇)

. For

𝐱 \in 𝑉

𝑞 (𝑇) 𝑢 (𝑇) 𝑝 (𝑇) 𝐱 =

𝑢 (𝑇) (𝑝 𝑞) (𝑇) 𝐱 =

𝟎

, so the first term lies in

ker 𝑞 (𝑇)

, and likewise the second in

ker 𝑝 (𝑇)

𝑉 =

ker 𝑝 (𝑇) +

ker 𝑞 (𝑇)

. If

𝐱

is in both kernels, the identity gives

𝐱 = 𝟎

. Invariance is Proposition 7.2.30. For

𝑇^{2} = id

, take

𝑝 = 𝑥 - 1

𝑞 = 𝑥 + 1

, coprime since

1 =

((𝑥 + 1) - (𝑥 - 1)) / 2

By Corollary 7.1.48 there are $𝑢, 𝑣 \in 𝐹 [𝑥]$ with $𝑢 𝑝 + 𝑣 𝑞 = 1$ . By Proposition 7.2.27(a),

id = 𝑢 (𝑇) 𝑝 (𝑇) + 𝑣 (𝑇) 𝑞 (𝑇) .

(7.2.4)

Let $𝐱 \in 𝑉$ , and set $𝐲 = 𝑣 (𝑇) 𝑞 (𝑇) 𝐱$ and $𝐳 = 𝑢 (𝑇) 𝑝 (𝑇) 𝐱$ , so that $𝐱 = 𝐲 + 𝐳$ . Since polynomials in $𝑇$ commute (Proposition 7.2.27(b)),

𝑝 (𝑇) 𝐲 = 𝑣 (𝑇) 𝑝 (𝑇) 𝑞 (𝑇) 𝐱 = 𝟎, 𝑞 (𝑇) 𝐳 = 𝑢 (𝑇) 𝑝 (𝑇) 𝑞 (𝑇) 𝐱 = 𝟎 .

So $𝐲 \in ker 𝑝 (𝑇)$ and $𝐳 \in ker 𝑞 (𝑇)$ , and $𝑉 =$ $ker 𝑝 (𝑇) +$ $ker 𝑞 (𝑇)$ . If $𝐱 \in$ $ker 𝑝 (𝑇) \cap$ $ker 𝑞 (𝑇)$ , then (7.2.4) gives $𝐱 =$ $𝑢 (𝑇) 𝑝 (𝑇) 𝐱 +$ $𝑣 (𝑇) 𝑞 (𝑇) 𝐱 =$ $𝟎$ . So the sum is direct (Theorem 4.4.7). Both kernels are $𝑇$ -invariant by Proposition 7.2.30(a).

If $𝑇^{2} = id$ and $2 \neq 0$ , take $𝑝 = 𝑥 - 1$ and $𝑞 = 𝑥 + 1$ . Then $𝑝 𝑞 = 𝑥^{2} - 1$ , so $𝑝 (𝑇) 𝑞 (𝑇) =$ $𝑇^{2} -$ $id =$ $0$ , and $2^{- 1} 𝑞 - 2^{- 1} 𝑝 = 1$ , so $𝑝$ and $𝑞$ are coprime (Corollary 7.1.48). Hence $𝑉 =$ $ker (𝑇 - id) \oplus$ $ker (𝑇 + id) =$ $𝐸_{1} \oplus$ $𝐸_{- 1}$ . Over $𝔽_{2}$ this fails: the swap of Example 7.2.17 satisfies $𝜎^{2} = id$ , but $𝐸_{1}$ is a line in $𝔽_{2}^{2}$ and $𝐸_{- 1} = 𝐸_{1}$ .

Exercise 7.2.56.standard Let

𝑍 (𝑥, 𝑦, 𝑧) =

(𝑧, 𝑥, 𝑦)

be the cyclic permutation of coordinates on

𝐹^{3}

. Find all eigenvalues of

𝑍

, and an eigenvector for each, over

ℂ

, over

ℝ

, over

𝔽_{7}

, and over

𝔽_{3}

Solution.

Uses. Definition 7.2.6 (Eigenvalue, eigenvector), Proposition 7.2.28 (Polynomials act on eigenvectors by their values), Definition 7.2.13 (Eigenspace)

Hint 1.

𝑍^{3} = id

, so every eigenvalue is a cube root of

1

(Proposition 7.2.28). Conversely, for

𝜆^{3} = 1

, solve

𝑍 𝐯 = 𝜆 𝐯

starting from

𝑥 = 1

Hint 2. The equations are

𝑧 = 𝜆 𝑥

𝑥 = 𝜆 𝑦

𝑦 = 𝜆 𝑧

. Try

𝐯 = (1, 𝜆^{2}, 𝜆)

Sketch.

𝑍 𝐯 = 𝜆 𝐯

means

𝑧 = 𝜆 𝑥

𝑦 = 𝜆 𝑧 = 𝜆^{2} 𝑥

𝑥 = 𝜆 𝑦 = 𝜆^{3} 𝑥

, so

𝐯 = 𝑥 (1, 𝜆^{2}, 𝜆)

and

(𝜆^{3} - 1) 𝑥 = 0

. Eigenvalues are the cube roots of

1

𝐹

, each with eigenspace

span ((1, 𝜆^{2}, 𝜆))

ℂ

1, 𝜔, 𝜔^{2}

ℝ

1

𝔽_{7}

1, 2, 4

𝔽_{3}

1

only.

$𝑍 𝐯 = 𝜆 𝐯$ means $(𝑧, 𝑥, 𝑦) =$ $(𝜆 𝑥, 𝜆 𝑦, 𝜆 𝑧)$ , that is, $𝑧 = 𝜆 𝑥$ , $𝑥 = 𝜆 𝑦$ , $𝑦 = 𝜆 𝑧$ . Then $𝑦 = 𝜆 𝑧 = 𝜆^{2} 𝑥$ and $𝑥 = 𝜆 𝑦 = 𝜆^{3} 𝑥$ . So every solution is $𝐯 = 𝑥 (1, 𝜆^{2}, 𝜆)$ , and it is nonzero only if $𝑥 \neq 0$ and $𝜆^{3} = 1$ . Conversely, if $𝜆^{3} = 1$ , then

𝑍 (1, 𝜆^{2}, 𝜆) = (𝜆, 1, 𝜆^{2}) = (𝜆, 𝜆^{3}, 𝜆^{2}) = 𝜆 (1, 𝜆^{2}, 𝜆) .

So the eigenvalues are the cube roots of $1$ in $𝐹$ , and $𝐸_{𝜆} =$ $span ((1, 𝜆^{2}, 𝜆))$ for each.

Over $ℂ$ : $1, 𝜔, 𝜔^{2}$ with $𝜔 =$ $𝑒^{2 𝜋 𝑖 / 3} =$ $(- 1 + 𝑖 \sqrt{3}) / 2$ , and eigenvectors $(1, 1, 1)$ , $(1, 𝜔^{2}, 𝜔)$ , $(1, 𝜔, 𝜔^{2})$ .
Over $ℝ$ : only $1$ , since $𝑡 \mapsto 𝑡^{3}$ is injective on $ℝ$ . Eigenvector $(1, 1, 1)$ .
Over $𝔽_{7}$ : $2^{3} = 8 = 1$ and $4^{3} = 64 = 1$ , and $3^{3} = 6$ , $5^{3} = 6$ , $6^{3} = 6$ are not $1$ . So the eigenvalues are $1, 2, 4$ , with eigenvectors $(1, 1, 1)$ , $(1, 4, 2)$ , and $(1, 2, 4)$ .
Over $𝔽_{3}$ : $0^{3} = 0$ , $1^{3} = 1$ , $2^{3} = 8 = 2$ . Only $1$ , with $𝐸_{1} = span ((1, 1, 1))$ .

Over $ℂ$ and $𝔽_{7}$ there are $3$ distinct eigenvalues, so the three eigenvectors form a basis (Theorem 7.2.21). Over $𝔽_{3}$ there is only one eigenvector line.

Exercise 7.2.57.hard Let

dim 𝑉 = 𝑛

and

1 \leq 𝑘 < 𝑛

. Suppose

𝑇 \in L (𝑉)

leaves every

𝑘

-dimensional subspace of

𝑉

invariant. Show that

𝑇 = 𝜆 id

for some

𝜆 \in 𝐹

Solution.

Uses. Theorem 4.1.16 (Extending to a basis), Exercise 7.2.54, Proposition 7.2.7 (Invariant lines are spanned by eigenvectors), Exercise 2.1.48

Hint 1. Show that every line

span (𝐯)

is invariant. Intersections of invariant subspaces are invariant.

Hint 2. For

𝐯 \neq 𝟎

and

𝐰 \notin span (𝐯)

, build a

𝑘

-dimensional subspace containing

𝐯

but not

𝐰

by extending

𝐯, 𝐰

to a basis.

Hint 3. So

span (𝐯)

is the intersection of all

𝑘

-dimensional subspaces that contain

𝐯

. Finish with Exercise 2.1.48.

Sketch. Fix

𝐯 \neq 𝟎

. For

𝐰 \notin span (𝐯)

, extend

𝐯, 𝐰

to a basis

𝐯, 𝐰, 𝐛_{3}, \dots, 𝐛_{𝑛}

and let

𝑈 =

span (𝐯, 𝐛_{3}, \dots, 𝐛_{𝑘 + 1})

, of dimension

𝑘

, containing

𝐯

but not

𝐰

. So

span (𝐯)

is the intersection of the

𝑘

-dimensional subspaces containing

𝐯

, all invariant, hence invariant. Every vector is then an eigenvector or

𝟎

, so

𝑇

is scalar.

Fix $𝐯 \neq 𝟎$ , and let $𝑋$ be the intersection of all $𝑘$ -dimensional subspaces of $𝑉$ that contain $𝐯$ . There is at least one such subspace, since $𝑘 \geq 1$ (extend $𝐯$ to a basis and take the span of $𝐯$ and $𝑘 - 1$ further basis vectors). So $𝑋$ is a subspace containing $span (𝐯)$ . We show $𝑋 = span (𝐯)$ . Let $𝐰 \notin span (𝐯)$ . Then $𝐯, 𝐰$ is independent, and it extends to a basis $𝐯, 𝐰, 𝐛_{3}, \dots, 𝐛_{𝑛}$ of $𝑉$ (Theorem 4.1.16). Since $𝑘 - 1 \leq 𝑛 - 2$ , the list $𝐯, 𝐛_{3}, \dots, 𝐛_{𝑘 + 1}$ makes sense; its span $𝑈$ has dimension $𝑘$ and contains $𝐯$ , but not $𝐰$ , because $𝐰$ is not a combination of the other basis vectors. So $𝐰 \notin 𝑋$ , and $𝑋 = span (𝐯)$ .

Each of these $𝑘$ -dimensional subspaces is $𝑇$ -invariant by hypothesis, so $𝑋$ is $𝑇$ -invariant: if $𝐱 \in 𝑋$ , then $𝑇 𝐱$ lies in each of them, as in Exercise 7.2.54(a). So $𝑇 𝐯 \in span (𝐯)$ for every $𝐯 \neq 𝟎$ , and trivially for $𝐯 = 𝟎$ . By Exercise 2.1.48, $𝑇 = 𝜆 id$ for some $𝜆$ . (The converse is clear, and the hypothesis $𝑘 < 𝑛$ is needed: every operator leaves the one $𝑛$ -dimensional subspace $𝑉$ invariant.)

Exercise 7.2.58.routine Find

𝐴 \in ℚ^{2 \times 2}

with

𝐴 (1, 2) = 3 (1, 2)

and

𝐴 (1, 1) = - (1, 1)

. Is

𝐴

unique?

Solution.

Uses. Theorem 2.2.1 (Linear maps are determined by a basis), Proposition 2.5.28 (Inverse of a $2 \times 2$ matrix)

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

Hint 1.

(1, 2), (1, 1)

is a basis, so Theorem 2.2.1 gives exactly one such map.

Hint 2. With

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

, the conditions say

𝐴 𝑆 = 𝑆 diag (3, - 1)

. Solve for

𝐴

Sketch.

𝐴 𝑆 = 𝑆 𝐷

with

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

and

𝐷 = diag (3, - 1)

, so

𝐴 =

𝑆 𝐷 𝑆^{- 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})

. It is unique because a linear map is determined by its values on a basis.

The vectors $(1, 2)$ and $(1, 1)$ form a basis of $ℚ^{2}$ , so by Theorem 2.2.1 there is exactly one linear map with the prescribed values, and $𝐴$ is unique. To find it, let $𝑆 = (\begin{matrix} 1 & 1 \\ 2 & 1 \end{matrix})$ , whose columns are the two vectors, and $𝐷 = diag (3, - 1)$ . The conditions say $𝐴 𝑆 = 𝑆 𝐷$ , column by column. By Proposition 2.5.28, $det 𝑆 = - 1$ and $𝑆^{- 1} = (\begin{matrix} - 1 & 1 \\ 2 & - 1 \end{matrix})$ , so

𝐴 = 𝑆 𝐷 𝑆^{- 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: $𝐴 (1, 2) =$ $(- 5 + 8, - 8 + 14) =$ $(3, 6)$ and $𝐴 (1, 1) = (- 1, - 1)$ .

More practice

The practice bank has 36 exercises on this section’s topics.