2.1 Linear maps · Matrix Theory

In Section 1.6 we attached to each vector $𝐯$ of a space $𝑉$ with basis $ℬ︀$ its coordinate vector ${[𝐯]}_{ℬ︀} \in 𝐹^{𝑛}$ , and we saw that coordinates respect linear combinations: ${[𝑐_{1} 𝐯_{1} + \dots + 𝑐_{𝑘} 𝐯_{𝑘}]}_{ℬ︀} =$ $𝑐_{1} {[𝐯_{1}]}_{ℬ︀} +$ $\dots +$ $𝑐_{𝑘} {[𝐯_{𝑘}]}_{ℬ︀}$ (Proposition 1.6.19). Forming a combination in $𝑉$ and then taking coordinates gives the same answer as taking coordinates first and forming the combination in $𝐹^{𝑛}$ . A vector space has no structure besides addition and scalar multiplication, so the functions worth studying between vector spaces are the ones with this property. They are the linear maps, and from here on they are the main objects of the book.

2.1.1Definition and first properties

Definition 2.1.1 (Linear map). Let $𝑉$ and $𝑊$ be vector spaces over the same field $𝐹$ . A function $𝑇 : 𝑉 \to 𝑊$ is a linear map, or linear transformation, if it respects linear combinations: for every $𝑘 \geq 1$ , all $𝐯_{1}, \dots, 𝐯_{𝑘} \in 𝑉$ , and all $𝑎_{1}, \dots, 𝑎_{𝑘} \in 𝐹$ ,

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

A linear map from $𝑉$ to itself is also called a linear operator, or just an operator, on $𝑉$ .

Notation 2.1.2.

L (𝑉, 𝑊)

denotes the set of linear maps

𝑉 \to 𝑊

, and

L (𝑉) = L (𝑉, 𝑉)

the set of operators on

𝑉

. Other texts write

𝑇 𝐯

for

𝑇 (𝐯)

; we keep the parentheses in this section. When the field matters, we say that

𝑇

is $𝐹$ -linear.

Realize that the two sides of the defining equation are computed in different spaces. On the left, the linear combination is formed in $𝑉$ and then $𝑇$ is applied to the result. On the right, $𝑇$ is applied to each $𝐯_{𝑖}$ first and the combination is formed in $𝑊$ , with the same coefficients. Linearity says that the order does not matter: applying $𝑇$ commutes with forming linear combinations.

Intuition. The definition asks about combinations of every length

𝑘

, which is unpleasant to check directly. But a linear combination is built one operation at a time from scalar multiples and sums. So a function that respects scalar multiplication and addition separately should respect every linear combination: split off the last term

𝑎_{𝑘} 𝐯_{𝑘}

using addition, pull out its coefficient using scalar multiplication, and handle the shorter combination that remains in the same way. This is an induction on

𝑘

. Conversely,

𝑎 𝐯

and

𝐮 + 𝐯

are themselves linear combinations, of lengths

1

and

2

Theorem 2.1.3 (Checking linearity). Let

𝑉

and

𝑊

be vector spaces over

𝐹

, and let

𝑇 : 𝑉 \to 𝑊

be a function. The following are equivalent.

(i) $𝑇$ is linear.
(ii) $𝑇$ is additive and homogeneous: $𝑇 (𝐮 + 𝐯) =$ $𝑇 (𝐮) +$ $𝑇 (𝐯)$ and $𝑇 (𝑎 𝐯) = 𝑎 𝑇 (𝐯)$ for all $𝐮, 𝐯 \in 𝑉$ and all $𝑎 \in 𝐹$ .
(iii)For all $𝐮, 𝐯 \in 𝑉$ and all $𝑎 \in 𝐹$ , $𝑇 (𝑎 𝐮 + 𝐯) =$ $𝑎 𝑇 (𝐮) +$ $𝑇 (𝐯)$ .

Proof.routine

Uses. Definition 2.1.1 (Linear map), Definition 1.2.1 (Vector space)

Techniques. Induction on the length of a list (Technique 1.6.9)

Hint 1. Sums and scalar multiples are linear combinations of length

2

and

1

, so (i)

\Rightarrow

(iii) and (iii)

\Rightarrow

(ii) come down to choosing special cases. For (ii)

\Rightarrow

(i), induct on the length

𝑘

Hint 2. For (iii)

\Rightarrow

(ii), first take

𝑎 = 1

and

𝐮 = 𝐯 = 𝟎

to learn that

𝑇 (𝟎) = 𝟎

. Then

𝑎 = 1

gives additivity and

𝐯 = 𝟎

gives homogeneity.

Hint 3. For the induction step, write

𝑎_{1} 𝐯_{1} +

\dots +

𝑎_{𝑘 + 1} 𝐯_{𝑘 + 1}

𝐮 + 𝑎_{𝑘 + 1} 𝐯_{𝑘 + 1}

with

𝐮 =

𝑎_{1} 𝐯_{1} +

\dots +

𝑎_{𝑘} 𝐯_{𝑘}

. Apply additivity, then homogeneity to the last term and the induction hypothesis to

𝑇 (𝐮)

Sketch. (i)

\Rightarrow

(iii) is the definition with

𝑘 = 2

and coefficients

𝑎, 1

. In (iii), the choice

𝑎 = 1

𝐮 = 𝐯 = 𝟎

gives

𝑇 (𝟎) =

𝑇 (𝟎) +

𝑇 (𝟎)

, so

𝑇 (𝟎) = 𝟎

; then

𝑎 = 1

gives additivity and

𝐯 = 𝟎

gives homogeneity. For (ii)

\Rightarrow

(i), induct on

𝑘

: the case

𝑘 = 1

is homogeneity, and additivity splits a combination of length

𝑘 + 1

into one of length

𝑘

plus a single term.

We prove (i) $\Rightarrow$ (iii) $\Rightarrow$ (ii) $\Rightarrow$ (i).

(i) $\Rightarrow$ (iii). Let $𝐮, 𝐯 \in 𝑉$ and $𝑎 \in 𝐹$ . The definition of linearity with $𝑘 = 2$ , $𝐯_{1} = 𝐮$ , $𝐯_{2} = 𝐯$ , $𝑎_{1} = 𝑎$ , and $𝑎_{2} = 1$ gives $𝑇 (𝑎 𝐮 + 1 𝐯) =$ $𝑎 𝑇 (𝐮) +$ $1 𝑇 (𝐯)$ . Since $1 𝐱 = 𝐱$ for every vector $𝐱$ , in $𝑉$ and in $𝑊$ , this is (iii).

(iii) $\Rightarrow$ (ii). Taking $𝑎 = 1$ and $𝐮 = 𝐯 = 𝟎$ in (iii) gives

𝑇 (𝟎) = 𝑇 (1 \cdot 𝟎 + 𝟎) = 1 \cdot 𝑇 (𝟎) + 𝑇 (𝟎) = 𝑇 (𝟎) + 𝑇 (𝟎) .

Adding $- 𝑇 (𝟎)$ to both sides and regrouping, $𝟎 = 𝑇 (𝟎)$ . Now let $𝐮, 𝐯 \in 𝑉$ and $𝑎 \in 𝐹$ . With $𝑎 = 1$ , (iii) reads $𝑇 (𝐮 + 𝐯) =$ $𝑇 (𝐮) +$ $𝑇 (𝐯)$ , which is additivity. With $𝐯 = 𝟎$ , it reads

𝑇 (𝑎 𝐮) = 𝑇 (𝑎 𝐮 + 𝟎) = 𝑎 𝑇 (𝐮) + 𝑇 (𝟎) = 𝑎 𝑇 (𝐮),

which is homogeneity.

(ii) $\Rightarrow$ (i). We show by induction on $𝑘 \geq 1$ that $𝑇 (𝑎_{1} 𝐯_{1} + \dots + 𝑎_{𝑘} 𝐯_{𝑘}) =$ $𝑎_{1} 𝑇 (𝐯_{1}) +$ $\dots +$ $𝑎_{𝑘} 𝑇 (𝐯_{𝑘})$ for all $𝐯_{1}, \dots, 𝐯_{𝑘} \in 𝑉$ and $𝑎_{1}, \dots, 𝑎_{𝑘} \in 𝐹$ . For $𝑘 = 1$ this is homogeneity. Suppose it holds for some $𝑘 \geq 1$ , and let $𝐯_{1}, \dots, 𝐯_{𝑘 + 1} \in 𝑉$ and $𝑎_{1}, \dots, 𝑎_{𝑘 + 1} \in 𝐹$ . Put $𝐮 =$ $𝑎_{1} 𝐯_{1} +$ $\dots +$ $𝑎_{𝑘} 𝐯_{𝑘}$ . Then

\begin{aligned} 𝑇 (𝑎_{1} 𝐯_{1} + \dots + 𝑎_{𝑘 + 1} 𝐯_{𝑘 + 1}) & = 𝑇 (𝐮 + 𝑎_{𝑘 + 1} 𝐯_{𝑘 + 1}) \\ = 𝑇 (𝐮) + 𝑇 (𝑎_{𝑘 + 1} 𝐯_{𝑘 + 1}) \\ = 𝑎_{1} 𝑇 (𝐯_{1}) + \dots + 𝑎_{𝑘} 𝑇 (𝐯_{𝑘}) + 𝑎_{𝑘 + 1} 𝑇 (𝐯_{𝑘 + 1}), \end{aligned}

as desired. The second step is additivity, and the third applies the induction hypothesis to $𝑇 (𝐮)$ and homogeneity to the last term. □

From now on we verify linearity through (ii) or (iii), and we use (i) whenever we apply a linear map to a linear combination. Neither half of (ii) implies the other: complex conjugation is additive but not homogeneous over $ℂ$ (Non-example 2.1.14), and Exercise 2.1.38 gives a homogeneous map that is not additive.

Proposition 2.1.4 (Linear maps preserve $𝟎$ and negatives). Let $𝑇 \in L (𝑉, 𝑊)$ . Then $𝑇 (𝟎_{𝑉}) = 𝟎_{𝑊}$ , and for all $𝐮, 𝐯 \in 𝑉$ ,

𝑇 (- 𝐯) = - 𝑇 (𝐯) and 𝑇 (𝐮 - 𝐯) = 𝑇 (𝐮) - 𝑇 (𝐯) .

Proof (via homogeneity).routine

Uses. Theorem 2.1.3 (Checking linearity), Proposition 1.2.32 (The scalar zero), Proposition 1.2.34 (Negatives and the scalar $- 1$ )

Hint 1. Write

𝟎_{𝑉}

as a scalar multiple of some vector, so that homogeneity applies.

Hint 2.

𝟎_{𝑉} = 0 \cdot 𝟎_{𝑉}

, and

0 𝐰 = 𝟎_{𝑊}

for every

𝐰 \in 𝑊

. For negatives,

- 𝐯 = (- 1) 𝐯

Sketch.

𝑇 (𝟎) =

𝑇 (0 \cdot 𝟎) =

0 \cdot

𝑇 (𝟎) =

𝟎

, using

0 𝐱 = 𝟎

𝑉

and then in

𝑊

. Similarly

𝑇 (- 𝐯) =

𝑇 ((- 1) 𝐯) =

(- 1) 𝑇 (𝐯) =

- 𝑇 (𝐯)

, and the last identity follows by additivity.

By Proposition 1.2.32, $0 𝐱 = 𝟎$ for every vector $𝐱$ , in $𝑉$ and in $𝑊$ . So by homogeneity,

𝑇 (𝟎_{𝑉}) = 𝑇 (0 \cdot 𝟎_{𝑉}) = 0 \cdot 𝑇 (𝟎_{𝑉}) = 𝟎_{𝑊} .

By Proposition 1.2.34, $- 𝐱 = (- 1) 𝐱$ in $𝑉$ and in $𝑊$ , so homogeneity gives $𝑇 (- 𝐯) =$ $𝑇 ((- 1) 𝐯) =$ $(- 1) 𝑇 (𝐯) =$ $- 𝑇 (𝐯)$ . Finally, by additivity, $𝑇 (𝐮 - 𝐯) =$ $𝑇 (𝐮 + (- 𝐯)) =$ $𝑇 (𝐮) +$ $𝑇 (- 𝐯) =$ $𝑇 (𝐮) -$ $𝑇 (𝐯)$ . □

Proof (via additivity).routine

Uses. Theorem 2.1.3 (Checking linearity), Definition 1.2.1 (Vector space), Proposition 1.2.28 (Uniqueness of additive inverses)

Techniques. Uniqueness via two candidates (Technique 1.1.8)

Hint 1. Write

𝟎_{𝑉}

as a sum, so that additivity applies.

Hint 2.

𝑇 (𝟎) =

𝑇 (𝟎 + 𝟎) =

𝑇 (𝟎) +

𝑇 (𝟎)

. Now cancel. For negatives, show that

𝑇 (- 𝐯)

is an additive inverse of

𝑇 (𝐯)

Sketch.

𝑇 (𝟎) =

𝑇 (𝟎 + 𝟎) =

𝑇 (𝟎) +

𝑇 (𝟎)

, and adding

- 𝑇 (𝟎)

gives

𝟎 = 𝑇 (𝟎)

. Then

𝑇 (𝐯) +

𝑇 (- 𝐯) =

𝑇 (𝐯 - 𝐯) =

𝑇 (𝟎) =

𝟎

, so

𝑇 (- 𝐯) = - 𝑇 (𝐯)

by uniqueness of additive inverses.

By additivity, $𝑇 (𝟎_{𝑉}) =$ $𝑇 (𝟎_{𝑉} + 𝟎_{𝑉}) =$ $𝑇 (𝟎_{𝑉}) +$ $𝑇 (𝟎_{𝑉})$ . Writing $𝐰 = 𝑇 (𝟎_{𝑉})$ , we have $𝐰 = 𝐰 + 𝐰$ , so

𝟎_{𝑊} = 𝐰 + (- 𝐰) = (𝐰 + 𝐰) + (- 𝐰) = 𝐰 + (𝐰 + (- 𝐰)) = 𝐰 + 𝟎_{𝑊} = 𝐰 .

Next, for $𝐯 \in 𝑉$ , additivity gives $𝑇 (𝐯) +$ $𝑇 (- 𝐯) =$ $𝑇 (𝐯 + (- 𝐯)) =$ $𝑇 (𝟎_{𝑉}) =$ $𝟎_{𝑊}$ . So $𝑇 (- 𝐯)$ is an additive inverse of $𝑇 (𝐯)$ , and by Proposition 1.2.28 it equals $- 𝑇 (𝐯)$ . The identity $𝑇 (𝐮 - 𝐯) =$ $𝑇 (𝐮) -$ $𝑇 (𝐯)$ follows from additivity as in the first proof. □

The first proof used only homogeneity and the second only additivity. So a function satisfying either condition alone sends $𝟎$ to $𝟎$ ; we use this for additive maps in Exercise 2.1.36 and Exercise 2.1.37. In practice Proposition 2.1.4 is a quick test in the contrapositive: a function with $𝑇 (𝟎) \neq 𝟎$ is not linear. It is only a necessary condition. The map $ℝ^{2} \to ℝ$ , $(𝑥, 𝑦) \mapsto 𝑥 𝑦$ , sends $𝟎$ to $0$ , but it sends $(1, 1)$ to $1$ and $2 (1, 1) = (2, 2)$ to $4 \neq 2 \cdot 1$ , so it is not homogeneous.

Exercise 2.1.5.routine Let $𝑊$ be a vector space over $𝐹$ , and regard $𝐹$ as a vector space over itself.

(a)Show that for every $𝐰 \in 𝑊$ , the map $𝐹 \to 𝑊$ , $𝑥 \mapsto 𝑥 𝐰$ , is linear.
(b)Show that every $𝑇 \in L (𝐹, 𝑊)$ has this form, with $𝐰 = 𝑇 (1)$ . In particular, the linear maps $𝐹 \to 𝐹$ are exactly the maps $𝑥 \mapsto 𝑎 𝑥$ with $𝑎 \in 𝐹$ .

Solution.

Uses. Theorem 2.1.3 (Checking linearity), Definition 1.2.1 (Vector space)

Hint 1. (a) is two of the vector space axioms of

𝑊

Hint 2. (b) Every

𝑥 \in 𝐹

is a scalar multiple of the single vector

1 \in 𝐹

𝑥 = 𝑥 \cdot 1

. Apply homogeneity.

Sketch. (a)

(𝑥 + 𝑦) 𝐰 =

𝑥 𝐰 +

𝑦 𝐰

and

(𝑎 𝑥) 𝐰 = 𝑎 (𝑥 𝐰)

. (b)

𝑇 (𝑥) =

𝑇 (𝑥 \cdot 1) =

𝑥 𝑇 (1)

(a) For $𝑥, 𝑦, 𝑎 \in 𝐹$ , the vector space axioms of $𝑊$ give $(𝑥 + 𝑦) 𝐰 =$ $𝑥 𝐰 +$ $𝑦 𝐰$ and $(𝑎 𝑥) 𝐰 = 𝑎 (𝑥 𝐰)$ . So the map is additive and homogeneous, hence linear by Theorem 2.1.3.

(b) Let $𝑇 \in L (𝐹, 𝑊)$ and $𝐰 = 𝑇 (1)$ . For $𝑥 \in 𝐹$ , the element $𝑥 = 𝑥 \cdot 1$ is the scalar $𝑥$ times the vector $1 \in 𝐹$ , so homogeneity gives $𝑇 (𝑥) =$ $𝑇 (𝑥 \cdot 1) =$ $𝑥 𝑇 (1) =$ $𝑥 𝐰$ . Taking $𝑊 = 𝐹$ , every $𝑇 \in L (𝐹)$ is $𝑥 \mapsto 𝑎 𝑥$ with $𝑎 = 𝑇 (1)$ , and these maps are linear by (a).

2.1.2Examples

Many familiar operations are linear maps. In each example below, linearity is checked with Theorem 2.1.3.

Example 2.1.6 (Zero, identity, and scaling maps). Let $𝑉$ and $𝑊$ be vector spaces over $𝐹$ .

The zero map $𝑉 \to 𝑊$ , $𝐯 \mapsto 𝟎_{𝑊}$ , is linear, since both sides of (iii) in Theorem 2.1.3 are $𝟎_{𝑊}$ (by Proposition 1.2.33). We denote it by $0$ ; the context tells it apart from $0 \in 𝐹$ and $𝟎_{𝑊}$ .
The identity map ${id}_{𝑉} : 𝑉 \to 𝑉$ , $𝐯 \mapsto 𝐯$ , is linear, since both sides of (iii) are $𝑎 𝐮 + 𝐯$ .
For $𝜆 \in 𝐹$ , the scaling map $𝑉 \to 𝑉$ , $𝐯 \mapsto 𝜆 𝐯$ , is linear:
$𝜆 (𝑎 𝐮 + 𝐯) = (𝜆 𝑎) 𝐮 + 𝜆 𝐯 = (𝑎 𝜆) 𝐮 + 𝜆 𝐯 = 𝑎 (𝜆 𝐮) + 𝜆 𝐯 .$
The middle step is the commutativity of multiplication in $𝐹$ .

Example 2.1.7 (Maps $𝐹^{𝑛} \to 𝐹^{𝑚}$ given by linear formulas). Let $𝐴 \in 𝐹^{𝑚 \times 𝑛}$ (Definition 1.2.7) have entries $𝑎_{𝑖 𝑗}$ , and define $𝑇_{𝐴} : 𝐹^{𝑛} \to 𝐹^{𝑚}$ by

𝑇_{𝐴} (𝑥_{1}, \dots, 𝑥_{𝑛}) = (𝑎_{11} 𝑥_{1} + \dots + 𝑎_{1 𝑛} 𝑥_{𝑛}, \dots, 𝑎_{𝑚 1} 𝑥_{1} + \dots + 𝑎_{𝑚 𝑛} 𝑥_{𝑛}) .

That is, the $𝑖$ -th coordinate of $𝑇_{𝐴} (𝐱)$ is $\sum_{𝑗 = 1}^{𝑛} 𝑎_{𝑖 𝑗} 𝑥_{𝑗}$ . This map is linear. For $𝐱, 𝐲 \in 𝐹^{𝑛}$ and $𝑐 \in 𝐹$ , the $𝑖$ -th coordinate of $𝑇_{𝐴} (𝑐 𝐱 + 𝐲)$ is

\sum_{𝑗 = 1}^{𝑛} 𝑎_{𝑖 𝑗} (𝑐 𝑥_{𝑗} + 𝑦_{𝑗}) = 𝑐 \sum_{𝑗 = 1}^{𝑛} 𝑎_{𝑖 𝑗} 𝑥_{𝑗} + \sum_{𝑗 = 1}^{𝑛} 𝑎_{𝑖 𝑗} 𝑦_{𝑗},

which is the $𝑖$ -th coordinate of $𝑐 𝑇_{𝐴} (𝐱) +$ $𝑇_{𝐴} (𝐲)$ . For example, $𝑇 : ℝ^{3} \to ℝ^{2}$ , $𝑇 (𝑥, 𝑦, 𝑧) =$ $(2 𝑥 - 𝑦 + 3 𝑧, 𝑥 + 5 𝑧)$ , is $𝑇_{𝐴}$ with

𝐴 = (\begin{matrix} 2 & - 1 & 3 \\ 1 & 0 & 5 \end{matrix}) .

Over $𝔽_{2}$ , the map $(𝑥, 𝑦, 𝑧) \mapsto$ $(𝑥 + 𝑦, 𝑦 + 𝑧)$ is $𝑇_{𝐴}$ with $𝐴 = (\begin{matrix} 1 & 1 & 0 \\ 0 & 1 & 1 \end{matrix})$ . The coordinate projections $(𝑥_{1}, \dots, 𝑥_{𝑛}) \mapsto 𝑥_{𝑖}$ are special cases too.

In Section 2.3 we write $𝑇_{𝐴} (𝐱)$ as a matrix–vector product $𝐴 𝐱$ (Definition 2.3.8, Definition 2.3.15) and show that every linear map $𝐹^{𝑛} \to 𝐹^{𝑚}$ is $𝑇_{𝐴}$ for some $𝐴$ (Proposition 2.3.17).

Example 2.1.8 (Coordinates). Let $ℬ︀ = 𝐛_{1}, \dots, 𝐛_{𝑛}$ be a basis of $𝑉$ . The coordinate map $𝑉 \to 𝐹^{𝑛}$ , $𝐯 \mapsto {[𝐯]}_{ℬ︀}$ (Definition 1.6.14), is linear; this is Proposition 1.6.19. In the other direction, for any list $𝐯_{1}, \dots, 𝐯_{𝑛}$ in $𝑉$ , basis or not, the map

𝐹^{𝑛} \to 𝑉, (𝑎_{1}, \dots, 𝑎_{𝑛}) \mapsto 𝑎_{1} 𝐯_{1} + \dots + 𝑎_{𝑛} 𝐯_{𝑛},

is linear, because $\sum_{𝑖} (𝑐 𝑎_{𝑖} + 𝑑_{𝑖}) 𝐯_{𝑖} =$ $𝑐 \sum_{𝑖} 𝑎_{𝑖} 𝐯_{𝑖} +$ $\sum_{𝑖} 𝑑_{𝑖} 𝐯_{𝑖}$ by the vector space axioms. For the basis $ℬ︀$ , the two maps undo each other: ${[𝑎_{1} 𝐛_{1} + \dots + 𝑎_{𝑛} 𝐛_{𝑛}]}_{ℬ︀} =$ $(𝑎_{1}, \dots, 𝑎_{𝑛})$ by the definition of coordinates, and $𝐯 =$ $𝑐_{1} 𝐛_{1} +$ $\dots +$ $𝑐_{𝑛} 𝐛_{𝑛}$ where $(𝑐_{1}, \dots, 𝑐_{𝑛}) =$ ${[𝐯]}_{ℬ︀}$ . So once a basis is chosen, $𝑉$ and $𝐹^{𝑛}$ correspond through a pair of linear maps; Section 4.5 develops this.

Example 2.1.9 (Maps of polynomials). Let $𝑝 =$ $𝑎_{0} +$ $𝑎_{1} 𝑥 +$ $\dots +$ $𝑎_{𝑛} 𝑥^{𝑛} \in$ $𝐹 [𝑥]$ (Definition 1.2.13).

Evaluation at $𝑡 \in 𝐹$ is ${ev}_{𝑡} : 𝐹 [𝑥] \to 𝐹$ , ${ev}_{𝑡} (𝑝) =$ $𝑝 (𝑡) =$ $𝑎_{0} +$ $𝑎_{1} 𝑡 +$ $\dots +$ $𝑎_{𝑛} 𝑡^{𝑛}$ .
Multiplication by $𝑥$ is $𝑀 : 𝐹 [𝑥] \to 𝐹 [𝑥]$ , $𝑀 (𝑝) =$ $𝑥 𝑝 =$ $𝑎_{0} 𝑥 +$ $𝑎_{1} 𝑥^{2} +$ $\dots +$ $𝑎_{𝑛} 𝑥^{𝑛 + 1}$ .
Differentiation is $𝐷 : 𝐹 [𝑥] \to 𝐹 [𝑥]$ ,
$𝐷 (𝑝) = 𝑎_{1} + 2 𝑎_{2} 𝑥 + 3 𝑎_{3} 𝑥^{2} + \dots + 𝑛 𝑎_{𝑛} 𝑥^{𝑛 - 1},$
where $𝑘 𝑎_{𝑘}$ means $𝑎_{𝑘} + \dots + 𝑎_{𝑘}$ ( $𝑘$ terms). For $𝐹 = ℝ$ this is the derivative of calculus, and over any field we take the formula as the definition (the formal derivative). We also write $𝑝^{'}$ for $𝐷 (𝑝)$ .

All three maps are linear, with the same proof. Let $𝑞 =$ $𝑏_{0} +$ $𝑏_{1} 𝑥 +$ $\dots +$ $𝑏_{𝑛} 𝑥^{𝑛}$ (padding with zero coefficients, we may use the same $𝑛$ for $𝑝$ and $𝑞$ ) and $𝑐 \in 𝐹$ . Polynomials are added and scaled coefficientwise, so $𝑐 𝑝 +$ $𝑞 =$ $\sum_{𝑘} (𝑐 𝑎_{𝑘} + 𝑏_{𝑘}) 𝑥^{𝑘}$ . Each of the three maps sends the coefficient of $𝑥^{𝑘}$ to a fixed place and multiplies it by a factor that does not depend on the polynomial ( $𝑡^{𝑘}$ , $1$ , or $𝑘$ ). For $𝐷$ ,

𝐷 (𝑐 𝑝 + 𝑞) = \sum_{𝑘 = 1}^{𝑛} 𝑘 (𝑐 𝑎_{𝑘} + 𝑏_{𝑘}) 𝑥^{𝑘 - 1} = 𝑐 \sum_{𝑘 = 1}^{𝑛} 𝑘 𝑎_{𝑘} 𝑥^{𝑘 - 1} + \sum_{𝑘 = 1}^{𝑛} 𝑘 𝑏_{𝑘} 𝑥^{𝑘 - 1} = 𝑐 𝐷 (𝑝) + 𝐷 (𝑞),

where $𝑘 (𝑐 𝑎_{𝑘} + 𝑏_{𝑘}) =$ $𝑐 (𝑘 𝑎_{𝑘}) +$ $𝑘 𝑏_{𝑘}$ because $𝑘 𝑦$ is a sum of $𝑘$ copies of $𝑦$ . The computations for ${ev}_{𝑡}$ and $𝑀$ are the same with $𝑡^{𝑘}$ and $𝑥^{𝑘 + 1}$ in place of $𝑘 𝑥^{𝑘 - 1}$ .

Over a finite field the formal derivative behaves unlike the one from calculus. Over $𝔽_{2}$ , for example, $𝐷 (𝑥^{2}) = 2 𝑥 = 0$ , so a nonconstant polynomial can have derivative $0$ (Exercise 2.1.34).

Example 2.1.10 (Integration). Let $𝐶 [0, 1]$ be the real vector space of continuous functions $[0, 1] \to ℝ$ (Section 1.2). The map

𝐶 [0, 1] \to ℝ, 𝑓 \mapsto \int_{0}^{1} 𝑓 (𝑡) d 𝑡,

is linear: $\int_{0}^{1} (𝑐 𝑓 + 𝑔) =$ $𝑐 \int_{0}^{1} 𝑓 +$ $\int_{0}^{1} 𝑔$ is the linearity of the integral. Likewise $𝐽 : 𝐶 [0, 1] \to$ $𝐶 [0, 1]$ , $(𝐽 𝑓) (𝑥) =$ $\int_{0}^{𝑥} 𝑓 (𝑡) d 𝑡$ , is linear. Here $𝐽 𝑓$ is continuous (even differentiable) by the fundamental theorem of calculus, and $𝐽 (𝑐 𝑓 + 𝑔) =$ $𝑐 𝐽 𝑓 +$ $𝐽 𝑔$ because the two functions agree at every $𝑥 \in [0, 1]$ .

Linear maps of the plane

Rotations, projections, and reflections of $ℝ^{2}$ that fix the origin are linear. In each case we derive a coordinate formula from the geometry. The formula has the shape of Example 2.1.7, so the map is linear.

Example 2.1.11 (Rotations, projections, and reflections). Rotation. Let $𝑅_{𝜃} : ℝ^{2} \to ℝ^{2}$ rotate the plane counterclockwise about $𝟎$ by the angle $𝜃$ . Write a point in polar form, $(𝑥, 𝑦) =$ $(𝑟 cos 𝜑, 𝑟 sin 𝜑)$ . Rotation adds $𝜃$ to the angle, so by the angle addition formulas,

\begin{aligned} 𝑅_{𝜃} (𝑥, 𝑦) & = (𝑟 cos (𝜑 + 𝜃), 𝑟 sin (𝜑 + 𝜃)) \\ = (𝑟 cos 𝜑 cos 𝜃 - 𝑟 sin 𝜑 sin 𝜃, 𝑟 sin 𝜑 cos 𝜃 + 𝑟 cos 𝜑 sin 𝜃) \\ = (𝑥 cos 𝜃 - 𝑦 sin 𝜃, 𝑥 sin 𝜃 + 𝑦 cos 𝜃) . \end{aligned}

Projection. Let $ℓ$ be the line through $𝟎$ in the direction of the unit vector $𝐮 = (cos 𝜃, sin 𝜃)$ . The projection ${Pr}_{ℓ} : ℝ^{2} \to ℝ^{2}$ sends $𝐯 = (𝑥, 𝑦)$ to the foot $𝑡 𝐮$ of the perpendicular from $𝐯$ to $ℓ$ . Two vectors $(𝑎, 𝑏)$ and $(𝑐, 𝑑)$ are perpendicular when $𝑎 𝑐 + 𝑏 𝑑 = 0$ , so $𝑡$ is determined by

(𝑥 - 𝑡 cos 𝜃) cos 𝜃 + (𝑦 - 𝑡 sin 𝜃) sin 𝜃 = 0, that is, 𝑡 = 𝑥 cos 𝜃 + 𝑦 sin 𝜃,

using ${cos}^{2} 𝜃 + {sin}^{2} 𝜃 = 1$ . Hence

\begin{aligned} {Pr}_{ℓ} (𝑥, 𝑦) & = (𝑥 cos 𝜃 + 𝑦 sin 𝜃) (cos 𝜃, sin 𝜃) \\ = (𝑥 {cos}^{2} 𝜃 + 𝑦 sin 𝜃 cos 𝜃, 𝑥 sin 𝜃 cos 𝜃 + 𝑦 {sin}^{2} 𝜃) . \end{aligned}

Reflection. The reflection $𝑆_{ℓ}$ across $ℓ$ sends $𝐯$ to the point $𝑆_{ℓ} (𝐯)$ for which ${Pr}_{ℓ} (𝐯)$ is the midpoint of $𝐯$ and $𝑆_{ℓ} (𝐯)$ . So $𝑆_{ℓ} (𝐯) =$ $2 {Pr}_{ℓ} (𝐯) -$ $𝐯$ , and with the double angle formulas,

𝑆_{ℓ} (𝑥, 𝑦) = (𝑥 cos 2 𝜃 + 𝑦 sin 2 𝜃, 𝑥 sin 2 𝜃 - 𝑦 cos 2 𝜃) .

All three formulas have the form of Example 2.1.7, so $𝑅_{𝜃}$ , ${Pr}_{ℓ}$ , and $𝑆_{ℓ}$ are linear. For instance, if $ℓ$ is the $𝑥$ -axis ( $𝜃 = 0$ ), then ${Pr}_{ℓ} (𝑥, 𝑦) = (𝑥, 0)$ and $𝑆_{ℓ} (𝑥, 𝑦) =$ $(𝑥, - 𝑦)$ , and the rotation by a right angle is $𝑅_{𝜋 / 2} (𝑥, 𝑦) =$ $(- 𝑦, 𝑥)$ . Orthogonal projections in any inner product space are the subject of Section 6.4.

Intuition (What a linear map does to the grid). Let $𝑇 : ℝ^{2} \to ℝ^{2}$ be linear. Since $(𝑥, 𝑦) =$ $𝑥 𝐞_{1} +$ $𝑦 𝐞_{2}$ , linearity gives

𝑇 (𝑥, 𝑦) = 𝑥 𝑇 (𝐞_{1}) + 𝑦 𝑇 (𝐞_{2}) .

So $𝑇$ is determined by the two vectors $𝑇 (𝐞_{1})$ and $𝑇 (𝐞_{2})$ (Section 2.2 proves the analogue for every basis), and the grid point $(𝑥, 𝑦)$ goes to $𝑥$ steps along $𝑇 (𝐞_{1})$ followed by $𝑦$ steps along $𝑇 (𝐞_{2})$ . The square grid becomes a grid of parallelograms spanned by $𝑇 (𝐞_{1})$ and $𝑇 (𝐞_{2})$ . To be precise about lines, apply $𝑇$ to the line through $𝐩$ in the direction $𝐝$ :

𝑇 (𝐩 + 𝑡 𝐝) = 𝑇 (𝐩) + 𝑡 𝑇 (𝐝) (𝑡 \in ℝ) .

This is the line through $𝑇 (𝐩)$ in the direction $𝑇 (𝐝)$ , or the single point $𝑇 (𝐩)$ if $𝑇 (𝐝) = 𝟎$ . Reading off this formula:

lines go to lines, or collapse to points;
parallel lines, which share a direction $𝐝$ , go to lines that share the direction $𝑇 (𝐝)$ (they may coincide, or all collapse to points if $𝑇 (𝐝) = 𝟎$ );
evenly spaced points $𝐩, 𝐩 + 𝐝, 𝐩 + 2 𝐝, \dots$ go to evenly spaced points;
the origin stays where it is.

Drag the tips of $𝑇 (𝐞_{1})$ and $𝑇 (𝐞_{2})$ below (or focus one and use the arrow keys) and watch the grid follow. The panel lists the four numbers in $𝑇 (𝑥, 𝑦) =$ $(𝑎 𝑥 + 𝑏 𝑦, 𝑐 𝑥 + 𝑑 𝑦)$ as an array $𝐴$ , the notation of Section 2.3; the handles labelled $𝐴 𝐞_{1}$ and $𝐴 𝐞_{2}$ are $𝑇 (𝐞_{1}) = (𝑎, 𝑐)$ and $𝑇 (𝐞_{2}) = (𝑏, 𝑑)$ . The determinant readout belongs to Section 5.1 and can be ignored for now.

A linear map of the plane. Grid lines go to evenly spaced parallel lines, the unit square goes to the parallelogram spanned by

𝑇 (𝐞_{1})

and

𝑇 (𝐞_{2})

, and the origin stays fixed.

2.1.3Non-examples

The quickest way to see that a map is not linear is to find that it moves the origin (Proposition 2.1.4). The simplest maps that do this are the affine maps.

Non-example 2.1.12 (Affine maps). Let $𝑎, 𝑏 \in 𝐹$ . The map $𝑓 : 𝐹 \to 𝐹$ , $𝑓 (𝑥) = 𝑎 𝑥 + 𝑏$ , is linear if and only if $𝑏 = 0$ . Indeed, if $𝑏 \neq 0$ , then $𝑓 (0) = 𝑏 \neq 0$ , so $𝑓$ is not linear by Proposition 2.1.4; if $𝑏 = 0$ , then $𝑓 (𝑥) = 𝑎 𝑥$ is linear by Exercise 2.1.5. So the “linear functions” of school algebra, whose graphs are lines, are linear maps only when the line passes through the origin.

More generally, for $𝐿 \in L (𝑉, 𝑊)$ and $𝐛 \in 𝑊$ , the map $𝑇 : 𝑉 \to 𝑊$ , $𝑇 (𝐯) = 𝐿 (𝐯) + 𝐛$ , is called affine. Since $𝑇 (𝟎) = 𝐛$ , it is linear only if $𝐛 = 𝟎$ .

Intuition. The failure of an affine map is more specific than “

𝑇 (𝟎) \neq 𝟎

”. Compute both sides of the defining equation of linearity for

𝑇 = 𝐿 + 𝐛

: the left side picks up one copy of

𝐛

, and the right side picks up

𝑐_{1} + \dots + 𝑐_{𝑘}

copies. So, when

𝐛 \neq 𝟎

𝑇

respects exactly those combinations whose coefficients add up to

1

Proposition 2.1.13 (Which combinations an affine map respects). Let $𝐿 \in L (𝑉, 𝑊)$ , $𝐛 \in 𝑊$ , and $𝑇 (𝐯) = 𝐿 (𝐯) + 𝐛$ . For all $𝐯_{1}, \dots, 𝐯_{𝑘} \in 𝑉$ and $𝑐_{1}, \dots, 𝑐_{𝑘} \in 𝐹$ ,

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

In particular, if $𝐛 \neq 𝟎$ , then $𝑇 (\sum_{𝑖} 𝑐_{𝑖} 𝐯_{𝑖}) =$ $\sum_{𝑖} 𝑐_{𝑖} 𝑇 (𝐯_{𝑖})$ holds if and only if $𝑐_{1} + \dots + 𝑐_{𝑘} = 1$ . So $𝑇$ is linear if and only if $𝐛 = 𝟎$ .

Proof.routine

Uses. Definition 2.1.1 (Linear map), Proposition 1.2.35 (Zero products)

Hint 1. Expand both

𝑇 (\sum_{𝑖} 𝑐_{𝑖} 𝐯_{𝑖})

and

\sum_{𝑖} 𝑐_{𝑖} 𝑇 (𝐯_{𝑖})

using the formula for

𝑇

and the linearity of

𝐿

Hint 2. The terms involving

𝐿

cancel. What remains is

𝐛 - (𝑐_{1} + \dots + 𝑐_{𝑘}) 𝐛

. When is

𝑠 𝐛 = 𝟎

for a nonzero

𝐛

Sketch.

𝑇 (\sum 𝑐_{𝑖} 𝐯_{𝑖}) =

\sum 𝑐_{𝑖} 𝐿 (𝐯_{𝑖}) +

𝐛

and

\sum 𝑐_{𝑖} 𝑇 (𝐯_{𝑖}) =

\sum 𝑐_{𝑖} 𝐿 (𝐯_{𝑖}) +

(\sum 𝑐_{𝑖}) 𝐛

; subtract. If

𝐛 \neq 𝟎

, then

(1 - \sum 𝑐_{𝑖}) 𝐛 = 𝟎

forces

\sum 𝑐_{𝑖} = 1

. The combination

0 \cdot 𝐯

has coefficient sum

0

, so

𝑇

is not linear.

By the linearity of $𝐿$ ,

𝑇 (𝑐_{1} 𝐯_{1} + \dots + 𝑐_{𝑘} 𝐯_{𝑘}) = 𝑐_{1} 𝐿 (𝐯_{1}) + \dots + 𝑐_{𝑘} 𝐿 (𝐯_{𝑘}) + 𝐛,

while, by distributivity in $𝑊$ ,

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

Subtracting gives the displayed identity. If $𝐛 \neq 𝟎$ , then by Proposition 1.2.35, $(1 - 𝑐_{1} - \dots - 𝑐_{𝑘}) 𝐛 =$ $𝟎$ if and only if $𝑐_{1} + \dots + 𝑐_{𝑘} = 1$ . In particular, the combination $0 \cdot 𝐯$ (with $𝑘 = 1$ and $𝑐_{1} = 0$ ) is not respected, so $𝑇$ is not linear. If $𝐛 = 𝟎$ , then $𝑇 = 𝐿$ is linear. □

Combinations whose coefficients add up to $1$ are called affine combinations. Examples are the points $(1 - 𝑡) 𝐮 + 𝑡 𝐯$ of the line through $𝐮$ and $𝐯$ , and in particular the midpoint $\frac{1}{2} 𝐮 + \frac{1}{2} 𝐯$ when $1 + 1 \neq 0$ . An affine map respects these, which is why an affine map of $ℝ^{2}$ , like a linear one, still sends lines to lines or points, parallel lines to lines with a common direction, and evenly spaced points to evenly spaced points. What it does not respect is anything that refers to the origin, such as the scalar multiple $2 𝐯$ , whose single coefficient is $2$ . Compare the first two panels of Figure 2.1.1: the affine map produces the same grid as the linear one, translated. Affine maps and affine combinations are the subject of Section 4.7.

Figure 2.1.1. The square grid under three maps of

ℝ^{2}

; the faint lines are the original axes, and the shaded cell is the image of the unit square. Left: the linear map

𝑇 (𝑥, 𝑦) =

(1.5 𝑥 - 0.5 𝑦, 0.5 𝑥 + 𝑦)

. Middle: the affine map

𝐯 \mapsto 𝑇 (𝐯) + (1, 0.5)

, which moves the origin to the red point. Right:

(𝑥, 𝑦) \mapsto

(𝑥, 𝑦 + 𝑥^{2} / 4)

, which fixes the origin but bends the horizontal grid lines, so it is not linear either.

Non-example 2.1.14 (Complex conjugation). Let $𝑐 : ℂ \to ℂ$ be complex conjugation, $𝑐 (𝑎 + 𝑏 𝑖) =$ $𝑎 + 𝑏 𝑖 =$ $𝑎 -$ $𝑏 𝑖$ for $𝑎, 𝑏 \in ℝ$ . Regard $ℂ$ as a vector space over $ℝ$ (Section 1.2), with $𝑎 + 𝑏 𝑖$ corresponding to $(𝑎, 𝑏) \in ℝ^{2}$ . Then $𝑐$ is the map $(𝑎, 𝑏) \mapsto (𝑎, - 𝑏)$ , the reflection in the real axis, which is $ℝ$ -linear by Example 2.1.11.

Now regard $ℂ$ as a vector space over $ℂ$ . Conjugation is still additive, since $𝑧 + 𝑤 = 𝑧 + 𝑤$ . But it is not homogeneous: $𝑐 (𝑖 \cdot 1) = 𝑖 = - 𝑖$ , while $𝑖 \cdot 𝑐 (1) = 𝑖$ . So conjugation is $ℝ$ -linear but not $ℂ$ -linear. Whether a function is linear depends on the field of scalars, and not only on the sets and the function.

Formulas that are visibly nonlinear fail homogeneity at once. The map $ℝ \to ℝ$ , $𝑥 \mapsto 𝑥^{2}$ , sends $2 \cdot 1$ to $4 \neq 2 \cdot 1^{2}$ , and $ℝ \to ℝ$ , $𝑥 \mapsto sin 𝑥$ , sends $𝜋 = 2 \cdot 𝜋 / 2$ to $0 \neq 2 sin (𝜋 / 2)$ . As a rule of thumb, in a coordinate formula for a linear map, each output coordinate is a sum of constant multiples of the input coordinates: no constant terms, no products of coordinates, and no other functions applied to them. For maps $𝐹^{𝑛} \to 𝐹^{𝑚}$ , Proposition 2.3.17 makes this precise.

Exercise 2.1.15.routine Decide which of the following maps are linear.

(a) $𝑇 : ℝ^{2} \to ℝ^{2}$ , $𝑇 (𝑥, 𝑦) =$ $(𝑦, 𝑥 - 3 𝑦)$ .
(b) $𝑇 : ℝ^{2} \to ℝ^{2}$ , $𝑇 (𝑥, 𝑦) =$ $(𝑥 + 1, 𝑦)$ .
(c) $𝑇 : ℝ \to ℝ^{2}$ , $𝑇 (𝑥) = (𝑥, | 𝑥 |)$ .
(d) $𝑇 : 𝔽_{2}^{2} \to 𝔽_{2}$ , $𝑇 (𝑥, 𝑦) = 𝑥 𝑦$ .

Solution.

Uses. Example 2.1.7 (Maps $𝐹^{𝑛} \to 𝐹^{𝑚}$ given by linear formulas), Proposition 2.1.4 (Linear maps preserve $𝟎$ and negatives), Theorem 2.1.3 (Checking linearity)

Hint 1. Test

𝑇 (𝟎)

first, then

𝑇 (- 𝐯)

𝑇 (2 𝐯)

. These catch most nonlinear formulas.

Hint 2. In (d) there are only two scalars, so homogeneity says very little. Test additivity on

𝐞_{1}

and

𝐞_{2}

Sketch. (a) Linear: it is

𝑇_{𝐴}

with

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

. (b) Not linear:

𝑇 (𝟎) = (1, 0)

. (c) Not linear:

𝑇 (- 1) =

(- 1, 1) \neq

- 𝑇 (1)

. (d) Homogeneous but not additive:

𝑇 (1, 0) +

𝑇 (0, 1) =

0 \neq

1 =

𝑇 (1, 1)

(a) Linear: $𝑇 = 𝑇_{𝐴}$ for $𝐴 = (\begin{matrix} 0 & 1 \\ 1 & - 3 \end{matrix})$ , so Example 2.1.7 applies.

(b) Not linear: $𝑇 (0, 0) = (1, 0) \neq 𝟎$ , contradicting Proposition 2.1.4.

(c) Not linear: $𝑇 (- 1) = (- 1, 1)$ , while $(- 1) 𝑇 (1) = (- 1, - 1)$ , so $𝑇$ is not homogeneous. (Note that $𝑇 (0) = 𝟎$ , so the first test does not detect this one.)

(d) Not linear. Homogeneity does hold: the only scalars are $0$ and $1$ , and $𝑇 (0 𝐯) =$ $𝑇 (0, 0) =$ $0 =$ $0 𝑇 (𝐯)$ and $𝑇 (1 𝐯) = 1 𝑇 (𝐯)$ . But additivity fails: $𝑇 (1, 0) +$ $𝑇 (0, 1) =$ $0 +$ $0 =$ $0$ , while $𝑇 (1, 1) = 1$ .

2.1.4What linear maps preserve

Intuition. A vector space is a set with an addition and a scalar multiplication, and nothing else. Every notion we have defined so far (linear combination, span, linear relation, linear dependence) is defined from these two operations. A linear map respects both, so it carries each notion in

𝑉

to the corresponding notion in

𝑊

, at least in one direction. The direction matters: a linear map may send different vectors to the same vector, so it can destroy independence, but it never creates independence where there was none.

For a subset $𝐴 \subseteq 𝑉$ we write $𝑇 (𝐴) =$ ${𝑇 (𝐚) | 𝐚 \in 𝐴}$ , and for a list $ℒ︀ = 𝐯_{1}, \dots, 𝐯_{𝑘}$ in $𝑉$ we write $𝑇 (ℒ︀)$ for the list $𝑇 (𝐯_{1}),$ $\dots,$ $𝑇 (𝐯_{𝑘})$ in $𝑊$ .

Proposition 2.1.16 (Linear maps preserve spans and linear relations). Let $𝑇 \in L (𝑉, 𝑊)$ , and let $ℒ︀ = 𝐯_{1}, \dots, 𝐯_{𝑘}$ be a list in $𝑉$ .

(a) $im (𝑇 ↾_{span ℒ︀}) =$ $span 𝑇 (ℒ︀)$ . In particular, if $ℒ︀$ spans $𝑉$ , then $𝑇 (ℒ︀)$ spans $im 𝑇$ .
(b)If $𝑎_{1} 𝐯_{1} +$ $\dots +$ $𝑎_{𝑘} 𝐯_{𝑘} =$ $𝟎$ , then $𝑎_{1} 𝑇 (𝐯_{1}) +$ $\dots +$ $𝑎_{𝑘} 𝑇 (𝐯_{𝑘}) =$ $𝟎$ . Consequently, if $ℒ︀$ is linearly dependent, then so is $𝑇 (ℒ︀)$ . Equivalently, if $𝑇 (ℒ︀)$ is linearly independent, then so is $ℒ︀$ .

Proof.routine

Uses. Definition 1.3.10 (Span), Definition 2.1.1 (Linear map), Proposition 2.1.4 (Linear maps preserve $𝟎$ and negatives), Theorem 1.5.5 (Independence is decided at zero)

Techniques. Double containment (Technique 1.3.14)

Hint 1. For (a), prove the two inclusions separately. An element of

span ℒ︀

is a combination

\sum 𝑎_{𝑖} 𝐯_{𝑖}

; what is its image?

Hint 2. For (b), apply

𝑇

to both sides of the relation. For the statement about dependence, use the description of dependence by a nontrivial representation of

𝟎

(Theorem 1.5.5).

Sketch.

𝑇 (\sum 𝑎_{𝑖} 𝐯_{𝑖}) =

\sum 𝑎_{𝑖} 𝑇 (𝐯_{𝑖})

shows both that images of elements of

span ℒ︀

lie in

span 𝑇 (ℒ︀)

, and that every element of

span 𝑇 (ℒ︀)

is such an image. Applying

𝑇

to a nontrivial relation

\sum 𝑎_{𝑖} 𝐯_{𝑖} = 𝟎

gives a relation among the

𝑇 (𝐯_{𝑖})

with the same, not all zero, coefficients.

(a) If $𝑘 = 0$ , then $span ℒ︀ = {𝟎}$ and $span 𝑇 (ℒ︀) = {𝟎}$ , and $𝑇 ({𝟎}) = {𝟎}$ by Proposition 2.1.4. Let $𝑘 \geq 1$ . We prove the two inclusions.

( $\subseteq$ ) An element of $im (𝑇 ↾_{span ℒ︀})$ has the form $𝑇 (𝑎_{1} 𝐯_{1} + \dots + 𝑎_{𝑘} 𝐯_{𝑘})$ with $𝑎_{𝑖} \in 𝐹$ . By linearity it equals $𝑎_{1} 𝑇 (𝐯_{1}) +$ $\dots +$ $𝑎_{𝑘} 𝑇 (𝐯_{𝑘})$ , which lies in $span 𝑇 (ℒ︀)$ .

( $\supseteq$ ) An element of $span 𝑇 (ℒ︀)$ has the form $𝑎_{1} 𝑇 (𝐯_{1}) +$ $\dots +$ $𝑎_{𝑘} 𝑇 (𝐯_{𝑘})$ , which by linearity equals $𝑇 (𝑎_{1} 𝐯_{1} + \dots + 𝑎_{𝑘} 𝐯_{𝑘})$ . Since $𝑎_{1} 𝐯_{1} +$ $\dots +$ $𝑎_{𝑘} 𝐯_{𝑘} \in$ $span ℒ︀$ , this lies in $im (𝑇 ↾_{span ℒ︀})$ .

If $ℒ︀$ spans $𝑉$ , then $im 𝑇 =$ $im (𝑇 ↾_{span ℒ︀}) =$ $span 𝑇 (ℒ︀)$ , so $𝑇 (ℒ︀)$ spans $im 𝑇$ .

(b) Applying $𝑇$ to both sides of $𝑎_{1} 𝐯_{1} +$ $\dots +$ $𝑎_{𝑘} 𝐯_{𝑘} =$ $𝟎$ and using linearity and Proposition 2.1.4,

𝑎_{1} 𝑇 (𝐯_{1}) + \dots + 𝑎_{𝑘} 𝑇 (𝐯_{𝑘}) = 𝑇 (𝑎_{1} 𝐯_{1} + \dots + 𝑎_{𝑘} 𝐯_{𝑘}) = 𝑇 (𝟎) = 𝟎 .

If $ℒ︀$ is linearly dependent, then by Theorem 1.5.5 there are scalars $𝑎_{1}, \dots, 𝑎_{𝑘}$ , not all $0$ , with $\sum_{𝑖} 𝑎_{𝑖} 𝐯_{𝑖} = 𝟎$ . By what we just showed, the same scalars give $\sum_{𝑖} 𝑎_{𝑖} 𝑇 (𝐯_{𝑖}) =$ $𝟎$ , a nontrivial representation of $𝟎$ by $𝑇 (ℒ︀)$ . So $𝑇 (ℒ︀)$ is linearly dependent, again by Theorem 1.5.5. The last sentence of (b) is the contrapositive. □

The converse of the last statement fails. If $ℓ$ is the $𝑥$ -axis, then ${Pr}_{ℓ}$ sends the independent list $𝐞_{1}, 𝐞_{2}$ to the dependent list $(1, 0), (0, 0)$ , and the zero map sends every list to a list of zero vectors. The maps that preserve independence are identified in Section 2.6: they are the injective ones. Part (a) with $𝑘 = 1$ is the statement about lines from the grid picture: for $𝐯 \neq 𝟎$ , the line $span (𝐯)$ through $𝟎$ goes to the line $span (𝑇 (𝐯))$ , or to ${𝟎}$ if $𝑇 (𝐯) = 𝟎$ .

2.1.5Composition

Linear maps can be applied one after another. Recall that for functions $𝑆 : 𝑈 \to 𝑉$ and $𝑇 : 𝑉 \to 𝑊$ , the composition $𝑇 \circ 𝑆 : 𝑈 \to 𝑊$ is $𝐮 \mapsto 𝑇 (𝑆 (𝐮))$ (Section A.1).

Proposition 2.1.17 (Compositions of linear maps are linear). Let

𝑈

𝑉

, and

𝑊

be vector spaces over

𝐹

. If

𝑆 \in L (𝑈, 𝑉)

and

𝑇 \in L (𝑉, 𝑊)

, then

𝑇 \circ 𝑆 \in L (𝑈, 𝑊)

Proof.routine

Uses. Definition 2.1.1 (Linear map)

Hint 1. Take a linear combination in

𝑈

and follow it through

𝑆

and then through

𝑇

Hint 2. Use the linearity of

𝑆

to move the combination outside

𝑆

, then the linearity of

𝑇

to move it outside

𝑇

Sketch.

(𝑇 \circ 𝑆) (\sum 𝑎_{𝑖} 𝐮_{𝑖}) =

𝑇 (\sum 𝑎_{𝑖} 𝑆 (𝐮_{𝑖})) =

\sum 𝑎_{𝑖} 𝑇 (𝑆 (𝐮_{𝑖}))

Let $𝐮_{1}, \dots, 𝐮_{𝑘} \in 𝑈$ and $𝑎_{1}, \dots, 𝑎_{𝑘} \in 𝐹$ . Then

\begin{aligned} (𝑇 \circ 𝑆) (𝑎_{1} 𝐮_{1} + \dots + 𝑎_{𝑘} 𝐮_{𝑘}) & = 𝑇 (𝑆 (𝑎_{1} 𝐮_{1} + \dots + 𝑎_{𝑘} 𝐮_{𝑘})) & (definition of \circ) \\ = 𝑇 (𝑎_{1} 𝑆 (𝐮_{1}) + \dots + 𝑎_{𝑘} 𝑆 (𝐮_{𝑘})) & (linearity of 𝑆) \\ = 𝑎_{1} 𝑇 (𝑆 (𝐮_{1})) + \dots + 𝑎_{𝑘} 𝑇 (𝑆 (𝐮_{𝑘})) & (linearity of 𝑇) \\ = 𝑎_{1} (𝑇 \circ 𝑆) (𝐮_{1}) + \dots + 𝑎_{𝑘} (𝑇 \circ 𝑆) (𝐮_{𝑘}) & (definition of \circ), \end{aligned}

as desired. □

Notation 2.1.18. For linear maps we usually write

𝑇 𝑆

for

𝑇 \circ 𝑆

. For

𝑇 \in L (𝑉)

and an integer

𝑘 \geq 1

𝑇^{𝑘} = 𝑇 𝑇 \dots 𝑇

(

𝑘

factors), and

𝑇^{0} = {id}_{𝑉}

. Composition is associative (Proposition 2.1.24), so

𝑇^{𝑘}

needs no parentheses, and

𝑇^{𝑗} 𝑇^{𝑘} = 𝑇^{𝑗 + 𝑘}

Example 2.1.19 (Composites).

(a) $𝐷^{2} = 𝐷 𝐷$ on $𝐹 [𝑥]$ is the second formal derivative,
$𝐷^{2} (𝑎_{0} + 𝑎_{1} 𝑥 + \dots + 𝑎_{𝑛} 𝑥^{𝑛}) = 2 𝑎_{2} + 6 𝑎_{3} 𝑥 + \dots + 𝑛 (𝑛 - 1) 𝑎_{𝑛} 𝑥^{𝑛 - 2} .$
Each application of $𝐷$ sends constants to $0$ and lowers the degree of a nonconstant polynomial by at least one (by exactly one over $ℝ$ or $ℂ$ ; over $𝔽_{2}$ , $𝐷 (𝑥^{2} + 𝑥) = 1$ ). So $𝐷^{𝑛 + 1} (𝑝) = 0$ whenever $deg 𝑝 \leq 𝑛$ .
(b)Order matters. Let $𝑃 (𝑥, 𝑦) = (𝑥, 0)$ be the projection onto the $𝑥$ -axis and $𝑅 (𝑥, 𝑦) =$ $(- 𝑦, 𝑥)$ the rotation by a right angle (Example 2.1.11). Then
$(𝑃 𝑅) (𝑥, 𝑦) = 𝑃 (- 𝑦, 𝑥) = (- 𝑦, 0), (𝑅 𝑃) (𝑥, 𝑦) = 𝑅 (𝑥, 0) = (0, 𝑥) .$
So $𝑃 𝑅 \neq 𝑅 𝑃$ : for instance $(𝑃 𝑅) (𝐞_{1}) = 𝟎$ , while $(𝑅 𝑃) (𝐞_{1}) = 𝐞_{2}$ .

Composition of linear maps is not commutative: even when $𝑇 𝑆$ and $𝑆 𝑇$ are both defined, they need not be equal. For matrices this becomes the noncommutativity of matrix multiplication (Section 2.4).

Exercise 2.1.20.routine Let $𝑆 : ℝ^{3} \to ℝ^{2}$ and $𝑇 : ℝ^{2} \to ℝ^{3}$ be given by

𝑆 (𝑥, 𝑦, 𝑧) = (𝑥 + 𝑦, 𝑦 + 𝑧), 𝑇 (𝑢, 𝑣) = (𝑢 - 𝑣, 2 𝑣, 𝑢) .

Compute $𝑆 𝑇$ and $𝑇 𝑆$ .

Solution.

Uses. Proposition 2.1.17 (Compositions of linear maps are linear)

Hint.

𝑆 𝑇

means “first

𝑇

, then

𝑆

”, so it is a map

ℝ^{2} \to ℝ^{2}

. Substitute the coordinates of

𝑇 (𝑢, 𝑣)

into the formula for

𝑆

Sketch.

𝑆 𝑇 (𝑢, 𝑣) =

(𝑢 + 𝑣, 𝑢 + 2 𝑣)

and

𝑇 𝑆 (𝑥, 𝑦, 𝑧) =

(𝑥 - 𝑧, 2 𝑦 + 2 𝑧, 𝑥 + 𝑦)

$𝑆 𝑇 : ℝ^{2} \to ℝ^{2}$ is

(𝑆 𝑇) (𝑢, 𝑣) = 𝑆 (𝑢 - 𝑣, 2 𝑣, 𝑢) = ((𝑢 - 𝑣) + 2 𝑣, 2 𝑣 + 𝑢) = (𝑢 + 𝑣, 𝑢 + 2 𝑣),

and $𝑇 𝑆 : ℝ^{3} \to ℝ^{3}$ is

(𝑇 𝑆) (𝑥, 𝑦, 𝑧) = 𝑇 (𝑥 + 𝑦, 𝑦 + 𝑧) = ((𝑥 + 𝑦) - (𝑦 + 𝑧), 2 (𝑦 + 𝑧), 𝑥 + 𝑦) = (𝑥 - 𝑧, 2 𝑦 + 2 𝑧, 𝑥 + 𝑦) .

The two composites have different domains, so here the question whether $𝑆 𝑇 = 𝑇 𝑆$ does not even arise.

2.1.6The vector space $L (𝑉, 𝑊)$

Functions with values in a vector space can be added and scaled value by value. For linear maps this makes $L (𝑉, 𝑊)$ a vector space, so everything we prove about vector spaces applies to spaces of linear maps.

Definition 2.1.21 (Pointwise operations). Let $𝑋$ be a set and $𝑊$ a vector space over $𝐹$ . For functions $𝑓, 𝑔 : 𝑋 \to 𝑊$ and $𝑐 \in 𝐹$ , define $𝑓 + 𝑔 : 𝑋 \to 𝑊$ and $𝑐 𝑓 : 𝑋 \to 𝑊$ by

(𝑓 + 𝑔) (𝑥) = 𝑓 (𝑥) + 𝑔 (𝑥), (𝑐 𝑓) (𝑥) = 𝑐 𝑓 (𝑥) (𝑥 \in 𝑋) .

We write $𝑊^{𝑋}$ for the set of all functions $𝑋 \to 𝑊$ , with these operations.

Lemma 2.1.22 (Functions into a vector space). Let

𝑋

be a set and

𝑊

a vector space over

𝐹

. Then

𝑊^{𝑋}

is a vector space over

𝐹

. Its zero vector is the zero function

𝑥 \mapsto 𝟎_{𝑊}

, and the additive inverse of

𝑓

𝑥 \mapsto - 𝑓 (𝑥)

Proof.routine

Uses. Definition 1.2.1 (Vector space), Definition 2.1.21 (Pointwise operations)

Techniques. Proving from axioms (Technique 1.1.7)

Hint 1. Each axiom is an equality between two functions

𝑋 \to 𝑊

. Two functions are equal when they agree at every point.

Hint 2. Evaluate both sides of an axiom at

𝑥 \in 𝑋

. You get the same axiom in

𝑊

, applied to the vectors

𝑓 (𝑥)

𝑔 (𝑥)

, ….

Sketch. Each vector space axiom for

𝑊^{𝑋}

, evaluated at a point

𝑥

, becomes the same axiom in

𝑊

for the values at

𝑥

. For example

(𝑓 + 𝑔) (𝑥) =

𝑓 (𝑥) +

𝑔 (𝑥) =

𝑔 (𝑥) +

𝑓 (𝑥) =

(𝑔 + 𝑓) (𝑥)

The operations produce functions $𝑋 \to 𝑊$ , so $𝑊^{𝑋}$ is closed under them. Each vector space axiom asserts an equality between functions $𝑋 \to 𝑊$ , and two such functions are equal when their values agree at every $𝑥 \in 𝑋$ . At a point $𝑥$ , each axiom becomes the corresponding axiom of $𝑊$ . For example, for $𝑓, 𝑔, ℎ \in 𝑊^{𝑋}$ , $𝑎, 𝑏 \in 𝐹$ , and $𝑥 \in 𝑋$ :

\begin{aligned} (𝑓 + 𝑔) (𝑥) & = 𝑓 (𝑥) + 𝑔 (𝑥) = 𝑔 (𝑥) + 𝑓 (𝑥) = (𝑔 + 𝑓) (𝑥), \\ ((𝑓 + 𝑔) + ℎ) (𝑥) & = (𝑓 (𝑥) + 𝑔 (𝑥)) + ℎ (𝑥) = 𝑓 (𝑥) + (𝑔 (𝑥) + ℎ (𝑥)) = (𝑓 + (𝑔 + ℎ)) (𝑥), \\ (𝑎 (𝑓 + 𝑔)) (𝑥) & = 𝑎 (𝑓 (𝑥) + 𝑔 (𝑥)) = 𝑎 𝑓 (𝑥) + 𝑎 𝑔 (𝑥) = (𝑎 𝑓 + 𝑎 𝑔) (𝑥), \\ ((𝑎 + 𝑏) 𝑓) (𝑥) & = (𝑎 + 𝑏) 𝑓 (𝑥) = 𝑎 𝑓 (𝑥) + 𝑏 𝑓 (𝑥) = (𝑎 𝑓 + 𝑏 𝑓) (𝑥), \end{aligned}

and similarly $((𝑎 𝑏) 𝑓) (𝑥) =$ $(𝑎 𝑏) 𝑓 (𝑥) =$ $𝑎 (𝑏 𝑓 (𝑥)) =$ $(𝑎 (𝑏 𝑓)) (𝑥)$ and $(1 𝑓) (𝑥) =$ $1 𝑓 (𝑥) =$ $𝑓 (𝑥)$ . Let $𝑧$ be the zero function and $- 𝑓$ the function $𝑥 \mapsto - 𝑓 (𝑥)$ . Then $(𝑓 + 𝑧) (𝑥) =$ $𝑓 (𝑥) +$ $𝟎_{𝑊} =$ $𝑓 (𝑥)$ and $(𝑓 + (- 𝑓)) (𝑥) =$ $𝑓 (𝑥) +$ $(- 𝑓 (𝑥)) =$ $𝟎_{𝑊} =$ $𝑧 (𝑥)$ , so $𝑧$ is an additive identity and $- 𝑓$ is an additive inverse of $𝑓$ . □

For $𝑊 = 𝐹$ this is the vector space $𝐹^{𝑋}$ of Section 1.2. Now $L (𝑉, 𝑊)$ is a subset of $𝑊^{𝑉}$ .

Intuition. We could check the vector space axioms for

L (𝑉, 𝑊)

one at a time, by evaluating at each

𝐯

. That would repeat the proof of Lemma 2.1.22 word for word. Passing through

𝑊^{𝑉}

separates the two things that are true. The axioms hold for all functions

𝑉 \to 𝑊

, linear or not; what is special about linear maps is only that the operations of

𝑊^{𝑉}

do not destroy linearity. The subspace criterion asks for exactly that.

Proposition 2.1.23 (

L (𝑉, 𝑊)

is a vector space). Let

𝑉

and

𝑊

be vector spaces over

𝐹

. If

𝑆, 𝑇 \in L (𝑉, 𝑊)

and

𝑐 \in 𝐹

, then

𝑆 + 𝑇

and

𝑐 𝑇

are linear. Hence

L (𝑉, 𝑊)

is a subspace of

𝑊^{𝑉}

, and in particular it is a vector space over

𝐹

. Its zero vector is the zero map, and

- 𝑇

is the map

𝐯 \mapsto - 𝑇 (𝐯)

Proof.routine

Uses. Lemma 2.1.22 (Functions into a vector space), Theorem 1.4.3 (Subspace criterion), Theorem 2.1.3 (Checking linearity), Example 2.1.6 (Zero, identity, and scaling maps)

Hint 1. By Lemma 2.1.22 and the subspace criterion, it suffices to show that the zero map is linear and that

𝑆 + 𝑇

and

𝑐 𝑇

are linear.

Hint 2. Check (iii) of Theorem 2.1.3 for

𝑆 + 𝑇

: expand

(𝑆 + 𝑇) (𝑎 𝐮 + 𝐯)

by the definition of

𝑆 + 𝑇

, use linearity of

𝑆

and of

𝑇

, and regroup.

Hint 3. For

𝑐 𝑇

you need to move

𝑐

past

𝑎

. Which field axiom allows that?

Sketch. The zero map is linear. For

𝑆 + 𝑇

(𝑆 + 𝑇) (𝑎 𝐮 + 𝐯) =

𝑎 𝑆 (𝐮) +

𝑆 (𝐯) +

𝑎 𝑇 (𝐮) +

𝑇 (𝐯) =

𝑎 (𝑆 + 𝑇) (𝐮) +

(𝑆 + 𝑇) (𝐯)

. For

𝑐 𝑇

𝑐 (𝑎 𝑇 (𝐮) + 𝑇 (𝐯)) =

𝑎 (𝑐 𝑇 (𝐮)) +

𝑐 𝑇 (𝐯)

since

𝑐 𝑎 = 𝑎 𝑐

. So

L (𝑉, 𝑊)

is a nonempty subset of the vector space

𝑊^{𝑉}

, closed under both operations.

By Lemma 2.1.22, $𝑊^{𝑉}$ is a vector space, and $L (𝑉, 𝑊) \subseteq 𝑊^{𝑉}$ . We check the conditions of Theorem 1.4.3. The zero map is linear (Example 2.1.6), so $L (𝑉, 𝑊)$ is nonempty and contains the zero vector of $𝑊^{𝑉}$ .

Let $𝑆, 𝑇 \in L (𝑉, 𝑊)$ , $𝑐 \in 𝐹$ , $𝐮, 𝐯 \in 𝑉$ , and $𝑎 \in 𝐹$ . Using the definition of $𝑆 + 𝑇$ , then the linearity of $𝑆$ and $𝑇$ , then the vector space axioms of $𝑊$ to regroup,

\begin{aligned} (𝑆 + 𝑇) (𝑎 𝐮 + 𝐯) & = 𝑆 (𝑎 𝐮 + 𝐯) + 𝑇 (𝑎 𝐮 + 𝐯) \\ = (𝑎 𝑆 (𝐮) + 𝑆 (𝐯)) + (𝑎 𝑇 (𝐮) + 𝑇 (𝐯)) \\ = 𝑎 (𝑆 (𝐮) + 𝑇 (𝐮)) + (𝑆 (𝐯) + 𝑇 (𝐯)) = 𝑎 (𝑆 + 𝑇) (𝐮) + (𝑆 + 𝑇) (𝐯) . \end{aligned}

Similarly,

\begin{aligned} (𝑐 𝑇) (𝑎 𝐮 + 𝐯) & = 𝑐 (𝑎 𝑇 (𝐮) + 𝑇 (𝐯)) = (𝑐 𝑎) 𝑇 (𝐮) + 𝑐 𝑇 (𝐯) \\ = (𝑎 𝑐) 𝑇 (𝐮) + 𝑐 𝑇 (𝐯) = 𝑎 (𝑐 𝑇) (𝐮) + (𝑐 𝑇) (𝐯), \end{aligned}

where the step $𝑐 𝑎 = 𝑎 𝑐$ is the commutativity of multiplication in $𝐹$ . By Theorem 2.1.3, $𝑆 + 𝑇$ and $𝑐 𝑇$ are linear. So $L (𝑉, 𝑊)$ is a subspace of $𝑊^{𝑉}$ , hence a vector space, with the same zero vector and the same additive inverses as $𝑊^{𝑉}$ . □

Linear maps now have three operations: addition, scalar multiplication, and composition. The next proposition records how they interact.

Proposition 2.1.24 (Algebra of composition). Let

𝑈

𝑉

𝑊

, and

𝑋

be vector spaces over

𝐹

, and let

𝑎 \in 𝐹

(a) $(𝑅 𝑆) 𝑇 = 𝑅 (𝑆 𝑇)$ for $𝑇 \in L (𝑈, 𝑉)$ , $𝑆 \in L (𝑉, 𝑊)$ , and $𝑅 \in L (𝑊, 𝑋)$ .
(b) ${id}_{𝑊} 𝑇 = 𝑇 = 𝑇 {id}_{𝑉}$ for $𝑇 \in L (𝑉, 𝑊)$ .
(c) $(𝑆_{1} + 𝑆_{2}) 𝑇 =$ $𝑆_{1} 𝑇 +$ $𝑆_{2} 𝑇$ and $𝑆 (𝑇_{1} + 𝑇_{2}) =$ $𝑆 𝑇_{1} +$ $𝑆 𝑇_{2}$ for $𝑇,$ $𝑇_{1},$ $𝑇_{2} \in$ $L (𝑈, 𝑉)$ and $𝑆,$ $𝑆_{1},$ $𝑆_{2} \in$ $L (𝑉, 𝑊)$ .
(d) $(𝑎 𝑆) 𝑇 =$ $𝑎 (𝑆 𝑇) =$ $𝑆 (𝑎 𝑇)$ for $𝑇 \in L (𝑈, 𝑉)$ and $𝑆 \in L (𝑉, 𝑊)$ .

Proof.routine

Uses. Proposition 2.1.23 ( $L (𝑉, 𝑊)$ is a vector space), Definition 2.1.21 (Pointwise operations), Theorem 2.1.3 (Checking linearity)

Hint 1. Two maps are equal when they agree at every vector. Evaluate both sides of each identity at

𝐮

Hint 2. Most identities follow from the definitions alone. In

𝑆 (𝑇_{1} + 𝑇_{2}) =

𝑆 𝑇_{1} +

𝑆 𝑇_{2}

and

𝑆 (𝑎 𝑇) = 𝑎 (𝑆 𝑇)

you need a property of

𝑆

Sketch. Evaluate at

𝐮

. Associativity and the identity laws hold for all functions.

((𝑆_{1} + 𝑆_{2}) 𝑇) (𝐮) =

𝑆_{1} (𝑇 (𝐮)) +

𝑆_{2} (𝑇 (𝐮))

by the definition of

𝑆_{1} + 𝑆_{2}

(𝑆 (𝑇_{1} + 𝑇_{2})) (𝐮) =

𝑆 (𝑇_{1} (𝐮) + 𝑇_{2} (𝐮))

splits by additivity of

𝑆

, and

𝑆 (𝑎 𝑇 (𝐮)) =

𝑎 𝑆 (𝑇 (𝐮))

by homogeneity of

𝑆

In each case both sides are maps with the same domain and codomain, so it suffices to show that they agree at every vector $𝐮$ of the domain.

(a) $((𝑅 𝑆) 𝑇) (𝐮) =$ $(𝑅 𝑆) (𝑇 (𝐮)) =$ $𝑅 (𝑆 (𝑇 (𝐮))) =$ $𝑅 ((𝑆 𝑇) (𝐮)) =$ $(𝑅 (𝑆 𝑇)) (𝐮)$ .

(b) $({id}_{𝑊} 𝑇) (𝐮) =$ ${id}_{𝑊} (𝑇 (𝐮)) =$ $𝑇 (𝐮)$ and $(𝑇 {id}_{𝑉}) (𝐮) =$ $𝑇 ({id}_{𝑉} (𝐮)) =$ $𝑇 (𝐮)$ .

(c) By the definition of the sum of two maps,

((𝑆_{1} + 𝑆_{2}) 𝑇) (𝐮) = (𝑆_{1} + 𝑆_{2}) (𝑇 (𝐮)) = 𝑆_{1} (𝑇 (𝐮)) + 𝑆_{2} (𝑇 (𝐮)) = (𝑆_{1} 𝑇 + 𝑆_{2} 𝑇) (𝐮) .

By the definition of $𝑇_{1} + 𝑇_{2}$ and then additivity of $𝑆$ ,

(𝑆 (𝑇_{1} + 𝑇_{2})) (𝐮) = 𝑆 (𝑇_{1} (𝐮) + 𝑇_{2} (𝐮)) = 𝑆 (𝑇_{1} (𝐮)) + 𝑆 (𝑇_{2} (𝐮)) = (𝑆 𝑇_{1} + 𝑆 𝑇_{2}) (𝐮) .

(d) $((𝑎 𝑆) 𝑇) (𝐮) =$ $𝑎 𝑆 (𝑇 (𝐮)) =$ $(𝑎 (𝑆 𝑇)) (𝐮)$ by the definition of $𝑎 𝑆$ , and by the definition of $𝑎 𝑇$ and homogeneity of $𝑆$ ,

(𝑆 (𝑎 𝑇)) (𝐮) = 𝑆 (𝑎 𝑇 (𝐮)) = 𝑎 𝑆 (𝑇 (𝐮)) = (𝑎 (𝑆 𝑇)) (𝐮) .

□

Only the second identity in (c) and the last one in (d) used linearity, in both cases of the left factor $𝑆$ . For functions that are not linear they can fail: for $𝑠 (𝑥) = 𝑥^{2}$ on $ℝ$ , $(𝑠 \circ ({id}_{ℝ} + {id}_{ℝ})) (1) =$ $𝑠 (2) =$ $4$ , while $(𝑠 \circ {id}_{ℝ} + 𝑠 \circ {id}_{ℝ}) (1) =$ $2$ . Parts (c) and (d) also say that composition is linear in each factor: for fixed $𝑇 \in L (𝑈, 𝑉)$ , the map $L (𝑉, 𝑊) \to L (𝑈, 𝑊)$ , $𝑆 \mapsto 𝑆 𝑇$ , is linear, and for fixed $𝑆 \in L (𝑉, 𝑊)$ , so is $L (𝑈, 𝑉) \to L (𝑈, 𝑊)$ , $𝑇 \mapsto 𝑆 𝑇$ .

Exercise 2.1.25.routine Let

𝑇 : 𝐹^{2} \to 𝐹^{2}

𝑇 (𝑥, 𝑦) =

(𝑦, 𝑥 - 𝑦)

. Show that

𝑇^{2} + 𝑇 = {id}_{𝐹^{2}}

, and deduce that

𝑇 (𝑇 + {id}_{𝐹^{2}}) =

(𝑇 + {id}_{𝐹^{2}}) 𝑇 =

{id}_{𝐹^{2}}

Solution.

Uses. Proposition 2.1.24 (Algebra of composition), Notation 2.1.18

Hint 1. Compute

𝑇^{2} (𝑥, 𝑦) =

𝑇 (𝑇 (𝑥, 𝑦))

and add

𝑇 (𝑥, 𝑦)

Hint 2. For the second part, expand

𝑇 (𝑇 + id)

with Proposition 2.1.24.

Sketch.

𝑇^{2} (𝑥, 𝑦) =

(𝑥 - 𝑦, 2 𝑦 - 𝑥)

, and adding

(𝑦, 𝑥 - 𝑦)

gives

(𝑥, 𝑦)

. Then

𝑇 (𝑇 + id) =

𝑇^{2} +

𝑇 =

id

and

(𝑇 + id) 𝑇 =

𝑇^{2} +

𝑇 =

id

by parts (b) and (c) of Proposition 2.1.24.

We have $𝑇^{2} (𝑥, 𝑦) =$ $𝑇 (𝑦, 𝑥 - 𝑦) =$ $(𝑥 - 𝑦, 𝑦 - (𝑥 - 𝑦)) =$ $(𝑥 - 𝑦, 2 𝑦 - 𝑥)$ , so

(𝑇^{2} + 𝑇) (𝑥, 𝑦) = (𝑥 - 𝑦 + 𝑦, 2 𝑦 - 𝑥 + 𝑥 - 𝑦) = (𝑥, 𝑦) .

That is, $𝑇^{2} + 𝑇 = id$ . By parts (b) and (c) of Proposition 2.1.24, $𝑇 (𝑇 + id) =$ $𝑇^{2} +$ $𝑇 id =$ $𝑇^{2} +$ $𝑇 =$ $id$ and $(𝑇 + id) 𝑇 =$ $𝑇^{2} +$ $id 𝑇 =$ $𝑇^{2} +$ $𝑇 =$ $id$ . So applying $𝑇 + id$ undoes $𝑇$ , in either order; invertibility is the subject of Section 2.5.

Assembling linear maps

The results of this section and the last give a way to prove linearity without computing.

Technique 2.1.26 (Assemble linear maps from linear pieces). To show that a map is linear, write it as a sum, scalar multiple, or composite of maps already known to be linear. By Proposition 2.1.23 and Proposition 2.1.17 the result is linear, with no computation.

Reach for it when a map is given by a formula built out of evaluations, derivatives, integrals, coordinates, projections, or other maps from the examples above. It also handles whole families at once. For instance, for $𝑇 \in L (𝑉)$ and $𝑎_{0}, \dots, 𝑎_{𝑛} \in 𝐹$ , every $𝑎_{0} {id}_{𝑉} +$ $𝑎_{1} 𝑇 +$ $\dots +$ $𝑎_{𝑛} 𝑇^{𝑛}$ is linear. Conversely, a formula that involves a product of two inputs, a constant term, or a nonlinear function of a coordinate is a candidate for a counterexample; test $𝑇 (𝟎)$ and $𝑇 (2 𝐯)$ first. The canonical example is the reflection $𝑆_{ℓ} = 2 {Pr}_{ℓ} - {id}_{ℝ^{2}}$ (Example 2.1.27).

Used in 7 places and 1 practice exercise

Exercise 2.1.29, Exercise 2.1.42, Exercise 2.1.47, Theorem 2.2.1, Theorem 2.3.20, Exercise 2.8.5, Proposition 4.6.7; 1 practice exercise.

Example 2.1.27 (Maps assembled from linear pieces).

(a)By Example 2.1.11, $𝑆_{ℓ} (𝐯) =$ $2 {Pr}_{ℓ} (𝐯) -$ $𝐯$ for every $𝐯$ . That is, $𝑆_{ℓ} = 2 {Pr}_{ℓ} - {id}_{ℝ^{2}}$ in $L (ℝ^{2})$ . Since ${Pr}_{ℓ}$ is linear, so is $𝑆_{ℓ}$ , and the double angle formulas are not needed for this.
(b)The map $𝜑 : 𝐹 [𝑥] \to 𝐹$ , $𝜑 (𝑝) =$ $𝑝 (1) -$ $3 𝑝^{'} (0)$ , is linear, because $𝜑 = {ev}_{1} - 3 {ev}_{0} 𝐷$ with ${ev}_{𝑡}$ and $𝐷$ from Example 2.1.9.
(c)On the real vector space of infinitely differentiable functions $ℝ \to ℝ$ (a subspace of $ℝ^{ℝ}$ ), differentiation $𝐷 : 𝑓 \mapsto 𝑓^{'}$ is linear, so $𝑓 \mapsto 𝑓^{″} - 3 𝑓^{'} + 2 𝑓$ is linear, being $𝐷^{2} - 3 𝐷 + 2 id$ . This is why the differential equation $𝑦^{″} - 3 𝑦^{'} + 2 𝑦 = 0$ is called linear: it asks for the vectors that a linear map sends to $𝟎$ . By Section 2.6 such vectors always form a subspace.

Exercise 2.1.28.routine Fix

𝐯_{0} \in 𝑉

. Show that the map

L (𝑉, 𝑊) \to 𝑊

𝑇 \mapsto 𝑇 (𝐯_{0})

, is linear.

Solution.

Uses. Definition 2.1.21 (Pointwise operations), Proposition 2.1.23 ( $L (𝑉, 𝑊)$ is a vector space), Theorem 2.1.3 (Checking linearity)

Hint. The vectors of

L (𝑉, 𝑊)

are maps, and they are added and scaled pointwise. Evaluate

𝑎 𝑆 + 𝑇

𝐯_{0}

Sketch.

(𝑎 𝑆 + 𝑇) (𝐯_{0}) =

𝑎 𝑆 (𝐯_{0}) +

𝑇 (𝐯_{0})

by the definition of the operations.

Write $𝐸 (𝑇) = 𝑇 (𝐯_{0})$ . For $𝑆, 𝑇 \in L (𝑉, 𝑊)$ and $𝑎 \in 𝐹$ , the definition of the operations on $L (𝑉, 𝑊)$ gives

𝐸 (𝑎 𝑆 + 𝑇) = (𝑎 𝑆 + 𝑇) (𝐯_{0}) = 𝑎 𝑆 (𝐯_{0}) + 𝑇 (𝐯_{0}) = 𝑎 𝐸 (𝑆) + 𝐸 (𝑇) .

So $𝐸$ is linear by Theorem 2.1.3. Note that the linearity of $𝑆$ and $𝑇$ played no role; the same computation shows that evaluation at a point is linear on $𝑊^{𝑋}$ for any set $𝑋$ .

Aside (Additive maps that are not linear).Beyond 21-242 By Exercise 2.1.36 and Exercise 2.1.37, an additive map between vector spaces over

𝔽_{𝑝}

or over

ℚ

is automatically linear, because every scalar can be reached from

1

by addition, subtraction, and division by positive integers. Over

ℝ

this is false, but no formula exhibits a counterexample. Regard

ℝ

as a vector space over

ℚ

. It has a basis (a Hamel basis), by Zorn’s lemma (Section 15.1). Pick a basis vector

𝐛

, and let

𝑇 : ℝ \to ℝ

send each real number to its coordinate along

𝐛

. Then

𝑇

ℚ

-linear, hence additive, and it takes only rational values. If

𝑇

were

ℝ

-linear, then

𝑇 (𝑥) = 𝑥 𝑇 (1)

for all

𝑥

(Exercise 2.1.5); since

𝑇

takes only rational values, this forces

𝑇 (1) = 0

, and then

𝑇 = 0

, contradicting

𝑇 (𝐛) = 1

. Such a map is wildly discontinuous: an additive map

ℝ \to ℝ

that is continuous is linear (Exercise 2.1.51).

Exercises

Exercise 2.1.29.routine Decide which of the following maps are linear. Here $𝐹$ is any field, $𝐴 \in 𝐹^{2 \times 2}$ has entries $𝑎_{𝑖 𝑗}$ , and $𝐹^{2 \times 2}$ is added and scaled entrywise.

(a) $𝐹^{2 \times 2} \to 𝐹$ , $𝐴 \mapsto 𝑎_{11} + 𝑎_{22}$ .
(b) $𝐹^{2 \times 2} \to 𝐹$ , $𝐴 \mapsto$ $𝑎_{11} 𝑎_{22} -$ $𝑎_{12} 𝑎_{21}$ .
(c) $𝐹^{2 \times 2} \to 𝐹^{2 \times 2}$ , the map that swaps the entries $𝑎_{12}$ and $𝑎_{21}$ .
(d) $𝐹 [𝑥] \to 𝐹 [𝑥]$ , $𝑝 \mapsto 𝑝 + 𝑝^{'}$ .
(e) $ℂ \to ℂ$ , $𝑧 \mapsto 𝑖 𝑧$ , with $ℂ$ regarded first as a vector space over $ℂ$ , then as one over $ℝ$ .

Solution.

Uses. Example 2.1.7 (Maps $𝐹^{𝑛} \to 𝐹^{𝑚}$ given by linear formulas), Example 2.1.9 (Maps of polynomials), Non-example 2.1.14 (Complex conjugation)

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

Hint 1. Entrywise operations make

𝐹^{2 \times 2}

look like

𝐹^{4}

. Which of the formulas are sums of constant multiples of the entries?

Hint 2. For (b), try two matrices with a single nonzero entry each, on the diagonal. For (e) over

ℂ

, compare

𝑇 (𝑖 \cdot 1)

with

𝑖 \cdot 𝑇 (1)

Sketch. (a), (c), and (d) are linear: (a) and (c) are linear formulas in the entries, and (d) is

id + 𝐷

. (b) is not additive: the two diagonal matrices with a single

1

each map to

0

, while their sum, the identity matrix, maps to

1

. (e) is

ℝ

-linear, being

(𝑎, 𝑏) \mapsto (𝑏, 𝑎)

in coordinates, but not

ℂ

-linear:

𝑇 (𝑖) = 1

while

𝑖 𝑇 (1) = - 1

(a) Linear. The entries of $𝑐 𝐴 + 𝐵$ are $𝑐 𝑎_{𝑖 𝑗} + 𝑏_{𝑖 𝑗}$ , so the map sends $𝑐 𝐴 + 𝐵$ to $(𝑐 𝑎_{11} + 𝑏_{11}) +$ $(𝑐 𝑎_{22} + 𝑏_{22}) =$ $𝑐 (𝑎_{11} + 𝑎_{22}) +$ $(𝑏_{11} + 𝑏_{22})$ .

(b) Not linear, over any field. Let $𝐴$ have $𝑎_{11} = 1$ and all other entries $0$ , and let $𝐵$ have $𝑏_{22} = 1$ and all other entries $0$ . Both map to $0$ , but $𝐴 + 𝐵$ has diagonal entries $1, 1$ and off-diagonal entries $0$ , so it maps to $1 \neq 0$ . The map is not additive.

(c) Linear. Listing the entries as $(𝑎_{11}, 𝑎_{12}, 𝑎_{21}, 𝑎_{22})$ , the map is $(𝑎, 𝑏, 𝑐, 𝑑) \mapsto$ $(𝑎, 𝑐, 𝑏, 𝑑)$ , a linear formula as in Example 2.1.7 (each output coordinate is one input coordinate), and the operations of $𝐹^{2 \times 2}$ are those of $𝐹^{4}$ in these coordinates.

(d) Linear: the map is ${id}_{𝐹 [𝑥]} + 𝐷$ , a sum of linear maps (Technique 2.1.26).

(e) Write $𝑇 (𝑧) = 𝑖 𝑧$ . For $𝑧 = 𝑎 + 𝑏 𝑖$ , $𝑇 (𝑧) =$ $𝑖 (𝑎 - 𝑏 𝑖) =$ $𝑏 +$ $𝑎 𝑖$ . Over $ℝ$ , in the coordinates $(𝑎, 𝑏)$ this is $(𝑎, 𝑏) \mapsto (𝑏, 𝑎)$ , which is linear (it is the reflection across the line $𝑦 = 𝑥$ ). Over $ℂ$ it is not homogeneous: $𝑇 (𝑖 \cdot 1) =$ $𝑇 (𝑖) =$ $𝑖 \cdot$ $(- 𝑖) =$ $1$ , while $𝑖 \cdot 𝑇 (1) = 𝑖 \cdot 𝑖 = - 1$ .

Exercise 2.1.30.standard How many functions

𝔽_{2}^{2} \to 𝔽_{2}

are there? How many of them send

𝟎

0

? How many are linear? List the linear ones.

Solution.

Uses. Example 2.1.7 (Maps $𝐹^{𝑛} \to 𝐹^{𝑚}$ given by linear formulas), Proposition 2.1.4 (Linear maps preserve $𝟎$ and negatives)

Hint 1. A function

𝔽_{2}^{2} \to 𝔽_{2}

is a choice of one of two values at each of the four points of

𝔽_{2}^{2}

Hint 2. A linear

𝑇

satisfies

𝑇 (𝑥, 𝑦) =

𝑥 𝑇 (𝐞_{1}) +

𝑦 𝑇 (𝐞_{2})

. How many choices are there for the pair

(𝑇 (𝐞_{1}), 𝑇 (𝐞_{2}))

, and does each choice give a linear map?

Sketch.

2^{4} = 16

functions, of which

2^{3} = 8

send

𝟎

0

. A linear map is

(𝑥, 𝑦) \mapsto 𝛼 𝑥 + 𝛽 𝑦

with

(𝛼, 𝛽) =

(𝑇 (𝐞_{1}), 𝑇 (𝐞_{2}))

, and every such formula is linear, so there are

4

0

𝑥

𝑦

, and

𝑥 + 𝑦

$𝔽_{2}^{2}$ has $4$ elements and $𝔽_{2}$ has $2$ , so there are $2^{4} = 16$ functions $𝔽_{2}^{2} \to 𝔽_{2}$ . Those with $𝑇 (𝟎) = 0$ are free only at the other three points, so there are $2^{3} = 8$ of them.

If $𝑇$ is linear, then since $(𝑥, 𝑦) =$ $𝑥 𝐞_{1} +$ $𝑦 𝐞_{2}$ ,

𝑇 (𝑥, 𝑦) = 𝑥 𝑇 (𝐞_{1}) + 𝑦 𝑇 (𝐞_{2}) = 𝛼 𝑥 + 𝛽 𝑦, where 𝛼 = 𝑇 (𝐞_{1}), 𝛽 = 𝑇 (𝐞_{2}) .

So $𝑇$ is determined by $(𝛼, 𝛽) \in 𝔽_{2}^{2}$ , and there are at most $4$ linear maps. Conversely, each formula $(𝑥, 𝑦) \mapsto 𝛼 𝑥 + 𝛽 𝑦$ is linear by Example 2.1.7. So there are exactly $4$ linear maps: $(𝑥, 𝑦) \mapsto 0$ , $𝑥$ , $𝑦$ , and $𝑥 + 𝑦$ .

So half of the functions that fix $𝟎$ are not linear. They are $(𝑥, 𝑦) \mapsto 𝑥 𝑦$ , $𝑥 + 𝑥 𝑦$ , $𝑦 + 𝑥 𝑦$ , and $𝑥 + 𝑦 + 𝑥 𝑦$ .

Exercise 2.1.31.routine Let $𝑇 \in L (𝑉, 𝑊)$ and let $ℒ︀$ be a list in $𝑉$ . Prove or disprove each statement.

(a)If $ℒ︀$ spans $𝑉$ , then $𝑇 (ℒ︀)$ spans $𝑊$ .
(b)If $𝑇 (ℒ︀)$ spans $𝑊$ , then $ℒ︀$ spans $𝑉$ .
(c)If $𝑇 (ℒ︀)$ spans $𝑊$ , then $im 𝑇 = 𝑊$ .

Solution.

Uses. Proposition 2.1.16 (Linear maps preserve spans and linear relations)

Hint 1. Proposition 2.1.16 says that

𝑇 (ℒ︀)

spans

im (𝑇 ↾_{span ℒ︀})

. When is that all of

𝑊

Hint 2. For counterexamples, try the zero map and a coordinate projection

ℝ^{2} \to ℝ

Sketch. (a) False: the zero map

ℝ \to ℝ

and

ℒ︀ = 1

. (b) False:

(𝑥, 𝑦) \mapsto 𝑥

from

ℝ^{2}

ℝ

and

ℒ︀ = 𝐞_{1}

. (c) True:

𝑊 =

span 𝑇 (ℒ︀) =

im (𝑇 ↾_{span ℒ︀}) \subseteq

im 𝑇 \subseteq

𝑊

(a) False. Let $𝑇 : ℝ \to ℝ$ be the zero map and $ℒ︀$ the list $1$ , which spans $ℝ$ . Then $𝑇 (ℒ︀)$ is the list $0$ , which spans ${0} \neq ℝ$ .

(b) False. Let $𝑇 : ℝ^{2} \to ℝ$ , $𝑇 (𝑥, 𝑦) = 𝑥$ , and $ℒ︀ = 𝐞_{1}$ . Then $𝑇 (ℒ︀)$ is the list $1$ , which spans $ℝ$ , but $ℒ︀$ does not span $ℝ^{2}$ : every element of $span (𝐞_{1})$ has second coordinate $0$ .

(c) True. By Proposition 2.1.16, $𝑊 =$ $span 𝑇 (ℒ︀) =$ $im (𝑇 ↾_{span ℒ︀}) \subseteq$ $im 𝑇 \subseteq$ $𝑊$ , so $im 𝑇 = 𝑊$ .

Exercise 2.1.32.routine Fix

𝑞 \in 𝐹 [𝑥]

. For

𝑝 =

𝑎_{0} +

𝑎_{1} 𝑥 +

\dots +

𝑎_{𝑛} 𝑥^{𝑛}

, let

𝑝 (𝑞) =

𝑎_{0} +

𝑎_{1} 𝑞 +

\dots +

𝑎_{𝑛} 𝑞^{𝑛} \in

𝐹 [𝑥]

be the result of substituting

𝑞

for

𝑥

. Show that the map

𝐶_{𝑞} : 𝐹 [𝑥] \to

𝐹 [𝑥]

𝑝 \mapsto 𝑝 (𝑞)

, is linear. (For

𝑞 = 𝑥 + 1

, this is the shift

𝑝 (𝑥) \mapsto 𝑝 (𝑥 + 1)

Solution.

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

Hint 1. Write

𝑐 𝑝 + 𝑟

in terms of the coefficients of

𝑝

and

𝑟

, using the same

𝑛

for both.

Hint 2. Substituting

𝑞

into

\sum_{𝑘} (𝑐 𝑎_{𝑘} + 𝑏_{𝑘}) 𝑥^{𝑘}

gives

\sum_{𝑘} (𝑐 𝑎_{𝑘} + 𝑏_{𝑘}) 𝑞^{𝑘}

. Split it using the vector space axioms of

𝐹 [𝑥]

Sketch. With

𝑝 = \sum 𝑎_{𝑘} 𝑥^{𝑘}

and

𝑟 = \sum 𝑏_{𝑘} 𝑥^{𝑘}

𝑐 𝑝 +

𝑟 =

\sum (𝑐 𝑎_{𝑘} + 𝑏_{𝑘}) 𝑥^{𝑘}

, so

𝐶_{𝑞} (𝑐 𝑝 + 𝑟) =

\sum (𝑐 𝑎_{𝑘} + 𝑏_{𝑘}) 𝑞^{𝑘} =

𝑐 \sum 𝑎_{𝑘} 𝑞^{𝑘} +

\sum 𝑏_{𝑘} 𝑞^{𝑘}

Let $𝑝 =$ $\sum_{𝑘 = 0}^{𝑛} 𝑎_{𝑘} 𝑥^{𝑘}$ and $𝑟 =$ $\sum_{𝑘 = 0}^{𝑛} 𝑏_{𝑘} 𝑥^{𝑘}$ , padding with zero coefficients so that both use the same $𝑛$ . (Padding does not change $𝑝 (𝑞)$ , since the added terms are $0 \cdot 𝑞^{𝑘} = 0$ .) For $𝑐 \in 𝐹$ , $𝑐 𝑝 +$ $𝑟 =$ $\sum_{𝑘} (𝑐 𝑎_{𝑘} + 𝑏_{𝑘}) 𝑥^{𝑘}$ , so

𝐶_{𝑞} (𝑐 𝑝 + 𝑟) = \sum_{𝑘 = 0}^{𝑛} (𝑐 𝑎_{𝑘} + 𝑏_{𝑘}) 𝑞^{𝑘} = 𝑐 \sum_{𝑘 = 0}^{𝑛} 𝑎_{𝑘} 𝑞^{𝑘} + \sum_{𝑘 = 0}^{𝑛} 𝑏_{𝑘} 𝑞^{𝑘} = 𝑐 𝐶_{𝑞} (𝑝) + 𝐶_{𝑞} (𝑟),

by distributivity and the other vector space axioms of $𝐹 [𝑥]$ . By Theorem 2.1.3, $𝐶_{𝑞}$ is linear.

Exercise 2.1.33.standard Now substitute in the other order. For

𝑞 \in ℝ [𝑥]

, let

𝐿_{𝑞} : ℝ [𝑥] \to ℝ [𝑥]

𝑝 \mapsto 𝑞 (𝑝)

, the result of substituting

𝑝

for

𝑥

𝑞

(notation of Exercise 2.1.32). Find all

𝑞

for which

𝐿_{𝑞}

is linear.

Solution.

Uses. Theorem 2.1.3 (Checking linearity), Example 2.1.6 (Zero, identity, and scaling maps)

Hint 1. Apply

𝐿_{𝑞}

to the polynomial

𝑥

, and to the polynomial

2 𝑥

. What does homogeneity say?

Hint 2.

𝐿_{𝑞} (𝑥) = 𝑞

and

𝐿_{𝑞} (2 𝑥) = 𝑞 (2 𝑥)

. If

𝑞 = \sum 𝑏_{𝑘} 𝑥^{𝑘}

, then

𝑞 (2 𝑥) =

\sum 2^{𝑘} 𝑏_{𝑘} 𝑥^{𝑘}

. Compare coefficients with

2 𝑞

Sketch.

𝐿_{𝑞}

is linear exactly when

𝑞 = 𝑐 𝑥

for some

𝑐 \in ℝ

. If

𝐿_{𝑞}

is linear,

𝑞 (2 𝑥) =

𝐿_{𝑞} (2 𝑥) =

2 𝐿_{𝑞} (𝑥) =

2 𝑞

, so

(2^{𝑘} - 2) 𝑏_{𝑘} = 0

for every

𝑘

, and

𝑏_{𝑘} = 0

unless

𝑘 = 1

. Conversely

𝐿_{𝑐 𝑥} (𝑝) = 𝑐 𝑝

is a scaling map.

The answer is $𝑞 = 𝑐 𝑥$ with $𝑐 \in ℝ$ .

If $𝑞 = 𝑐 𝑥$ , then $𝐿_{𝑞} (𝑝) = 𝑐 𝑝$ , which is linear (Example 2.1.6). Conversely, suppose $𝐿_{𝑞}$ is linear, and write $𝑞 =$ $𝑏_{0} +$ $𝑏_{1} 𝑥 +$ $\dots +$ $𝑏_{𝑛} 𝑥^{𝑛}$ . Then $𝐿_{𝑞} (𝑥) = 𝑞$ and

𝐿_{𝑞} (2 𝑥) = 𝑞 (2 𝑥) = \sum_{𝑘 = 0}^{𝑛} 𝑏_{𝑘} {(2 𝑥)}^{𝑘} = \sum_{𝑘 = 0}^{𝑛} 2^{𝑘} 𝑏_{𝑘} 𝑥^{𝑘} .

By homogeneity, $𝐿_{𝑞} (2 𝑥) =$ $2 𝐿_{𝑞} (𝑥) =$ $\sum_{𝑘} 2 𝑏_{𝑘} 𝑥^{𝑘}$ . Comparing coefficients, $(2^{𝑘} - 2) 𝑏_{𝑘} = 0$ for every $𝑘$ . For $𝑘 \neq 1$ we have $2^{𝑘} - 2 \neq 0$ (it is $- 1$ for $𝑘 = 0$ and at least $2$ for $𝑘 \geq 2$ ), so $𝑏_{𝑘} = 0$ . Hence $𝑞 = 𝑏_{1} 𝑥$ .

The test used the scalar $2$ , which is $0$ in $𝔽_{2}$ , so over $𝔽_{2}$ the condition $𝐿_{𝑞} (2 𝑥) =$ $2 𝐿_{𝑞} (𝑥)$ says nothing. Indeed over $𝔽_{2}$ the answer is different: for $𝑞 = 𝑥^{2}$ , the map $𝑝 \mapsto 𝑝^{2}$ turns out to be linear (try it).

Exercise 2.1.34.routine Find all

𝑝 \in 𝔽_{2} [𝑥]

with

𝐷 (𝑝) = 0

, where

𝐷

is the formal derivative. Do the same over

𝔽_{3}

and over

ℝ

Solution.

Uses. Example 2.1.9 (Maps of polynomials), Proposition 1.1.17 (No zero divisors), Proposition 1.1.24 ( $ℤ / 𝑛 ℤ$ is a field if and only if $𝑛$ is prime)

Hint 1.

𝐷 (\sum 𝑎_{𝑘} 𝑥^{𝑘}) =

\sum 𝑘 𝑎_{𝑘} 𝑥^{𝑘 - 1}

, so

𝐷 (𝑝) = 0

if and only if

𝑘 𝑎_{𝑘} = 0

for every

𝑘 \geq 1

Hint 2. In

𝔽_{𝑞}

for a prime

𝑞

𝑘 𝑎_{𝑘} = (𝑘 \cdot 1) 𝑎_{𝑘}

, and

𝑘 \cdot 1 = 0

exactly when

𝑞

divides

𝑘

Sketch.

𝐷 (𝑝) = 0

if and only if

𝑘 𝑎_{𝑘} = 0

for all

𝑘 \geq 1

. Over

𝔽_{2}

this means

𝑎_{𝑘} = 0

for odd

𝑘

, so

𝑝

is a polynomial in

𝑥^{2}

. Over

𝔽_{3}

𝑝

is a polynomial in

𝑥^{3}

. Over

ℝ

𝑝

is constant.

Let $𝑝 =$ $𝑎_{0} +$ $𝑎_{1} 𝑥 +$ $\dots +$ $𝑎_{𝑛} 𝑥^{𝑛}$ . Then $𝐷 (𝑝) =$ $\sum_{𝑘 = 1}^{𝑛} 𝑘 𝑎_{𝑘} 𝑥^{𝑘 - 1}$ , which is $0$ if and only if $𝑘 𝑎_{𝑘} = 0$ for every $1 \leq 𝑘 \leq 𝑛$ . Here $𝑘 𝑎_{𝑘} = (𝑘 \cdot 1) 𝑎_{𝑘}$ , where $𝑘 \cdot 1$ is the sum of $𝑘$ ones in $𝐹$ . By Proposition 1.1.17, $𝑘 𝑎_{𝑘} = 0$ if and only if $𝑘 \cdot 1 = 0$ or $𝑎_{𝑘} = 0$ .

In $𝔽_{𝑞}$ for a prime $𝑞$ , the sum of $𝑘$ ones is $𝑘 mod 𝑞$ , which is $0$ exactly when $𝑞$ divides $𝑘$ . So over $𝔽_{2}$ , $𝐷 (𝑝) = 0$ if and only if $𝑎_{𝑘} = 0$ for every odd $𝑘$ , that is, $𝑝 =$ $𝑏_{0} +$ $𝑏_{1} 𝑥^{2} +$ $𝑏_{2} 𝑥^{4} +$ $\dots$ for some $𝑏_{𝑖} \in 𝔽_{2}$ . For example $𝐷 (𝑥^{2} + 1) = 2 𝑥 = 0$ . Over $𝔽_{3}$ , $𝐷 (𝑝) = 0$ if and only if $𝑎_{𝑘} = 0$ whenever $3$ does not divide $𝑘$ , that is, $𝑝 =$ $𝑏_{0} +$ $𝑏_{1} 𝑥^{3} +$ $𝑏_{2} 𝑥^{6} +$ $\dots$ . Over $ℝ$ , $𝑘 \cdot 1 = 𝑘 \neq 0$ for $𝑘 \geq 1$ , so $𝐷 (𝑝) = 0$ if and only if $𝑝 = 𝑎_{0}$ is constant.

Exercise 2.1.35.routine Find the error in the following argument.

Claim. Every additive map $𝑇 : ℝ \to ℝ$ is linear.

Proof. Let $𝑐, 𝑥 \in ℝ$ . Then

𝑇 (𝑐 𝑥) = 𝑇 (𝑥 + 𝑥 + \dots + 𝑥) = 𝑇 (𝑥) + 𝑇 (𝑥) + \dots + 𝑇 (𝑥) = 𝑐 𝑇 (𝑥),

so $𝑇$ is homogeneous. Since it is also additive, it is linear.

Solution.

Uses. Theorem 2.1.3 (Checking linearity)

Hint. How many terms does the sum

𝑥 + 𝑥 + \dots + 𝑥

have? For which

𝑐

does that make sense?

Sketch. Writing

𝑐 𝑥

as a sum of

𝑐

copies of

𝑥

only makes sense for positive integers

𝑐

. The argument proves

𝑇 (𝑛 𝑥) = 𝑛 𝑇 (𝑥)

for

𝑛 = 1, 2, 3, \dots

and nothing about

𝑐 = 1 / 2

𝑐 = \sqrt{2}

The first step writes $𝑐 𝑥$ as a sum of $𝑐$ copies of $𝑥$ , which only makes sense when $𝑐$ is a positive integer. For such $𝑐 = 𝑛$ the argument is correct: by induction on $𝑛$ , using additivity, $𝑇 (𝑛 𝑥) =$ $𝑇 ((𝑛 - 1) 𝑥 + 𝑥) =$ $𝑇 ((𝑛 - 1) 𝑥) +$ $𝑇 (𝑥) =$ $𝑛 𝑇 (𝑥)$ . But it says nothing about $𝑐 = 1 / 2$ or $𝑐 = \sqrt{2}$ , and homogeneity requires $𝑇 (𝑐 𝑥) = 𝑐 𝑇 (𝑥)$ for every real $𝑐$ .

The gap is not a technicality. The conclusion $𝑇 (𝑐 𝑥) = 𝑐 𝑇 (𝑥)$ can be rescued for rational $𝑐$ with a further idea (Exercise 2.1.37), but over $ℝ$ the claim is false: there are additive maps $ℝ \to ℝ$ that are not linear (Aside (Additive maps that are not linear)).

Exercise 2.1.36.standard Let $𝑉$ and $𝑊$ be vector spaces over $𝐹$ , and let $𝑇 : 𝑉 \to 𝑊$ be additive. Prove or disprove that $𝑇$ is linear when

(a) $𝐹 = 𝔽_{𝑝}$ for a prime $𝑝$ ;
(b) $𝐹 = ℂ$ .

Solution.

Uses. Proposition 2.1.4 (Linear maps preserve $𝟎$ and negatives), Non-example 2.1.14 (Complex conjugation), Definition 1.1.19 (Integers modulo $𝑛$ )

Hint 1. In

𝔽_{𝑝}

, every element is

1 + 1 + \dots + 1

for some number of terms.

Hint 2. If

𝑐 = 1 + \dots + 1

(

𝑛

terms), then

𝑐 𝐯 = 𝐯 + \dots + 𝐯

. Apply additivity

𝑛 - 1

times. Do not forget

𝑐 = 0

Hint 3. For (b), look at the non-examples in this section.

Sketch. (a) True. Each

𝑐 \in 𝔽_{𝑝}

𝑛 \cdot 1

for some

0 \leq 𝑛 < 𝑝

, so

𝑐 𝐯

is a sum of

𝑛

copies of

𝐯

and additivity gives

𝑇 (𝑐 𝐯) = 𝑐 𝑇 (𝐯)

; the case

𝑛 = 0

𝑇 (𝟎) = 𝟎

, which holds for additive maps. (b) False: complex conjugation

ℂ \to ℂ

is additive but not

ℂ

-linear.

(a) True. Let $𝑐 \in$ $𝔽_{𝑝} =$ ${0, 1, \dots, 𝑝 - 1}$ , and let $𝑛$ be $𝑐$ regarded as an integer. In $𝔽_{𝑝}$ the sum of $𝑛$ ones is $𝑛 mod 𝑝 = 𝑐$ . If $𝑛 = 0$ , then $𝑇 (0 𝐯) =$ $𝑇 (𝟎) =$ $𝟎 =$ $0 𝑇 (𝐯)$ , where $𝑇 (𝟎) = 𝟎$ holds for additive maps by the second proof of Proposition 2.1.4. If $𝑛 \geq 1$ , distributivity and $1 𝐯 = 𝐯$ give $𝑐 𝐯 =$ $(1 + \dots + 1) 𝐯 =$ $𝐯 +$ $\dots +$ $𝐯$ ( $𝑛$ terms), and additivity, applied $𝑛 - 1$ times, gives

𝑇 (𝑐 𝐯) = 𝑇 (𝐯 + \dots + 𝐯) = 𝑇 (𝐯) + \dots + 𝑇 (𝐯) = (1 + \dots + 1) 𝑇 (𝐯) = 𝑐 𝑇 (𝐯) .

So $𝑇$ is homogeneous, hence linear.

(b) False. Complex conjugation $ℂ \to ℂ$ is additive but not $ℂ$ -linear (Non-example 2.1.14).

Exercise 2.1.37.standard Let

𝑉

and

𝑊

be vector spaces over

ℚ

, and let

𝑇 : 𝑉 \to 𝑊

be additive. Show that

𝑇

is linear.

Solution.

Uses. Proposition 2.1.4 (Linear maps preserve $𝟎$ and negatives), Exercise 2.1.35

Hint 1. First show

𝑇 (𝑛 𝐯) = 𝑛 𝑇 (𝐯)

for every integer

𝑛

, positive, zero, or negative.

Hint 2. For

𝑐 = 𝑚 / 𝑛

with

𝑛 \geq 1

, let

𝐰 = 𝑐 𝐯

. Then

𝑛 𝐰 = 𝑚 𝐯

. Apply

𝑇

and use the integer case on both sides.

Sketch. For integers

𝑛 \geq 0

𝑇 (𝑛 𝐯) = 𝑛 𝑇 (𝐯)

by induction, and

𝑇 (- 𝐮) = - 𝑇 (𝐮)

extends this to

𝑛 < 0

. If

𝑐 = 𝑚 / 𝑛

, then

𝑛 (𝑐 𝐯) = 𝑚 𝐯

, so

𝑛 𝑇 (𝑐 𝐯) =

𝑚 𝑇 (𝐯)

; divide by

𝑛

Integers. We first show $𝑇 (𝑛 𝐯) = 𝑛 𝑇 (𝐯)$ for every $𝐯 \in 𝑉$ and every integer $𝑛$ . For $𝑛 = 0$ this is $𝑇 (𝟎) = 𝟎$ , which holds for additive maps by the second proof of Proposition 2.1.4. If it holds for some $𝑛 \geq 0$ , then $𝑇 ((𝑛 + 1) 𝐯) =$ $𝑇 (𝑛 𝐯 + 𝐯) =$ $𝑛 𝑇 (𝐯) +$ $𝑇 (𝐯) =$ $(𝑛 + 1) 𝑇 (𝐯)$ . So it holds for all $𝑛 \geq 0$ . For $𝑛 < 0$ , the second proof of Proposition 2.1.4 also shows $𝑇 (- 𝐮) = - 𝑇 (𝐮)$ for additive $𝑇$ , so $𝑇 (𝑛 𝐯) =$ $𝑇 (- ((- 𝑛) 𝐯)) =$ $- ((- 𝑛) 𝑇 (𝐯)) =$ $𝑛 𝑇 (𝐯)$ .

Rationals. Let $𝑐 = 𝑚 / 𝑛$ with $𝑚, 𝑛 \in ℤ$ and $𝑛 \geq 1$ , and put $𝐰 = 𝑐 𝐯$ . Then $𝑛 𝐰 =$ $(𝑛 𝑐) 𝐯 =$ $𝑚 𝐯$ , so by the integer case

𝑛 𝑇 (𝐰) = 𝑇 (𝑛 𝐰) = 𝑇 (𝑚 𝐯) = 𝑚 𝑇 (𝐯) .

Multiplying by $1 / 𝑛 \in ℚ$ gives $𝑇 (𝑐 𝐯) =$ $𝑇 (𝐰) =$ $(𝑚 / 𝑛) 𝑇 (𝐯) =$ $𝑐 𝑇 (𝐯)$ . So $𝑇$ is homogeneous, hence linear.

Exercise 2.1.38.standard Let

𝐹

be any field. Find a function

𝑇 : 𝐹^{2} \to 𝐹

that is homogeneous but not additive.

Solution.

Uses. Proposition 1.1.17 (No zero divisors)

Hint 1. Homogeneity only relates

𝑇 (𝐯)

to the values of

𝑇

on the line

span (𝐯)

. It says nothing about how different lines through

𝟎

interact.

Hint 2. Let

𝑇

be a linear function on the

𝑥

-axis and

0

off the

𝑥

-axis.

Sketch.

𝑇 (𝑥, 𝑦) = 𝑥

𝑦 = 0

, and

𝑇 (𝑥, 𝑦) = 0

𝑦 \neq 0

. Scaling by

𝑐 \neq 0

keeps a point on or off the

𝑥

-axis, which gives homogeneity. But

𝑇 (1, 1) +

𝑇 (0, - 1) =

0 \neq

1 =

𝑇 (1, 0)

Define $𝑇 (𝑥, 𝑦) = 𝑥$ if $𝑦 = 0$ , and $𝑇 (𝑥, 𝑦) = 0$ if $𝑦 \neq 0$ .

Homogeneity. Let $𝑐 \in 𝐹$ and $(𝑥, 𝑦) \in 𝐹^{2}$ . If $𝑐 = 0$ , then $𝑇 (𝑐 (𝑥, 𝑦)) =$ $𝑇 (0, 0) =$ $0 =$ $𝑐 𝑇 (𝑥, 𝑦)$ . Let $𝑐 \neq 0$ . If $𝑦 = 0$ , then $𝑇 (𝑐 𝑥, 0) =$ $𝑐 𝑥 =$ $𝑐 𝑇 (𝑥, 0)$ . If $𝑦 \neq 0$ , then $𝑐 𝑦 \neq 0$ by Proposition 1.1.17, so $𝑇 (𝑐 𝑥, 𝑐 𝑦) =$ $0 =$ $𝑐 \cdot$ $0 =$ $𝑐 𝑇 (𝑥, 𝑦)$ .

Additivity fails. $𝑇 (1, 1) +$ $𝑇 (0, - 1) =$ $0 +$ $0 =$ $0$ , since $1 \neq 0$ and $- 1 \neq 0$ , while $𝑇 ((1, 1) + (0, - 1)) =$ $𝑇 (1, 0) =$ $1$ .

Homogeneity is a condition on each line through $𝟎$ separately, and additivity is what ties different lines together.

Exercise 2.1.39.standard Let $𝑇 : 𝑉 \to 𝑊$ satisfy $𝑇 (𝟎) = 𝟎$ and

𝑇 (𝑎 𝐮 + 𝑏 𝐯) = 𝑎 𝑇 (𝐮) + 𝑏 𝑇 (𝐯) for all 𝐮, 𝐯 \in 𝑉 and all 𝑎, 𝑏 \in 𝐹 with 𝑎 + 𝑏 = 1 .

(a)Show that if $1 + 1 \neq 0$ in $𝐹$ , then $𝑇$ is linear.
(b)Show that the conclusion of (a) can fail over $𝔽_{2}$ .

Solution.

Uses. Theorem 2.1.3 (Checking linearity), Exercise 2.1.15

Hint 1. For homogeneity, use the pair of coefficients

𝑎

and

1 - 𝑎

, with

𝟎

as the second vector.

Hint 2. For additivity, write

𝐮 +

𝐯 =

(1 + 1) (ℎ 𝐮 + ℎ 𝐯)

with

ℎ = {(1 + 1)}^{- 1}

, and note that

ℎ + ℎ = 1

Hint 3. For (b), what does the hypothesis say when the only scalars are

0

and

1

Sketch. (a)

𝑇 (𝑎 𝐮) =

𝑇 (𝑎 𝐮 + (1 - 𝑎) 𝟎) =

𝑎 𝑇 (𝐮)

. With

ℎ = {(1 + 1)}^{- 1}

, homogeneity and the hypothesis with

𝑎 = 𝑏 = ℎ

give

𝑇 (𝐮 + 𝐯) =

(1 + 1) 𝑇 (ℎ 𝐮 + ℎ 𝐯) =

(1 + 1) (ℎ 𝑇 (𝐮) + ℎ 𝑇 (𝐯)) =

𝑇 (𝐮) +

𝑇 (𝐯)

. (b) Over

𝔽_{2}

𝑎 + 𝑏 = 1

forces

{𝑎, 𝑏} = {0, 1}

, and the hypothesis is empty. So

(𝑥, 𝑦) \mapsto 𝑥 𝑦

𝔽_{2}^{2}

satisfies it but is not linear.

(a) Homogeneity. For $𝑎 \in 𝐹$ and $𝐮 \in 𝑉$ , the coefficients $𝑎$ and $1 - 𝑎$ add up to $1$ , so

𝑇 (𝑎 𝐮) = 𝑇 (𝑎 𝐮 + (1 - 𝑎) 𝟎) = 𝑎 𝑇 (𝐮) + (1 - 𝑎) 𝑇 (𝟎) = 𝑎 𝑇 (𝐮) .

Additivity. Since $1 + 1 \neq 0$ , it has an inverse $ℎ = {(1 + 1)}^{- 1}$ , and $ℎ + ℎ = (1 + 1) ℎ = 1$ . For $𝐮, 𝐯 \in 𝑉$ we have $(1 + 1) (ℎ 𝐮 + ℎ 𝐯) =$ $𝐮 +$ $𝐯$ , so by homogeneity and then the hypothesis with $𝑎 = 𝑏 = ℎ$ ,

𝑇 (𝐮 + 𝐯) = (1 + 1) 𝑇 (ℎ 𝐮 + ℎ 𝐯) = (1 + 1) (ℎ 𝑇 (𝐮) + ℎ 𝑇 (𝐯)) = 𝑇 (𝐮) + 𝑇 (𝐯) .

By Theorem 2.1.3, $𝑇$ is linear.

(b) In $𝔽_{2}$ , $𝑎 + 𝑏 = 1$ holds only for $(𝑎, 𝑏) = (1, 0)$ and $(0, 1)$ , and then the hypothesis reads $𝑇 (𝐮) = 𝑇 (𝐮)$ or $𝑇 (𝐯) = 𝑇 (𝐯)$ , which is always true. So every $𝑇$ with $𝑇 (𝟎) = 𝟎$ satisfies the hypotheses. For example $𝑇 : 𝔽_{2}^{2} \to 𝔽_{2}$ , $𝑇 (𝑥, 𝑦) = 𝑥 𝑦$ , does, and it is not linear (Exercise 2.1.15).

Exercise 2.1.40.routine

(a)Find $𝑆, 𝑇 \in L (ℝ^{2})$ with $𝑆 𝑇 = 0$ but $𝑇 𝑆 \neq 0$ .
(b)Find $𝑇 \in L (ℝ^{2})$ with $𝑇 \neq 0$ but $𝑇^{2} = 0$ .

Solution.

Uses. Example 2.1.7 (Maps $𝐹^{𝑛} \to 𝐹^{𝑚}$ given by linear formulas)

Hint 1.

𝑆 𝑇 = 0

means that

𝑆

sends every vector of the form

𝑇 (𝐯)

𝟎

. Choose

𝑇

with all its values on a line, and

𝑆

that kills that line.

Hint 2. Try

𝑆 (𝑥, 𝑦) = (𝑥, 0)

, the projection onto the

𝑥

-axis, and a

𝑇

whose values lie on the

𝑦

-axis.

Sketch.

𝑆 (𝑥, 𝑦) = (𝑥, 0)

and

𝑇 (𝑥, 𝑦) = (0, 𝑥)

: then

𝑆 𝑇 (𝑥, 𝑦) =

𝑆 (0, 𝑥) =

𝟎

, while

𝑇 𝑆 (𝑥, 𝑦) =

(0, 𝑥)

. The same

𝑇

answers (b):

𝑇^{2} (𝑥, 𝑦) =

𝑇 (0, 𝑥) =

𝟎

Let $𝑆 (𝑥, 𝑦) = (𝑥, 0)$ and $𝑇 (𝑥, 𝑦) = (0, 𝑥)$ ; both are linear by Example 2.1.7.

(a) $(𝑆 𝑇) (𝑥, 𝑦) =$ $𝑆 (0, 𝑥) =$ $(0, 0)$ for all $(𝑥, 𝑦)$ , so $𝑆 𝑇 = 0$ . But $(𝑇 𝑆) (𝑥, 𝑦) =$ $𝑇 (𝑥, 0) =$ $(0, 𝑥)$ , so $(𝑇 𝑆) (𝐞_{1}) =$ $𝐞_{2} \neq$ $𝟎$ and $𝑇 𝑆 \neq 0$ .

(b) The same $𝑇$ works: $𝑇 (𝐞_{1}) = 𝐞_{2} \neq 𝟎$ , and $𝑇^{2} (𝑥, 𝑦) =$ $𝑇 (0, 𝑥) =$ $(0, 0)$ . Geometrically, $𝑇$ moves everything onto the $𝑦$ -axis and sends the $𝑦$ -axis to $𝟎$ .

Exercise 2.1.41.routine Find the error in the following argument.

Claim. If $𝑃 \in L (𝑉)$ satisfies $𝑃^{2} = 𝑃$ , then $𝑃 = 0$ or $𝑃 = {id}_{𝑉}$ .

Proof. Let $𝐯 \in 𝑉$ . Then $𝑃 (𝑃 (𝐯) - 𝐯) =$ $𝑃^{2} (𝐯) -$ $𝑃 (𝐯) =$ $𝟎$ . Since $𝑃$ is linear, either $𝑃 = 0$ or $𝑃 (𝐯) - 𝐯 = 𝟎$ . If $𝑃 \neq 0$ , the second alternative holds for every $𝐯$ , so $𝑃 = {id}_{𝑉}$ .

Solution.

Uses. Example 2.1.11 (Rotations, projections, and reflections), Proposition 2.1.4 (Linear maps preserve $𝟎$ and negatives)

Hint 1. Test the claim on a projection from Example 2.1.11.

Hint 2. Which step would be valid for numbers but not for linear maps? Can a linear map send a nonzero vector to

𝟎

without being

0

Sketch. The step “

𝑃 (𝐰) = 𝟎

implies

𝑃 = 0

𝐰 = 𝟎

” is false. The projection

𝑃 (𝑥, 𝑦) = (𝑥, 0)

satisfies

𝑃^{2} = 𝑃

and is neither

0

nor

id

; there

𝑃 (𝑃 (𝐯) - 𝐯) =

𝑃 (0, - 𝑦) =

𝟎

with

(0, - 𝑦) \neq 𝟎

for

𝑦 \neq 0

The claim is false: $𝑃 (𝑥, 𝑦) = (𝑥, 0)$ (the projection ${Pr}_{ℓ}$ of Example 2.1.11 onto the $𝑥$ -axis) satisfies $𝑃^{2} (𝑥, 𝑦) =$ $𝑃 (𝑥, 0) =$ $(𝑥, 0) =$ $𝑃 (𝑥, 𝑦)$ , and $𝑃$ is neither $0$ nor ${id}_{ℝ^{2}}$ .

The error is the step “since $𝑃$ is linear, either $𝑃 = 0$ or $𝑃 (𝐯) - 𝐯 = 𝟎$ ”. It treats $𝑃 (𝐰) = 𝟎$ like a product of numbers equal to $0$ , but a linear map can send a nonzero vector to $𝟎$ without being the zero map. In the example, $𝑃 (𝐯) - 𝐯 = (0, - 𝑦)$ , which is nonzero when $𝑦 \neq 0$ , and $𝑃 (0, - 𝑦) = 𝟎$ . The vectors that a linear map sends to $𝟎$ are studied in Section 2.6.

Exercise 2.1.42.standard Let $ℓ$ be the line through $𝟎$ at angle $𝜃$ , and let $𝑃 = {Pr}_{ℓ}$ and $𝑆 = 𝑆_{ℓ}$ be as in Example 2.1.11.

(a)Show from the formula for $𝑃$ that $𝑃^{2} = 𝑃$ .
(b)Using only $𝑆 = 2 𝑃 - id$ , (a), and Proposition 2.1.24, show that $𝑆^{2} = id$ and $𝑆 𝑃 = 𝑃 𝑆 = 𝑃$ . Interpret these identities geometrically.

Solution.

Uses. Example 2.1.11 (Rotations, projections, and reflections), Example 2.1.27 (Maps assembled from linear pieces), Proposition 2.1.24 (Algebra of composition)

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

Hint 1. Write

𝑃 (𝑥, 𝑦) =

𝑠 (cos 𝜃, sin 𝜃)

with

𝑠 =

𝑥 cos 𝜃 +

𝑦 sin 𝜃

, and apply

𝑃

to the point

𝑠 (cos 𝜃, sin 𝜃)

Hint 2. Expand

(2 𝑃 - id) (2 𝑃 - id)

as you would expand

{(2 𝑝 - 1)}^{2}

, but justify each step by a part of Proposition 2.1.24.

Sketch. (a)

𝑃

applied to

𝑠 (cos 𝜃, sin 𝜃)

gives

𝑠 ({cos}^{2} 𝜃 + {sin}^{2} 𝜃) (cos 𝜃, sin 𝜃) =

𝑃 (𝑥, 𝑦)

. (b)

𝑆^{2} =

4 𝑃^{2} -

2 𝑃 -

2 𝑃 +

id =

id

and

𝑆 𝑃 =

2 𝑃^{2} -

𝑃 =

𝑃 =

𝑃 𝑆

. Reflecting twice returns every point; a point and its mirror image have the same foot on

ℓ

; points of

ℓ

are fixed by

𝑆

(a) Let $𝐮 = (cos 𝜃, sin 𝜃)$ and $𝑠 =$ $𝑥 cos 𝜃 +$ $𝑦 sin 𝜃$ , so $𝑃 (𝑥, 𝑦) = 𝑠 𝐮$ . By linearity of $𝑃$ , $𝑃 (𝑠 𝐮) = 𝑠 𝑃 (𝐮)$ , and by the formula,

𝑃 (𝐮) = (cos 𝜃 \cdot cos 𝜃 + sin 𝜃 \cdot sin 𝜃) 𝐮 = 𝐮 .

So $𝑃 (𝑃 (𝑥, 𝑦)) =$ $𝑠 𝐮 =$ $𝑃 (𝑥, 𝑦)$ , that is, $𝑃^{2} = 𝑃$ .

(b) Using parts (c), (d), and (b) of Proposition 2.1.24,

\begin{aligned} 𝑆^{2} = (2 𝑃 - id) (2 𝑃 - id) & = (2 𝑃) (2 𝑃) - (2 𝑃) id - id (2 𝑃) + id id \\ = 4 𝑃^{2} - 2 𝑃 - 2 𝑃 + id = 4 𝑃 - 4 𝑃 + id = id, \end{aligned}

where we used $𝑃^{2} = 𝑃$ . Similarly,

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

Geometrically, $𝑆^{2} = id$ says that reflecting twice returns every point to where it started. $𝑃 𝑆 = 𝑃$ says that a point and its mirror image have the same foot of perpendicular on $ℓ$ . $𝑆 𝑃 = 𝑃$ says that $𝑆$ fixes every point of $ℓ$ , since the values of $𝑃$ are exactly the points of $ℓ$ .

Exercise 2.1.43.standard For an angle $𝛼$ , let $𝑅_{𝛼}$ be the rotation by $𝛼$ and $𝑆_{𝛼}$ the reflection across the line through $𝟎$ at angle $𝛼$ (Example 2.1.11). Using the formulas, show that

𝑅_{𝛼} 𝑅_{𝛽} = 𝑅_{𝛼 + 𝛽} and 𝑆_{𝛼} 𝑆_{𝛽} = 𝑅_{2 (𝛼 - 𝛽)} .

Interpret the second identity geometrically.

Solution.

Uses. Example 2.1.11 (Rotations, projections, and reflections), Proposition 2.1.17 (Compositions of linear maps are linear)

Hint 1. Substitute the formula for

𝑅_{𝛽} (𝑥, 𝑦)

into the formula for

𝑅_{𝛼}

, and collect the coefficients of

𝑥

and

𝑦

Hint 2. The coefficients come out as

cos 𝛼 cos 𝛽 -

sin 𝛼 sin 𝛽

and similar expressions. Recognize them with the angle addition formulas. For the reflections, the angles are

2 𝛼

and

2 𝛽

Sketch. The first coordinate of

𝑅_{𝛼} 𝑅_{𝛽} (𝑥, 𝑦)

𝑥 (cos 𝛼 cos 𝛽 - sin 𝛼 sin 𝛽) -

𝑦 (sin 𝛼 cos 𝛽 + cos 𝛼 sin 𝛽) =

𝑥 cos (𝛼 + 𝛽) -

𝑦 sin (𝛼 + 𝛽)

, and the second is similar. For reflections, the coefficients become

cos 2 𝛼 cos 2 𝛽 +

sin 2 𝛼 sin 2 𝛽 =

cos (2 𝛼 - 2 𝛽)

and

sin 2 𝛼 cos 2 𝛽 -

cos 2 𝛼 sin 2 𝛽 =

sin (2 𝛼 - 2 𝛽)

Rotations. Write $(𝑢, 𝑣) =$ $𝑅_{𝛽} (𝑥, 𝑦) =$ $(𝑥 cos 𝛽 - 𝑦 sin 𝛽, 𝑥 sin 𝛽 + 𝑦 cos 𝛽)$ . Then $𝑅_{𝛼} 𝑅_{𝛽} (𝑥, 𝑦) =$ $(𝑢 cos 𝛼 - 𝑣 sin 𝛼, 𝑢 sin 𝛼 + 𝑣 cos 𝛼)$ . The first coordinate is

𝑥 (cos 𝛼 cos 𝛽 - sin 𝛼 sin 𝛽) - 𝑦 (cos 𝛼 sin 𝛽 + sin 𝛼 cos 𝛽) = 𝑥 cos (𝛼 + 𝛽) - 𝑦 sin (𝛼 + 𝛽),

and the second is

𝑥 (sin 𝛼 cos 𝛽 + cos 𝛼 sin 𝛽) + 𝑦 (cos 𝛼 cos 𝛽 - sin 𝛼 sin 𝛽) = 𝑥 sin (𝛼 + 𝛽) + 𝑦 cos (𝛼 + 𝛽) .

So $𝑅_{𝛼} 𝑅_{𝛽} = 𝑅_{𝛼 + 𝛽}$ .

Reflections. Write $(𝑢, 𝑣) =$ $𝑆_{𝛽} (𝑥, 𝑦) =$ $(𝑥 cos 2 𝛽 + 𝑦 sin 2 𝛽, 𝑥 sin 2 𝛽 - 𝑦 cos 2 𝛽)$ . Then $𝑆_{𝛼} 𝑆_{𝛽} (𝑥, 𝑦) =$ $(𝑢 cos 2 𝛼 + 𝑣 sin 2 𝛼, 𝑢 sin 2 𝛼 - 𝑣 cos 2 𝛼)$ . The first coordinate is

𝑥 (cos 2 𝛼 cos 2 𝛽 + sin 2 𝛼 sin 2 𝛽) + 𝑦 (cos 2 𝛼 sin 2 𝛽 - sin 2 𝛼 cos 2 𝛽) = 𝑥 cos (2 𝛼 - 2 𝛽) - 𝑦 sin (2 𝛼 - 2 𝛽),

and the second is

𝑥 (sin 2 𝛼 cos 2 𝛽 - cos 2 𝛼 sin 2 𝛽) + 𝑦 (sin 2 𝛼 sin 2 𝛽 + cos 2 𝛼 cos 2 𝛽) = 𝑥 sin (2 𝛼 - 2 𝛽) + 𝑦 cos (2 𝛼 - 2 𝛽) .

So $𝑆_{𝛼} 𝑆_{𝛽} =$ $𝑅_{2 (𝛼 - 𝛽)}$ .

Geometrically: reflecting across one line and then across another is a rotation about their intersection point by twice the angle from the first line to the second. In particular, reflecting twice across the same line ( $𝛼 = 𝛽$ ) gives $𝑅_{0} = id$ . Also, two reflections never compose to a reflection: $𝑅_{𝜑}$ sends $𝐞_{1}, 𝐞_{2}$ to $(cos 𝜑, sin 𝜑),$ $(- sin 𝜑, cos 𝜑)$ , and $𝑆_{𝛾}$ sends them to $(cos 2 𝛾, sin 2 𝛾),$ $(sin 2 𝛾, - cos 2 𝛾)$ , so $𝑅_{𝜑} = 𝑆_{𝛾}$ would force $cos 𝜑 =$ $cos 2 𝛾 =$ $- cos 2 𝛾$ and $sin 𝜑 =$ $sin 2 𝛾 =$ $- sin 2 𝛾$ , hence $cos 𝜑 = sin 𝜑 = 0$ , which is impossible.

Exercise 2.1.44.routine Fix

𝐯_{0} \in 𝑉

and

𝐰_{0} \in 𝑊

. Show that

{𝑇 \in L (𝑉, 𝑊) | 𝑇 (𝐯_{0}) = 𝐰_{0}}

is a subspace of

L (𝑉, 𝑊)

if and only if

𝐰_{0} = 𝟎

Solution.

Uses. Definition 1.4.1 (Subspace), Theorem 1.4.3 (Subspace criterion), Proposition 2.1.23 ( $L (𝑉, 𝑊)$ is a vector space)

Hint 1. A subspace contains the zero vector. What is the zero vector of

L (𝑉, 𝑊)

, and where does it send

𝐯_{0}

Hint 2. If

𝐰_{0} = 𝟎

, check closure directly, or note that the set is where the linear map

𝑇 \mapsto 𝑇 (𝐯_{0})

of Exercise 2.1.28 takes the value

𝟎

Sketch. The zero map sends

𝐯_{0}

𝟎

, so it lies in the set only if

𝐰_{0} = 𝟎

. If

𝐰_{0} = 𝟎

, then

(𝑐 𝑆 + 𝑇) (𝐯_{0}) =

𝑐 𝑆 (𝐯_{0}) +

𝑇 (𝐯_{0}) =

𝟎

for

𝑆, 𝑇

in the set.

Call the set $𝑈$ . The zero vector of $L (𝑉, 𝑊)$ is the zero map (Proposition 2.1.23), which sends $𝐯_{0}$ to $𝟎$ .

If $𝐰_{0} \neq 𝟎$ , the zero map is not in $𝑈$ , so $𝑈$ is not a subspace.

If $𝐰_{0} = 𝟎$ , the zero map is in $𝑈$ . For $𝑆, 𝑇 \in 𝑈$ and $𝑐 \in 𝐹$ ,

(𝑐 𝑆 + 𝑇) (𝐯_{0}) = 𝑐 𝑆 (𝐯_{0}) + 𝑇 (𝐯_{0}) = 𝑐 𝟎 + 𝟎 = 𝟎,

Taking $𝑐 = 1$ shows that $𝑈$ is closed under addition, and taking $𝑇 = 0$ shows that it is closed under scalar multiplication. So $𝑈$ is a subspace by Theorem 1.4.3.

Exercise 2.1.45.standard Fix $𝐴 \in L (𝑉)$ , and let $𝐶 (𝐴) =$ ${𝑇 \in L (𝑉) | 𝐴 𝑇 = 𝑇 𝐴}$ be the set of operators that commute with $𝐴$ . Show that

(a) $𝐶 (𝐴)$ is a subspace of $L (𝑉)$ ;
(b)if $𝑆, 𝑇 \in 𝐶 (𝐴)$ , then $𝑆 𝑇 \in 𝐶 (𝐴)$ ;
(c) $𝑎_{0} {id}_{𝑉} +$ $𝑎_{1} 𝐴 +$ $\dots +$ $𝑎_{𝑛} 𝐴^{𝑛} \in$ $𝐶 (𝐴)$ for all $𝑎_{0}, \dots, 𝑎_{𝑛} \in 𝐹$ .

Solution.

Uses. Proposition 2.1.24 (Algebra of composition), Theorem 1.4.3 (Subspace criterion), Notation 2.1.18

Hint 1. Every step is an application of Proposition 2.1.24. For (a), expand

𝐴 (𝑐 𝑆 + 𝑇)

and

(𝑐 𝑆 + 𝑇) 𝐴

Hint 2. For (b), move

𝐴

across

𝑆 𝑇

one factor at a time, using associativity.

Hint 3. For (c),

{id}_{𝑉}

and

𝐴

commute with

𝐴

. Use (b) and induction for the powers, then (a).

Sketch. (a)

𝐴 0 = 0 = 0 𝐴

, and

𝐴 (𝑐 𝑆 + 𝑇) =

𝑐 𝐴 𝑆 +

𝐴 𝑇 =

𝑐 𝑆 𝐴 +

𝑇 𝐴 =

(𝑐 𝑆 + 𝑇) 𝐴

. (b)

𝐴 𝑆 𝑇 =

𝑆 𝐴 𝑇 =

𝑆 𝑇 𝐴

. (c)

{id}_{𝑉}, 𝐴 \in 𝐶 (𝐴)

, so every

𝐴^{𝑘} \in 𝐶 (𝐴)

by (b), and linear combinations stay in

𝐶 (𝐴)

by (a).

(a) The zero map commutes with $𝐴$ : $(𝐴 0) (𝐯) =$ $𝐴 (𝟎) =$ $𝟎 =$ $(0 𝐴) (𝐯)$ by Proposition 2.1.4. If $𝑆, 𝑇 \in 𝐶 (𝐴)$ and $𝑐 \in 𝐹$ , then by parts (c) and (d) of Proposition 2.1.24,

𝐴 (𝑐 𝑆 + 𝑇) = 𝑐 (𝐴 𝑆) + 𝐴 𝑇 = 𝑐 (𝑆 𝐴) + 𝑇 𝐴 = (𝑐 𝑆 + 𝑇) 𝐴 .

With $𝑐 = 1$ this shows that $𝐶 (𝐴)$ is closed under addition, and with $𝑇 = 0$ that it is closed under scalar multiplication. So $𝐶 (𝐴)$ is a subspace by Theorem 1.4.3.

(b) By associativity, $𝐴 (𝑆 𝑇) =$ $(𝐴 𝑆) 𝑇 =$ $(𝑆 𝐴) 𝑇 =$ $𝑆 (𝐴 𝑇) =$ $𝑆 (𝑇 𝐴) =$ $(𝑆 𝑇) 𝐴$ .

(c) ${id}_{𝑉} 𝐴 = 𝐴 = 𝐴 {id}_{𝑉}$ , and $𝐴 𝐴 = 𝐴 𝐴$ , so ${id}_{𝑉}, 𝐴 \in 𝐶 (𝐴)$ . If $𝐴^{𝑘} \in 𝐶 (𝐴)$ , then $𝐴^{𝑘 + 1} =$ $𝐴^{𝑘} 𝐴 \in$ $𝐶 (𝐴)$ by (b), so by induction every $𝐴^{𝑘} \in 𝐶 (𝐴)$ . Since $𝐶 (𝐴)$ is a subspace, it contains $𝑎_{0} {id}_{𝑉} +$ $𝑎_{1} 𝐴 +$ $\dots +$ $𝑎_{𝑛} 𝐴^{𝑛}$ .

Exercise 2.1.46.standard Let $𝑀 \in L (𝐹 [𝑥])$ be multiplication by $𝑥$ and $𝐷$ the formal derivative (Example 2.1.9).

(a)Show that $𝐷 𝑀 -$ $𝑀 𝐷 =$ ${id}_{𝐹 [𝑥]}$ .
(b)Show that $𝐷 𝑀^{𝑘} -$ $𝑀^{𝑘} 𝐷 =$ $𝑘 𝑀^{𝑘 - 1}$ for every $𝑘 \geq 1$ , where $𝑘 𝑀^{𝑘 - 1}$ means $(𝑘 \cdot 1) 𝑀^{𝑘 - 1}$ . What does this identity say when both sides are applied to a polynomial $𝑝$ ?

Solution.

Uses. Example 2.1.9 (Maps of polynomials), Proposition 2.1.24 (Algebra of composition), Notation 2.1.18

Hint 1. (a) Apply both sides to

𝑝 = \sum_{𝑗} 𝑎_{𝑗} 𝑥^{𝑗}

and compare coefficients.

Hint 2. (b) Induct on

𝑘

. First check, using only Proposition 2.1.24, that

𝐷 𝑀^{𝑘 + 1} -

𝑀^{𝑘 + 1} 𝐷 =

(𝐷 𝑀^{𝑘} - 𝑀^{𝑘} 𝐷) 𝑀 +

𝑀^{𝑘} (𝐷 𝑀 - 𝑀 𝐷)

Sketch. (a)

𝐷 𝑀 (𝑝) =

\sum (𝑗 + 1) 𝑎_{𝑗} 𝑥^{𝑗}

and

𝑀 𝐷 (𝑝) =

\sum 𝑗 𝑎_{𝑗} 𝑥^{𝑗}

, so the difference is

𝑝

. (b) By the identity in the hint, the induction hypothesis, and (a),

𝐷 𝑀^{𝑘 + 1} -

𝑀^{𝑘 + 1} 𝐷 =

𝑘 𝑀^{𝑘 - 1} 𝑀 +

𝑀^{𝑘} =

(𝑘 + 1) 𝑀^{𝑘}

. Applied to

𝑝

, (b) is the product rule

𝐷 (𝑥^{𝑘} 𝑝) =

𝑘 𝑥^{𝑘 - 1} 𝑝 +

𝑥^{𝑘} 𝐷 (𝑝)

(a) Let $𝑝 =$ $\sum_{𝑗 = 0}^{𝑛} 𝑎_{𝑗} 𝑥^{𝑗}$ . Then $𝑀 (𝑝) =$ $\sum_{𝑗 = 0}^{𝑛} 𝑎_{𝑗} 𝑥^{𝑗 + 1}$ , so $(𝐷 𝑀) (𝑝) =$ $\sum_{𝑗 = 0}^{𝑛} (𝑗 + 1) 𝑎_{𝑗} 𝑥^{𝑗}$ , while $(𝑀 𝐷) (𝑝) =$ $𝑥 \sum_{𝑗 = 1}^{𝑛} 𝑗 𝑎_{𝑗} 𝑥^{𝑗 - 1} =$ $\sum_{𝑗 = 0}^{𝑛} 𝑗 𝑎_{𝑗} 𝑥^{𝑗}$ (the term $𝑗 = 0$ is $0$ ). Since $(𝑗 + 1) 𝑎_{𝑗} -$ $𝑗 𝑎_{𝑗} =$ $𝑎_{𝑗}$ ,

(𝐷 𝑀 - 𝑀 𝐷) (𝑝) = \sum_{𝑗 = 0}^{𝑛} 𝑎_{𝑗} 𝑥^{𝑗} = 𝑝 .

(b) We induct on $𝑘$ . For $𝑘 = 1$ this is (a), since $𝑀^{0} = id$ . Suppose $𝐷 𝑀^{𝑘} -$ $𝑀^{𝑘} 𝐷 =$ $𝑘 𝑀^{𝑘 - 1}$ for some $𝑘 \geq 1$ . By Proposition 2.1.24,

(𝐷 𝑀^{𝑘} - 𝑀^{𝑘} 𝐷) 𝑀 + 𝑀^{𝑘} (𝐷 𝑀 - 𝑀 𝐷) = 𝐷 𝑀^{𝑘 + 1} - 𝑀^{𝑘} 𝐷 𝑀 + 𝑀^{𝑘} 𝐷 𝑀 - 𝑀^{𝑘 + 1} 𝐷 = 𝐷 𝑀^{𝑘 + 1} - 𝑀^{𝑘 + 1} 𝐷 .

By the induction hypothesis and (a), the left side is $𝑘 𝑀^{𝑘 - 1} 𝑀 +$ $𝑀^{𝑘} id =$ $𝑘 𝑀^{𝑘} +$ $𝑀^{𝑘} =$ $(𝑘 + 1) 𝑀^{𝑘}$ , as desired.

Applied to $𝑝$ , (b) says $𝐷 (𝑥^{𝑘} 𝑝) -$ $𝑥^{𝑘} 𝐷 (𝑝) =$ $𝑘 𝑥^{𝑘 - 1} 𝑝$ , that is, $𝐷 (𝑥^{𝑘} 𝑝) =$ $𝑘 𝑥^{𝑘 - 1} 𝑝 +$ $𝑥^{𝑘} 𝐷 (𝑝)$ : the product rule for the product of $𝑥^{𝑘}$ and $𝑝$ . Exercise 2.1.49 proves the product rule in general.

Exercise 2.1.47.standard Regard $ℂ$ as a vector space over $ℝ$ .

(a)Show that for every $ℝ$ -linear $𝑇 : ℂ \to ℂ$ there are unique $𝛼, 𝛽 \in ℂ$ with $𝑇 (𝑧) = 𝛼 𝑧 + 𝛽 𝑧$ for all $𝑧 \in ℂ$ .
(b)Show that such a $𝑇$ is $ℂ$ -linear if and only if $𝛽 = 0$ , if and only if $𝑇 (𝑖) = 𝑖 𝑇 (1)$ .

Solution.

Uses. Non-example 2.1.14 (Complex conjugation), Exercise 2.1.5, Proposition 2.1.17 (Compositions of linear maps are linear), Proposition 2.1.23 ( $L (𝑉, 𝑊)$ is a vector space)

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

Hint 1. An

ℝ

-linear map

𝑇 : ℂ \to ℂ

satisfies

𝑇 (𝑎 + 𝑏 𝑖) =

𝑎 𝑇 (1) +

𝑏 𝑇 (𝑖)

for

𝑎, 𝑏 \in ℝ

. So it is determined by

𝑇 (1)

and

𝑇 (𝑖)

Hint 2. The map

𝑧 \mapsto 𝛼 𝑧 + 𝛽 𝑧

ℝ

-linear. Make its values at

1

and

𝑖

equal

𝑇 (1)

and

𝑇 (𝑖)

: solve

𝛼 + 𝛽 = 𝑇 (1)

and

(𝛼 - 𝛽) 𝑖 = 𝑇 (𝑖)

Sketch. Both

𝑇

and

𝑧 \mapsto 𝛼 𝑧 + 𝛽 𝑧

are

ℝ

-linear, so they agree everywhere as soon as they agree at

1

and

𝑖

, which means

𝛼 + 𝛽 = 𝑇 (1)

and

𝛼 - 𝛽 = - 𝑖 𝑇 (𝑖)

. This system has the unique solution

𝛼 =

(𝑇 (1) - 𝑖 𝑇 (𝑖)) / 2

𝛽 =

(𝑇 (1) + 𝑖 𝑇 (𝑖)) / 2

. Then

𝛽 = 0

if and only if

𝑇 (𝑖) = 𝑖 𝑇 (1)

, which holds when

𝑇

ℂ

-linear; and

𝛽 = 0

makes

𝑇 (𝑧) = 𝛼 𝑧

ℂ

-linear.

(a) For $𝑎, 𝑏 \in ℝ$ , $ℝ$ -linearity gives $𝑇 (𝑎 + 𝑏 𝑖) =$ $𝑎 𝑇 (1) +$ $𝑏 𝑇 (𝑖)$ , so an $ℝ$ -linear map $ℂ \to ℂ$ is determined by its values at $1$ and $𝑖$ . For $𝛼, 𝛽 \in ℂ$ , the map $𝑇_{𝛼, 𝛽} (𝑧) =$ $𝛼 𝑧 +$ $𝛽 𝑧$ is $ℝ$ -linear. Indeed, multiplication by a complex number is $ℂ$ -linear (Exercise 2.1.5), hence $ℝ$ -linear, and conjugation is $ℝ$ -linear (Non-example 2.1.14). So $𝑧 \mapsto 𝛽 𝑧$ , the composite of conjugation and multiplication by $𝛽$ , is $ℝ$ -linear by Proposition 2.1.17, and the sum of two $ℝ$ -linear maps is $ℝ$ -linear by Proposition 2.1.23. (Here $𝛽$ is not a scalar of the real vector space of $ℝ$ -linear maps, so $𝑧 \mapsto 𝛽 𝑧$ is a composite, not a scalar multiple.) Its values are $𝑇_{𝛼, 𝛽} (1) = 𝛼 + 𝛽$ and $𝑇_{𝛼, 𝛽} (𝑖) =$ $(𝛼 - 𝛽) 𝑖$ .

So $𝑇 = 𝑇_{𝛼, 𝛽}$ if and only if $𝛼 + 𝛽 = 𝑇 (1)$ and $(𝛼 - 𝛽) 𝑖 = 𝑇 (𝑖)$ , that is, $𝛼 - 𝛽 = - 𝑖 𝑇 (𝑖)$ . Adding and subtracting these two equations shows that they hold if and only if

𝛼 = \frac{𝑇 (1) - 𝑖 𝑇 (𝑖)}{2}, 𝛽 = \frac{𝑇 (1) + 𝑖 𝑇 (𝑖)}{2} .

So $𝛼$ and $𝛽$ exist and are unique.

(b) From the formula, $𝛽 = 0$ if and only if $𝑇 (1) + 𝑖 𝑇 (𝑖) = 0$ ; multiplying by $- 𝑖$ , this says $𝑇 (𝑖) = 𝑖 𝑇 (1)$ . If $𝑇$ is $ℂ$ -linear, then $𝑇 (𝑖) =$ $𝑇 (𝑖 \cdot 1) =$ $𝑖 𝑇 (1)$ . Conversely, if $𝛽 = 0$ , then $𝑇 (𝑧) = 𝛼 𝑧$ , which is $ℂ$ -linear by Exercise 2.1.5.

Exercise 2.1.48.hard Let

𝑇 \in L (𝑉)

satisfy

𝑇 (𝐯) \in span (𝐯)

for every

𝐯 \in 𝑉

. Show that there is a

𝜆 \in 𝐹

with

𝑇 = 𝜆 {id}_{𝑉}

Solution.

Uses. Proposition 1.2.35 (Zero products), Theorem 1.5.5 (Independence is decided at zero), Definition 2.1.1 (Linear map)

Hint 1. For each nonzero

𝐯

there is a scalar

𝜆_{𝐯}

with

𝑇 (𝐯) = 𝜆_{𝐯} 𝐯

, and it is unique. The task is to show that

𝜆_{𝐯}

does not depend on

𝐯

Hint 2. Take nonzero

𝐮, 𝐯

. If the list

𝐮, 𝐯

is linearly independent, expand

𝑇 (𝐮 + 𝐯)

in two ways and compare coefficients.

Hint 3. If

𝐮, 𝐯

is dependent, then

𝐯 = 𝑐 𝐮

for some

𝑐

; compute

𝑇 (𝐯)

directly.

Sketch. Write

𝑇 (𝐯) = 𝜆_{𝐯} 𝐯

for

𝐯 \neq 𝟎

. If

𝐮, 𝐯

is independent, then

𝜆_{𝐮 + 𝐯} (𝐮 + 𝐯) =

𝜆_{𝐮} 𝐮 +

𝜆_{𝐯} 𝐯

, and comparing coefficients gives

𝜆_{𝐮} = 𝜆_{𝐮 + 𝐯} = 𝜆_{𝐯}

. If

𝐯 = 𝑐 𝐮

, then

𝑇 (𝐯) =

𝑐 𝜆_{𝐮} 𝐮 =

𝜆_{𝐮} 𝐯

. So all

𝜆_{𝐯}

are equal.

If $𝑉 = {𝟎}$ , any $𝜆$ works. Otherwise, for each $𝐯 \neq 𝟎$ choose $𝜆_{𝐯} \in 𝐹$ with $𝑇 (𝐯) = 𝜆_{𝐯} 𝐯$ . It is unique: if $𝜆 𝐯 = 𝜇 𝐯$ , then $(𝜆 - 𝜇) 𝐯 = 𝟎$ , so $𝜆 = 𝜇$ by Proposition 1.2.35. We show $𝜆_{𝐮} = 𝜆_{𝐯}$ for all nonzero $𝐮, 𝐯 \in 𝑉$ .

Case 1: the list $𝐮, 𝐯$ is linearly independent. Then $𝐮 + 𝐯 \neq 𝟎$ (otherwise $1 𝐮 + 1 𝐯 = 𝟎$ would be a nontrivial relation), and

𝜆_{𝐮 + 𝐯} 𝐮 + 𝜆_{𝐮 + 𝐯} 𝐯 = 𝑇 (𝐮 + 𝐯) = 𝑇 (𝐮) + 𝑇 (𝐯) = 𝜆_{𝐮} 𝐮 + 𝜆_{𝐯} 𝐯 .

So $(𝜆_{𝐮 + 𝐯} - 𝜆_{𝐮}) 𝐮 +$ $(𝜆_{𝐮 + 𝐯} - 𝜆_{𝐯}) 𝐯 =$ $𝟎$ , and by Theorem 1.5.5 both coefficients are $0$ . Hence $𝜆_{𝐮} = 𝜆_{𝐮 + 𝐯} = 𝜆_{𝐯}$ .

Case 2: the list $𝐮, 𝐯$ is linearly dependent. By Theorem 1.5.5 there are $𝑎, 𝑏 \in 𝐹$ , not both $0$ , with $𝑎 𝐮 + 𝑏 𝐯 = 𝟎$ . If $𝑏 = 0$ , then $𝑎 \neq 0$ and $𝑎 𝐮 = 𝟎$ would force $𝐮 = 𝟎$ by Proposition 1.2.35; so $𝑏 \neq 0$ , and $𝐯 = 𝑐 𝐮$ with $𝑐 = - 𝑎 / 𝑏$ . Then

𝑇 (𝐯) = 𝑐 𝑇 (𝐮) = 𝑐 𝜆_{𝐮} 𝐮 = 𝜆_{𝐮} (𝑐 𝐮) = 𝜆_{𝐮} 𝐯,

so $𝜆_{𝐯} = 𝜆_{𝐮}$ by uniqueness.

So all the $𝜆_{𝐯}$ equal a common value $𝜆$ , and $𝑇 (𝐯) = 𝜆 𝐯$ for every $𝐯 \neq 𝟎$ . Since also $𝑇 (𝟎) = 𝟎 = 𝜆 𝟎$ , we get $𝑇 = 𝜆 {id}_{𝑉}$ .

Exercise 2.1.49.hard A linear map $𝑇 : 𝐹 [𝑥] \to 𝐹 [𝑥]$ is a derivation if $𝑇 (𝑝 𝑞) =$ $𝑇 (𝑝) 𝑞 +$ $𝑝 𝑇 (𝑞)$ for all $𝑝, 𝑞 \in 𝐹 [𝑥]$ .

(a)Show that the formal derivative $𝐷$ is a derivation (the product rule).
(b)Show that the derivations are exactly the maps $𝑝 \mapsto 𝑟 𝐷 (𝑝)$ for $𝑟 \in 𝐹 [𝑥]$ , and that $𝑟 = 𝑇 (𝑥)$ .

Solution.

Uses. Example 2.1.9 (Maps of polynomials), Proposition 2.1.17 (Compositions of linear maps are linear), Proposition 2.1.23 ( $L (𝑉, 𝑊)$ is a vector space)

Techniques. Check it on a basis (Technique 2.2.7)

Hint 1. For (a), fix

𝑞

. Both sides of

𝐷 (𝑝 𝑞) =

𝐷 (𝑝) 𝑞 +

𝑝 𝐷 (𝑞)

are linear in

𝑝

, so it is enough to check them for

𝑝 = 𝑥^{𝑗}

. Then fix

𝑝 = 𝑥^{𝑗}

and argue the same way in

𝑞

Hint 2. For (b), first find

𝑇 (1)

by applying the rule to

1 = 1 \cdot 1

. Then compute

𝑇 (𝑥^{𝑛})

by induction, writing

𝑥^{𝑛 + 1} = 𝑥^{𝑛} \cdot 𝑥

Hint 3. You should get

𝑇 (𝑥^{𝑛}) =

𝑛 𝑥^{𝑛 - 1} 𝑇 (𝑥)

. Two linear maps that agree on every

𝑥^{𝑛}

agree on every polynomial.

Sketch. (a) For a fixed

𝑞

𝑝 \mapsto 𝐷 (𝑝 𝑞)

𝑝 \mapsto 𝐷 (𝑝) 𝑞

, and

𝑝 \mapsto 𝑝 𝐷 (𝑞)

are linear (composites of

𝐷

and multiplication maps), so it suffices to take

𝑝 = 𝑥^{𝑗}

, and likewise

𝑞 = 𝑥^{𝑘}

; there

(𝑗 + 𝑘) 𝑥^{𝑗 + 𝑘 - 1} =

𝑗 𝑥^{𝑗 - 1} 𝑥^{𝑘} +

𝑥^{𝑗} 𝑘 𝑥^{𝑘 - 1}

. (b)

𝑇 (1) = 𝑇 (1) + 𝑇 (1)

gives

𝑇 (1) = 0

, and induction gives

𝑇 (𝑥^{𝑛}) =

𝑛 𝑥^{𝑛 - 1} 𝑟

with

𝑟 = 𝑇 (𝑥)

. So

𝑇

and

𝑝 \mapsto 𝑟 𝐷 (𝑝)

agree on monomials, hence everywhere. Conversely

𝑟 𝐷

is a derivation by (a).

For $𝑠 \in 𝐹 [𝑥]$ let $𝑀_{𝑠} : 𝐹 [𝑥] \to$ $𝐹 [𝑥]$ be multiplication by $𝑠$ , $𝑀_{𝑠} (𝑝) = 𝑠 𝑝$ . It is linear, since $𝑠 (𝑐 𝑝 + 𝑞) =$ $𝑐 (𝑠 𝑝) +$ $𝑠 𝑞$ in $𝐹 [𝑥]$ . We use one observation twice: if two linear maps $𝐹 [𝑥] \to 𝐹 [𝑥]$ agree on every monomial $𝑥^{𝑗}$ , they agree on every $𝑝 = \sum_{𝑗} 𝑎_{𝑗} 𝑥^{𝑗}$ , since by linearity both send $𝑝$ to $\sum_{𝑗} 𝑎_{𝑗}$ times the common value at $𝑥^{𝑗}$ . (This is the idea of Technique 2.2.7.)

(a) Fix $𝑞$ . The maps $𝑝 \mapsto$ $𝐷 (𝑝 𝑞) =$ $(𝐷 𝑀_{𝑞}) (𝑝)$ , $𝑝 \mapsto$ $𝐷 (𝑝) 𝑞 =$ $(𝑀_{𝑞} 𝐷) (𝑝)$ , and $𝑝 \mapsto$ $𝑝 𝐷 (𝑞) =$ $𝑀_{𝐷 (𝑞)} (𝑝)$ are linear, so $𝑝 \mapsto$ $𝐷 (𝑝) 𝑞 +$ $𝑝 𝐷 (𝑞)$ is linear too. Hence it suffices to prove $𝐷 (𝑥^{𝑗} 𝑞) =$ $𝐷 (𝑥^{𝑗}) 𝑞 +$ $𝑥^{𝑗} 𝐷 (𝑞)$ for every $𝑗 \geq 0$ . Now fix $𝑗$ ; by the same argument with the roles of $𝑝$ and $𝑞$ exchanged, it suffices to take $𝑞 = 𝑥^{𝑘}$ . For $𝑗, 𝑘 \geq 0$ ,

𝐷 (𝑥^{𝑗} 𝑥^{𝑘}) = (𝑗 + 𝑘) 𝑥^{𝑗 + 𝑘 - 1} = 𝑗 𝑥^{𝑗 - 1} 𝑥^{𝑘} + 𝑥^{𝑗} \cdot 𝑘 𝑥^{𝑘 - 1} = 𝐷 (𝑥^{𝑗}) 𝑥^{𝑘} + 𝑥^{𝑗} 𝐷 (𝑥^{𝑘}),

where a term with coefficient $0$ is read as $0$ (so that $𝑗 𝑥^{𝑗 - 1} = 0$ when $𝑗 = 0$ , matching $𝐷 (1) = 0$ ).

(b) Let $𝑇$ be a derivation and $𝑟 = 𝑇 (𝑥)$ . From $1 = 1 \cdot 1$ , $𝑇 (1) =$ $𝑇 (1) \cdot$ $1 +$ $1 \cdot$ $𝑇 (1) =$ $𝑇 (1) +$ $𝑇 (1)$ , so $𝑇 (1) = 0$ . We claim $𝑇 (𝑥^{𝑛}) =$ $𝑛 𝑥^{𝑛 - 1} 𝑟$ for all $𝑛 \geq 1$ . For $𝑛 = 1$ this is the definition of $𝑟$ . If it holds for $𝑛$ , then

𝑇 (𝑥^{𝑛 + 1}) = 𝑇 (𝑥^{𝑛} \cdot 𝑥) = 𝑇 (𝑥^{𝑛}) 𝑥 + 𝑥^{𝑛} 𝑇 (𝑥) = 𝑛 𝑥^{𝑛} 𝑟 + 𝑥^{𝑛} 𝑟 = (𝑛 + 1) 𝑥^{𝑛} 𝑟 .

So $𝑇 (𝑥^{𝑛}) =$ $𝑟 𝐷 (𝑥^{𝑛})$ for all $𝑛 \geq 0$ (for $𝑛 = 0$ both sides are $0$ ). The maps $𝑇$ and $𝑀_{𝑟} 𝐷$ are linear and agree on every monomial, so $𝑇 (𝑝) = 𝑟 𝐷 (𝑝)$ for all $𝑝$ .

Conversely, $𝑝 \mapsto 𝑟 𝐷 (𝑝)$ is linear, being $𝑀_{𝑟} 𝐷$ , and by (a) and commutativity of multiplication in $𝐹 [𝑥]$ ,

𝑟 𝐷 (𝑝 𝑞) = 𝑟 (𝐷 (𝑝) 𝑞 + 𝑝 𝐷 (𝑞)) = (𝑟 𝐷 (𝑝)) 𝑞 + 𝑝 (𝑟 𝐷 (𝑞)) .

So it is a derivation, with value $𝑟 𝐷 (𝑥) = 𝑟$ at $𝑥$ .

Exercise 2.1.50.challenge Let $𝔽$ be $ℝ$ or $ℂ$ , and let $𝑇 \in L (𝔽 [𝑥])$ commute with $𝐷$ , that is, $𝑇 𝐷 = 𝐷 𝑇$ . Show that there are scalars $𝑐_{0}, 𝑐_{1}, 𝑐_{2}, \dots \in 𝔽$ such that

𝑇 (𝑝) = 𝑐_{0} 𝑝 + 𝑐_{1} 𝐷 (𝑝) + 𝑐_{2} 𝐷^{2} (𝑝) + \dots + 𝑐_{𝑛} 𝐷^{𝑛} (𝑝) for every 𝑝 with deg 𝑝 \leq 𝑛 .

(The terms $𝑐_{𝑘} 𝐷^{𝑘} (𝑝)$ with $𝑘 > deg 𝑝$ are $0$ , so the right side does not depend on the choice of $𝑛 \geq deg 𝑝$ .) Conversely, show that every map of this form is linear and commutes with $𝐷$ .

Solution.

Uses. Example 2.1.9 (Maps of polynomials), Proposition 2.1.24 (Algebra of composition), Exercise 2.1.45

Hint 1. Work with

𝑒_{𝑛} = 𝑥^{𝑛} / 𝑛!

for

𝑛 \geq 0

. Then

𝐷 (𝑒_{𝑛}) = 𝑒_{𝑛 - 1}

for

𝑛 \geq 1

and

𝐷 (𝑒_{0}) = 0

, and every polynomial is a combination of the

𝑒_{𝑛}

in exactly one way.

Hint 2. Write

𝑇 (𝑒_{𝑛}) =

\sum_{𝑗} 𝑏_{𝑛, 𝑗} 𝑒_{𝑗}

. Apply

𝐷 𝑇 = 𝑇 𝐷

𝑒_{0}

to see that

𝑇 (𝑒_{0})

is constant, and to

𝑒_{𝑛}

to get

𝑏_{𝑛, 𝑗 + 1} =

𝑏_{𝑛 - 1, 𝑗}

Hint 3. Let

𝑐_{𝑘} = 𝑏_{𝑘, 0}

, the constant term of

𝑇 (𝑒_{𝑘})

. Show by induction on

𝑛

that

𝑇 (𝑒_{𝑛}) =

𝑐_{0} 𝑒_{𝑛} +

𝑐_{1} 𝑒_{𝑛 - 1} +

\dots +

𝑐_{𝑛} 𝑒_{0}

, and note that

𝑒_{𝑛 - 𝑘} =

𝐷^{𝑘} (𝑒_{𝑛})

Sketch. With

𝑒_{𝑛} = 𝑥^{𝑛} / 𝑛!

𝐷

shifts

𝑒_{𝑛}

𝑒_{𝑛 - 1}

. Writing

𝑇 (𝑒_{𝑛}) =

\sum_{𝑗} 𝑏_{𝑛, 𝑗} 𝑒_{𝑗}

, the relation

𝐷 𝑇 (𝑒_{𝑛}) =

𝑇 (𝑒_{𝑛 - 1})

gives

𝑏_{𝑛, 𝑗 + 1} =

𝑏_{𝑛 - 1, 𝑗}

, and

𝐷 𝑇 (𝑒_{0}) = 0

gives

𝑏_{0, 𝑗} = 0

for

𝑗 \geq 1

. By induction,

𝑏_{𝑛, 𝑗} = 𝑐_{𝑛 - 𝑗}

for

𝑗 \leq 𝑛

and

0

for

𝑗 > 𝑛

, where

𝑐_{𝑘} = 𝑏_{𝑘, 0}

. So

𝑇 (𝑒_{𝑛}) =

\sum_{𝑘 \leq 𝑛} 𝑐_{𝑘} 𝑒_{𝑛 - 𝑘} =

\sum_{𝑘 \leq 𝑛} 𝑐_{𝑘} 𝐷^{𝑘} (𝑒_{𝑛})

, and linearity extends this to all

𝑝

Setup. For $𝑛 \geq 0$ let $𝑒_{𝑛} =$ $𝑥^{𝑛} / 𝑛! \in$ $𝔽 [𝑥]$ ; dividing by $𝑛!$ is allowed since $𝑛! \neq 0$ in $ℝ$ and $ℂ$ . Then $𝐷 (𝑒_{0}) = 0$ and $𝐷 (𝑒_{𝑛}) =$ $𝑛 𝑥^{𝑛 - 1} / 𝑛! =$ $𝑒_{𝑛 - 1}$ for $𝑛 \geq 1$ , so $𝐷^{𝑘} (𝑒_{𝑛}) =$ $𝑒_{𝑛 - 𝑘}$ for $𝑘 \leq 𝑛$ and $𝐷^{𝑘} (𝑒_{𝑛}) = 0$ for $𝑘 > 𝑛$ . A polynomial $\sum_{𝑗} 𝑎_{𝑗} 𝑥^{𝑗}$ equals $\sum_{𝑗} (𝑗! 𝑎_{𝑗}) 𝑒_{𝑗}$ , and comparing coefficients of $𝑥^{𝑗}$ shows that this is the only way to write it as a combination of the $𝑒_{𝑗}$ .

For each $𝑛$ write $𝑇 (𝑒_{𝑛}) =$ $\sum_{𝑗 \geq 0} 𝑏_{𝑛, 𝑗} 𝑒_{𝑗}$ , with only finitely many $𝑏_{𝑛, 𝑗}$ nonzero, and let $𝑐_{𝑘} = 𝑏_{𝑘, 0}$ . We claim that for every $𝑛 \geq 0$ ,

𝑏_{𝑛, 𝑗} = 𝑐_{𝑛 - 𝑗} for 0 \leq 𝑗 \leq 𝑛, and 𝑏_{𝑛, 𝑗} = 0 for 𝑗 > 𝑛 .

(2.1.1)

The case $𝑛 = 0$ . Since $𝐷 𝑇 = 𝑇 𝐷$ , $𝐷 (𝑇 (𝑒_{0})) =$ $𝑇 (𝐷 (𝑒_{0})) =$ $𝑇 (0) =$ $0$ . But $𝐷 (𝑇 (𝑒_{0})) =$ $\sum_{𝑗 \geq 1} 𝑏_{0, 𝑗} 𝑒_{𝑗 - 1}$ , and by uniqueness of coefficients $𝑏_{0, 𝑗} = 0$ for all $𝑗 \geq 1$ . Also $𝑏_{0, 0} = 𝑐_{0}$ by definition.

The induction step. Let $𝑛 \geq 1$ and suppose the claim holds for $𝑛 - 1$ . Then

\sum_{𝑗 \geq 0} 𝑏_{𝑛, 𝑗 + 1} 𝑒_{𝑗} = 𝐷 (𝑇 (𝑒_{𝑛})) = 𝑇 (𝐷 (𝑒_{𝑛})) = 𝑇 (𝑒_{𝑛 - 1}) = \sum_{𝑗 \geq 0} 𝑏_{𝑛 - 1, 𝑗} 𝑒_{𝑗},

so $𝑏_{𝑛, 𝑗 + 1} =$ $𝑏_{𝑛 - 1, 𝑗}$ for all $𝑗 \geq 0$ . For $0 \leq 𝑗 \leq 𝑛 - 1$ this gives $𝑏_{𝑛, 𝑗 + 1} =$ $𝑐_{𝑛 - 1 - 𝑗} =$ $𝑐_{𝑛 - (𝑗 + 1)}$ , and for $𝑗 \geq 𝑛$ it gives $𝑏_{𝑛, 𝑗 + 1} = 0$ . Together with $𝑏_{𝑛, 0} = 𝑐_{𝑛}$ , this is (2.1.1) for $𝑛$ .

Conclusion. By (2.1.1),

𝑇 (𝑒_{𝑚}) = \sum_{𝑗 = 0}^{𝑚} 𝑐_{𝑚 - 𝑗} 𝑒_{𝑗} = \sum_{𝑘 = 0}^{𝑚} 𝑐_{𝑘} 𝑒_{𝑚 - 𝑘} = \sum_{𝑘 = 0}^{𝑛} 𝑐_{𝑘} 𝐷^{𝑘} (𝑒_{𝑚}) for every 𝑛 \geq 𝑚,

since $𝑒_{𝑚 - 𝑘} =$ $𝐷^{𝑘} (𝑒_{𝑚})$ for $𝑘 \leq 𝑚$ and $𝐷^{𝑘} (𝑒_{𝑚}) = 0$ for $𝑘 > 𝑚$ . Now let $deg 𝑝 \leq 𝑛$ , and write $𝑝 =$ $\sum_{𝑚 = 0}^{𝑛} 𝑎_{𝑚} 𝑒_{𝑚}$ . By linearity of $𝑇$ and of each $𝐷^{𝑘}$ ,

𝑇 (𝑝) = \sum_{𝑚 = 0}^{𝑛} 𝑎_{𝑚} \sum_{𝑘 = 0}^{𝑛} 𝑐_{𝑘} 𝐷^{𝑘} (𝑒_{𝑚}) = \sum_{𝑘 = 0}^{𝑛} 𝑐_{𝑘} 𝐷^{𝑘} (\sum_{𝑚 = 0}^{𝑛} 𝑎_{𝑚} 𝑒_{𝑚}) = \sum_{𝑘 = 0}^{𝑛} 𝑐_{𝑘} 𝐷^{𝑘} (𝑝) .

Converse. Given scalars $𝑐_{0}, 𝑐_{1}, \dots$ , define $𝑇 (𝑝) =$ $\sum_{𝑘 = 0}^{𝑛} 𝑐_{𝑘} 𝐷^{𝑘} (𝑝)$ for any $𝑛 \geq deg 𝑝$ . For $𝑝, 𝑞 \in 𝔽 [𝑥]$ and $𝑎 \in 𝔽$ , choose one $𝑛$ at least $deg 𝑝$ and $deg 𝑞$ ; then $𝑇 (𝑎 𝑝 + 𝑞) =$ $\sum_{𝑘 \leq 𝑛} 𝑐_{𝑘} 𝐷^{𝑘} (𝑎 𝑝 + 𝑞) =$ $𝑎 𝑇 (𝑝) +$ $𝑇 (𝑞)$ , because each $𝐷^{𝑘}$ is linear. Since $deg 𝐷 (𝑝) \leq deg 𝑝$ ,

𝑇 (𝐷 (𝑝)) = \sum_{𝑘 = 0}^{𝑛} 𝑐_{𝑘} 𝐷^{𝑘 + 1} (𝑝) = 𝐷 (\sum_{𝑘 = 0}^{𝑛} 𝑐_{𝑘} 𝐷^{𝑘} (𝑝)) = 𝐷 (𝑇 (𝑝)) .

So $𝑇$ is linear and commutes with $𝐷$ . (Compare Exercise 2.1.45: when $𝑐_{𝑘} = 0$ for large $𝑘$ , $𝑇$ is a polynomial in $𝐷$ .)

Exercise 2.1.51.hardBeyond 21-242 Let

𝑇 : ℝ \to ℝ

be additive and continuous. Show that

𝑇 (𝑥) = 𝑥 𝑇 (1)

for every

𝑥 \in ℝ

, so that

𝑇

is linear.

Solution.

Uses. Exercise 2.1.37, every real number is a limit of a sequence of rational numbers

Hint 1. Regard

ℝ

as a vector space over

ℚ

. By Exercise 2.1.37,

𝑇

ℚ

-linear. What is

𝑇 (𝑞)

for rational

𝑞

Hint 2. Approximate a real

𝑥

by rationals and use continuity.

Sketch.

𝑇

ℚ

-linear, so

𝑇 (𝑞) =

𝑇 (𝑞 \cdot 1) =

𝑞 𝑇 (1)

for

𝑞 \in ℚ

. If

𝑞_{𝑛} \to 𝑥

with

𝑞_{𝑛} \in ℚ

, continuity gives

𝑇 (𝑥) =

lim 𝑇 (𝑞_{𝑛}) =

lim 𝑞_{𝑛} 𝑇 (1) =

𝑥 𝑇 (1)

Regard $ℝ$ as a vector space over $ℚ$ . Since $𝑇$ is additive, Exercise 2.1.37 shows that $𝑇$ is $ℚ$ -linear, so $𝑇 (𝑞) =$ $𝑇 (𝑞 \cdot 1) =$ $𝑞 𝑇 (1)$ for every $𝑞 \in ℚ$ . Let $𝑥 \in ℝ$ , and choose rationals $𝑞_{𝑛}$ with $𝑞_{𝑛} \to 𝑥$ . Since $𝑇$ is continuous,

𝑇 (𝑥) = lim_{𝑛 \to \infty} 𝑇 (𝑞_{𝑛}) = lim_{𝑛 \to \infty} 𝑞_{𝑛} 𝑇 (1) = 𝑥 𝑇 (1) .

So $𝑇$ is the map $𝑥 \mapsto 𝑎 𝑥$ with $𝑎 = 𝑇 (1)$ , which is $ℝ$ -linear by Exercise 2.1.5.

More practice

The practice bank has 52 exercises on this section’s topics.