2.2 Linear maps are determined by a basis

Suppose $𝑇 : ℝ^{2} \to ℝ^{2}$ is linear, with $𝑇 (1, 1) = (2, 0)$ and $𝑇 (1, - 1) = (0, 4)$ . Then $𝑇$ is already pinned down. For instance, $(3, 1) =$ $2 (1, 1) +$ $(1, - 1)$ , so linearity forces

𝑇 (3, 1) = 2 𝑇 (1, 1) + 𝑇 (1, - 1) = (4, 0) + (0, 4) = (4, 4),

and the same computation works at every vector, because every vector of $ℝ^{2}$ is a linear combination of $(1, 1)$ and $(1, - 1)$ . Two questions remain. Is there a linear map with these two values at all? And if we wrote $(3, 1)$ as a different combination, could the computation give a different answer? This section answers both questions for every basis of every vector space: a linear map may send the vectors of a basis anywhere, and once it has done so, it is determined. This is what will let us record a linear map by finitely many vectors, and in Section 2.3 by an array of scalars.

2.2.1Existence and uniqueness

Intuition (Where the formula comes from). Let $ℬ︀ = 𝐯_{1}, \dots, 𝐯_{𝑛}$ be a basis of $𝑉$ , and suppose $𝑇 : 𝑉 \to 𝑊$ is linear. Every $𝐯 \in 𝑉$ is a linear combination $𝐯 =$ $𝑎_{1} 𝐯_{1} +$ $\dots +$ $𝑎_{𝑛} 𝐯_{𝑛}$ , and linearity gives

𝑇 (𝐯) = 𝑎_{1} 𝑇 (𝐯_{1}) + \dots + 𝑎_{𝑛} 𝑇 (𝐯_{𝑛}) .

So the $𝑛$ vectors $𝑇 (𝐯_{1}),$ $\dots,$ $𝑇 (𝐯_{𝑛})$ determine $𝑇$ . Now turn this around. Given vectors $𝐰_{1}, \dots, 𝐰_{𝑛} \in 𝑊$ , a linear map with $𝑇 (𝐯_{𝑖}) = 𝐰_{𝑖}$ , if there is one, can only be

𝑇 (𝑎_{1} 𝐯_{1} + \dots + 𝑎_{𝑛} 𝐯_{𝑛}) = 𝑎_{1} 𝐰_{1} + \dots + 𝑎_{𝑛} 𝐰_{𝑛} .

Read this as a recipe: to find $𝑇 (𝐯)$ , write $𝐯$ in terms of the basis, then replace each $𝐯_{𝑖}$ by $𝐰_{𝑖}$ . For the recipe to define a function $𝑉 \to 𝑊$ , two things must hold. Every $𝐯$ must have a representation, or the recipe says nothing about $𝐯$ ; spanning provides this. And no $𝐯$ may have two different representations, or the recipe could produce two different answers; linear independence provides this. A basis provides both, and then it remains to check that the function we get is linear.

Theorem 2.2.1 (Linear maps are determined by a basis). Let $𝑉$ and $𝑊$ be vector spaces over $𝐹$ , let $ℬ︀ = 𝐯_{1}, \dots, 𝐯_{𝑛}$ be a basis of $𝑉$ , and let $𝐰_{1}, \dots, 𝐰_{𝑛}$ be any list in $𝑊$ . Then there is exactly one linear map $𝑇 : 𝑉 \to 𝑊$ such that $𝑇 (𝐯_{𝑖}) = 𝐰_{𝑖}$ for $𝑖 = 1, \dots, 𝑛$ . It is given by

𝑇 (𝑎_{1} 𝐯_{1} + \dots + 𝑎_{𝑛} 𝐯_{𝑛}) = 𝑎_{1} 𝐰_{1} + \dots + 𝑎_{𝑛} 𝐰_{𝑛} for all 𝑎_{1}, \dots, 𝑎_{𝑛} \in 𝐹 .

(2.2.1)

That is, $𝑇 (𝐯) =$ $𝑐_{1} 𝐰_{1} +$ $\dots +$ $𝑐_{𝑛} 𝐰_{𝑛}$ , where $(𝑐_{1}, \dots, 𝑐_{𝑛}) =$ ${[𝐯]}_{ℬ︀}$ .

In other words, to specify a linear map out of $𝑉$ it suffices to say what it does to a basis of $𝑉$ , and every choice is allowed.

Proof, part 1 of 2: uniqueness.routine

Uses. Definition 1.3.10 (Span), Definition 2.1.1 (Linear map)

Techniques. Uniqueness via two candidates (Technique 1.1.8)

Hint 1. Suppose

𝑆

and

𝑇

both work. To compare them at an arbitrary

𝐯 \in 𝑉

, write

𝐯

in terms of the basis.

Hint 2. Apply the linearity of

𝑆

and of

𝑇

𝐯 =

𝑎_{1} 𝐯_{1} +

\dots +

𝑎_{𝑛} 𝐯_{𝑛}

. Both give

𝑎_{1} 𝐰_{1} + \dots + 𝑎_{𝑛} 𝐰_{𝑛}

Sketch. Since

ℬ︀

spans

𝑉

, every

𝐯

is some

\sum_{𝑖} 𝑎_{𝑖} 𝐯_{𝑖}

. Linearity gives

𝑆 (𝐯) =

\sum_{𝑖} 𝑎_{𝑖} 𝐰_{𝑖} =

𝑇 (𝐯)

. The same computation shows that any such map satisfies (2.2.1).

Suppose $𝑆, 𝑇 \in L (𝑉, 𝑊)$ both satisfy $𝑆 (𝐯_{𝑖}) =$ $𝐰_{𝑖} =$ $𝑇 (𝐯_{𝑖})$ for every $𝑖$ . We will show that $𝑆 (𝐯) = 𝑇 (𝐯)$ for every $𝐯 \in 𝑉$ . Let $𝐯 \in 𝑉$ . Since $ℬ︀$ spans $𝑉$ , there are scalars $𝑎_{1}, \dots, 𝑎_{𝑛}$ with $𝐯 =$ $𝑎_{1} 𝐯_{1} +$ $\dots +$ $𝑎_{𝑛} 𝐯_{𝑛}$ . Then

\begin{aligned} 𝑆 (𝐯) & = 𝑎_{1} 𝑆 (𝐯_{1}) + \dots + 𝑎_{𝑛} 𝑆 (𝐯_{𝑛}) & (linearity of 𝑆) \\ = 𝑎_{1} 𝐰_{1} + \dots + 𝑎_{𝑛} 𝐰_{𝑛} \\ = 𝑎_{1} 𝑇 (𝐯_{1}) + \dots + 𝑎_{𝑛} 𝑇 (𝐯_{𝑛}) = 𝑇 (𝐯) & (linearity of 𝑇) . \end{aligned}

So $𝑆 = 𝑇$ . The first two lines show more: every linear map that sends each $𝐯_{𝑖}$ to $𝐰_{𝑖}$ satisfies (2.2.1).

Intuition. The uniqueness proof did more than prove uniqueness: it found the only possible candidate, the map given by (2.2.1). This is a common way to organize an existence-and-uniqueness proof. First assume that a solution exists and work out what it must be, which proves uniqueness; then check that the forced candidate really is a solution, which proves existence. Here the check has three parts: (2.2.1) defines a function

𝑉 \to 𝑊

, this function sends

𝐯_{𝑖}

𝐰_{𝑖}

, and it is linear. The uniqueness half used only that

ℬ︀

spans

𝑉

. Independence enters in the first part of the check.

Proof (directly), part 2 of 2: existence.routine

Uses. Definition 1.6.1 (Basis), Definition 1.5.1 (Linear independence), Definition 1.6.14 (Coordinates), Theorem 2.1.3 (Checking linearity)

Hint 1. Define

𝑇 (𝐯)

by (2.2.1), using the coordinates of

𝐯

. Why is this a single, well-defined vector of

𝑊

Hint 2. What are the coordinates of

𝐯_{𝑖}

itself?

Hint 3. For linearity, if

𝐮 = \sum_{𝑖} 𝑏_{𝑖} 𝐯_{𝑖}

and

𝐯 = \sum_{𝑖} 𝑐_{𝑖} 𝐯_{𝑖}

, then

𝑎 𝐮 +

𝐯 =

\sum_{𝑖} (𝑎 𝑏_{𝑖} + 𝑐_{𝑖}) 𝐯_{𝑖}

, and by uniqueness of representations this is the representation of

𝑎 𝐮 + 𝐯

Sketch. Each

𝐯

has exactly one coordinate list

(𝑐_{1}, \dots, 𝑐_{𝑛})

(at least one by spanning, at most one by independence), so

𝑇 (𝐯) =

\sum_{𝑖} 𝑐_{𝑖} 𝐰_{𝑖}

is a well-defined function. The coordinates of

𝐯_{𝑖}

are

𝐞_{𝑖}

, so

𝑇 (𝐯_{𝑖}) = 𝐰_{𝑖}

. The coordinates of

𝑎 𝐮 + 𝐯

are

𝑎 {[𝐮]}_{ℬ︀} + {[𝐯]}_{ℬ︀}

, which gives linearity.

$𝑇$ is a function $𝑉 \to 𝑊$ . Let $𝐯 \in 𝑉$ . Since $ℬ︀$ is a basis, $𝐯$ can be written as a linear combination of $ℬ︀$ in exactly one way (Definition 1.6.1): in at least one way because $ℬ︀$ spans $𝑉$ , and in at most one way because $ℬ︀$ is linearly independent (Definition 1.5.1). Let $(𝑐_{1}, \dots, 𝑐_{𝑛}) =$ ${[𝐯]}_{ℬ︀}$ be the coefficients (Definition 1.6.14), and define

𝑇 (𝐯) = 𝑐_{1} 𝐰_{1} + \dots + 𝑐_{𝑛} 𝐰_{𝑛} .

This assigns to each $𝐯 \in 𝑉$ exactly one vector of $𝑊$ . If $𝐯 =$ $𝑎_{1} 𝐯_{1} +$ $\dots +$ $𝑎_{𝑛} 𝐯_{𝑛}$ is any representation, then $𝑎_{𝑖} = 𝑐_{𝑖}$ for every $𝑖$ , so $𝑇$ satisfies (2.2.1).

$𝑇 (𝐯_{𝑖}) = 𝐰_{𝑖}$ . Since $𝐯_{𝑖} =$ $0 𝐯_{1} +$ $\dots +$ $1 𝐯_{𝑖} +$ $\dots +$ $0 𝐯_{𝑛}$ , (2.2.1) gives $𝑇 (𝐯_{𝑖}) =$ $0 𝐰_{1} +$ $\dots +$ $1 𝐰_{𝑖} +$ $\dots +$ $0 𝐰_{𝑛} =$ $𝐰_{𝑖}$ .

$𝑇$ is linear. Let $𝐮, 𝐯 \in 𝑉$ and $𝑎 \in 𝐹$ , and write $𝐮 = \sum_{𝑖} 𝑏_{𝑖} 𝐯_{𝑖}$ and $𝐯 = \sum_{𝑖} 𝑐_{𝑖} 𝐯_{𝑖}$ . By the vector space axioms,

𝑎 𝐮 + 𝐯 = \sum_{𝑖 = 1}^{𝑛} (𝑎 𝑏_{𝑖} + 𝑐_{𝑖}) 𝐯_{𝑖},

and since (2.2.1) holds for every representation,

𝑇 (𝑎 𝐮 + 𝐯) = \sum_{𝑖 = 1}^{𝑛} (𝑎 𝑏_{𝑖} + 𝑐_{𝑖}) 𝐰_{𝑖} = 𝑎 \sum_{𝑖 = 1}^{𝑛} 𝑏_{𝑖} 𝐰_{𝑖} + \sum_{𝑖 = 1}^{𝑛} 𝑐_{𝑖} 𝐰_{𝑖} = 𝑎 𝑇 (𝐮) + 𝑇 (𝐯) .

By Theorem 2.1.3, $𝑇$ is linear. □

The recipe “take coordinates, then form the combination with the $𝐰_{𝑖}$ ” is a composite of two maps that we already know to be linear. Seeing it that way gives a second proof of existence with no computation at all.

Proof (via coordinates), part 2 of 2: existence.routine

Uses. Definition 1.6.14 (Coordinates), Proposition 1.6.19 (Coordinates respect linear combinations), Example 2.1.8 (Coordinates), Proposition 2.1.17 (Compositions of linear maps are linear)

Techniques. Assemble linear maps from linear pieces (Technique 2.1.26)

Hint 1. The coordinate map

𝑉 \to 𝐹^{𝑛}

is linear, and so is

𝐹^{𝑛} \to 𝑊

(𝑎_{1}, \dots, 𝑎_{𝑛}) \mapsto

\sum_{𝑖} 𝑎_{𝑖} 𝐰_{𝑖}

Hint 2. Compose them, and compute the composite at

𝑎_{1} 𝐯_{1} + \dots + 𝑎_{𝑛} 𝐯_{𝑛}

Sketch. Let

𝐶 (𝐯) = {[𝐯]}_{ℬ︀}

and

𝐿 (𝑎_{1}, \dots, 𝑎_{𝑛}) =

\sum_{𝑖} 𝑎_{𝑖} 𝐰_{𝑖}

. Both are linear, so

𝑇 = 𝐿 𝐶

is linear, and

𝑇 (\sum_{𝑖} 𝑎_{𝑖} 𝐯_{𝑖}) =

𝐿 (𝑎_{1}, \dots, 𝑎_{𝑛}) =

\sum_{𝑖} 𝑎_{𝑖} 𝐰_{𝑖}

Let $𝐶 : 𝑉 \to 𝐹^{𝑛}$ be the coordinate map $𝐯 \mapsto {[𝐯]}_{ℬ︀}$ , which is linear by Proposition 1.6.19, and let $𝐿 : 𝐹^{𝑛} \to 𝑊$ be the map $(𝑎_{1}, \dots, 𝑎_{𝑛}) \mapsto$ $𝑎_{1} 𝐰_{1} +$ $\dots +$ $𝑎_{𝑛} 𝐰_{𝑛}$ , which is linear by Example 2.1.8. By Proposition 2.1.17, $𝑇 = 𝐿 𝐶$ is linear. For scalars $𝑎_{1}, \dots, 𝑎_{𝑛}$ , the definition of coordinates gives ${[𝑎_{1} 𝐯_{1} + \dots + 𝑎_{𝑛} 𝐯_{𝑛}]}_{ℬ︀} =$ $(𝑎_{1}, \dots, 𝑎_{𝑛})$ , so

𝑇 (𝑎_{1} 𝐯_{1} + \dots + 𝑎_{𝑛} 𝐯_{𝑛}) = 𝐿 (𝑎_{1}, \dots, 𝑎_{𝑛}) = 𝑎_{1} 𝐰_{1} + \dots + 𝑎_{𝑛} 𝐰_{𝑛},

which is (2.2.1). In particular, taking $𝑎_{𝑖} = 1$ and the other coefficients $0$ , $𝑇 (𝐯_{𝑖}) = 𝐰_{𝑖}$ . □

In the second proof both halves of “basis” are hidden inside $𝐶$ : the coordinate vector ${[𝐯]}_{ℬ︀}$ is defined only because every vector has exactly one representation.

Nothing is assumed about $𝐰_{1}, \dots, 𝐰_{𝑛}$ . They may be linearly dependent, or even all equal to $𝟎$ , which gives the zero map. If $𝑛 = 0$ , then $ℬ︀$ is the empty list, $𝑉 = {𝟎}$ , and (2.2.1) with its empty sums says that the only linear map ${𝟎} \to 𝑊$ is the zero map.

Remark 2.2.2 (Where the two halves of “basis” were used). The proof of uniqueness used only that

ℬ︀

spans

𝑉

. The proof of existence used both halves: spanning so that the recipe applies to every

𝐯

, and independence so that it gives only one answer. Section 2.2.3 and Section 2.2.4 show what each half achieves on its own. A spanning list still gives uniqueness, but existence needs a condition on the

𝐰_{𝑖}

. An independent list still gives existence, at least on its span, but if it does not span

𝑉

, uniqueness fails (unless

𝑊 = {𝟎}

Geometrically, Theorem 2.2.1 says the following. A basis $𝐛_{1}, 𝐛_{2}$ of $ℝ^{2}$ cuts the plane into a grid of parallelograms, and the grid point $𝑎_{1} 𝐛_{1} + 𝑎_{2} 𝐛_{2}$ is reached by $𝑎_{1}$ steps along $𝐛_{1}$ and $𝑎_{2}$ steps along $𝐛_{2}$ . The linear map with $𝑇 (𝐛_{𝑖}) = 𝐰_{𝑖}$ sends it to the point reached by the same steps along $𝐰_{1}$ and $𝐰_{2}$ (Figure 2.2.1). This is the picture from Section 2.1, where $𝐛_{1}, 𝐛_{2}$ was the standard basis.

Figure 2.2.1. Left: the grid of

ℬ︀

-coordinates for the basis

ℬ︀ = 𝐛_{1}, 𝐛_{2}

ℝ^{2}

. The vector

𝐯

with

{[𝐯]}_{ℬ︀} = (2, 1)

is two steps along

𝐛_{1}

and one along

𝐛_{2}

. Right: the linear map

𝑇

with

𝑇 (𝐛_{1}) = 𝐰_{1}

and

𝑇 (𝐛_{2}) = 𝐰_{2}

carries the grid onto the grid built from

𝐰_{1}, 𝐰_{2}

, and

𝑇 (𝐯) = 2 𝐰_{1} + 𝐰_{2}

is reached by the same steps.

In practice, Theorem 2.2.1 is used most often to build maps, by choosing a basis in which the desired map is easy to describe.

Example 2.2.3 (Building maps from a convenient basis). Let $𝐹$ be a field with $1 + 1 \neq 0$ , and write $2 = 1 + 1$ . In $𝐹^{2}$ let $𝐛_{1} = (1, 1)$ and $𝐛_{2} = (1, - 1)$ . This is a basis of $𝐹^{2}$ . Indeed, $(𝑥, 𝑦) =$ $𝑎 𝐛_{1} +$ $𝑏 𝐛_{2}$ means $𝑎 + 𝑏 = 𝑥$ and $𝑎 - 𝑏 = 𝑦$ . Adding and subtracting these equations gives $2 𝑎 = 𝑥 + 𝑦$ and $2 𝑏 = 𝑥 - 𝑦$ , and since $2 \neq 0$ ,

𝑎 = \frac{𝑥 + 𝑦}{2}, 𝑏 = \frac{𝑥 - 𝑦}{2} .

Conversely these values satisfy both equations, so every $(𝑥, 𝑦)$ is a combination of $𝐛_{1}, 𝐛_{2}$ in exactly one way.

By Theorem 2.2.1 there are unique linear maps $𝑃, 𝑆 : 𝐹^{2} \to 𝐹^{2}$ with

𝑃 (𝐛_{1}) = 𝐛_{1}, 𝑃 (𝐛_{2}) = 𝟎, 𝑆 (𝐛_{1}) = 𝐛_{1}, 𝑆 (𝐛_{2}) = - 𝐛_{2},

and (2.2.1) gives their formulas:

\begin{aligned} 𝑃 (𝑥, 𝑦) & = \frac{𝑥 + 𝑦}{2} 𝐛_{1} = (\frac{𝑥 + 𝑦}{2}, \frac{𝑥 + 𝑦}{2}), \\ 𝑆 (𝑥, 𝑦) & = \frac{𝑥 + 𝑦}{2} 𝐛_{1} - \frac{𝑥 - 𝑦}{2} 𝐛_{2} = (𝑦, 𝑥) . \end{aligned}

Over $ℝ$ , $𝐛_{1}$ spans the line $ℓ$ given by $𝑦 = 𝑥$ , and $𝐛_{2}$ is perpendicular to $ℓ$ . So $𝑃$ fixes $ℓ$ pointwise and kills the perpendicular direction: it is the projection ${Pr}_{ℓ}$ . Similarly $𝑆$ fixes $ℓ$ and flips the perpendicular direction: it is the reflection $𝑆_{ℓ}$ . The formulas agree with those of Example 2.1.11 for $𝜃 = 𝜋 / 4$ , and this time no trigonometry was needed.

Over $𝔽_{2}$ the construction breaks down. There $𝐛_{2} =$ $(1, - 1) =$ $(1, 1) =$ $𝐛_{1}$ , so $𝐛_{1}, 𝐛_{2}$ is not a basis, and indeed no linear map can send the same vector both to $𝐛_{1}$ and to $𝟎$ .

Technique 2.2.4 (Construct a map by choosing it on a basis). To produce a linear map with some property, choose a basis of the domain that is adapted to the property, decide where each basis vector should go, and let Theorem 2.2.1 extend the choice to a linear map. Linearity then costs nothing, and the property itself can usually be checked on the basis (Technique 2.2.7). If a formula is wanted, write a general vector in the chosen basis and apply (2.2.1).

Reach for it when the property is easy to state on special vectors (vectors to be fixed, sent to $𝟎$ , swapped, or scaled), or when all you need is that some linear map with a given behaviour exists, as when building a counterexample. The canonical example is the projection and reflection of Example 2.2.3. Another is that a subspace is the image of one linear map and the set of vectors sent to $𝟎$ by another (Exercise 2.2.35).

Used in 25 places and 6 practice exercises

Proposition 2.2.11, Proposition 2.2.14, Exercise 2.2.28, Exercise 2.2.35, Exercise 2.2.40, Theorem 2.7.8, Theorem 2.7.4, Theorem 2.7.2, Exercise 2.7.29, Exercise 2.8.7, Problem 2.3, Problem 2.15, Proposition 3.5.9, Exercise 3.5.40, Proposition 4.1.19, Proposition 4.4.18, Exercise 4.4.42, Theorem 4.5.6, Theorem 4.5.8, Theorem 4.7.39, Problem 4.9, Problem 4.10, Exercise 7.2.58, Corollary 8.4.7, Proposition 8.4.19; 6 practice exercises.

Exercise 2.2.5.routine Show that there is exactly one linear map

𝑇 : ℝ^{2} \to ℝ

with

𝑇 (1, 1) = 1

and

𝑇 (1, 2) = 0

. Find

𝑇 (𝑥, 𝑦)

and

𝑇 (3, 5)

Solution.

Uses. Theorem 2.2.1 (Linear maps are determined by a basis), Definition 1.6.1 (Basis)

Hint 1. First check that

(1, 1), (1, 2)

is a basis of

ℝ^{2}

by solving

(𝑥, 𝑦) =

𝑎 (1, 1) +

𝑏 (1, 2)

for

𝑎

and

𝑏

Hint 2. Then

𝑇 (𝑥, 𝑦) =

𝑎 𝑇 (1, 1) +

𝑏 𝑇 (1, 2) =

𝑎

Sketch. Solving gives

𝑎 = 2 𝑥 - 𝑦

and

𝑏 = 𝑦 - 𝑥

, uniquely, so the list is a basis and

𝑇 (𝑥, 𝑦) =

𝑎 =

2 𝑥 -

𝑦

. Then

𝑇 (3, 5) = 1

The equations $𝑎 + 𝑏 = 𝑥$ and $𝑎 + 2 𝑏 = 𝑦$ give $𝑏 = 𝑦 - 𝑥$ by subtraction, and then $𝑎 = 𝑥 - 𝑏 = 2 𝑥 - 𝑦$ . Conversely these values satisfy both equations, so every $(𝑥, 𝑦)$ is a combination of $(1, 1), (1, 2)$ in exactly one way, and the list is a basis of $ℝ^{2}$ . By Theorem 2.2.1 there is exactly one such $𝑇$ , and by (2.2.1),

𝑇 (𝑥, 𝑦) = (2 𝑥 - 𝑦) \cdot 1 + (𝑦 - 𝑥) \cdot 0 = 2 𝑥 - 𝑦 .

Check: $𝑇 (1, 1) = 1$ and $𝑇 (1, 2) = 0$ . Finally, $𝑇 (3, 5) = 6 - 5 = 1$ .

2.2.2Checking on a basis

The proof of uniqueness in Theorem 2.2.1 used only that $ℬ︀$ spans $𝑉$ , so it proves more.

Corollary 2.2.6 (Agreeing on a spanning list). Let

𝑆, 𝑇 \in L (𝑉, 𝑊)

, and let

𝐯_{1}, \dots, 𝐯_{𝑘}

be a list that spans

𝑉

. If

𝑆 (𝐯_{𝑖}) = 𝑇 (𝐯_{𝑖})

for every

𝑖

, then

𝑆 = 𝑇

. In particular, two linear maps that agree on a basis are equal.

Proof (by computation).routine

Uses. Definition 1.3.10 (Span), Definition 2.1.1 (Linear map)

Hint 1. Take any

𝐯 \in 𝑉

and write it as a combination of the spanning list.

Hint 2. This is the uniqueness half of the proof of Theorem 2.2.1, with a spanning list in place of a basis.

Sketch. Write

𝐯 = \sum_{𝑖} 𝑎_{𝑖} 𝐯_{𝑖}

. Then

𝑆 (𝐯) =

\sum_{𝑖} 𝑎_{𝑖} 𝑆 (𝐯_{𝑖}) =

\sum_{𝑖} 𝑎_{𝑖} 𝑇 (𝐯_{𝑖}) =

𝑇 (𝐯)

Let $𝐯 \in 𝑉$ . Since $𝐯_{1}, \dots, 𝐯_{𝑘}$ spans $𝑉$ , we can write $𝐯 =$ $𝑎_{1} 𝐯_{1} +$ $\dots +$ $𝑎_{𝑘} 𝐯_{𝑘}$ . By linearity of $𝑆$ and $𝑇$ ,

𝑆 (𝐯) = \sum_{𝑖 = 1}^{𝑘} 𝑎_{𝑖} 𝑆 (𝐯_{𝑖}) = \sum_{𝑖 = 1}^{𝑘} 𝑎_{𝑖} 𝑇 (𝐯_{𝑖}) = 𝑇 (𝐯) .

As $𝐯$ was arbitrary, $𝑆 = 𝑇$ . □

The second proof does not compute with a general vector at all. It looks at the set of vectors on which $𝑆$ and $𝑇$ agree, and shows that this set is a subspace containing the spanning list.

Proof (via a subspace).routine

Uses. Theorem 1.4.3 (Subspace criterion), Proposition 1.4.14 (The span is the smallest subspace containing a list), Proposition 2.1.4 (Linear maps preserve $𝟎$ and negatives)

Hint 1. Consider

𝐸 =

{𝐯 \in 𝑉 | 𝑆 (𝐯) = 𝑇 (𝐯)}

. Which vectors do we know lie in

𝐸

Hint 2. Show that

𝐸

is a subspace. What is the smallest subspace containing

𝐯_{1}, \dots, 𝐯_{𝑘}

Sketch.

𝐸 =

{𝐯 | 𝑆 (𝐯) = 𝑇 (𝐯)}

contains

𝟎

and is closed under addition and scalar multiplication, by linearity. It contains every

𝐯_{𝑖}

, so it contains

span (𝐯_{1}, \dots, 𝐯_{𝑘}) =

𝑉

Let $𝐸 =$ ${𝐯 \in 𝑉 | 𝑆 (𝐯) = 𝑇 (𝐯)}$ . We check that $𝐸$ is a subspace of $𝑉$ with Theorem 1.4.3. It contains $𝟎$ , since $𝑆 (𝟎) = 𝟎 = 𝑇 (𝟎)$ by Proposition 2.1.4. If $𝐮, 𝐯 \in 𝐸$ and $𝑎 \in 𝐹$ , then

𝑆 (𝑎 𝐮 + 𝐯) = 𝑎 𝑆 (𝐮) + 𝑆 (𝐯) = 𝑎 𝑇 (𝐮) + 𝑇 (𝐯) = 𝑇 (𝑎 𝐮 + 𝐯),

so $𝑎 𝐮 + 𝐯 \in 𝐸$ . By hypothesis, $𝐸$ contains $𝐯_{1}, \dots, 𝐯_{𝑘}$ . Since the span of a list is the smallest subspace containing it (Proposition 1.4.14), $𝑉 =$ $span (𝐯_{1}, \dots, 𝐯_{𝑘}) \subseteq$ $𝐸$ . So $𝐸 = 𝑉$ , that is, $𝑆 = 𝑇$ . □

The second proof applies to more than equality. Whenever the vectors with some property form a subspace, the property holds on all of $𝑉$ as soon as it holds on a spanning list. (In the language of Section 2.6, $𝐸$ is the kernel of $𝑆 - 𝑇$ .)

Technique 2.2.7 (Check it on a basis). To prove that two linear maps $𝑆, 𝑇 : 𝑉 \to 𝑊$ are equal, it is enough to check $𝑆 (𝐯_{𝑖}) = 𝑇 (𝐯_{𝑖})$ for the vectors of a basis of $𝑉$ , or of any spanning list (Corollary 2.2.6). Many statements can be put in this form. An identity such as $𝑃^{2} = 𝑃$ , $𝑆 𝑇 = 𝑇 𝑆$ , or $𝑇 = 𝑎 id + 𝑏 𝑅$ is an equation between linear maps, because sums, scalar multiples, and composites of linear maps are linear (Proposition 2.1.23, Proposition 2.1.17). A containment $𝑇 (𝑉) \subseteq 𝑈$ , for a subspace $𝑈$ of $𝑊$ , holds as soon as $𝑇 (𝐯_{𝑖}) \in 𝑈$ for every vector of a spanning list, because $𝑇 (𝑉)$ is the span of the $𝑇 (𝐯_{𝑖})$ (Proposition 2.1.16).

Reach for it when the basis vectors are easy to compute with, typically for a basis on which the maps involved act simply, and when a computation at a general vector would be messy. Before using it, make sure that both sides really are linear in the vector. A condition such as “ $𝑇 (𝐯)$ is a multiple of $𝐯$ ” is not an equation between linear maps, and it can hold on a basis without holding everywhere (Exercise 2.2.25). The canonical example is Corollary 2.2.6 itself.

Used in 47 places and 12 practice exercises

Exercise 2.1.49, Exercise 2.2.9, Proposition 2.2.14, Exercise 2.2.23, Exercise 2.2.25, Exercise 2.2.28, Exercise 2.2.29, Exercise 2.2.30, Exercise 2.2.33, Exercise 2.2.34, Exercise 2.2.36, Exercise 2.2.39, Exercise 2.2.40, Exercise 2.3.38, Proposition 2.4.28, Exercise 2.4.49, Exercise 2.4.50, Proposition 2.6.18, Proposition 2.6.22, Exercise 2.6.26, Theorem 2.7.8, Theorem 2.7.4, Theorem 2.7.2, Exercise 2.8.6, Exercise 2.8.8, Problem 2.3, Problem 2.4, Problem 2.5, Problem 2.6, Problem 2.7, Lemma 3.2.6, Proposition 3.5.9, Exercise 3.5.40, Exercise 4.1.55, Theorem 4.5.8, Proposition 4.5.35, Exercise 4.5.47, Proposition 6.1.22, Theorem 6.6.3, Corollary 7.1.12, Proposition 7.2.4, Proposition 7.2.27, Proposition 7.4.7, Proposition 7.4.26, Proposition 7.5.2, Exercise 7.6.48, Problem 7.9; 12 practice exercises.

Example 2.2.8 (Identities checked on a basis). Let $𝑃$ and $𝑆$ be the projection and reflection of Example 2.2.3. We claim that

𝑃^{2} = 𝑃, 𝑆^{2} = id, 𝑆 = 2 𝑃 - id, 𝑆 𝑃 = 𝑃 𝑆 = 𝑃 .

Each side of each identity is a linear map $𝐹^{2} \to 𝐹^{2}$ (Proposition 2.1.17, Proposition 2.1.23), so by Corollary 2.2.6 it suffices to compare the two sides at $𝐛_{1}$ and at $𝐛_{2}$ . Every map involved sends $𝐛_{1}$ to $𝐛_{1}$ (for $2 𝑃 - id$ , because $2 𝐛_{1} - 𝐛_{1} = 𝐛_{1}$ ), so both sides of each identity send $𝐛_{1}$ to $𝐛_{1}$ . At $𝐛_{2}$ ,

\begin{aligned} 𝑃^{2} (𝐛_{2}) & = 𝑃 (𝟎) = 𝟎 = 𝑃 (𝐛_{2}), \\ 𝑆^{2} (𝐛_{2}) & = 𝑆 (- 𝐛_{2}) = 𝐛_{2}, \\ (2 𝑃 - id) (𝐛_{2}) & = 𝟎 - 𝐛_{2} = 𝑆 (𝐛_{2}), \\ 𝑆 𝑃 (𝐛_{2}) & = 𝑆 (𝟎) = 𝟎 = 𝑃 (𝐛_{2}), \\ 𝑃 𝑆 (𝐛_{2}) & = 𝑃 (- 𝐛_{2}) = 𝟎 = 𝑃 (𝐛_{2}) . \end{aligned}

Exercise 2.1.42 proves the same identities over $ℝ$ from the coordinate formulas. In the basis $𝐛_{1}, 𝐛_{2}$ each map does something simple to each basis vector, and the identities become one-line checks.

Exercise 2.2.9.routine Let

𝐯_{1}, \dots, 𝐯_{𝑛}

be a basis of

𝑉

, and let

𝑇 \in L (𝑉)

be the linear map with

𝑇 (𝐯_{𝑖}) = 𝐯_{𝑖} + 𝐯_{1}

for every

𝑖

(so

𝑇 (𝐯_{1}) = 2 𝐯_{1}

). Show that

𝑇^{2} = 3 𝑇 - 2 {id}_{𝑉}

Solution.

Uses. Corollary 2.2.6 (Agreeing on a spanning list), Proposition 2.1.23 ( $L (𝑉, 𝑊)$ is a vector space), Proposition 2.1.17 (Compositions of linear maps are linear)

Techniques. Check it on a basis (Technique 2.2.7)

Hint 1. Both sides are linear maps, so it suffices to compare them at each

𝐯_{𝑖}

Hint 2. Compute

𝑇^{2} (𝐯_{𝑖}) =

𝑇 (𝐯_{𝑖} + 𝐯_{1}) =

𝑇 (𝐯_{𝑖}) +

𝑇 (𝐯_{1})

Sketch.

𝑇^{2} (𝐯_{𝑖}) =

(𝐯_{𝑖} + 𝐯_{1}) +

2 𝐯_{1} =

𝐯_{𝑖} +

3 𝐯_{1}

, and

(3 𝑇 - 2 id) (𝐯_{𝑖}) =

3 𝐯_{𝑖} +

3 𝐯_{1} -

2 𝐯_{𝑖}

is the same vector.

Both $𝑇^{2}$ and $3 𝑇 - 2 {id}_{𝑉}$ are linear (Proposition 2.1.17, Proposition 2.1.23), so by Corollary 2.2.6 it suffices to show that they agree at each $𝐯_{𝑖}$ . For every $𝑖$ ,

𝑇^{2} (𝐯_{𝑖}) = 𝑇 (𝐯_{𝑖} + 𝐯_{1}) = 𝑇 (𝐯_{𝑖}) + 𝑇 (𝐯_{1}) = (𝐯_{𝑖} + 𝐯_{1}) + 2 𝐯_{1} = 𝐯_{𝑖} + 3 𝐯_{1},

and

(3 𝑇 - 2 {id}_{𝑉}) (𝐯_{𝑖}) = 3 (𝐯_{𝑖} + 𝐯_{1}) - 2 𝐯_{𝑖} = 𝐯_{𝑖} + 3 𝐯_{1} .

(For $𝑖 = 1$ both computations give $4 𝐯_{1}$ .) So $𝑇^{2} = 3 𝑇 - 2 {id}_{𝑉}$ .

2.2.3Spanning lists and linear relations

Theorem 2.2.1 needs both halves of “basis”. Here is what happens with a list that spans but is not independent.

Non-example 2.2.10 (A spanning list that is not independent). The list $𝐞_{1}, 𝐞_{2}, 𝐞_{1} + 𝐞_{2}$ spans $ℝ^{2}$ . Try to prescribe $𝑇 (𝐞_{1}) = 𝐰_{1}$ , $𝑇 (𝐞_{2}) = 𝐰_{2}$ , and $𝑇 (𝐞_{1} + 𝐞_{2}) = 𝐰_{3}$ for a linear map $𝑇 : ℝ^{2} \to 𝑊$ .

Uniqueness survives, since its proof used only spanning: any such $𝑇$ is $𝑇 (𝑥, 𝑦) =$ $𝑥 𝐰_{1} +$ $𝑦 𝐰_{2}$ . Existence fails unless $𝐰_{3} = 𝐰_{1} + 𝐰_{2}$ , because a linear map satisfies $𝑇 (𝐞_{1} + 𝐞_{2}) =$ $𝑇 (𝐞_{1}) +$ $𝑇 (𝐞_{2})$ . In terms of the recipe: the vector $(1, 1)$ has the two representations $1 \cdot (𝐞_{1} + 𝐞_{2})$ and $1 \cdot 𝐞_{1} + 1 \cdot 𝐞_{2}$ , and the recipe turns them into $𝐰_{3}$ and $𝐰_{1} + 𝐰_{2}$ .

The obstruction is a linear relation. If $𝑎_{1} 𝐯_{1} +$ $\dots +$ $𝑎_{𝑘} 𝐯_{𝑘} =$ $𝟎$ , then every linear map satisfies $𝑎_{1} 𝑇 (𝐯_{1}) +$ $\dots +$ $𝑎_{𝑘} 𝑇 (𝐯_{𝑘}) =$ $𝟎$ (Proposition 2.1.16). So prescribed values must satisfy every linear relation that the vectors satisfy. For a spanning list, this is the only obstruction.

Proposition 2.2.11 (Prescribing values on a spanning list). Let $ℒ︀ = 𝐯_{1}, \dots, 𝐯_{𝑘}$ be a list that spans $𝑉$ , and let $𝐰_{1}, \dots, 𝐰_{𝑘} \in 𝑊$ . There is a linear map $𝑇 : 𝑉 \to 𝑊$ with $𝑇 (𝐯_{𝑖}) = 𝐰_{𝑖}$ for every $𝑖$ if and only if, for all $𝑎_{1}, \dots, 𝑎_{𝑘} \in 𝐹$ ,

𝑎_{1} 𝐯_{1} + \dots + 𝑎_{𝑘} 𝐯_{𝑘} = 𝟎 ⟹ 𝑎_{1} 𝐰_{1} + \dots + 𝑎_{𝑘} 𝐰_{𝑘} = 𝟎 .

(2.2.2)

In that case $𝑇$ is unique.

In words, the list $𝐰_{1}, \dots, 𝐰_{𝑘}$ must satisfy every linear relation that $ℒ︀$ satisfies. If $ℒ︀$ is linearly independent, its only relation is the trivial one (Theorem 1.5.5), so the condition holds automatically, and we recover Theorem 2.2.1. There are two natural ways to prove the converse direction: repair the recipe directly, or throw away vectors until a basis remains.

Proof (directly).standard

Uses. Proposition 2.1.16 (Linear maps preserve spans and linear relations), Corollary 2.2.6 (Agreeing on a spanning list), Theorem 2.1.3 (Checking linearity), Definition 1.3.10 (Span)

Hint 1. Necessity is Proposition 2.1.16. For sufficiency, define

𝑇 (𝐯) =

\sum_{𝑖} 𝑎_{𝑖} 𝐰_{𝑖}

for any representation

𝐯 = \sum_{𝑖} 𝑎_{𝑖} 𝐯_{𝑖}

. Why does the choice of representation not matter?

Hint 2. If

\sum_{𝑖} 𝑎_{𝑖} 𝐯_{𝑖} =

\sum_{𝑖} 𝑏_{𝑖} 𝐯_{𝑖}

, then

\sum_{𝑖} (𝑎_{𝑖} - 𝑏_{𝑖}) 𝐯_{𝑖} =

𝟎

. Apply (2.2.2).

Hint 3. Once

𝑇

is well defined, linearity and

𝑇 (𝐯_{𝑖}) = 𝐰_{𝑖}

follow because you may compute

𝑇

from any representation you like.

Sketch. Necessity: apply

𝑇

to a relation. Sufficiency: set

𝑇 (\sum_{𝑖} 𝑎_{𝑖} 𝐯_{𝑖}) =

\sum_{𝑖} 𝑎_{𝑖} 𝐰_{𝑖}

. Two representations of the same vector differ by a relation of

ℒ︀

, which (2.2.2) turns into a relation of the

𝐰_{𝑖}

, so the two values agree. Since any representation may be used,

𝑇 (𝐯_{𝑖}) = 𝐰_{𝑖}

and linearity follow as in Theorem 2.2.1. Uniqueness is Corollary 2.2.6.

Necessity. If $𝑇$ is linear with $𝑇 (𝐯_{𝑖}) = 𝐰_{𝑖}$ and $\sum_{𝑖} 𝑎_{𝑖} 𝐯_{𝑖} = 𝟎$ , then $\sum_{𝑖} 𝑎_{𝑖} 𝐰_{𝑖} =$ $\sum_{𝑖} 𝑎_{𝑖} 𝑇 (𝐯_{𝑖}) =$ $𝟎$ by Proposition 2.1.16(b).

Sufficiency. Assume (2.2.2). Let $𝐯 \in 𝑉$ . Since $ℒ︀$ spans $𝑉$ , $𝐯$ has at least one representation $𝐯 =$ $𝑎_{1} 𝐯_{1} +$ $\dots +$ $𝑎_{𝑘} 𝐯_{𝑘}$ . We claim that the vector $𝑎_{1} 𝐰_{1} + \dots + 𝑎_{𝑘} 𝐰_{𝑘}$ is the same for all representations of $𝐯$ . Indeed, if also $𝐯 =$ $𝑏_{1} 𝐯_{1} +$ $\dots +$ $𝑏_{𝑘} 𝐯_{𝑘}$ , then subtracting,

(𝑎_{1} - 𝑏_{1}) 𝐯_{1} + \dots + (𝑎_{𝑘} - 𝑏_{𝑘}) 𝐯_{𝑘} = 𝟎,

so by (2.2.2), $\sum_{𝑖} (𝑎_{𝑖} - 𝑏_{𝑖}) 𝐰_{𝑖} =$ $𝟎$ , that is, $\sum_{𝑖} 𝑎_{𝑖} 𝐰_{𝑖} =$ $\sum_{𝑖} 𝑏_{𝑖} 𝐰_{𝑖}$ . So we may define $𝑇 : 𝑉 \to 𝑊$ by

𝑇 (𝑎_{1} 𝐯_{1} + \dots + 𝑎_{𝑘} 𝐯_{𝑘}) = 𝑎_{1} 𝐰_{1} + \dots + 𝑎_{𝑘} 𝐰_{𝑘},

computed from any representation. Using the representation $𝐯_{𝑖} =$ $0 𝐯_{1} +$ $\dots +$ $1 𝐯_{𝑖} +$ $\dots +$ $0 𝐯_{𝑘}$ gives $𝑇 (𝐯_{𝑖}) = 𝐰_{𝑖}$ . For linearity, let $𝐮 = \sum_{𝑖} 𝑎_{𝑖} 𝐯_{𝑖}$ , $𝐯 = \sum_{𝑖} 𝑏_{𝑖} 𝐯_{𝑖}$ , and $𝑐 \in 𝐹$ . Then $𝑐 𝐮 +$ $𝐯 =$ $\sum_{𝑖} (𝑐 𝑎_{𝑖} + 𝑏_{𝑖}) 𝐯_{𝑖}$ is a representation of $𝑐 𝐮 + 𝐯$ , so

𝑇 (𝑐 𝐮 + 𝐯) = \sum_{𝑖 = 1}^{𝑘} (𝑐 𝑎_{𝑖} + 𝑏_{𝑖}) 𝐰_{𝑖} = 𝑐 𝑇 (𝐮) + 𝑇 (𝐯) .

By Theorem 2.1.3, $𝑇$ is linear.

Uniqueness. Two such maps agree on the spanning list $ℒ︀$ , so they are equal by Corollary 2.2.6. □

Proof (by reducing to a basis).standard

Uses. Theorem 1.6.10 (Every spanning list contains a basis), Theorem 2.2.1 (Linear maps are determined by a basis), Proposition 2.1.16 (Linear maps preserve spans and linear relations), Corollary 2.2.6 (Agreeing on a spanning list)

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

Hint 1. Some sublist of

ℒ︀

is a basis of

𝑉

(Theorem 1.6.10). Define

𝑇

on that basis.

Hint 2. Each remaining

𝐯_{𝑗}

is a combination of the basis vectors. That expression is a linear relation among the vectors of

ℒ︀

; use (2.2.2) to compute

𝑇 (𝐯_{𝑗})

Sketch. Let

𝐮_{1}, \dots, 𝐮_{𝑛}

be a sublist of

ℒ︀

that is a basis of

𝑉

, and let

𝑇

send each

𝐮_{𝑟}

to its prescribed value. For any

𝑗

, write

𝐯_{𝑗} = \sum_{𝑟} 𝑐_{𝑟} 𝐮_{𝑟}

; this is a relation of

ℒ︀

, so the prescribed values satisfy it, and

𝑇 (𝐯_{𝑗}) =

\sum_{𝑟} 𝑐_{𝑟} 𝑇 (𝐮_{𝑟})

is the prescribed value

𝐰_{𝑗}

Necessity and uniqueness are as in the first proof. For sufficiency, assume (2.2.2). By Theorem 1.6.10, some sublist $𝐯_{𝑖_{1}}, \dots, 𝐯_{𝑖_{𝑛}}$ of $ℒ︀$ , with $𝑖_{1} < \dots < 𝑖_{𝑛}$ , is a basis of $𝑉$ ; write $𝐮_{𝑟} = 𝐯_{𝑖_{𝑟}}$ . By Theorem 2.2.1 there is a linear map $𝑇 : 𝑉 \to 𝑊$ with $𝑇 (𝐮_{𝑟}) = 𝐰_{𝑖_{𝑟}}$ for $𝑟 = 1, \dots, 𝑛$ . We show that $𝑇 (𝐯_{𝑗}) = 𝐰_{𝑗}$ for every $𝑗$ .

Fix $𝑗$ . Since $𝐮_{1}, \dots, 𝐮_{𝑛}$ spans $𝑉$ , there are scalars with $𝐯_{𝑗} =$ $𝑐_{1} 𝐮_{1} +$ $\dots +$ $𝑐_{𝑛} 𝐮_{𝑛}$ . The equation

𝐯_{𝑗} - 𝑐_{1} 𝐯_{𝑖_{1}} - \dots - 𝑐_{𝑛} 𝐯_{𝑖_{𝑛}} = 𝟎

is a linear relation among the vectors of $ℒ︀$ (collect the coefficient of each $𝐯_{𝑖}$ ). By (2.2.2), the same combination of the $𝐰_{𝑖}$ vanishes: $𝐰_{𝑗} =$ $𝑐_{1} 𝐰_{𝑖_{1}} +$ $\dots +$ $𝑐_{𝑛} 𝐰_{𝑖_{𝑛}}$ . Hence

𝑇 (𝐯_{𝑗}) = 𝑐_{1} 𝑇 (𝐮_{1}) + \dots + 𝑐_{𝑛} 𝑇 (𝐮_{𝑛}) = 𝑐_{1} 𝐰_{𝑖_{1}} + \dots + 𝑐_{𝑛} 𝐰_{𝑖_{𝑛}} = 𝐰_{𝑗},

as desired. □

The first proof is the general method for defining a function by a formula that refers to a choice: show that the value does not depend on the choice. The second proof reduces to a case already handled, and it shows more than the statement does.

Remark 2.2.12 (Checking fewer relations). The second proof used (2.2.2) only for the

𝑘 - 𝑛

relations

𝐯_{𝑗} =

𝑐_{1} 𝐮_{1} +

\dots +

𝑐_{𝑛} 𝐮_{𝑛}

that express each vector of

ℒ︀

outside the chosen basis in terms of the basis. So: if a sublist

𝐮_{1}, \dots, 𝐮_{𝑛}

ℒ︀

is a basis of

𝑉

, and for each remaining

𝐯_{𝑗}

the prescribed values satisfy the relation expressing

𝐯_{𝑗}

in terms of

𝐮_{1}, \dots, 𝐮_{𝑛}

, then

𝑇

exists. Every other relation of

ℒ︀

is then satisfied automatically, by Proposition 2.1.16. In examples this is how existence is checked: one relation per redundant vector.

2.2.4Prescribing values on any list

Now the other half: a list that is independent but does not span.

Non-example 2.2.13 (An independent list that does not span). The list

𝐞_{1}

is linearly independent in

ℝ^{2}

but does not span it. Prescribe

𝑇 (𝐞_{1}) = 1

for a linear map

𝑇 : ℝ^{2} \to ℝ

. Existence survives:

(𝑥, 𝑦) \mapsto 𝑥

works. Uniqueness fails:

(𝑥, 𝑦) \mapsto 𝑥 + 𝑐 𝑦

works for every

𝑐 \in ℝ

. The recipe says nothing about

𝐞_{2}

, which has no representation in terms of

𝐞_{1}

, so

𝑇 (𝐞_{2})

can be anything.

In general, given vectors $𝐯_{1}, \dots, 𝐯_{𝑘} \in 𝑉$ and $𝐰_{1}, \dots, 𝐰_{𝑘} \in 𝑊$ , we can first define a map on $span (𝐯_{1}, \dots, 𝐯_{𝑘})$ with Proposition 2.2.11, and then extend it to $𝑉$ by choosing values on vectors that complete a basis. The extension step needs such vectors to exist.

Proposition 2.2.14 (Prescribing values on any list). Let $ℒ︀ = 𝐯_{1}, \dots, 𝐯_{𝑘}$ be a list in $𝑉$ , and let $𝐰_{1}, \dots, 𝐰_{𝑘} \in 𝑊$ . Suppose that $𝑉$ has a basis $𝐮_{1},$ $\dots,$ $𝐮_{𝑚},$ $𝐱_{1},$ $\dots,$ $𝐱_{𝑟}$ in which $𝐮_{1}, \dots, 𝐮_{𝑚}$ is a basis of $span ℒ︀$ .

(a)There is a linear map $𝑇 : 𝑉 \to 𝑊$ with $𝑇 (𝐯_{𝑖}) = 𝐰_{𝑖}$ for every $𝑖$ if and only if (2.2.2) holds.
(b)If such a $𝑇$ exists, it is unique if and only if $ℒ︀$ spans $𝑉$ or $𝑊 = {𝟎}$ .

Proof.standard

Uses. Proposition 2.2.11 (Prescribing values on a spanning list), Proposition 2.1.16 (Linear maps preserve spans and linear relations), Theorem 2.2.1 (Linear maps are determined by a basis), Corollary 2.2.6 (Agreeing on a spanning list), Proposition 1.4.14 (The span is the smallest subspace containing a list)

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

Hint 1.

span ℒ︀

is a vector space spanned by

ℒ︀

, so Proposition 2.2.11 gives a linear map

𝑆 : span ℒ︀ \to 𝑊

. Extend it with Theorem 2.2.1.

Hint 2. Let

𝑇

agree with

𝑆

𝐮_{1}, \dots, 𝐮_{𝑚}

and send the

𝐱_{𝑗}

anywhere. Why does

𝑇

agree with

𝑆

on all of

span ℒ︀

Hint 3. For (b), if

ℒ︀

does not span

𝑉

, then

𝑟 \geq 1

. Change the value at

𝐱_{1}

Sketch. Necessity in (a) is Proposition 2.1.16. Conversely, Proposition 2.2.11 applied to

span ℒ︀

gives

𝑆

with

𝑆 (𝐯_{𝑖}) = 𝐰_{𝑖}

; let

𝑇

agree with

𝑆

on the

𝐮

’s and vanish on the

𝐱

’s. Then

𝑇 = 𝑆

span (𝐮_{1}, \dots, 𝐮_{𝑚}) =

span ℒ︀

. For (b), spanning gives uniqueness by Corollary 2.2.6; if

ℒ︀

does not span, then

𝑟 \geq 1

, and changing

𝑇 (𝐱_{1})

to a nonzero vector gives a second map.

(a) If $𝑇$ exists, (2.2.2) holds by Proposition 2.1.16(b). Conversely, assume (2.2.2). The subspace $span ℒ︀$ is a vector space (Proposition 1.4.14), and $ℒ︀$ spans it. So Proposition 2.2.11, applied with $span ℒ︀$ in place of $𝑉$ , gives a linear map $𝑆 : span ℒ︀ \to 𝑊$ with $𝑆 (𝐯_{𝑖}) = 𝐰_{𝑖}$ for every $𝑖$ . By Theorem 2.2.1 there is a linear map $𝑇 : 𝑉 \to 𝑊$ with

𝑇 (𝐮_{𝑗}) = 𝑆 (𝐮_{𝑗}) (𝑗 = 1, \dots, 𝑚), 𝑇 (𝐱_{𝑙}) = 𝟎 (𝑙 = 1, \dots, 𝑟) .

Every $𝐯 \in span ℒ︀$ is a combination $𝐯 =$ $𝑐_{1} 𝐮_{1} +$ $\dots +$ $𝑐_{𝑚} 𝐮_{𝑚}$ , and then $𝑇 (𝐯) =$ $\sum_{𝑗} 𝑐_{𝑗} 𝑆 (𝐮_{𝑗}) =$ $𝑆 (𝐯)$ . In particular $𝑇 (𝐯_{𝑖}) =$ $𝑆 (𝐯_{𝑖}) =$ $𝐰_{𝑖}$ .

(b) If $ℒ︀$ spans $𝑉$ , two such maps agree on a spanning list and are equal by Corollary 2.2.6. If $𝑊 = {𝟎}$ , there is only one function $𝑉 \to 𝑊$ at all. Conversely, suppose that $ℒ︀$ does not span $𝑉$ and that $𝑊$ contains some $𝐰 \neq 𝟎$ . Then $𝑟 \geq 1$ : if $𝑟 = 0$ , then $𝐮_{1}, \dots, 𝐮_{𝑚}$ would be a basis of $𝑉$ , and $𝑉 =$ $span (𝐮_{1}, \dots, 𝐮_{𝑚}) =$ $span ℒ︀$ . Let $𝑇$ be the map built in (a), and let $𝑇^{'}$ be the linear map with $𝑇^{'} (𝐮_{𝑗}) =$ $𝑆 (𝐮_{𝑗})$ for all $𝑗$ , $𝑇^{'} (𝐱_{1}) = 𝐰$ , and $𝑇^{'} (𝐱_{𝑙}) = 𝟎$ for $𝑙 \geq 2$ . The argument in (a) shows that $𝑇^{'} (𝐯_{𝑖}) = 𝐰_{𝑖}$ for every $𝑖$ , but $𝑇^{'} (𝐱_{1}) =$ $𝐰 \neq$ $𝟎 =$ $𝑇 (𝐱_{1})$ . So $𝑇$ is not unique. □

The hypothesis on $𝑉$ always holds when $𝑉$ is spanned by a finite list. By Theorem 1.6.10, $span ℒ︀$ has a basis taken from $ℒ︀$ , and in Section 4.1 we will see that every linearly independent list in such a $𝑉$ extends to a basis (Theorem 4.1.16). In $𝐹^{𝑛}$ one can also extend by hand, adding standard basis vectors one at a time. Without any finiteness assumption, the hypothesis still holds if one accepts the axiom of choice (Section 15.1).

Putting Proposition 2.2.11, Remark 2.2.12, and Proposition 2.2.14 together gives a procedure.

Algorithm 2.2.15 (Deciding whether a linear map with prescribed values exists)

Input: Lists $𝐯_{1}, \dots, 𝐯_{𝑘}$ in $𝑉$ and $𝐰_{1}, \dots, 𝐰_{𝑘}$ in $𝑊$ .
Output: Whether some $𝑇 \in L (𝑉, 𝑊)$ has $𝑇 (𝐯_{𝑖}) = 𝐰_{𝑖}$ for every $𝑖$ , and whether it is unique.

Go through

𝐯_{1}, \dots, 𝐯_{𝑘}

in order, and discard each vector that is a linear combination of the vectors kept before it. The kept vectors form a basis

ℬ︀

span (𝐯_{1}, \dots, 𝐯_{𝑘})

For each discarded

𝐯_{𝑗}

, write

𝐯_{𝑗} = \sum 𝑐_{𝑟} 𝐮_{𝑟}

with the

𝐮_{𝑟}

ℬ︀

, and check that the corresponding

𝐰

’s satisfy the same equation. If one check fails, no

𝑇

exists. Otherwise a linear map with the prescribed values exists on

span (𝐯_{1}, \dots, 𝐯_{𝑘})

Suppose every check passed. If

𝐯_{1}, \dots, 𝐯_{𝑘}

spans

𝑉

, then

𝑇

exists and is unique. Otherwise extend

ℬ︀

to a basis

ℬ︀, 𝐱_{1}, \dots, 𝐱_{𝑟}

𝑉

. Then

𝑇

exists, its values at

𝐱_{1}, \dots, 𝐱_{𝑟}

can be chosen freely, and it is not unique unless

𝑊 = {𝟎}

In step 1, the kept vectors span the same subspace as the whole list, since each discarded vector is a combination of kept ones. They are linearly independent: otherwise, by the linear dependence lemma (Lemma 1.5.10), some kept vector would be a combination of the kept vectors before it, and it would have been discarded. Applied to $span (𝐯_{1}, \dots, 𝐯_{𝑘})$ , Remark 2.2.12 justifies step 2, and Proposition 2.2.14 justifies step 3.

Example 2.2.16 (Deciding existence and uniqueness).

(a)Is there a linear $𝑇 : ℝ^{2} \to ℝ^{2}$ with
$𝑇 (1, 2) = (1, 1), 𝑇 (3, 1) = (2, - 1), 𝑇 (4, 3) = (3, 1) ?$
Keep $(1, 2)$ , and keep $(3, 1)$ , which is not a multiple of $(1, 2)$ . The two kept vectors form a basis of $ℝ^{2}$ : the equations $𝑎 + 3 𝑏 = 𝑥$ and $2 𝑎 + 𝑏 = 𝑦$ have the unique solution $𝑏 = (2 𝑥 - 𝑦) / 5$ , $𝑎 = (3 𝑦 - 𝑥) / 5$ . Then $(4, 3) = (1, 2) + (3, 1)$ is discarded, and the check asks whether $(3, 1) =$ $(1, 1) +$ $(2, - 1) =$ $(3, 0)$ . It is not, so no such $𝑇$ exists.
(b)Change the last value to $𝑇 (4, 3) = (3, 0)$ . Now the check passes, and the list spans $ℝ^{2}$ , so exactly one $𝑇$ exists. With the coordinates from (a),
$𝑇 (𝑥, 𝑦) = \frac{3 𝑦 - 𝑥}{5} (1, 1) + \frac{2 𝑥 - 𝑦}{5} (2, - 1) = (\frac{3 𝑥 + 𝑦}{5}, \frac{4 𝑦 - 3 𝑥}{5}) .$
As a check, $𝑇 (1, 2) = (1, 1)$ , $𝑇 (3, 1) = (2, - 1)$ , and $𝑇 (4, 3) = (3, 0)$ .
(c)Let $𝐹$ be any field. Is there a linear $𝑇 : 𝐹^{3} \to 𝐹^{2}$ with
$𝑇 (1, 1, 0) = (1, 0), 𝑇 (0, 1, 1) = (0, 1), 𝑇 (1, 2, 1) = (1, 1) ?$
Keep $𝐮_{1} = (1, 1, 0)$ and $𝐮_{2} = (0, 1, 1)$ , which is not a multiple of $𝐮_{1}$ (its first coordinate is $0$ and its second is not). Discard $(1, 2, 1) = 𝐮_{1} + 𝐮_{2}$ ; the check $(1, 0) + (0, 1) = (1, 1)$ passes. The list does not span $𝐹^{3}$ : every vector $𝑎 𝐮_{1} +$ $𝑏 𝐮_{2} =$ $(𝑎, 𝑎 + 𝑏, 𝑏)$ of the span has middle coordinate equal to the sum of the other two, and $𝐞_{1}$ does not. Adding $𝐞_{1}$ gives a basis of $𝐹^{3}$ , because
$(𝑥, 𝑦, 𝑧) = 𝑎 𝐮_{1} + 𝑏 𝐮_{2} + 𝑐 𝐞_{1} = (𝑎 + 𝑐, 𝑎 + 𝑏, 𝑏)$
holds exactly when $𝑏 = 𝑧$ , $𝑎 = 𝑦 - 𝑧$ , and $𝑐 = 𝑥 - 𝑦 + 𝑧$ . So for each $𝐜 \in 𝐹^{2}$ there is exactly one linear $𝑇$ with the prescribed values and $𝑇 (𝐞_{1}) = 𝐜$ , namely
$𝑇 (𝑥, 𝑦, 𝑧) = (𝑦 - 𝑧) (1, 0) + 𝑧 (0, 1) + (𝑥 - 𝑦 + 𝑧) 𝐜 = (𝑦 - 𝑧, 𝑧) + (𝑥 - 𝑦 + 𝑧) 𝐜 .$
Different choices of $𝐜$ give different maps, so $𝑇$ exists but is not unique.

2.2.5Maps out of $𝐹^{𝑛}$

For $𝑉 = 𝐹^{𝑛}$ with the standard basis, Theorem 2.2.1 takes a particularly concrete form.

Corollary 2.2.17 (Linear maps out of $𝐹^{𝑛}$ ). Let $𝑊$ be a vector space over $𝐹$ and $𝐰_{1}, \dots, 𝐰_{𝑛} \in 𝑊$ . There is exactly one linear map $𝑇 : 𝐹^{𝑛} \to 𝑊$ with $𝑇 (𝐞_{𝑗}) = 𝐰_{𝑗}$ for $𝑗 = 1, \dots, 𝑛$ , namely

𝑇 (𝑥_{1}, \dots, 𝑥_{𝑛}) = 𝑥_{1} 𝐰_{1} + \dots + 𝑥_{𝑛} 𝐰_{𝑛} .

In particular, every $𝑇 \in L (𝐹^{𝑛}, 𝑊)$ satisfies $𝑇 (𝐱) =$ $𝑥_{1} 𝑇 (𝐞_{1}) +$ $\dots +$ $𝑥_{𝑛} 𝑇 (𝐞_{𝑛})$ for all $𝐱 =$ $(𝑥_{1}, \dots, 𝑥_{𝑛}) \in$ $𝐹^{𝑛}$ .

Proof.routine

Uses. Definition 1.6.8 (Standard bases), Theorem 2.2.1 (Linear maps are determined by a basis)

Hint 1. Apply Theorem 2.2.1 to the standard basis.

Hint 2. What are the coordinates of

(𝑥_{1}, \dots, 𝑥_{𝑛})

in the standard basis?

Sketch.

𝐞_{1}, \dots, 𝐞_{𝑛}

is a basis of

𝐹^{𝑛}

, and

(𝑥_{1}, \dots, 𝑥_{𝑛}) =

𝑥_{1} 𝐞_{1} +

\dots +

𝑥_{𝑛} 𝐞_{𝑛}

. Apply (2.2.1).

The standard basis

𝐞_{1}, \dots, 𝐞_{𝑛}

is a basis of

𝐹^{𝑛}

(Definition 1.6.8), and

(𝑥_{1}, \dots, 𝑥_{𝑛}) =

𝑥_{1} 𝐞_{1} +

\dots +

𝑥_{𝑛} 𝐞_{𝑛}

. So Theorem 2.2.1 gives exactly one linear

𝑇

with

𝑇 (𝐞_{𝑗}) = 𝐰_{𝑗}

, and (2.2.1) is the displayed formula. For an arbitrary

𝑇 \in L (𝐹^{𝑛}, 𝑊)

, apply this with

𝐰_{𝑗} = 𝑇 (𝐞_{𝑗})

. □

Take $𝑊 = 𝐹^{𝑚}$ , and let $𝑎_{𝑖 𝑗}$ be the $𝑖$ -th coordinate of $𝑇 (𝐞_{𝑗})$ . Comparing $𝑖$ -th coordinates in $𝑇 (𝐱) =$ $𝑥_{1} 𝑇 (𝐞_{1}) +$ $\dots +$ $𝑥_{𝑛} 𝑇 (𝐞_{𝑛})$ shows that the $𝑖$ -th coordinate of $𝑇 (𝐱)$ is $𝑎_{𝑖 1} 𝑥_{1} +$ $\dots +$ $𝑎_{𝑖 𝑛} 𝑥_{𝑛}$ . So every linear map $𝐹^{𝑛} \to 𝐹^{𝑚}$ is given by a linear formula as in Example 2.1.7, and the $𝑚 𝑛$ coefficients are the coordinates of $𝑇 (𝐞_{1}),$ $\dots,$ $𝑇 (𝐞_{𝑛})$ . Writing these $𝑛$ vectors side by side, as the columns of an $𝑚 \times 𝑛$ array, is the starting point of Section 2.3.

The corollary also explains the rotation formula of Example 2.1.11. Rotating $𝐞_{1}$ by $𝜃$ gives $(cos 𝜃, sin 𝜃)$ , and rotating $𝐞_{2}$ gives $(- sin 𝜃, cos 𝜃)$ . Once we know that $𝑅_{𝜃}$ is linear, Corollary 2.2.17 gives $𝑅_{𝜃} (𝑥, 𝑦) =$ $𝑥 (cos 𝜃, sin 𝜃) +$ $𝑦 (- sin 𝜃, cos 𝜃)$ without the angle addition formulas. (In Example 2.1.11 linearity was deduced from the formula, so this is a consistency check rather than an independent proof.)

Taken together, existence and uniqueness say that linear maps out of $𝑉$ correspond exactly to lists of vectors in $𝑊$ .

Corollary 2.2.18 (Linear maps correspond to lists of vectors). Let $ℬ︀ = 𝐯_{1}, \dots, 𝐯_{𝑛}$ be a basis of $𝑉$ . The map

𝑇 \mapsto (𝑇 (𝐯_{1}), \dots, 𝑇 (𝐯_{𝑛}))

is a bijection from $L (𝑉, 𝑊)$ onto the set of lists of length $𝑛$ in $𝑊$ . It respects the vector space operations: the list of $𝑆 + 𝑇$ is the entrywise sum of the lists of $𝑆$ and $𝑇$ , and the list of $𝑐 𝑇$ is $𝑐$ times the list of $𝑇$ .

Proof.routine

Uses. Theorem 2.2.1 (Linear maps are determined by a basis), Definition 2.1.21 (Pointwise operations)

Hint 1. Surjectivity is the existence half of Theorem 2.2.1. What is injectivity?

Hint 2. For the operations, use the definition of

𝑆 + 𝑇

and

𝑐 𝑇

(Definition 2.1.21).

Sketch. Existence in Theorem 2.2.1 says every list is the list of some

𝑇

; uniqueness says different maps have different lists.

(𝑆 + 𝑇) (𝐯_{𝑖}) =

𝑆 (𝐯_{𝑖}) +

𝑇 (𝐯_{𝑖})

and

(𝑐 𝑇) (𝐯_{𝑖}) =

𝑐 𝑇 (𝐯_{𝑖})

by definition.

Given a list

𝐰_{1}, \dots, 𝐰_{𝑛}

𝑊

, the existence part of Theorem 2.2.1 gives

𝑇 \in L (𝑉, 𝑊)

whose list is

𝐰_{1}, \dots, 𝐰_{𝑛}

, so the map is surjective. If

𝑆

and

𝑇

have the same list, then

𝑆 = 𝑇

by the uniqueness part, so the map is injective. Finally, by Definition 2.1.21,

(𝑆 + 𝑇) (𝐯_{𝑖}) =

𝑆 (𝐯_{𝑖}) +

𝑇 (𝐯_{𝑖})

and

(𝑐 𝑇) (𝐯_{𝑖}) =

𝑐 𝑇 (𝐯_{𝑖})

for each

𝑖

. □

Example 2.2.19 (Counting linear maps over $𝔽_{𝑝}$ ). Let $𝑝$ be prime. The space $𝔽_{𝑝}^{𝑚}$ has $𝑝^{𝑚}$ elements, so there are ${(𝑝^{𝑚})}^{𝑛}$ lists of length $𝑛$ in it. By Corollary 2.2.18, applied with the standard basis of $𝔽_{𝑝}^{𝑛}$ ,

| L (𝔽_{𝑝}^{𝑛}, 𝔽_{𝑝}^{𝑚}) | = 𝑝^{𝑚 𝑛} .

For $𝑛 = 2$ , $𝑚 = 1$ , and $𝑝 = 2$ this gives the $4$ linear maps $𝔽_{2}^{2} \to 𝔽_{2}$ found in Exercise 2.1.30. For comparison, there are ${(𝑝^{𝑚})}^{𝑝^{𝑛}}$ functions $𝔽_{𝑝}^{𝑛} \to 𝔽_{𝑝}^{𝑚}$ in all; for $𝔽_{2}^{2} \to 𝔽_{2}^{2}$ that is $256$ functions, of which only $16$ are linear.

2.2.6Maps out of $𝐹 [𝑥]$

No finite list spans $𝐹 [𝑥]$ (Section 1.6), so Theorem 2.2.1 does not apply to it. But the monomials $1, 𝑥, 𝑥^{2}, \dots$ behave like a basis: every polynomial is a combination of finitely many of them, and its coefficients are uniquely determined. The proof of Theorem 2.2.1 goes through unchanged.

Proposition 2.2.20 (Linear maps out of $𝐹 [𝑥]$ ). Let $𝑊$ be a vector space over $𝐹$ , and let $𝐰_{0}, 𝐰_{1}, 𝐰_{2}, \dots$ be any sequence of vectors in $𝑊$ . There is exactly one linear map $𝑇 : 𝐹 [𝑥] \to 𝑊$ with $𝑇 (𝑥^{𝑘}) = 𝐰_{𝑘}$ for every $𝑘 \geq 0$ , namely

𝑇 (𝑎_{0} + 𝑎_{1} 𝑥 + \dots + 𝑎_{𝑑} 𝑥^{𝑑}) = 𝑎_{0} 𝐰_{0} + 𝑎_{1} 𝐰_{1} + \dots + 𝑎_{𝑑} 𝐰_{𝑑} .

Proof.routine

Uses. Definition 1.2.13 (Polynomials), Notation 1.2.15, Theorem 2.1.3 (Checking linearity)

Hint 1. Imitate the proof of Theorem 2.2.1. What plays the role of the coordinates of a polynomial?

Hint 2. The coefficients of

𝑝

are determined by

𝑝

(Definition 1.2.13). Check that writing

𝑝

with extra zero terms does not change the right side.

Sketch. Define

𝑇 (𝑝) =

\sum_{𝑘} 𝑎_{𝑘} 𝐰_{𝑘}

over the coefficients

𝑎_{𝑘}

𝑝

, a finite sum; extra zero terms contribute

0 𝐰_{𝑘} = 𝟎

, so

𝑇

is well defined. The operations on

𝐹 [𝑥]

are coefficientwise, which gives linearity, and

𝑥^{𝑘}

has a single coefficient

1

, which gives

𝑇 (𝑥^{𝑘}) = 𝐰_{𝑘}

. A linear

𝑇^{'}

with

𝑇^{'} (𝑥^{𝑘}) = 𝐰_{𝑘}

satisfies

𝑇^{'} (𝑝) =

\sum_{𝑘} 𝑎_{𝑘} 𝑇^{'} (𝑥^{𝑘}) =

𝑇 (𝑝)

Let $𝑝 \in 𝐹 [𝑥]$ have coefficients $(𝑎_{0}, 𝑎_{1}, \dots)$ , of which only finitely many are nonzero (Definition 1.2.13). Choose $𝑑$ with $𝑎_{𝑘} = 0$ for $𝑘 > 𝑑$ , so that $𝑝 =$ $𝑎_{0} +$ $𝑎_{1} 𝑥 +$ $\dots +$ $𝑎_{𝑑} 𝑥^{𝑑}$ (Notation 1.2.15), and define

𝑇 (𝑝) = 𝑎_{0} 𝐰_{0} + 𝑎_{1} 𝐰_{1} + \dots + 𝑎_{𝑑} 𝐰_{𝑑} .

This does not depend on $𝑑$ , since increasing $𝑑$ only adds terms $0 𝐰_{𝑘} = 𝟎$ , and the coefficients are determined by $𝑝$ . So $𝑇$ is a function $𝐹 [𝑥] \to 𝑊$ .

Let $𝑝, 𝑞 \in 𝐹 [𝑥]$ have coefficients $(𝑎_{𝑘})$ and $(𝑏_{𝑘})$ , let $𝑐 \in 𝐹$ , and choose one $𝑑$ that works for both. The operations on $𝐹 [𝑥]$ are coefficientwise, so $𝑐 𝑝 + 𝑞$ has coefficients $(𝑐 𝑎_{𝑘} + 𝑏_{𝑘})$ , and

𝑇 (𝑐 𝑝 + 𝑞) = \sum_{𝑘 = 0}^{𝑑} (𝑐 𝑎_{𝑘} + 𝑏_{𝑘}) 𝐰_{𝑘} = 𝑐 \sum_{𝑘 = 0}^{𝑑} 𝑎_{𝑘} 𝐰_{𝑘} + \sum_{𝑘 = 0}^{𝑑} 𝑏_{𝑘} 𝐰_{𝑘} = 𝑐 𝑇 (𝑝) + 𝑇 (𝑞) .

By Theorem 2.1.3, $𝑇$ is linear. The polynomial $𝑥^{𝑘}$ has coefficient $1$ in position $𝑘$ and $0$ elsewhere, so $𝑇 (𝑥^{𝑘}) = 𝐰_{𝑘}$ . Finally, if $𝑇^{'}$ is linear and $𝑇^{'} (𝑥^{𝑘}) = 𝐰_{𝑘}$ for every $𝑘$ , then $𝑇^{'} (𝑝) =$ $\sum_{𝑘 = 0}^{𝑑} 𝑎_{𝑘} 𝑇^{'} (𝑥^{𝑘}) =$ $\sum_{𝑘 = 0}^{𝑑} 𝑎_{𝑘} 𝐰_{𝑘} =$ $𝑇 (𝑝)$ , so $𝑇^{'} = 𝑇$ . □

For example, the maps of Example 2.1.9 are the linear maps with $𝑥^{𝑘} \mapsto 𝑡^{𝑘}$ (evaluation at $𝑡$ ), $𝑥^{𝑘} \mapsto 𝑥^{𝑘 + 1}$ (multiplication by $𝑥$ ), and $𝑥^{𝑘} \mapsto 𝑘 𝑥^{𝑘 - 1}$ (differentiation, with $1 \mapsto 0$ ), and the substitution map $𝑝 \mapsto 𝑝 (𝑞)$ of Exercise 2.1.32 is $𝑥^{𝑘} \mapsto 𝑞^{𝑘}$ . So Proposition 2.2.20 gives a second proof that these maps are linear. Its uniqueness half says that two linear maps out of $𝐹 [𝑥]$ that agree on every monomial are equal, which is Technique 2.2.7 with the monomials in place of a basis; Exercise 2.1.49 used exactly this. The same argument works for any family of vectors such that every vector is a combination of them with uniquely determined coefficients, all but finitely many of which are $0$ . Such a family is a basis in the sense of Section 15.1.

Exercises

Exercise 2.2.21.routine

(a)Show that there is exactly one linear map $𝑇 : 𝔽_{5}^{2} \to 𝔽_{5}^{2}$ with $𝑇 (1, 2) = (1, 0)$ and $𝑇 (2, 1) = (0, 1)$ , and find $𝑇 (𝑥, 𝑦)$ .
(b)Show that there is no linear map $𝑇 : 𝔽_{3}^{2} \to 𝔽_{3}^{2}$ with $𝑇 (1, 2) = (1, 0)$ and $𝑇 (2, 1) = (0, 1)$ .

Solution.

Uses. Theorem 2.2.1 (Linear maps are determined by a basis), Definition 1.6.1 (Basis), Proposition 2.1.16 (Linear maps preserve spans and linear relations)

Hint 1. (a) Solve

(𝑥, 𝑦) =

𝑎 (1, 2) +

𝑏 (2, 1)

for

𝑎

and

𝑏

𝔽_{5}

. You will need

3^{- 1}

𝔽_{5}

Hint 2. (b) Compare

(2, 1)

with

2 \cdot (1, 2)

𝔽_{3}^{2}

Sketch. (a) Solving gives

𝑏 = 4 𝑥 + 3 𝑦

and

𝑎 = 3 𝑥 + 4 𝑦

, uniquely, so the list is a basis and

𝑇 (𝑥, 𝑦) =

𝑎 (1, 0) +

𝑏 (0, 1) =

(3 𝑥 + 4 𝑦, 4 𝑥 + 3 𝑦)

. (b) In

𝔽_{3}^{2}

(2, 1) = 2 (1, 2)

, so a linear

𝑇

would have

𝑇 (2, 1) =

(2, 0) \neq

(0, 1)

(a) Consider the equations $𝑎 + 2 𝑏 = 𝑥$ and $2 𝑎 + 𝑏 = 𝑦$ over $𝔽_{5}$ . Subtracting twice the first from the second gives $- 3 𝑏 = 𝑦 - 2 𝑥$ , that is, $3 𝑏 = 2 𝑥 - 𝑦$ . Since $3 \cdot 2 = 6 = 1$ in $𝔽_{5}$ , $3^{- 1} = 2$ , so

𝑏 = 2 (2 𝑥 - 𝑦) = 4 𝑥 - 2 𝑦 = 4 𝑥 + 3 𝑦, 𝑎 = 𝑥 - 2 𝑏 = 𝑥 - 8 𝑥 - 6 𝑦 = 3 𝑥 + 4 𝑦 .

Conversely, these values satisfy both equations: $𝑎 +$ $2 𝑏 =$ $11 𝑥 +$ $10 𝑦 =$ $𝑥$ and $2 𝑎 +$ $𝑏 =$ $10 𝑥 +$ $11 𝑦 =$ $𝑦$ in $𝔽_{5}$ . So every vector has exactly one representation, and $(1, 2), (2, 1)$ is a basis of $𝔽_{5}^{2}$ . By Theorem 2.2.1 there is exactly one such $𝑇$ , and by (2.2.1)

𝑇 (𝑥, 𝑦) = 𝑎 (1, 0) + 𝑏 (0, 1) = (3 𝑥 + 4 𝑦, 4 𝑥 + 3 𝑦) .

Check: $𝑇 (1, 2) =$ $(11, 10) =$ $(1, 0)$ and $𝑇 (2, 1) =$ $(10, 11) =$ $(0, 1)$ in $𝔽_{5}^{2}$ .

(b) In $𝔽_{3}^{2}$ , $2 \cdot$ $(1, 2) =$ $(2, 4) =$ $(2, 1)$ . A linear $𝑇$ would therefore satisfy $𝑇 (2, 1) =$ $2 𝑇 (1, 2) =$ $(2, 0)$ , which is not $(0, 1)$ . So no such $𝑇$ exists. Here $(1, 2), (2, 1)$ is linearly dependent, and the prescribed values do not satisfy the relation $(2, 1) - 2 \cdot (1, 2) = 𝟎$ .

Exercise 2.2.22.routine For each of the following, decide whether a linear map with the given values exists, and if so, whether it is unique.

(a) $𝑇 : ℝ^{2} \to ℝ^{2}$ with $𝑇 (1, 2) = (1, 0)$ and $𝑇 (2, 4) = (2, 1)$ .
(b) $𝑇 : ℝ^{3} \to ℝ^{2}$ with $𝑇 (1, 0, 1) = (1, 1)$ , $𝑇 (0, 1, 1) = (2, 0)$ , and $𝑇 (1, 1, 2) = (3, 1)$ .
(c) $𝑇 : ℝ^{2} \to ℝ$ with $𝑇 (1, 1) = 2$ , $𝑇 (1, - 1) = 0$ , and $𝑇 (3, 1) = 4$ .
(d) $𝑇 : 𝔽_{2}^{3} \to 𝔽_{2}$ with $𝑇 (1, 1, 0) =$ $𝑇 (0, 1, 1) =$ $𝑇 (1, 0, 1) =$ $1$ .

Solution.

Uses. Algorithm 2.2.15 (Deciding whether a linear map with prescribed values exists), Proposition 2.2.14 (Prescribing values on any list), Proposition 2.1.16 (Linear maps preserve spans and linear relations)

Hint 1. Follow Algorithm 2.2.15: look for a linear relation among the given vectors, then ask whether they span the domain.

Hint 2. (a)

(2, 4) = 2 (1, 2)

. (b)

(1, 1, 2)

is the sum of the other two vectors; do these span

ℝ^{3}

? (c)

(3, 1) =

2 (1, 1) +

(1, - 1)

. (d) Add the three vectors in

𝔽_{2}^{3}

Sketch. (a) No:

𝑇 (2, 4)

would be

2 𝑇 (1, 2) = (2, 0)

. (b) Exists, since the relation is respected, but is not unique, since the vectors span only a plane. (c) Exists and is unique:

𝑇 (𝑥, 𝑦) = 𝑥 + 𝑦

. (d) No: the three vectors add up to

𝟎

𝔽_{2}^{3}

, but the values add up to

1

(a) No. Since $(2, 4) = 2 (1, 2)$ , a linear $𝑇$ would have $𝑇 (2, 4) =$ $2 𝑇 (1, 2) =$ $(2, 0) \neq$ $(2, 1)$ .

(b) Exists, but is not unique. Let $𝐮_{1} = (1, 0, 1)$ and $𝐮_{2} = (0, 1, 1)$ . They are independent, since $𝑎 𝐮_{1} +$ $𝑏 𝐮_{2} =$ $(𝑎, 𝑏, 𝑎 + 𝑏)$ is $𝟎$ only for $𝑎 = 𝑏 = 0$ . The third vector is $(1, 1, 2) = 𝐮_{1} + 𝐮_{2}$ , and the values satisfy the same relation: $(1, 1) + (2, 0) = (3, 1)$ . The list does not span $ℝ^{3}$ , since every vector $(𝑎, 𝑏, 𝑎 + 𝑏)$ of the span has third coordinate equal to the sum of the first two, and $𝐞_{3}$ does not. Adding $𝐞_{3}$ gives a basis: $(𝑥, 𝑦, 𝑧) =$ $𝑎 𝐮_{1} +$ $𝑏 𝐮_{2} +$ $𝑐 𝐞_{3}$ exactly when $𝑎 = 𝑥$ , $𝑏 = 𝑦$ , and $𝑐 = 𝑧 - 𝑥 - 𝑦$ . By Proposition 2.2.14, a map exists and is not unique. Explicitly, for each $𝐜 \in ℝ^{2}$ ,

𝑇 (𝑥, 𝑦, 𝑧) = 𝑥 (1, 1) + 𝑦 (2, 0) + (𝑧 - 𝑥 - 𝑦) 𝐜

has the prescribed values and $𝑇 (𝐞_{3}) = 𝐜$ .

(c) Exists and is unique. The vectors $(1, 1), (1, - 1)$ form a basis of $ℝ^{2}$ (Example 2.2.3), and $(3, 1) =$ $2 (1, 1) +$ $(1, - 1)$ . The values satisfy the same relation, $2 \cdot 2 + 0 = 4$ , so by Remark 2.2.12 a map exists, and it is unique because the list spans $ℝ^{2}$ . Explicitly, $𝑇 (𝑥, 𝑦) = 𝑥 + 𝑦$ , which gives $𝑇 (1, 1) = 2$ , $𝑇 (1, - 1) = 0$ , and $𝑇 (3, 1) = 4$ .

(d) No. In $𝔽_{2}^{3}$ ,

(1, 1, 0) + (0, 1, 1) + (1, 0, 1) = (2, 2, 2) = 𝟎,

so by Proposition 2.1.16 a linear $𝑇$ would satisfy $𝑇 (1, 1, 0) +$ $𝑇 (0, 1, 1) +$ $𝑇 (1, 0, 1) =$ $0$ . But $1 + 1 + 1 = 1$ in $𝔽_{2}$ .

Exercise 2.2.23.routine

(a)Show that the unique linear map $𝑇 : ℝ [𝑥] \to ℝ$ with $𝑇 (𝑥^{𝑛}) = 1 / (𝑛 + 1)$ for every $𝑛 \geq 0$ is $𝑝 \mapsto \int_{0}^{1} 𝑝 (𝑡) d 𝑡$ .
(b)Let $𝐹$ be a field. Show that the unique linear map $𝑈 : 𝐹 [𝑥] \to 𝐹$ with $𝑈 (𝑥^{𝑛}) = 𝑛$ for every $𝑛 \geq 0$ (where $𝑛$ means $𝑛 \cdot 1 \in 𝐹$ ) is $𝑝 \mapsto 𝑝^{'} (1)$ . Compute $𝑈 (𝑥^{2} + 𝑥)$ over $ℝ$ , over $𝔽_{2}$ , and over $𝔽_{3}$ .

Solution.

Uses. Proposition 2.2.20 (Linear maps out of $𝐹 [𝑥]$ ), Example 2.1.10 (Integration), Example 2.1.9 (Maps of polynomials), Proposition 2.1.17 (Compositions of linear maps are linear), Proposition 1.2.18 (Degree and evaluation versus the operations)

Techniques. Check it on a basis (Technique 2.2.7)

Hint 1. In each part, show that the proposed map is linear and has the right values on monomials; then Proposition 2.2.20 says it is the map in question.

Hint 2.

\int_{0}^{1} 𝑡^{𝑛} d 𝑡 =

1 / (𝑛 + 1)

, and

𝑝 \mapsto 𝑝^{'} (1)

is the composite of differentiation and evaluation at

1

Sketch. (a)

𝑝 \mapsto \int_{0}^{1} 𝑝

is linear and sends

𝑥^{𝑛}

1 / (𝑛 + 1)

. (b)

𝑝 \mapsto 𝑝^{'} (1)

is linear, being evaluation at

1

after

𝐷

, and sends

𝑥^{𝑛}

𝑛 \cdot 1^{𝑛 - 1} = 𝑛

. By uniqueness in Proposition 2.2.20, these are

𝑇

and

𝑈

𝑈 (𝑥^{2} + 𝑥) = 2 + 1 = 3

, which is

3

1

, and

0

ℝ

𝔽_{2}

, and

𝔽_{3}

(a) Let $𝐼 (𝑝) =$ $\int_{0}^{1} 𝑝 (𝑡) d 𝑡$ . Evaluation respects the operations of $ℝ [𝑥]$ , $(𝑐 𝑝 + 𝑞) (𝑡) =$ $𝑐 𝑝 (𝑡) +$ $𝑞 (𝑡)$ (Proposition 1.2.18), so by the linearity of the integral (Example 2.1.10), $𝐼 (𝑐 𝑝 + 𝑞) =$ $𝑐 𝐼 (𝑝) +$ $𝐼 (𝑞)$ , and $𝐼$ is linear. Since $𝐼 (𝑥^{𝑛}) =$ $\int_{0}^{1} 𝑡^{𝑛} d 𝑡 =$ $1 / (𝑛 + 1)$ , the maps $𝐼$ and $𝑇$ are linear and agree on every monomial, so $𝐼 = 𝑇$ by the uniqueness part of Proposition 2.2.20.

(b) Let $𝑈^{'} (𝑝) = 𝑝^{'} (1)$ , that is, $𝑈^{'} = {ev}_{1} \circ 𝐷$ with ${ev}_{1}$ the evaluation at $1$ and $𝐷$ the formal derivative. Both are linear (Example 2.1.9), so $𝑈^{'}$ is linear (Proposition 2.1.17). We have $𝑈^{'} (1) = {ev}_{1} (0) = 0$ , and $𝑈^{'} (𝑥^{𝑛}) =$ ${ev}_{1} (𝑛 𝑥^{𝑛 - 1}) =$ $𝑛 \cdot$ $1^{𝑛 - 1} =$ $𝑛$ for $𝑛 \geq 1$ . So $𝑈^{'}$ and $𝑈$ agree on every monomial, and $𝑈^{'} = 𝑈$ by Proposition 2.2.20.

Finally, $𝑈 (𝑥^{2} + 𝑥) =$ $2 +$ $1 =$ $3 \cdot$ $1$ . This is $3$ over $ℝ$ , $1$ over $𝔽_{2}$ , and $0$ over $𝔽_{3}$ . Over $𝔽_{3}$ , indeed, $𝐷 (𝑥^{2} + 𝑥) = 2 𝑥 + 1$ and $2 + 1 = 0$ .

Exercise 2.2.24.routine Find the error in the following argument.

Claim. Let $𝐯_{1}, \dots, 𝐯_{𝑘}$ span $𝑉$ , and let $𝐰_{1}, \dots, 𝐰_{𝑘} \in 𝑊$ be arbitrary. Then there is a linear map $𝑇 : 𝑉 \to 𝑊$ with $𝑇 (𝐯_{𝑖}) = 𝐰_{𝑖}$ for every $𝑖$ .

Proof. For each $𝐯 \in 𝑉$ , choose scalars with $𝐯 =$ $𝑎_{1} 𝐯_{1} +$ $\dots +$ $𝑎_{𝑘} 𝐯_{𝑘}$ , which is possible because the list spans $𝑉$ , and set $𝑇 (𝐯) =$ $𝑎_{1} 𝐰_{1} +$ $\dots +$ $𝑎_{𝑘} 𝐰_{𝑘}$ . Choosing $𝑎_{𝑖} = 1$ and $𝑎_{𝑗} = 0$ for $𝑗 \neq 𝑖$ shows $𝑇 (𝐯_{𝑖}) = 𝐰_{𝑖}$ . If $𝐮 = \sum_{𝑖} 𝑎_{𝑖} 𝐯_{𝑖}$ and $𝐯 = \sum_{𝑖} 𝑏_{𝑖} 𝐯_{𝑖}$ , then $𝑐 𝐮 +$ $𝐯 =$ $\sum_{𝑖} (𝑐 𝑎_{𝑖} + 𝑏_{𝑖}) 𝐯_{𝑖}$ , so $𝑇 (𝑐 𝐮 + 𝐯) =$ $\sum_{𝑖} (𝑐 𝑎_{𝑖} + 𝑏_{𝑖}) 𝐰_{𝑖} =$ $𝑐 𝑇 (𝐮) +$ $𝑇 (𝐯)$ . So $𝑇$ is linear.

Solution.

Uses. Proposition 2.2.11 (Prescribing values on a spanning list), Non-example 2.2.10 (A spanning list that is not independent)

Hint 1. Does

𝑇 (𝐯)

depend on which scalars were chosen?

Hint 2. Try

𝑉 = 𝑊 = ℝ

, the spanning list

1, 1

, and the values

0, 1

Sketch.

𝑇 (𝐯)

depends on the chosen representation, and the proof silently uses other representations: the trivial one for

𝐯_{𝑖}

, and

\sum_{𝑖} (𝑐 𝑎_{𝑖} + 𝑏_{𝑖}) 𝐯_{𝑖}

for

𝑐 𝐮 + 𝐯

. Counterexample: the list

1, 1

ℝ

with values

0, 1

. The claim becomes true under (2.2.2).

The definition of $𝑇$ involves a choice of scalars for each $𝐯$ , and when $𝐯$ has several representations, different choices can give different values. The proof then treats $𝑇 (𝐯)$ as if it could be computed from any representation. It computes $𝑇 (𝐯_{𝑖})$ from the representation with $𝑎_{𝑖} = 1$ , which need not be the one chosen for $𝐯_{𝑖}$ , and it computes $𝑇 (𝑐 𝐮 + 𝐯)$ from the representation $\sum_{𝑖} (𝑐 𝑎_{𝑖} + 𝑏_{𝑖}) 𝐯_{𝑖}$ , which need not be the one chosen for $𝑐 𝐮 + 𝐯$ . Both steps are valid only if the value does not depend on the representation, and that is exactly what can fail.

The claim is false. In $𝑉 = 𝑊 = ℝ$ , the list $1, 1$ spans $ℝ$ . With $𝐰_{1} = 0$ and $𝐰_{2} = 1$ , a linear $𝑇$ would need $𝑇 (1) = 0$ and $𝑇 (1) = 1$ . The argument becomes correct under the extra hypothesis (2.2.2), which is exactly what makes the value independent of the representation; the first proof of Proposition 2.2.11 is the repaired version.

Exercise 2.2.25.routine Let $ℬ︀ = 𝐯_{1}, \dots, 𝐯_{𝑛}$ be a basis of $𝑉$ , and let $𝑇 \in L (𝑉, 𝑊)$ . Prove or disprove each statement.

(a)If $𝑊 = 𝑉$ and $𝑇 (𝐯_{𝑖}) \in$ $span (𝐯_{𝑖})$ for every $𝑖$ , then $𝑇 (𝐯) \in span (𝐯)$ for every $𝐯 \in 𝑉$ .
(b)If $𝑈$ is a subspace of $𝑊$ and $𝑇 (𝐯_{𝑖}) \in 𝑈$ for every $𝑖$ , then $𝑇 (𝑉) \subseteq 𝑈$ .
(c)If $𝑇 (𝐯_{𝑖}) \neq 𝟎$ for every $𝑖$ , then $𝑇 (𝐯) \neq 𝟎$ for every nonzero $𝐯 \in 𝑉$ .
(d)If $𝑇 (𝐯_{1}) =$ $𝑇 (𝐯_{2}) =$ $\dots =$ $𝑇 (𝐯_{𝑛})$ , then $𝑇 (𝑉) \subseteq$ $span (𝑇 (𝐯_{1}))$ .

Solution.

Uses. Proposition 2.1.16 (Linear maps preserve spans and linear relations), 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. Technique 2.2.7 applies to equations between linear maps and to statements that reduce to them. Which of these statements are of that kind?

Hint 2. For (a) and (c), try

𝑇 (𝑥, 𝑦) =

(𝑥, 2 𝑦)

and

𝑇 (𝑥, 𝑦) = 𝑥 - 𝑦

ℝ^{2}

with the standard basis.

Sketch. (a) False:

𝑇 (𝑥, 𝑦) =

(𝑥, 2 𝑦)

scales

𝐞_{1}

and

𝐞_{2}

, but

𝑇 (1, 1) = (1, 2)

. (b) True:

𝑇 (𝑉) =

span 𝑇 (ℬ︀) \subseteq

𝑈

. (c) False:

(𝑥, 𝑦) \mapsto 𝑥 - 𝑦

kills

(1, 1)

. (d) True:

𝑇 (\sum_{𝑖} 𝑎_{𝑖} 𝐯_{𝑖}) =

(\sum_{𝑖} 𝑎_{𝑖}) 𝑇 (𝐯_{1})

(a) False. On $ℝ^{2}$ with the standard basis, $𝑇 (𝑥, 𝑦) =$ $(𝑥, 2 𝑦)$ satisfies $𝑇 (𝐞_{1}) = 𝐞_{1}$ and $𝑇 (𝐞_{2}) = 2 𝐞_{2}$ . But $𝑇 (1, 1) = (1, 2)$ , which is not in $span ((1, 1))$ , since multiples of $(1, 1)$ have equal coordinates. The condition “ $𝑇 (𝐯) \in span (𝐯)$ ” is not an equation between linear maps, so it does not pass from a basis to all vectors.

(b) True. By Proposition 2.1.16, $𝑇 (𝑉) =$ $𝑇 (span ℬ︀) =$ $span (𝑇 (𝐯_{1}), \dots, 𝑇 (𝐯_{𝑛}))$ . Since $𝑈$ is a subspace containing every $𝑇 (𝐯_{𝑖})$ , it contains their span (Proposition 1.4.14).

(c) False. Let $𝑇 : ℝ^{2} \to ℝ$ , $𝑇 (𝑥, 𝑦) = 𝑥 - 𝑦$ . Then $𝑇 (𝐞_{1}) = 1$ and $𝑇 (𝐞_{2}) = - 1$ are nonzero, but $𝑇 (1, 1) = 0$ .

(d) True. Let $𝐰$ be the common value. For $𝐯 = \sum_{𝑖} 𝑎_{𝑖} 𝐯_{𝑖}$ , linearity gives $𝑇 (𝐯) =$ $\sum_{𝑖} 𝑎_{𝑖} 𝐰 =$ $(𝑎_{1} + \dots + 𝑎_{𝑛}) 𝐰 \in$ $span (𝐰)$ .

Exercise 2.2.26.standard

(a)Find the error in the following argument.
Claim. There is a linear map $𝑇 : ℝ^{2} \to ℝ$ with $𝑇 (1, 2) = 1$ , $𝑇 (2, 1) = 1$ , and $𝑇 (1, 1) = 1$ .
Proof. Any two of the vectors $(1, 2)$ , $(2, 1)$ , $(1, 1)$ form a basis of $ℝ^{2}$ . So by Theorem 2.2.1, for any two of the three prescribed values there is a linear map that takes them. Hence there is a linear map that takes all three.
(b)Find all $𝑐 \in ℝ$ for which some linear $𝑇 : ℝ^{2} \to ℝ$ satisfies $𝑇 (1, 2) = 𝑇 (2, 1) = 1$ and $𝑇 (1, 1) = 𝑐$ .

Solution.

Uses. Theorem 2.2.1 (Linear maps are determined by a basis)

Hint 1. The theorem gives three maps, one for each pair of vectors. Why should they be the same map?

Hint 2. Write

(1, 1)

as a combination of

(1, 2)

and

(2, 1)

Sketch. The three maps given by the theorem need not coincide, and the last sentence does not follow. The linear map with

𝑇 (1, 2) = 𝑇 (2, 1) = 1

is unique, namely

𝑇 (𝑥, 𝑦) =

(𝑥 + 𝑦) / 3

, so

𝑇 (1, 1)

is forced to be

2 / 3

. Hence

𝑐 = 2 / 3

only, and the claim is false.

(a) The argument produces, for each of the three pairs, a linear map that takes the two prescribed values on that pair. These are three maps, and nothing in the argument shows that they are the same map, or that any one of them takes all three values. The first sentence is true, but the conclusion does not follow from it, and (b) shows that the claim is false.

(b) The vectors $(1, 2), (2, 1)$ form a basis of $ℝ^{2}$ : the equations $𝑎 + 2 𝑏 = 𝑥$ and $2 𝑎 + 𝑏 = 𝑦$ have the unique solution $𝑎 = (2 𝑦 - 𝑥) / 3$ , $𝑏 = (2 𝑥 - 𝑦) / 3$ . By Theorem 2.2.1, there is exactly one linear $𝑇_{0}$ with $𝑇_{0} (1, 2) =$ $𝑇_{0} (2, 1) =$ $1$ , namely

𝑇_{0} (𝑥, 𝑦) = 𝑎 + 𝑏 = \frac{𝑥 + 𝑦}{3} .

Any $𝑇$ with $𝑇 (1, 2) = 𝑇 (2, 1) = 1$ equals $𝑇_{0}$ , so $𝑇 (1, 1) =$ $𝑇_{0} (1, 1) =$ $2 / 3$ . Hence $𝑐 = 2 / 3$ is the only possible value, and for it $𝑇_{0}$ works. In particular no linear map has $𝑇 (1, 1) = 1$ . Equivalently, $(1, 1) =$ $\frac{1}{3} (1, 2) +$ $\frac{1}{3} (2, 1)$ , and the prescribed values violate this relation unless $𝑐 = 2 / 3$ .

Exercise 2.2.27.standard Regard $ℂ^{2}$ as a vector space over $ℂ$ .

(a)Show that there is exactly one $ℂ$ -linear map $𝑇 : ℂ^{2} \to ℂ^{2}$ with $𝑇 (1, 𝑖) = (1, 0)$ and $𝑇 (1, - 𝑖) = (0, 1)$ , and find $𝑇 (𝑧_{1}, 𝑧_{2})$ .
(b)Show that there is no $ℂ$ -linear map $𝑇 : ℂ^{2} \to ℂ^{2}$ with $𝑇 (1, 𝑖) = (1, 0)$ and $𝑇 (𝑖, - 1) = (0, 1)$ .
(c)Now regard $ℂ^{2}$ as a vector space over $ℝ$ (Proposition 1.2.19). Show that there is an $ℝ$ -linear map $𝑇 : ℂ^{2} \to ℂ^{2}$ with $𝑇 (1, 𝑖) = (1, 0)$ and $𝑇 (𝑖, - 1) = (0, 1)$ .

Solution.

Uses. Theorem 2.2.1 (Linear maps are determined by a basis), Definition 1.6.1 (Basis), Proposition 1.2.19 (Restriction of scalars), Non-example 2.1.14 (Complex conjugation)

Hint 1. (a) Solve

(𝑧_{1}, 𝑧_{2}) =

𝑎 (1, 𝑖) +

𝑏 (1, - 𝑖)

for

𝑎, 𝑏 \in ℂ

Hint 2. (b) Over

ℂ

, how is

(𝑖, - 1)

related to

(1, 𝑖)

Hint 3. (c) Over

ℝ

the scalar

𝑖

is not available. Extend

(1, 𝑖), (𝑖, - 1)

to a basis of

ℂ^{2}

over

ℝ

by adding

(0, 1)

and

(0, 𝑖)

Sketch. (a)

𝑎 = (𝑧_{1} - 𝑖 𝑧_{2}) / 2

and

𝑏 = (𝑧_{1} + 𝑖 𝑧_{2}) / 2

, uniquely, so

𝑇 (𝑧_{1}, 𝑧_{2}) =

((𝑧_{1} - 𝑖 𝑧_{2}) / 2, (𝑧_{1} + 𝑖 𝑧_{2}) / 2)

. (b)

(𝑖, - 1) = 𝑖 (1, 𝑖)

, so

𝑇 (𝑖, - 1)

would be

(𝑖, 0)

. (c)

(1, 𝑖),

(𝑖, - 1),

(0, 1),

(0, 𝑖)

is a basis over

ℝ

; apply Theorem 2.2.1 over

ℝ

(a) The equation $(𝑧_{1}, 𝑧_{2}) =$ $𝑎 (1, 𝑖) +$ $𝑏 (1, - 𝑖)$ means $𝑎 + 𝑏 = 𝑧_{1}$ and $𝑖 (𝑎 - 𝑏) = 𝑧_{2}$ , that is, $𝑎 - 𝑏 = - 𝑖 𝑧_{2}$ . Adding and subtracting,

𝑎 = \frac{𝑧_{1} - 𝑖 𝑧_{2}}{2}, 𝑏 = \frac{𝑧_{1} + 𝑖 𝑧_{2}}{2},

and these values satisfy both equations. So $(1, 𝑖), (1, - 𝑖)$ is a basis of $ℂ^{2}$ , and by Theorem 2.2.1 there is exactly one such $𝑇$ :

𝑇 (𝑧_{1}, 𝑧_{2}) = 𝑎 (1, 0) + 𝑏 (0, 1) = (\frac{𝑧_{1} - 𝑖 𝑧_{2}}{2}, \frac{𝑧_{1} + 𝑖 𝑧_{2}}{2}) .

Check: at $(1, 𝑖)$ , $𝑧_{1} - 𝑖 𝑧_{2} = 1 - 𝑖^{2} = 2$ and $𝑧_{1} + 𝑖 𝑧_{2} = 1 + 𝑖^{2} = 0$ , so $𝑇 (1, 𝑖) = (1, 0)$ ; similarly $𝑇 (1, - 𝑖) = (0, 1)$ .

(b) Since $𝑖 \cdot 𝑖 = - 1$ , we have $(𝑖, - 1) = 𝑖 (1, 𝑖)$ . A $ℂ$ -linear $𝑇$ would satisfy $𝑇 (𝑖, - 1) =$ $𝑖 𝑇 (1, 𝑖) =$ $(𝑖, 0)$ , which is not $(0, 1)$ .

(c) Over $ℝ$ we show that $(1, 𝑖),$ $(𝑖, - 1),$ $(0, 1),$ $(0, 𝑖)$ is a basis of $ℂ^{2}$ . For $𝑎, 𝑏, 𝑐, 𝑑 \in ℝ$ ,

𝑎 (1, 𝑖) + 𝑏 (𝑖, - 1) + 𝑐 (0, 1) + 𝑑 (0, 𝑖) = (𝑎 + 𝑏 𝑖, (𝑐 - 𝑏) + (𝑎 + 𝑑) 𝑖) .

This equals a given $(𝑧_{1}, 𝑧_{2})$ exactly when $𝑎 + 𝑏 𝑖 = 𝑧_{1}$ and $(𝑐 - 𝑏) +$ $(𝑎 + 𝑑) 𝑖 =$ $𝑧_{2}$ . Comparing real and imaginary parts, the first equation determines $𝑎$ and $𝑏$ , and then the second determines $𝑐$ and $𝑑$ . So there is exactly one solution, and the list is a basis of $ℂ^{2}$ over $ℝ$ . By Theorem 2.2.1 over the field $ℝ$ , there is an $ℝ$ -linear $𝑇$ with $𝑇 (1, 𝑖) = (1, 0)$ , $𝑇 (𝑖, - 1) = (0, 1)$ , and, say, $𝑇 (0, 1) =$ $𝑇 (0, 𝑖) =$ $𝟎$ . By (b), $𝑇$ is not $ℂ$ -linear. As with complex conjugation (Non-example 2.1.14), whether a map is linear depends on the field of scalars.

Exercise 2.2.28.standard Let

𝑈 =

{(𝑥, 𝑦, 𝑧) \in ℝ^{3} | 𝑥 + 𝑦 + 𝑧 = 0}

. Show that there is exactly one linear map

𝑇 : ℝ^{3} \to ℝ^{3}

with

𝑇 (𝐮) = 𝐮

for every

𝐮 \in 𝑈

and

𝑇 (1, 1, 1) = 𝟎

. Find a formula for

𝑇

, and show that

𝑇^{2} = 𝑇

Solution.

Uses. Theorem 2.2.1 (Linear maps are determined by a basis), Corollary 2.2.6 (Agreeing on a spanning list)

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

Hint 1. Find a basis of

ℝ^{3}

consisting of two vectors of

𝑈

and

(1, 1, 1)

. Try

(1, - 1, 0)

and

(0, 1, - 1)

Hint 2. Express

(𝑥, 𝑦, 𝑧)

in that basis. The coefficient of

(1, 1, 1)

(𝑥 + 𝑦 + 𝑧) / 3

Hint 3. For

𝑇^{2} = 𝑇

, compare both sides on your basis.

Sketch.

𝐮_{1} = (1, - 1, 0)

and

𝐮_{2} = (0, 1, - 1)

span

𝑈

, and with

𝐮_{3} = (1, 1, 1)

they form a basis of

ℝ^{3}

, in which

(𝑥, 𝑦, 𝑧)

has

𝐮_{3}

-coefficient

𝑠 = (𝑥 + 𝑦 + 𝑧) / 3

. The conditions force

𝐮_{1} \mapsto 𝐮_{1}

𝐮_{2} \mapsto 𝐮_{2}

𝐮_{3} \mapsto 𝟎

, so

𝑇 (𝐯) = 𝐯 - 𝑠 𝐮_{3}

. Both

𝑇^{2}

and

𝑇

fix

𝐮_{1}, 𝐮_{2}

and kill

𝐮_{3}

Let $𝐮_{1} = (1, - 1, 0)$ , $𝐮_{2} = (0, 1, - 1)$ , and $𝐮_{3} = (1, 1, 1)$ . For $(𝑥, 𝑦, 𝑧) \in ℝ^{3}$ ,

𝑎 𝐮_{1} + 𝑏 𝐮_{2} + 𝑐 𝐮_{3} = (𝑎 + 𝑐, - 𝑎 + 𝑏 + 𝑐, - 𝑏 + 𝑐) .

Setting this equal to $(𝑥, 𝑦, 𝑧)$ and adding the three coordinates gives $3 𝑐 = 𝑥 + 𝑦 + 𝑧$ . The first and third coordinates then give $𝑎 = 𝑥 - 𝑐$ and $𝑏 = 𝑐 - 𝑧$ , and the second holds automatically, since $- 𝑎 +$ $𝑏 +$ $𝑐 =$ $3 𝑐 -$ $𝑥 -$ $𝑧 =$ $𝑦$ . So the unique solution is

𝑐 = \frac{𝑥 + 𝑦 + 𝑧}{3}, 𝑎 = 𝑥 - 𝑐, 𝑏 = 𝑐 - 𝑧,

and $𝐮_{1}, 𝐮_{2}, 𝐮_{3}$ is a basis of $ℝ^{3}$ . Both $𝐮_{1}$ and $𝐮_{2}$ lie in $𝑈$ , and if $(𝑥, 𝑦, 𝑧) \in 𝑈$ then $𝑐 = 0$ , so $𝑈 = span (𝐮_{1}, 𝐮_{2})$ .

Uniqueness. Any $𝑇$ with the required properties satisfies $𝑇 (𝐮_{1}) = 𝐮_{1}$ , $𝑇 (𝐮_{2}) = 𝐮_{2}$ , and $𝑇 (𝐮_{3}) = 𝟎$ , and by Theorem 2.2.1 only one linear map does.

Existence. Let $𝑇$ be that linear map. For $𝐮 = 𝑎 𝐮_{1} + 𝑏 𝐮_{2} \in 𝑈$ , $𝑇 (𝐮) =$ $𝑎 𝐮_{1} +$ $𝑏 𝐮_{2} =$ $𝐮$ , and $𝑇 (1, 1, 1) = 𝟎$ . For a general vector $𝐯 =$ $(𝑥, 𝑦, 𝑧) =$ $𝑎 𝐮_{1} +$ $𝑏 𝐮_{2} +$ $𝑐 𝐮_{3}$ ,

𝑇 (𝐯) = 𝑎 𝐮_{1} + 𝑏 𝐮_{2} = 𝐯 - 𝑐 𝐮_{3} = (𝑥 - 𝑠, 𝑦 - 𝑠, 𝑧 - 𝑠), where 𝑠 = \frac{𝑥 + 𝑦 + 𝑧}{3} .

So $𝑇$ subtracts from each coordinate the average of the three coordinates.

$𝑇^{2} = 𝑇$ . Both sides are linear. For $𝑖 = 1, 2$ , $𝑇^{2} (𝐮_{𝑖}) =$ $𝑇 (𝐮_{𝑖}) =$ $𝐮_{𝑖}$ , and $𝑇^{2} (𝐮_{3}) =$ $𝑇 (𝟎) =$ $𝟎 =$ $𝑇 (𝐮_{3})$ . By Corollary 2.2.6, $𝑇^{2} = 𝑇$ .

Exercise 2.2.29.standard Let $𝐯_{1}, \dots, 𝐯_{𝑛}$ be a basis of $𝑉$ with $𝑛 \geq 1$ , and let $𝑇 \in L (𝑉)$ be the linear map with $𝑇 (𝐯_{𝑖}) = 𝐯_{𝑖 + 1}$ for $𝑖 < 𝑛$ and $𝑇 (𝐯_{𝑛}) = 𝐯_{1}$ . Let $𝐬 = 𝐯_{1} + \dots + 𝐯_{𝑛}$ and $𝑁 =$ ${id}_{𝑉} +$ $𝑇 +$ $𝑇^{2} +$ $\dots +$ $𝑇^{𝑛 - 1}$ .

(a)Show that $𝑇^{𝑛} = {id}_{𝑉}$ .
(b)Show that $𝑁 (𝑎_{1} 𝐯_{1} + \dots + 𝑎_{𝑛} 𝐯_{𝑛}) =$ $(𝑎_{1} + \dots + 𝑎_{𝑛}) 𝐬$ for all scalars $𝑎_{𝑖}$ .
(c)Show that $𝑇 𝑁 = 𝑁 𝑇 = 𝑁$ .

Solution.

Uses. Corollary 2.2.6 (Agreeing on a spanning list), Proposition 2.1.24 (Algebra of composition), Proposition 2.1.23 ( $L (𝑉, 𝑊)$ is a vector space)

Techniques. Check it on a basis (Technique 2.2.7)

Hint 1. All three statements are identities between linear maps, or follow from one. Check them on

𝐯_{1}, \dots, 𝐯_{𝑛}

Hint 2.

𝑇

moves each basis vector one step around the cycle

𝐯_{1} \to

𝐯_{2} \to

\dots \to

𝐯_{𝑛} \to

𝐯_{1}

. Where is

𝐯_{𝑖}

after

𝑘

steps?

Hint 3.

𝑁 (𝐯_{𝑖})

is the sum of the vectors that

𝐯_{𝑖}

visits in its first

𝑛 - 1

steps, together with

𝐯_{𝑖}

itself.

Sketch. With indices read mod

𝑛

𝑇^{𝑘} (𝐯_{𝑖}) =

𝐯_{𝑖 + 𝑘}

. So

𝑇^{𝑛} (𝐯_{𝑖}) = 𝐯_{𝑖}

, and

𝑁 (𝐯_{𝑖}) =

𝐯_{𝑖} +

𝐯_{𝑖 + 1} +

\dots +

𝐯_{𝑖 + 𝑛 - 1} =

𝐬

, since these are all

𝑛

basis vectors once each. Linearity gives (b). Since

𝑇 (𝐬) = 𝐬

𝑇 𝑁

and

𝑁 𝑇

also send every

𝐯_{𝑖}

𝐬

For an integer $𝑗$ , write $𝐯_{𝑗}$ for $𝐯_{𝑟}$ , where $𝑟 \in {1, \dots, 𝑛}$ and $𝑟 \equiv 𝑗$ (mod $𝑛$ ). With this convention $𝑇 (𝐯_{𝑗}) = 𝐯_{𝑗 + 1}$ for every integer $𝑗$ , and by induction on $𝑘$ , $𝑇^{𝑘} (𝐯_{𝑖}) =$ $𝐯_{𝑖 + 𝑘}$ for all $𝑘 \geq 0$ . All maps below are linear by Proposition 2.1.17 and Proposition 2.1.23.

(a) $𝑇^{𝑛} (𝐯_{𝑖}) =$ $𝐯_{𝑖 + 𝑛} =$ $𝐯_{𝑖}$ for each $𝑖$ . So $𝑇^{𝑛}$ and ${id}_{𝑉}$ agree on a basis, and $𝑇^{𝑛} = {id}_{𝑉}$ by Corollary 2.2.6.

(b) For each $𝑖$ ,

𝑁 (𝐯_{𝑖}) = \sum_{𝑘 = 0}^{𝑛 - 1} 𝑇^{𝑘} (𝐯_{𝑖}) = \sum_{𝑘 = 0}^{𝑛 - 1} 𝐯_{𝑖 + 𝑘} = 𝐬,

because the $𝑛$ consecutive integers $𝑖, 𝑖 + 1, \dots, 𝑖 + 𝑛 - 1$ cover each residue mod $𝑛$ exactly once. By linearity, $𝑁 (\sum_{𝑖} 𝑎_{𝑖} 𝐯_{𝑖}) =$ $\sum_{𝑖} 𝑎_{𝑖} 𝐬 =$ $(𝑎_{1} + \dots + 𝑎_{𝑛}) 𝐬$ .

(c) $𝑇 (𝐬) =$ $\sum_{𝑖} 𝑇 (𝐯_{𝑖}) =$ $\sum_{𝑖} 𝐯_{𝑖 + 1} =$ $𝐬$ , the same $𝑛$ vectors in another order. So $𝑇 𝑁 (𝐯_{𝑖}) =$ $𝑇 (𝐬) =$ $𝐬 =$ $𝑁 (𝐯_{𝑖})$ , and $𝑁 𝑇 (𝐯_{𝑖}) =$ $𝑁 (𝐯_{𝑖 + 1}) =$ $𝐬 =$ $𝑁 (𝐯_{𝑖})$ . By Corollary 2.2.6, $𝑇 𝑁 = 𝑁 𝑇 = 𝑁$ . (Alternatively, by Proposition 2.1.24 and (a), $𝑇 𝑁 =$ $𝑇 +$ $𝑇^{2} +$ $\dots +$ $𝑇^{𝑛} =$ $𝑇 +$ $\dots +$ $𝑇^{𝑛 - 1} +$ ${id}_{𝑉} =$ $𝑁$ , and likewise $𝑁 𝑇 =$ $𝑇 +$ $𝑇^{2} +$ $\dots +$ $𝑇^{𝑛} =$ $𝑁$ .)

Exercise 2.2.30.standard Let

𝑅 \in L (ℝ^{2})

be the rotation

𝑅 (𝑥, 𝑦) =

(- 𝑦, 𝑥)

. Show that

𝑇 \in L (ℝ^{2})

satisfies

𝑇 𝑅 = 𝑅 𝑇

if and only if

𝑇 = 𝑎 id + 𝑏 𝑅

for some

𝑎, 𝑏 \in ℝ

. Interpret the result using complex numbers.

Solution.

Uses. Corollary 2.2.6 (Agreeing on a spanning list), Proposition 2.1.24 (Algebra of composition), Exercise 2.1.5

Techniques. Check it on a basis (Technique 2.2.7)

Hint 1.

𝐞_{1}

and

𝑅 (𝐞_{1}) = 𝐞_{2}

form a basis of

ℝ^{2}

, so

𝑇

is determined by

𝑇 (𝐞_{1})

and

𝑇 (𝑅 (𝐞_{1}))

Hint 2. If

𝑇 𝑅 = 𝑅 𝑇

, then

𝑇 (𝑅 (𝐞_{1})) =

𝑅 (𝑇 (𝐞_{1}))

. Write

𝑇 (𝐞_{1}) = (𝑎, 𝑏)

and compare

𝑇

with

𝑎 id + 𝑏 𝑅

𝐞_{1}

and

𝐞_{2}

Sketch. If

𝑇 𝑅 = 𝑅 𝑇

and

𝑇 (𝐞_{1}) = (𝑎, 𝑏)

, then

𝑇 (𝐞_{2}) =

𝑇 (𝑅 𝐞_{1}) =

𝑅 (𝑎, 𝑏) =

(- 𝑏, 𝑎)

. The map

𝑎 id + 𝑏 𝑅

also sends

𝐞_{1} \mapsto (𝑎, 𝑏)

and

𝐞_{2} \mapsto (- 𝑏, 𝑎)

, so it equals

𝑇

. Conversely

𝑎 id + 𝑏 𝑅

commutes with

𝑅

by Proposition 2.1.24. Under

(𝑥, 𝑦) \leftrightarrow 𝑥 + 𝑦 𝑖

𝑅

is multiplication by

𝑖

and

𝑎 id + 𝑏 𝑅

is multiplication by

𝑎 + 𝑏 𝑖

Suppose $𝑇 𝑅 = 𝑅 𝑇$ , and let $𝑇 (𝐞_{1}) = (𝑎, 𝑏)$ . Since $𝑅 (𝐞_{1}) = 𝐞_{2}$ ,

𝑇 (𝐞_{2}) = 𝑇 (𝑅 (𝐞_{1})) = 𝑅 (𝑇 (𝐞_{1})) = 𝑅 (𝑎, 𝑏) = (- 𝑏, 𝑎) .

On the other hand, $(𝑎 id + 𝑏 𝑅) (𝐞_{1}) =$ $(𝑎, 0) +$ $𝑏 (0, 1) =$ $(𝑎, 𝑏)$ and $(𝑎 id + 𝑏 𝑅) (𝐞_{2}) =$ $(0, 𝑎) +$ $𝑏 (- 1, 0) =$ $(- 𝑏, 𝑎)$ . So $𝑇$ and $𝑎 id + 𝑏 𝑅$ are linear maps that agree on the basis $𝐞_{1}, 𝐞_{2}$ , and $𝑇 = 𝑎 id + 𝑏 𝑅$ by Corollary 2.2.6.

Conversely, by Proposition 2.1.24, $(𝑎 id + 𝑏 𝑅) 𝑅 =$ $𝑎 𝑅 +$ $𝑏 𝑅^{2} =$ $𝑅 (𝑎 id + 𝑏 𝑅)$ .

Identify $(𝑥, 𝑦) \in ℝ^{2}$ with $𝑥 + 𝑦 𝑖 \in ℂ$ . Then $𝑅 (𝑥, 𝑦) =$ $(- 𝑦, 𝑥)$ corresponds to $- 𝑦 +$ $𝑥 𝑖 =$ $𝑖 (𝑥 + 𝑦 𝑖)$ , so $𝑅$ is multiplication by $𝑖$ , and $𝑎 id + 𝑏 𝑅$ is multiplication by $𝑎 + 𝑏 𝑖$ . So the $ℝ$ -linear maps of the plane that commute with the quarter turn are exactly the multiplications by complex numbers, which are the $ℂ$ -linear maps $ℂ \to ℂ$ (Exercise 2.1.5).

Exercise 2.2.31.standard Let $𝑝$ be prime.

(a)How many $𝑇 \in L (𝔽_{𝑝}^{2})$ satisfy $𝑇 (1, 1) \in$ $span ((1, 1))$ ?
(b)How many $𝑇 \in L (𝔽_{𝑝}^{2})$ satisfy both $𝑇 (1, 1) \in$ $span ((1, 1))$ and $𝑇 (1, - 1) \in$ $span ((1, - 1))$ ? (The answer depends on $𝑝$ .)

Solution.

Uses. Corollary 2.2.18 (Linear maps correspond to lists of vectors), Example 2.2.3 (Building maps from a convenient basis)

Hint 1. By Corollary 2.2.18, choosing

𝑇

is the same as freely choosing the images of the vectors of a basis. Pick a basis that contains

(1, 1)

Hint 2. For (b), when is

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

a basis of

𝔽_{𝑝}^{2}

? What happens when

𝑝 = 2

Sketch. (a)

(1, 1), (1, 0)

is a basis;

𝑇 (1, 1)

has

𝑝

allowed values and

𝑇 (1, 0)

has

𝑝^{2}

, so there are

𝑝^{3}

maps. (b) For odd

𝑝

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

is a basis, giving

𝑝 \cdot 𝑝 = 𝑝^{2}

. For

𝑝 = 2

(1, - 1) = (1, 1)

, the two conditions coincide, and the answer is

2^{3} = 8

(a) The list $(1, 1), (1, 0)$ is a basis of $𝔽_{𝑝}^{2}$ , since $(𝑥, 𝑦) =$ $𝑎 (1, 1) +$ $𝑏 (1, 0)$ exactly when $𝑎 = 𝑦$ and $𝑏 = 𝑥 - 𝑦$ . By Corollary 2.2.18, $𝑇 \mapsto$ $(𝑇 (1, 1), 𝑇 (1, 0))$ is a bijection from $L (𝔽_{𝑝}^{2})$ onto the pairs of vectors in $𝔽_{𝑝}^{2}$ . The condition restricts $𝑇 (1, 1)$ to $span ((1, 1)) =$ ${(𝑐, 𝑐) | 𝑐 \in 𝔽_{𝑝}}$ , which has $𝑝$ elements, and leaves $𝑇 (1, 0)$ free among $𝑝^{2}$ vectors. So there are $𝑝 \cdot 𝑝^{2} = 𝑝^{3}$ such maps.

(b) If $𝑝$ is odd, then $1 + 1 \neq 0$ in $𝔽_{𝑝}$ , so $(1, 1), (1, - 1)$ is a basis of $𝔽_{𝑝}^{2}$ (Example 2.2.3). The conditions restrict $𝑇 (1, 1)$ to the $𝑝$ vectors $(𝑐, 𝑐)$ and $𝑇 (1, - 1)$ to the $𝑝$ vectors $(𝑐, - 𝑐)$ , which are distinct for distinct $𝑐$ . So there are $𝑝^{2}$ such maps.

If $𝑝 = 2$ , then $- 1 = 1$ in $𝔽_{2}$ , so $(1, - 1) = (1, 1)$ and the second condition is the first one again. By (a) the answer is $2^{3} = 8$ .

Exercise 2.2.32.standard Let $ℒ︀ = 𝐯_{1}, \dots, 𝐯_{𝑛}$ be a list in $𝑉$ .

(a)Suppose $𝑊 \neq {𝟎}$ , and that for every list $𝐰_{1}, \dots, 𝐰_{𝑛}$ in $𝑊$ there is a linear map $𝑇 : 𝑉 \to 𝑊$ with $𝑇 (𝐯_{𝑖}) = 𝐰_{𝑖}$ for every $𝑖$ . Show that $ℒ︀$ is linearly independent.
(b)Conversely, show that if $ℒ︀$ is linearly independent, then for every vector space $𝑊$ and every $𝐰_{1}, \dots, 𝐰_{𝑛} \in 𝑊$ there is a linear map $span ℒ︀ \to 𝑊$ sending each $𝐯_{𝑖}$ to $𝐰_{𝑖}$ .

Solution.

Uses. Theorem 1.5.5 (Independence is decided at zero), Proposition 1.2.35 (Zero products), Theorem 2.2.1 (Linear maps are determined by a basis), Proposition 1.4.14 (The span is the smallest subspace containing a list)

Hint 1. (a) Suppose

𝑎_{1} 𝐯_{1} +

\dots +

𝑎_{𝑛} 𝐯_{𝑛} =

𝟎

. Choose the

𝐰_{𝑖}

so that applying

𝑇

to this relation isolates

𝑎_{𝑗}

Hint 2. Take

𝐰_{𝑗}

to be a nonzero vector of

𝑊

and all other

𝐰_{𝑖} = 𝟎

Hint 3. (b)

ℒ︀

is a basis of

span ℒ︀

Sketch. (a) If

\sum_{𝑖} 𝑎_{𝑖} 𝐯_{𝑖} = 𝟎

, pick

𝐰 \neq 𝟎

and let

𝑇

send

𝐯_{𝑗} \mapsto 𝐰

and the other

𝐯_{𝑖} \mapsto 𝟎

. Then

𝟎 =

𝑇 (\sum_{𝑖} 𝑎_{𝑖} 𝐯_{𝑖}) =

𝑎_{𝑗} 𝐰

, so

𝑎_{𝑗} = 0

. (b)

ℒ︀

spans

span ℒ︀

and is independent, so it is a basis; apply Theorem 2.2.1.

(a) Suppose $𝑎_{1} 𝐯_{1} +$ $\dots +$ $𝑎_{𝑛} 𝐯_{𝑛} =$ $𝟎$ , and fix $𝑗$ . Choose $𝐰 \in 𝑊$ with $𝐰 \neq 𝟎$ , and apply the hypothesis with $𝐰_{𝑗} = 𝐰$ and $𝐰_{𝑖} = 𝟎$ for $𝑖 \neq 𝑗$ . The resulting linear $𝑇$ gives

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

Since $𝐰 \neq 𝟎$ , Proposition 1.2.35 gives $𝑎_{𝑗} = 0$ . As $𝑗$ was arbitrary, every $𝑎_{𝑖} = 0$ , and $ℒ︀$ is linearly independent by Theorem 1.5.5. (The hypothesis $𝑊 \neq {𝟎}$ is needed: every list has the zero map into ${𝟎}$ .)

(b) The span of $ℒ︀$ is a subspace of $𝑉$ (Proposition 1.4.14), hence a vector space, and $ℒ︀$ spans it. If $ℒ︀$ is linearly independent, it is therefore a basis of $span ℒ︀$ , and Theorem 2.2.1 gives the map. To get a map on all of $𝑉$ , one needs to extend from $span ℒ︀$ to $𝑉$ , which is possible under the hypothesis of Proposition 2.2.14.

Exercise 2.2.33.standard Let $ℬ︀ = 𝐯_{1}, \dots, 𝐯_{𝑛}$ be a basis of $𝑉$ .

(a)Show that for each $𝑖$ there is exactly one linear map $𝜑_{𝑖} : 𝑉 \to 𝐹$ with $𝜑_{𝑖} (𝐯_{𝑖}) = 1$ and $𝜑_{𝑖} (𝐯_{𝑗}) = 0$ for $𝑗 \neq 𝑖$ , and that $𝜑_{𝑖} (𝐯)$ is the $𝑖$ -th coordinate of ${[𝐯]}_{ℬ︀}$ .
(b)Show that $𝐯 =$ $𝜑_{1} (𝐯) 𝐯_{1} +$ $\dots +$ $𝜑_{𝑛} (𝐯) 𝐯_{𝑛}$ for every $𝐯 \in 𝑉$ .
(c)Show that every $𝜑 \in L (𝑉, 𝐹)$ equals $𝜑 (𝐯_{1}) 𝜑_{1} +$ $\dots +$ $𝜑 (𝐯_{𝑛}) 𝜑_{𝑛}$ .

(The maps $𝜑_{1}, \dots, 𝜑_{𝑛}$ form the dual basis of $ℬ︀$ ; see Section 13.1.)

Solution.

Uses. Theorem 2.2.1 (Linear maps are determined by a basis), Definition 1.6.14 (Coordinates), Corollary 2.2.6 (Agreeing on a spanning list), Proposition 2.1.23 ( $L (𝑉, 𝑊)$ is a vector space)

Techniques. Check it on a basis (Technique 2.2.7)

Hint 1. (a) is Theorem 2.2.1 with

𝑊 = 𝐹

. Use (2.2.1) to evaluate

𝜑_{𝑖} (𝑎_{1} 𝐯_{1} + \dots + 𝑎_{𝑛} 𝐯_{𝑛})

Hint 2. (c) Both sides are linear maps

𝑉 \to 𝐹

. Compare them on the basis.

Sketch. (a)

𝜑_{𝑖} (\sum_{𝑗} 𝑎_{𝑗} 𝐯_{𝑗}) =

\sum_{𝑗} 𝑎_{𝑗} 𝜑_{𝑖} (𝐯_{𝑗}) =

𝑎_{𝑖}

. (b) is the definition of coordinates, rewritten with (a). (c) Both sides send

𝐯_{𝑘}

𝜑 (𝐯_{𝑘})

, since

𝜑_{𝑖} (𝐯_{𝑘})

1

for

𝑖 = 𝑘

and

0

otherwise.

(a) By Theorem 2.2.1 with $𝑊 = 𝐹$ , there is exactly one linear $𝜑_{𝑖}$ with the prescribed values on $ℬ︀$ . By (2.2.1),

𝜑_{𝑖} (𝑎_{1} 𝐯_{1} + \dots + 𝑎_{𝑛} 𝐯_{𝑛}) = \sum_{𝑗 = 1}^{𝑛} 𝑎_{𝑗} 𝜑_{𝑖} (𝐯_{𝑗}) = 𝑎_{𝑖},

so $𝜑_{𝑖} (𝐯)$ is the $𝑖$ -th coordinate of $𝐯$ (Definition 1.6.14).

(b) By definition of coordinates, $𝐯 =$ $𝑐_{1} 𝐯_{1} +$ $\dots +$ $𝑐_{𝑛} 𝐯_{𝑛}$ with $(𝑐_{1}, \dots, 𝑐_{𝑛}) =$ ${[𝐯]}_{ℬ︀}$ , and $𝑐_{𝑖} = 𝜑_{𝑖} (𝐯)$ by (a).

(c) Let $𝜓 =$ $𝜑 (𝐯_{1}) 𝜑_{1} +$ $\dots +$ $𝜑 (𝐯_{𝑛}) 𝜑_{𝑛}$ , which is linear by Proposition 2.1.23. For each $𝑘$ ,

𝜓 (𝐯_{𝑘}) = \sum_{𝑖 = 1}^{𝑛} 𝜑 (𝐯_{𝑖}) 𝜑_{𝑖} (𝐯_{𝑘}) = 𝜑 (𝐯_{𝑘}),

since only the term $𝑖 = 𝑘$ survives. So $𝜓$ and $𝜑$ agree on a basis, and $𝜓 = 𝜑$ by Corollary 2.2.6.

Exercise 2.2.34.standard Let $𝐯_{1}, \dots, 𝐯_{𝑛}$ be a basis of $𝑉$ and $𝐰_{1}, \dots, 𝐰_{𝑚}$ a basis of $𝑊$ . For $1 \leq 𝑖 \leq 𝑚$ and $1 \leq 𝑗 \leq 𝑛$ , let $𝐸_{𝑖 𝑗} \in L (𝑉, 𝑊)$ be the linear map with $𝐸_{𝑖 𝑗} (𝐯_{𝑗}) = 𝐰_{𝑖}$ and $𝐸_{𝑖 𝑗} (𝐯_{𝑘}) = 𝟎$ for $𝑘 \neq 𝑗$ . Show that the list of the $𝑚 𝑛$ maps

𝐸_{11}, 𝐸_{21}, \dots, 𝐸_{𝑚 1}, 𝐸_{12}, \dots, 𝐸_{𝑚 2}, \dots, 𝐸_{1 𝑛}, \dots, 𝐸_{𝑚 𝑛}

is a basis of $L (𝑉, 𝑊)$ . (Once dimension is defined in Section 4.1, this says $dim L (𝑉, 𝑊) = 𝑚 𝑛$ .)

Solution.

Uses. Theorem 2.2.1 (Linear maps are determined by a basis), Corollary 2.2.6 (Agreeing on a spanning list), Theorem 1.5.5 (Independence is decided at zero), Definition 2.1.21 (Pointwise operations)

Techniques. Check it on a basis (Technique 2.2.7)

Hint 1. Each

𝐸_{𝑖 𝑗}

exists and is unique by Theorem 2.2.1. For independence, apply a vanishing combination

\sum_{𝑖, 𝑗} 𝑐_{𝑖 𝑗} 𝐸_{𝑖 𝑗}

to a basis vector

𝐯_{𝑘}

Hint 2. For spanning, given

𝑇

, write

𝑇 (𝐯_{𝑗}) =

𝑎_{1 𝑗} 𝐰_{1} +

\dots +

𝑎_{𝑚 𝑗} 𝐰_{𝑚}

, and compare

𝑇

with

\sum_{𝑖, 𝑗} 𝑎_{𝑖 𝑗} 𝐸_{𝑖 𝑗}

on the basis of

𝑉

Sketch.

(\sum_{𝑖, 𝑗} 𝑐_{𝑖 𝑗} 𝐸_{𝑖 𝑗}) (𝐯_{𝑘}) =

\sum_{𝑖} 𝑐_{𝑖 𝑘} 𝐰_{𝑖}

. If the combination is

0

, independence of the

𝐰_{𝑖}

gives every

𝑐_{𝑖 𝑘} = 0

. Given

𝑇

with

𝑇 (𝐯_{𝑗}) =

\sum_{𝑖} 𝑎_{𝑖 𝑗} 𝐰_{𝑖}

, the map

\sum_{𝑖, 𝑗} 𝑎_{𝑖 𝑗} 𝐸_{𝑖 𝑗}

agrees with

𝑇

at every

𝐯_{𝑘}

, so it equals

𝑇

By Theorem 2.2.1 each $𝐸_{𝑖 𝑗}$ exists and is unique. For scalars $𝑐_{𝑖 𝑗}$ and each $𝑘$ , the operations on $L (𝑉, 𝑊)$ are pointwise (Definition 2.1.21), so

(\sum_{𝑖, 𝑗} 𝑐_{𝑖 𝑗} 𝐸_{𝑖 𝑗}) (𝐯_{𝑘}) = \sum_{𝑖, 𝑗} 𝑐_{𝑖 𝑗} 𝐸_{𝑖 𝑗} (𝐯_{𝑘}) = \sum_{𝑖 = 1}^{𝑚} 𝑐_{𝑖 𝑘} 𝐰_{𝑖},

(2.2.3)

because $𝐸_{𝑖 𝑗} (𝐯_{𝑘}) = 𝟎$ unless $𝑗 = 𝑘$ .

Independence. Suppose $\sum_{𝑖, 𝑗} 𝑐_{𝑖 𝑗} 𝐸_{𝑖 𝑗} =$ $0$ . Evaluating at $𝐯_{𝑘}$ and using (2.2.3) gives $\sum_{𝑖} 𝑐_{𝑖 𝑘} 𝐰_{𝑖} = 𝟎$ . Since $𝐰_{1}, \dots, 𝐰_{𝑚}$ is linearly independent, $𝑐_{𝑖 𝑘} = 0$ for every $𝑖$ (Theorem 1.5.5). This holds for every $𝑘$ , so all $𝑐_{𝑖 𝑗} = 0$ , and the list is linearly independent by Theorem 1.5.5.

Spanning. Let $𝑇 \in L (𝑉, 𝑊)$ . Since the $𝐰_{𝑖}$ span $𝑊$ , there are scalars $𝑎_{𝑖 𝑗}$ with $𝑇 (𝐯_{𝑗}) =$ $𝑎_{1 𝑗} 𝐰_{1} +$ $\dots +$ $𝑎_{𝑚 𝑗} 𝐰_{𝑚}$ for each $𝑗$ . By (2.2.3), $(\sum_{𝑖, 𝑗} 𝑎_{𝑖 𝑗} 𝐸_{𝑖 𝑗}) (𝐯_{𝑘}) =$ $\sum_{𝑖} 𝑎_{𝑖 𝑘} 𝐰_{𝑖} =$ $𝑇 (𝐯_{𝑘})$ for every $𝑘$ . So $𝑇 =$ $\sum_{𝑖, 𝑗} 𝑎_{𝑖 𝑗} 𝐸_{𝑖 𝑗}$ by Corollary 2.2.6.

The coefficients $𝑎_{𝑖 𝑗}$ in the last step are the entries of the matrix of $𝑇$ (Section 2.3).

Exercise 2.2.35.standard Let $𝑈$ be a subspace of $𝑉$ , and suppose that $𝑉$ has a basis $𝐮_{1},$ $\dots,$ $𝐮_{𝑚},$ $𝐱_{1},$ $\dots,$ $𝐱_{𝑟}$ in which $𝐮_{1}, \dots, 𝐮_{𝑚}$ is a basis of $𝑈$ . (By the results of Section 4.1, this holds whenever $𝑉$ is finite-dimensional.)

(a)Construct $𝑃 \in L (𝑉)$ with $𝑃 (𝑉) = 𝑈$ and $𝑃 (𝐮) = 𝐮$ for every $𝐮 \in 𝑈$ .
(b)Construct $𝑆 \in L (𝑉)$ with ${𝐯 \in 𝑉 | 𝑆 (𝐯) = 𝟎} =$ $𝑈$ .
(c)Carry out (a) and (b) for $𝑈 =$ $span ((1, 2, 3)) \subseteq$ $ℝ^{3}$ , and give formulas for $𝑃$ and $𝑆$ .

(In the language of Section 2.6, every such $𝑈$ is the image of a linear map and the kernel of another.)

Solution.

Uses. Theorem 2.2.1 (Linear maps are determined by a basis), Definition 1.6.14 (Coordinates), Corollary 1.5.17 (Sublists), Theorem 1.5.5 (Independence is decided at zero)

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

Hint 1. Decide what

𝑃

and

𝑆

should do to each basis vector, then apply Theorem 2.2.1.

Hint 2. Let

𝑃

fix each

𝐮_{𝑖}

and kill each

𝐱_{𝑗}

; let

𝑆

kill each

𝐮_{𝑖}

and fix each

𝐱_{𝑗}

Hint 3. For (b), compute

𝑆 (\sum_{𝑖} 𝑎_{𝑖} 𝐮_{𝑖} + \sum_{𝑗} 𝑏_{𝑗} 𝐱_{𝑗})

, and use that

𝐱_{1}, \dots, 𝐱_{𝑟}

is linearly independent. For (c), use the basis

(1, 2, 3), 𝐞_{2}, 𝐞_{3}

Sketch.

𝑃 : 𝐮_{𝑖} \mapsto 𝐮_{𝑖}

𝐱_{𝑗} \mapsto 𝟎

gives

𝑃 (\sum_{𝑖} 𝑎_{𝑖} 𝐮_{𝑖} + \sum_{𝑗} 𝑏_{𝑗} 𝐱_{𝑗}) =

\sum_{𝑖} 𝑎_{𝑖} 𝐮_{𝑖}

𝑆 : 𝐮_{𝑖} \mapsto 𝟎

𝐱_{𝑗} \mapsto 𝐱_{𝑗}

gives

𝑆 (𝐯) =

\sum_{𝑗} 𝑏_{𝑗} 𝐱_{𝑗}

, which is

𝟎

iff every

𝑏_{𝑗} = 0

iff

𝐯 \in 𝑈

. For

𝑈 = span ((1, 2, 3))

𝑃 (𝑥, 𝑦, 𝑧) =

(𝑥, 2 𝑥, 3 𝑥)

and

𝑆 (𝑥, 𝑦, 𝑧) =

(0, 𝑦 - 2 𝑥, 𝑧 - 3 𝑥)

Each $𝐯 \in 𝑉$ has exactly one representation $𝐯 =$ $\sum_{𝑖} 𝑎_{𝑖} 𝐮_{𝑖} +$ $\sum_{𝑗} 𝑏_{𝑗} 𝐱_{𝑗}$ .

(a) Let $𝑃$ be the linear map with $𝑃 (𝐮_{𝑖}) = 𝐮_{𝑖}$ and $𝑃 (𝐱_{𝑗}) = 𝟎$ (Theorem 2.2.1). Then $𝑃 (𝐯) =$ $\sum_{𝑖} 𝑎_{𝑖} 𝐮_{𝑖} \in$ $𝑈$ , so $𝑃 (𝑉) \subseteq 𝑈$ . If $𝐮 \in 𝑈$ , write $𝐮 = \sum_{𝑖} 𝑐_{𝑖} 𝐮_{𝑖}$ ; then $𝑃 (𝐮) =$ $\sum_{𝑖} 𝑐_{𝑖} 𝐮_{𝑖} =$ $𝐮$ . In particular every $𝐮 \in 𝑈$ equals $𝑃 (𝐮) \in 𝑃 (𝑉)$ , so $𝑃 (𝑉) = 𝑈$ .

(b) Let $𝑆$ be the linear map with $𝑆 (𝐮_{𝑖}) = 𝟎$ and $𝑆 (𝐱_{𝑗}) = 𝐱_{𝑗}$ . Then $𝑆 (𝐯) =$ $\sum_{𝑗} 𝑏_{𝑗} 𝐱_{𝑗}$ . If $𝑆 (𝐯) = 𝟎$ , then every $𝑏_{𝑗} = 0$ , because $𝐱_{1}, \dots, 𝐱_{𝑟}$ is linearly independent, being a sublist of a basis (Corollary 1.5.17, Theorem 1.5.5); so $𝐯 = \sum_{𝑖} 𝑎_{𝑖} 𝐮_{𝑖} \in 𝑈$ . Conversely, if $𝐯 = \sum_{𝑖} 𝑐_{𝑖} 𝐮_{𝑖} \in 𝑈$ , then $𝑆 (𝐯) =$ $\sum_{𝑖} 𝑐_{𝑖} 𝑆 (𝐮_{𝑖}) =$ $𝟎$ . So the vectors sent to $𝟎$ are exactly those of $𝑈$ .

(c) The single vector $(1, 2, 3)$ is a basis of $𝑈$ : it spans $𝑈$ , and it is linearly independent because it is nonzero. Together with $𝐞_{2}$ and $𝐞_{3}$ it forms a basis of $ℝ^{3}$ , since

(𝑥, 𝑦, 𝑧) = 𝑎 (1, 2, 3) + 𝑏 𝐞_{2} + 𝑐 𝐞_{3} ⟺ 𝑎 = 𝑥, 𝑏 = 𝑦 - 2 𝑥, 𝑐 = 𝑧 - 3 𝑥 .

So the constructions give

𝑃 (𝑥, 𝑦, 𝑧) = 𝑥 (1, 2, 3) = (𝑥, 2 𝑥, 3 𝑥), 𝑆 (𝑥, 𝑦, 𝑧) = (𝑦 - 2 𝑥) 𝐞_{2} + (𝑧 - 3 𝑥) 𝐞_{3} = (0, 𝑦 - 2 𝑥, 𝑧 - 3 𝑥) .

As a check, $𝑆 (𝑥, 𝑦, 𝑧) = 𝟎$ exactly when $𝑦 = 2 𝑥$ and $𝑧 = 3 𝑥$ , that is, when $(𝑥, 𝑦, 𝑧) =$ $𝑥 (1, 2, 3)$ .

Exercise 2.2.36.hard Let

𝐯_{1}, \dots, 𝐯_{𝑛}

be a basis of

𝑉

, and let

𝑇 \in L (𝑉)

satisfy

𝑇 (𝐯_{1}) = 𝟎

and

𝑇 (𝐯_{𝑘}) \in

span (𝐯_{1}, \dots, 𝐯_{𝑘 - 1})

for

2 \leq 𝑘 \leq 𝑛

. Show that

𝑇^{𝑛} = 0

Solution.

Uses. Proposition 2.1.16 (Linear maps preserve spans and linear relations), Proposition 1.4.14 (The span is the smallest subspace containing a list), Exercise 2.2.25

Techniques. Check it on a basis (Technique 2.2.7)

Hint 1. Let

𝑈_{𝑘} =

span (𝐯_{1}, \dots, 𝐯_{𝑘})

for

0 \leq 𝑘 \leq 𝑛

, with

𝑈_{0} = {𝟎}

. Show that

𝑇 (𝑈_{𝑘}) \subseteq 𝑈_{𝑘 - 1}

Hint 2. For

𝑇 (𝑈_{𝑘}) \subseteq 𝑈_{𝑘 - 1}

, check where

𝑇

sends the spanning list

𝐯_{1}, \dots, 𝐯_{𝑘}

𝑈_{𝑘}

(part (b) of Exercise 2.2.25).

Hint 3. Then show by induction on

𝑘

that

𝑇^{𝑘}

sends every vector of

𝑈_{𝑘}

𝟎

Sketch. With

𝑈_{𝑘} =

span (𝐯_{1}, \dots, 𝐯_{𝑘})

, each

𝑇 (𝐯_{𝑗})

with

𝑗 \leq 𝑘

lies in

𝑈_{𝑗 - 1} \subseteq 𝑈_{𝑘 - 1}

, so

𝑇 (𝑈_{𝑘}) \subseteq 𝑈_{𝑘 - 1}

. Each application of

𝑇

lowers the index by one, so

𝑇^{𝑘} (𝑈_{𝑘}) \subseteq

𝑈_{0} =

{𝟎}

, and

𝑘 = 𝑛

gives

𝑇^{𝑛} = 0

For $0 \leq 𝑘 \leq 𝑛$ , let $𝑈_{𝑘} =$ $span (𝐯_{1}, \dots, 𝐯_{𝑘})$ , so that $𝑈_{0} = {𝟎}$ , $𝑈_{0} \subseteq 𝑈_{1} \subseteq \dots \subseteq 𝑈_{𝑛} = 𝑉$ , and each $𝑈_{𝑘}$ is a subspace.

Claim: $𝑇 (𝑈_{𝑘}) \subseteq 𝑈_{𝑘 - 1}$ for $1 \leq 𝑘 \leq 𝑛$ . For $𝑗 \leq 𝑘$ , the hypothesis gives $𝑇 (𝐯_{𝑗}) \in$ $𝑈_{𝑗 - 1} \subseteq$ $𝑈_{𝑘 - 1}$ (for $𝑗 = 1$ , $𝑇 (𝐯_{1}) = 𝟎 \in 𝑈_{0}$ ). So $𝑈_{𝑘 - 1}$ is a subspace containing $𝑇 (𝐯_{1}),$ $\dots,$ $𝑇 (𝐯_{𝑘})$ , and by Proposition 2.1.16 and Proposition 1.4.14,

𝑇 (𝑈_{𝑘}) = span (𝑇 (𝐯_{1}), \dots, 𝑇 (𝐯_{𝑘})) \subseteq 𝑈_{𝑘 - 1} .

Claim: $𝑇^{𝑘} (𝐮) = 𝟎$ for every $𝐮 \in 𝑈_{𝑘}$ , for $0 \leq 𝑘 \leq 𝑛$ . We induct on $𝑘$ . For $𝑘 = 0$ , $𝑈_{0} = {𝟎}$ and $𝑇^{0} (𝟎) = 𝟎$ . Suppose the claim holds for some $𝑘 < 𝑛$ , and let $𝐮 \in 𝑈_{𝑘 + 1}$ . Then $𝑇 (𝐮) \in 𝑈_{𝑘}$ by the first claim, so $𝑇^{𝑘 + 1} (𝐮) =$ $𝑇^{𝑘} (𝑇 (𝐮)) =$ $𝟎$ .

Taking $𝑘 = 𝑛$ , $𝑇^{𝑛} (𝐯) = 𝟎$ for every $𝐯 \in 𝑈_{𝑛} = 𝑉$ , that is, $𝑇^{𝑛} = 0$ . (In the language of Section 2.3, the matrix of $𝑇$ in this basis is strictly upper triangular.)

Exercise 2.2.37.hard Let $ℒ︀ = 𝐯_{1}, \dots, 𝐯_{𝑘}$ be a list in $𝑉$ and $ℳ︀ = 𝐰_{1}, \dots, 𝐰_{𝑘}$ a list in $𝑊$ . Show that there is a bijective linear map $𝑆 : span ℒ︀ \to span ℳ︀$ with $𝑆 (𝐯_{𝑖}) = 𝐰_{𝑖}$ for every $𝑖$ if and only if $ℒ︀$ and $ℳ︀$ satisfy the same linear relations: for all $𝑎_{1}, \dots, 𝑎_{𝑘} \in 𝐹$ ,

𝑎_{1} 𝐯_{1} + \dots + 𝑎_{𝑘} 𝐯_{𝑘} = 𝟎 ⟺ 𝑎_{1} 𝐰_{1} + \dots + 𝑎_{𝑘} 𝐰_{𝑘} = 𝟎 .

Then decide whether such a map exists for $ℒ︀ =$ $(1, 0),$ $(0, 1),$ $(1, 1)$ and $ℳ︀ =$ $(1, 1),$ $(1, - 1),$ $(2, 0)$ in $ℝ^{2}$ .

Solution.

Uses. Proposition 2.2.11 (Prescribing values on a spanning list), Proposition 2.1.16 (Linear maps preserve spans and linear relations), Proposition 2.1.4 (Linear maps preserve $𝟎$ and negatives), Section A.1

Hint 1. For

(⟹)

, apply

𝑆

to a relation of

ℒ︀

; for the reverse implication between relations, use that

𝑆

is injective.

Hint 2. For

(⟸)

, Proposition 2.2.11 gives a linear

𝑆 : span ℒ︀ \to 𝑊

with

𝑆 (𝐯_{𝑖}) = 𝐰_{𝑖}

. What is its image? Why is it injective?

Hint 3. If

𝑆 (\sum_{𝑖} 𝑎_{𝑖} 𝐯_{𝑖}) =

𝑆 (\sum_{𝑖} 𝑏_{𝑖} 𝐯_{𝑖})

, then

\sum_{𝑖} (𝑎_{𝑖} - 𝑏_{𝑖}) 𝐰_{𝑖} =

𝟎

Sketch. If

𝑆

exists, relations of

ℒ︀

map to relations of

ℳ︀

, and if

\sum_{𝑖} 𝑎_{𝑖} 𝐰_{𝑖} = 𝟎

then

𝑆 (\sum_{𝑖} 𝑎_{𝑖} 𝐯_{𝑖}) =

𝟎 =

𝑆 (𝟎)

, so injectivity gives

\sum_{𝑖} 𝑎_{𝑖} 𝐯_{𝑖} = 𝟎

. Conversely, Proposition 2.2.11 gives

𝑆

span ℒ︀

; its image is

span ℳ︀

, and if

𝑆 (𝐮) = 𝑆 (𝐮^{'})

, the difference of coefficient lists is a relation of

ℳ︀

, hence of

ℒ︀

, so

𝐮 = 𝐮^{'}

. In the example both lists have exactly the relations

𝑎 = 𝑏 = - 𝑐

, so

𝑆

exists.

$(⟹)$ . Let $𝑆$ be such a map. If $\sum_{𝑖} 𝑎_{𝑖} 𝐯_{𝑖} = 𝟎$ , then $\sum_{𝑖} 𝑎_{𝑖} 𝐰_{𝑖} =$ $\sum_{𝑖} 𝑎_{𝑖} 𝑆 (𝐯_{𝑖}) =$ $𝑆 (\sum_{𝑖} 𝑎_{𝑖} 𝐯_{𝑖}) =$ $𝑆 (𝟎) =$ $𝟎$ (Proposition 2.1.4). If $\sum_{𝑖} 𝑎_{𝑖} 𝐰_{𝑖} = 𝟎$ , then $𝑆 (\sum_{𝑖} 𝑎_{𝑖} 𝐯_{𝑖}) =$ $𝟎 =$ $𝑆 (𝟎)$ , and since $𝑆$ is injective, $\sum_{𝑖} 𝑎_{𝑖} 𝐯_{𝑖} = 𝟎$ .

$(⟸)$ . The subspace $span ℒ︀$ is a vector space spanned by $ℒ︀$ , and every relation of $ℒ︀$ is a relation of $ℳ︀$ , which is (2.2.2). By Proposition 2.2.11, applied with $span ℒ︀$ in place of $𝑉$ , there is a linear $𝑆 : span ℒ︀ \to 𝑊$ with $𝑆 (𝐯_{𝑖}) = 𝐰_{𝑖}$ . By Proposition 2.1.16, $𝑆 (span ℒ︀) =$ $span (𝑆 (𝐯_{1}), \dots, 𝑆 (𝐯_{𝑘})) =$ $span ℳ︀$ , so $𝑆$ is a surjective linear map $span ℒ︀ \to span ℳ︀$ . It is injective: if $𝐮 = \sum_{𝑖} 𝑎_{𝑖} 𝐯_{𝑖}$ and $𝐮^{'} = \sum_{𝑖} 𝑏_{𝑖} 𝐯_{𝑖}$ satisfy $𝑆 (𝐮) = 𝑆 (𝐮^{'})$ , then

\sum_{𝑖 = 1}^{𝑘} (𝑎_{𝑖} - 𝑏_{𝑖}) 𝐰_{𝑖} = 𝑆 (𝐮) - 𝑆 (𝐮^{'}) = 𝟎,

which is a relation of $ℳ︀$ , hence of $ℒ︀$ : $\sum_{𝑖} (𝑎_{𝑖} - 𝑏_{𝑖}) 𝐯_{𝑖} =$ $𝟎$ , that is, $𝐮 = 𝐮^{'}$ . So $𝑆$ is bijective (Section A.1).

The example. For $ℒ︀$ : $𝑎 (1, 0) +$ $𝑏 (0, 1) +$ $𝑐 (1, 1) =$ $(𝑎 + 𝑐, 𝑏 + 𝑐)$ , which is $𝟎$ exactly when $𝑎 = 𝑏 = - 𝑐$ . For $ℳ︀$ : $𝑎 (1, 1) +$ $𝑏 (1, - 1) +$ $𝑐 (2, 0) =$ $(𝑎 + 𝑏 + 2 𝑐, 𝑎 - 𝑏)$ , which is $𝟎$ exactly when $𝑎 = 𝑏$ and $2 𝑎 + 2 𝑐 = 0$ , that is, when $𝑎 = 𝑏 = - 𝑐$ . The two lists have the same relations, so such an $𝑆$ exists. Here $span ℒ︀ = span ℳ︀ = ℝ^{2}$ , and $𝑆$ is the linear map with $𝑆 (𝐞_{1}) = (1, 1)$ and $𝑆 (𝐞_{2}) = (1, - 1)$ ; indeed $𝑆 (1, 1) = (2, 0)$ .

Exercise 2.2.38.hard Let

𝑛 \geq 2

, let

𝐬 =

𝐞_{1} +

\dots +

𝐞_{𝑛} \in

𝐹^{𝑛}

, and list the

𝑛 + 1

vectors

𝐯_{1} =

𝐞_{1},

\dots,

𝐯_{𝑛} =

𝐞_{𝑛},

𝐯_{𝑛 + 1} =

𝐬

. For which permutations

𝜎

{1, \dots, 𝑛 + 1}

is there a linear map

𝑇 : 𝐹^{𝑛} \to 𝐹^{𝑛}

with

𝑇 (𝐯_{𝑖}) = 𝐯_{𝜎 (𝑖)}

for every

𝑖

? The answer depends on whether

1 + 1 = 0

𝐹

Solution.

Uses. Proposition 2.2.11 (Prescribing values on a spanning list), Proposition 1.2.35 (Zero products), Definition 1.1.28 (Characteristic)

Hint 1. The list

𝐯_{1}, \dots, 𝐯_{𝑛 + 1}

spans

𝐹^{𝑛}

, so Proposition 2.2.11 applies. Find all linear relations among

𝐯_{1}, \dots, 𝐯_{𝑛 + 1}

Hint 2. Every relation is a multiple of

𝐬 - 𝐞_{1} - \dots - 𝐞_{𝑛} = 𝟎

. So

𝑇

exists if and only if

𝐯_{𝜎 (𝑛 + 1)} =

𝐯_{𝜎 (1)} +

\dots +

𝐯_{𝜎 (𝑛)}

Hint 3. If

𝜎 (𝑛 + 1) = 𝑘 \leq 𝑛

, then

𝐯_{𝜎 (1)},

\dots,

𝐯_{𝜎 (𝑛)}

are

𝐬

and the

𝐞_{𝑗}

with

𝑗 \neq 𝑘

, in some order.

Sketch.

\sum_{𝑖} 𝑎_{𝑖} 𝐞_{𝑖} +

𝑏 𝐬 =

𝟎

iff

𝑎_{𝑖} = - 𝑏

for all

𝑖

, so all relations are multiples of one, and

𝑇

exists iff

𝐯_{𝜎 (𝑛 + 1)} =

\sum_{𝑖 \leq 𝑛} 𝐯_{𝜎 (𝑖)}

. If

𝜎

fixes

𝑛 + 1

, the right side is

𝐬

: fine. If

𝜎 (𝑛 + 1) = 𝑘 \leq 𝑛

, the right side is

𝐬 +

\sum_{𝑗 \neq 𝑘} 𝐞_{𝑗} =

𝐞_{𝑘} +

2 \sum_{𝑗 \neq 𝑘} 𝐞_{𝑗}

, which equals

𝐞_{𝑘}

iff

1 + 1 = 0

. So: all

(𝑛 + 1)!

permutations if

1 + 1 = 0

, otherwise exactly the

𝑛!

that fix

𝑛 + 1

The relations. For scalars $𝑎_{1}, \dots, 𝑎_{𝑛}, 𝑏$ ,

𝑎_{1} 𝐞_{1} + \dots + 𝑎_{𝑛} 𝐞_{𝑛} + 𝑏 𝐬 = (𝑎_{1} + 𝑏, \dots, 𝑎_{𝑛} + 𝑏),

which is $𝟎$ exactly when $𝑎_{𝑖} = - 𝑏$ for every $𝑖$ . So the linear relations of the list are exactly the multiples, by $𝑏 \in 𝐹$ , of the relation $𝐬 - 𝐞_{1} - \dots - 𝐞_{𝑛} = 𝟎$ .

The criterion. The list contains the standard basis, so it spans $𝐹^{𝑛}$ . Given $𝜎$ , put $𝐰_{𝑖} = 𝐯_{𝜎 (𝑖)}$ . A list satisfies all multiples of a relation as soon as it satisfies the relation itself, so (2.2.2) holds exactly when $𝐰_{𝑛 + 1} -$ $𝐰_{1} -$ $\dots -$ $𝐰_{𝑛} =$ $𝟎$ . By Proposition 2.2.11, $𝑇$ exists if and only if

𝐯_{𝜎 (𝑛 + 1)} = 𝐯_{𝜎 (1)} + \dots + 𝐯_{𝜎 (𝑛)} .

(2.2.4)

Case $𝜎 (𝑛 + 1) = 𝑛 + 1$ . Then $𝜎$ permutes ${1, \dots, 𝑛}$ , and the right side of (2.2.4) is $𝐞_{1} +$ $\dots +$ $𝐞_{𝑛} =$ $𝐬 =$ $𝐯_{𝑛 + 1}$ . So $𝑇$ exists; it permutes the standard basis.

Case $𝜎 (𝑛 + 1) = 𝑘 \leq 𝑛$ . Then ${𝜎 (1), \dots, 𝜎 (𝑛)} =$ ${1, \dots, 𝑛 + 1} ∖ {𝑘}$ , so with $2 = 1 + 1$ the right side of (2.2.4) is

𝐬 + \sum_{𝑗 \leq 𝑛, 𝑗 \neq 𝑘} 𝐞_{𝑗} = 𝐞_{𝑘} + 2 \sum_{𝑗 \leq 𝑛, 𝑗 \neq 𝑘} 𝐞_{𝑗} .

Since $𝑛 \geq 2$ , there is some $𝑗 \neq 𝑘$ , and the vector $\sum_{𝑗 \neq 𝑘} 𝐞_{𝑗}$ is nonzero. By Proposition 1.2.35, (2.2.4) holds if and only if $2 = 0$ .

Answer. If $1 + 1 \neq 0$ in $𝐹$ , exactly the $𝑛!$ permutations with $𝜎 (𝑛 + 1) = 𝑛 + 1$ are realized by linear maps. If $1 + 1 = 0$ (characteristic $2$ , Definition 1.1.28), all $(𝑛 + 1)!$ permutations are. The reason is visible in the relation: when $1 + 1 = 0$ it reads $𝐞_{1} + \dots + 𝐞_{𝑛} + 𝐬 = 𝟎$ , which is symmetric in all $𝑛 + 1$ vectors.

Exercise 2.2.39.hard Find all linear maps

𝑓 : 𝐹 [𝑥] \to 𝐹 [𝑥]

such that

𝑓 (𝑝 𝑞) =

𝑓 (𝑝) 𝑓 (𝑞)

for all

𝑝, 𝑞 \in 𝐹 [𝑥]

. (Products of polynomials are studied in Section 7.1. You may use that multiplication of polynomials is associative and commutative, distributes over addition, satisfies

𝑐 (𝑝 𝑞) = (𝑐 𝑝) 𝑞

for

𝑐 \in 𝐹

, and has

𝑥^{𝑗} 𝑥^{𝑘} = 𝑥^{𝑗 + 𝑘}

Solution.

Uses. Proposition 2.2.20 (Linear maps out of $𝐹 [𝑥]$ ), Exercise 2.1.32, Proposition 1.1.17 (No zero divisors), Definition 1.2.16 (Degree)

Techniques. Check it on a basis (Technique 2.2.7)

Hint 1. Let

𝑒 = 𝑓 (1)

. Apply the condition to

1 = 1 \cdot 1

. What can

𝑒

be?

Hint 2. If

𝑒 = 1

, put

𝑞 = 𝑓 (𝑥)

. Compute

𝑓 (𝑥^{𝑛})

and compare

𝑓

with the substitution map

𝑝 \mapsto 𝑝 (𝑞)

on monomials.

Hint 3. To show that

𝑝 \mapsto 𝑝 (𝑞)

is multiplicative, fix

𝑟

and note that both sides of

(𝑝 𝑟) (𝑞) =

𝑝 (𝑞) 𝑟 (𝑞)

are linear in

𝑝

, so it suffices to take

𝑝 = 𝑥^{𝑗}

. Then fix

𝑗

and argue the same way in

𝑟

Sketch.

𝑒 = 𝑓 (1)

satisfies

𝑒^{2} = 𝑒

, and comparing degrees gives

𝑒 \in {0, 1}

. If

𝑒 = 0

, then

𝑓 (𝑝) = 𝑓 (𝑝) 𝑒 = 0

. If

𝑒 = 1

and

𝑞 = 𝑓 (𝑥)

, then

𝑓 (𝑥^{𝑛}) = 𝑞^{𝑛}

, so

𝑓

is the substitution map

𝐶_{𝑞} : 𝑝 \mapsto 𝑝 (𝑞)

by Proposition 2.2.20. Conversely

𝐶_{𝑞}

is linear, and multiplicative because both sides of

𝐶_{𝑞} (𝑝 𝑟) =

𝐶_{𝑞} (𝑝) 𝐶_{𝑞} (𝑟)

are linear in each of

𝑝

and

𝑟

and agree on monomials. Answer:

𝑓 = 0

𝑓 = 𝐶_{𝑞}

for some

𝑞 \in 𝐹 [𝑥]

For $𝑞 \in 𝐹 [𝑥]$ let $𝐶_{𝑞} : 𝐹 [𝑥] \to$ $𝐹 [𝑥]$ , $𝑝 \mapsto 𝑝 (𝑞)$ , be the substitution map, which is linear by Exercise 2.1.32 and satisfies $𝐶_{𝑞} (𝑥^{𝑛}) = 𝑞^{𝑛}$ . We show that the maps with the property are $𝑓 = 0$ and the maps $𝐶_{𝑞}$ .

The value $𝑒 = 𝑓 (1)$ . We have $𝑒 =$ $𝑓 (1 \cdot 1) =$ $𝑓 (1) 𝑓 (1) =$ $𝑒^{2}$ . Suppose $𝑒 \neq 0$ , with degree $𝑑$ and leading coefficient $𝑐 \neq 0$ (Definition 1.2.16). In $𝑒^{2}$ , the coefficient of $𝑥^{2 𝑑}$ is $𝑐^{2} \neq 0$ (Proposition 1.1.17) and all higher coefficients are $0$ , so $deg 𝑒^{2} = 2 𝑑$ . From $𝑒^{2} = 𝑒$ we get $2 𝑑 = 𝑑$ , so $𝑑 = 0$ , and $𝑒 = 𝑐$ is a nonzero constant with $𝑐^{2} = 𝑐$ , that is, $𝑐 (𝑐 - 1) = 0$ . Hence $𝑐 = 1$ . So $𝑒 = 0$ or $𝑒 = 1$ .

If $𝑒 = 0$ . For every $𝑝$ , $𝑓 (𝑝) =$ $𝑓 (𝑝 \cdot 1) =$ $𝑓 (𝑝) 𝑒 =$ $0$ , so $𝑓 = 0$ , which is linear and multiplicative.

If $𝑒 = 1$ . Let $𝑞 = 𝑓 (𝑥)$ . Then $𝑓 (𝑥^{𝑛}) = 𝑞^{𝑛}$ for every $𝑛 \geq 0$ , by induction: $𝑓 (𝑥^{0}) =$ $𝑓 (1) =$ $1 =$ $𝑞^{0}$ , and $𝑓 (𝑥^{𝑛 + 1}) =$ $𝑓 (𝑥^{𝑛} 𝑥) =$ $𝑓 (𝑥^{𝑛}) 𝑓 (𝑥) =$ $𝑞^{𝑛} 𝑞 =$ $𝑞^{𝑛 + 1}$ . So $𝑓$ and $𝐶_{𝑞}$ are linear maps that agree on every monomial, and $𝑓 = 𝐶_{𝑞}$ by Proposition 2.2.20.

Each $𝐶_{𝑞}$ is multiplicative. Fix $𝑟 \in 𝐹 [𝑥]$ . The maps $𝑝 \mapsto 𝐶_{𝑞} (𝑝 𝑟)$ and $𝑝 \mapsto$ $𝐶_{𝑞} (𝑝) 𝐶_{𝑞} (𝑟)$ are linear in $𝑝$ : the first is $𝐶_{𝑞}$ composed with multiplication by $𝑟$ , the second is multiplication by $𝐶_{𝑞} (𝑟)$ composed with $𝐶_{𝑞}$ , and multiplication by a fixed polynomial is linear by distributivity. By Proposition 2.2.20 it suffices to compare them at $𝑝 = 𝑥^{𝑗}$ . Now fix $𝑗$ . The maps $𝑟 \mapsto 𝐶_{𝑞} (𝑥^{𝑗} 𝑟)$ and $𝑟 \mapsto 𝑞^{𝑗} 𝐶_{𝑞} (𝑟)$ are linear in $𝑟$ for the same reason, so it suffices to compare them at $𝑟 = 𝑥^{𝑘}$ :

𝐶_{𝑞} (𝑥^{𝑗} 𝑥^{𝑘}) = 𝐶_{𝑞} (𝑥^{𝑗 + 𝑘}) = 𝑞^{𝑗 + 𝑘} = 𝑞^{𝑗} 𝑞^{𝑘} = 𝐶_{𝑞} (𝑥^{𝑗}) 𝐶_{𝑞} (𝑥^{𝑘}) .

So $𝐶_{𝑞} (𝑝 𝑟) =$ $𝐶_{𝑞} (𝑝) 𝐶_{𝑞} (𝑟)$ for all $𝑝, 𝑟$ .

Answer. The linear maps $𝑓$ with $𝑓 (𝑝 𝑞) =$ $𝑓 (𝑝) 𝑓 (𝑞)$ are the zero map and the substitution maps $𝑝 \mapsto 𝑝 (𝑞)$ , one for each $𝑞 \in 𝐹 [𝑥]$ . Different $𝑞$ give different maps, since $𝐶_{𝑞} (𝑥) = 𝑞$ .

Exercise 2.2.40.challenge Let $𝑛 \geq 1$ , and let $𝐯_{1}, \dots, 𝐯_{𝑛 + 1} \in 𝑉$ be such that any $𝑛$ of them, in their given order, form a basis of $𝑉$ .

(a)Show that $𝐯_{𝑛 + 1} =$ $𝑎_{1} 𝐯_{1} +$ $\dots +$ $𝑎_{𝑛} 𝐯_{𝑛}$ with every $𝑎_{𝑖} \neq 0$ .
(b)Show that if $𝑇 \in L (𝑉)$ satisfies $𝑇 (𝐯_{𝑖}) \in$ $span (𝐯_{𝑖})$ for $𝑖 = 1, \dots, 𝑛 + 1$ , then $𝑇 = 𝜆 {id}_{𝑉}$ for some $𝜆 \in 𝐹$ .
(c)Let $𝐰_{1}, \dots, 𝐰_{𝑛 + 1} \in 𝑊$ be such that any $𝑛$ of them form a basis of $𝑊$ . Show that there are a linear map $𝑇 : 𝑉 \to 𝑊$ and nonzero scalars $𝑐_{1}, \dots, 𝑐_{𝑛 + 1}$ with $𝑇 (𝐯_{𝑖}) = 𝑐_{𝑖} 𝐰_{𝑖}$ for every $𝑖$ , and that any two such maps are scalar multiples of each other.

Solution.

Uses. Theorem 1.5.5 (Independence is decided at zero), Definition 1.6.14 (Coordinates), Theorem 2.2.1 (Linear maps are determined by a basis), Corollary 2.2.6 (Agreeing on a spanning list)

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

Hint 1. (a) If

𝑎_{𝑗} = 0

, find a nontrivial linear relation among the

𝑛

vectors other than

𝐯_{𝑗}

Hint 2. (b) Write

𝑇 (𝐯_{𝑖}) = 𝜆_{𝑖} 𝐯_{𝑖}

and compute

𝑇 (𝐯_{𝑛 + 1})

in two ways. Compare coordinates in the basis

𝐯_{1}, \dots, 𝐯_{𝑛}

Hint 3. (c) Write

𝐰_{𝑛 + 1} =

\sum_{𝑖} 𝑏_{𝑖} 𝐰_{𝑖}

with all

𝑏_{𝑖} \neq 0

, and define

𝑇

on the basis

𝐯_{1}, \dots, 𝐯_{𝑛}

𝑇 (𝐯_{𝑖}) =

(𝑏_{𝑖} / 𝑎_{𝑖}) 𝐰_{𝑖}

. For uniqueness, argue as in (b).

Sketch. (a) If

𝑎_{𝑗} = 0

, then

𝐯_{𝑛 + 1} -

\sum_{𝑖 \neq 𝑗} 𝑎_{𝑖} 𝐯_{𝑖} =

𝟎

is a nontrivial relation among the

𝑛

vectors other than

𝐯_{𝑗}

, which form a basis. (b)

𝑇 (𝐯_{𝑛 + 1}) =

\sum_{𝑖} 𝑎_{𝑖} 𝜆_{𝑖} 𝐯_{𝑖} =

𝜆_{𝑛 + 1} \sum_{𝑖} 𝑎_{𝑖} 𝐯_{𝑖}

, so

𝑎_{𝑖} 𝜆_{𝑖} =

𝑎_{𝑖} 𝜆_{𝑛 + 1}

, and

𝑎_{𝑖} \neq 0

gives

𝜆_{𝑖} = 𝜆_{𝑛 + 1}

. (c) The map

𝐯_{𝑖} \mapsto

(𝑏_{𝑖} / 𝑎_{𝑖}) 𝐰_{𝑖}

sends

𝐯_{𝑛 + 1}

𝐰_{𝑛 + 1}

. If

𝑇^{'} (𝐯_{𝑖}) =

𝑐_{𝑖}^{'} 𝐰_{𝑖}

, comparing coordinates of

𝑇^{'} (𝐯_{𝑛 + 1})

forces

𝑐_{𝑖}^{'} =

𝑐_{𝑛 + 1}^{'} 𝑏_{𝑖} / 𝑎_{𝑖}

, so

𝑇^{'} = 𝑐_{𝑛 + 1}^{'} 𝑇

By hypothesis, deleting any one vector from $𝐯_{1}, \dots, 𝐯_{𝑛 + 1}$ leaves a basis of $𝑉$ ; in particular $𝐯_{1}, \dots, 𝐯_{𝑛}$ is a basis.

(a) Write $𝐯_{𝑛 + 1} =$ $𝑎_{1} 𝐯_{1} +$ $\dots +$ $𝑎_{𝑛} 𝐯_{𝑛}$ , with unique coefficients. Suppose $𝑎_{𝑗} = 0$ for some $𝑗$ . Then

1 \cdot 𝐯_{𝑛 + 1} - \sum_{𝑖 \leq 𝑛, 𝑖 \neq 𝑗} 𝑎_{𝑖} 𝐯_{𝑖} = 𝟎

is a linear relation, with the nonzero coefficient $1$ , among the vectors of the list obtained by deleting $𝐯_{𝑗}$ . That list is a basis, hence linearly independent, which contradicts Theorem 1.5.5. So every $𝑎_{𝑖} \neq 0$ .

(b) Write $𝑇 (𝐯_{𝑖}) = 𝜆_{𝑖} 𝐯_{𝑖}$ for $𝑖 = 1, \dots, 𝑛 + 1$ . Then

\sum_{𝑖 = 1}^{𝑛} 𝑎_{𝑖} 𝜆_{𝑖} 𝐯_{𝑖} = \sum_{𝑖 = 1}^{𝑛} 𝑎_{𝑖} 𝑇 (𝐯_{𝑖}) = 𝑇 (𝐯_{𝑛 + 1}) = 𝜆_{𝑛 + 1} 𝐯_{𝑛 + 1} = \sum_{𝑖 = 1}^{𝑛} 𝜆_{𝑛 + 1} 𝑎_{𝑖} 𝐯_{𝑖} .

By uniqueness of coordinates in the basis $𝐯_{1}, \dots, 𝐯_{𝑛}$ (Definition 1.6.14), $𝑎_{𝑖} 𝜆_{𝑖} =$ $𝑎_{𝑖} 𝜆_{𝑛 + 1}$ , and since $𝑎_{𝑖} \neq 0$ by (a), $𝜆_{𝑖} = 𝜆_{𝑛 + 1}$ . With $𝜆 = 𝜆_{𝑛 + 1}$ , $𝑇$ and $𝜆 {id}_{𝑉}$ agree on the basis $𝐯_{1}, \dots, 𝐯_{𝑛}$ , so $𝑇 = 𝜆 {id}_{𝑉}$ by Corollary 2.2.6.

(c) By (a) applied to the $𝐰_{𝑖}$ , $𝐰_{𝑛 + 1} =$ $𝑏_{1} 𝐰_{1} +$ $\dots +$ $𝑏_{𝑛} 𝐰_{𝑛}$ with every $𝑏_{𝑖} \neq 0$ .

Existence. By Theorem 2.2.1, let $𝑇 : 𝑉 \to 𝑊$ be the linear map with $𝑇 (𝐯_{𝑖}) =$ $(𝑏_{𝑖} / 𝑎_{𝑖}) 𝐰_{𝑖}$ for $𝑖 = 1, \dots, 𝑛$ . Then

𝑇 (𝐯_{𝑛 + 1}) = \sum_{𝑖 = 1}^{𝑛} 𝑎_{𝑖} (\frac{𝑏_{𝑖}}{𝑎_{𝑖}}) 𝐰_{𝑖} = \sum_{𝑖 = 1}^{𝑛} 𝑏_{𝑖} 𝐰_{𝑖} = 𝐰_{𝑛 + 1} .

So $𝑇 (𝐯_{𝑖}) = 𝑐_{𝑖} 𝐰_{𝑖}$ with $𝑐_{𝑖} = 𝑏_{𝑖} / 𝑎_{𝑖} \neq 0$ for $𝑖 \leq 𝑛$ and $𝑐_{𝑛 + 1} = 1$ .

Uniqueness up to scalars. Let $𝑇^{'} : 𝑉 \to 𝑊$ be linear with $𝑇^{'} (𝐯_{𝑖}) =$ $𝑐_{𝑖}^{'} 𝐰_{𝑖}$ , all $𝑐_{𝑖}^{'} \neq 0$ . Then

\sum_{𝑖 = 1}^{𝑛} 𝑎_{𝑖} 𝑐_{𝑖}^{'} 𝐰_{𝑖} = 𝑇^{'} (𝐯_{𝑛 + 1}) = 𝑐_{𝑛 + 1}^{'} 𝐰_{𝑛 + 1} = \sum_{𝑖 = 1}^{𝑛} 𝑐_{𝑛 + 1}^{'} 𝑏_{𝑖} 𝐰_{𝑖},

and comparing coordinates in the basis $𝐰_{1}, \dots, 𝐰_{𝑛}$ gives $𝑐_{𝑖}^{'} =$ $𝑐_{𝑛 + 1}^{'} 𝑏_{𝑖} / 𝑎_{𝑖}$ for $𝑖 \leq 𝑛$ . So $𝑇^{'} (𝐯_{𝑖}) =$ $𝑐_{𝑛 + 1}^{'} 𝑇 (𝐯_{𝑖})$ for $𝑖 \leq 𝑛$ , and $𝑇^{'} = 𝑐_{𝑛 + 1}^{'} 𝑇$ by Corollary 2.2.6. Every such map is therefore a nonzero multiple of $𝑇$ , and any two of them are multiples of each other.

In projective geometry, (c) says that any two frames of $𝑛 + 1$ points in general position are related by a projective transformation, which is unique.

More practice

The practice bank has 24 exercises on this section’s topics.