2.6 Kernel and image · Matrix Theory

In Section 2.5, a left inverse of $𝑇$ showed that the equation $𝑇 (𝐯) = 𝐰$ has at most one solution, and a right inverse showed that it has at least one. This section studies the two properties themselves. As properties of functions they are injectivity and surjectivity, and each is a statement about every $𝐰 \in 𝑊$ at once. For a linear map, linearity packs each of them into a single subspace: the kernel, which consists of the vectors sent to $𝟎$ , and the image, which consists of the vectors that are reached. In particular, whether $𝑇 (𝐯) = 𝐰$ has at most one solution for every $𝐰$ is decided by the one equation $𝑇 (𝐯) = 𝟎$ .

Throughout, $𝑈$ , $𝑉$ , and $𝑊$ are vector spaces over $𝐹$ , and $𝑇 \in L (𝑉, 𝑊)$ unless said otherwise.

2.6.1Injective, surjective, and bijective maps

The three notions below make sense for any function $𝑓 : 𝑋 \to 𝑌$ between sets, and Section A.1 treats them in that generality. For $𝑦 \in 𝑌$ , an element $𝑥 \in 𝑋$ with $𝑓 (𝑥) = 𝑦$ is a preimage of $𝑦$ . The definitions count preimages.

Definition 2.6.1 (Injective). A function

𝑓 : 𝑋 \to 𝑌

is injective if every

𝑦 \in 𝑌

has at most one preimage. Equivalently, for all

𝑥, 𝑥^{'} \in 𝑋

, if

𝑓 (𝑥) = 𝑓 (𝑥^{'})

, then

𝑥 = 𝑥^{'}

Definition 2.6.2 (Surjective). A function

𝑓 : 𝑋 \to 𝑌

is surjective if every

𝑦 \in 𝑌

has at least one preimage, that is, if for every

𝑦 \in 𝑌

there is some

𝑥 \in 𝑋

with

𝑓 (𝑥) = 𝑦

Definition 2.6.3 (Bijective). A function

𝑓 : 𝑋 \to 𝑌

is bijective if every

𝑦 \in 𝑌

has exactly one preimage, that is, if

𝑓

is both injective and surjective.

The two forms of injectivity say the same thing: if $𝑓 (𝑥) = 𝑓 (𝑥^{'}) = 𝑦$ , then $𝑥$ and $𝑥^{'}$ are both preimages of $𝑦$ , and “at most one” says that they are equal. The second form is the one we use in proofs. Its contrapositive reads: different inputs have different values. Other common names are one-to-one for injective, onto for surjective, and one-to-one correspondence for a bijection.

For a linear map $𝑇$ , the preimages of $𝐰$ are the solutions of the equation $𝑇 (𝐯) = 𝐰$ . So $𝑇$ is injective if and only if this equation has at most one solution for every $𝐰 \in 𝑊$ , surjective if and only if it has at least one, and bijective if and only if it has exactly one. In this language, Proposition 2.5.4 and Proposition 2.5.5 say that a left-invertible linear map is injective and a right-invertible one is surjective, so an invertible linear map is bijective. Conversely, a bijective linear map is invertible, because its inverse function turns out to be linear (Proposition 2.7.7). The converses for one-sided inverses are harder (Remark 2.5.7).

Example 2.6.4 (Injective, surjective, both, neither).

(a) $𝑇 : 𝐹 \to 𝐹^{2}$ , $𝑇 (𝑥) = (𝑥, 0)$ , is injective: if $(𝑥, 0) = (𝑥^{'}, 0)$ , then $𝑥 = 𝑥^{'}$ . It is not surjective, since $(0, 1)$ has no preimage (Example 2.5.3).
(b) $𝑃 : 𝐹^{2} \to 𝐹$ , $𝑃 (𝑥, 𝑦) = 𝑥$ , is surjective, since $𝑥 = 𝑃 (𝑥, 0)$ for every $𝑥 \in 𝐹$ . It is not injective, since $𝑃 (0, 1) = 0 = 𝑃 (0, 0)$ .
(c)The rotation $𝑅_{𝜃}$ of $ℝ^{2}$ is bijective, since it is invertible with inverse $𝑅_{- 𝜃}$ (Example 2.5.14).
(d)The projection ${Pr}_{ℓ}$ onto a line $ℓ$ through $𝟎$ in $ℝ^{2}$ is neither injective nor surjective (Non-example 2.5.15).
(e)On $𝐹 [𝑥]$ , multiplication by $𝑥$ , $𝑀 (𝑝) = 𝑥 𝑝$ (Example 2.1.9), is injective but not surjective. If $𝑝 = \sum_{𝑘} 𝑎_{𝑘} 𝑥^{𝑘}$ , then $𝑥 𝑝 =$ $\sum_{𝑘} 𝑎_{𝑘} 𝑥^{𝑘 + 1}$ has the coefficients of $𝑝$ , moved up by one place, so $𝑥 𝑝 = 𝑥 𝑞$ forces $𝑝 = 𝑞$ . Every value of $𝑀$ has constant term $0$ , so $1$ has no preimage. In the other direction, if $𝐹$ has characteristic $0$ , then differentiation $𝐷$ is surjective but not injective (Example 2.5.24).

In (e), an operator on $𝐹 [𝑥]$ is injective without being surjective, and another is surjective without being injective. For an operator on a finite-dimensional space this cannot happen (Corollary 4.2.18).

2.6.2Kernel and image

Checking injectivity straight from Definition 2.6.1 means comparing $𝑇 (𝐯)$ with $𝑇 (𝐯^{'})$ for every pair of vectors. For a linear map there is a shortcut, since $𝑇 (𝐯) = 𝑇 (𝐯^{'})$ if and only if $𝑇 (𝐯 - 𝐯^{'}) = 𝟎$ . Every coincidence of values is therefore recorded by a vector that $𝑇$ sends to $𝟎$ . Surjectivity, on the other hand, asks which vectors $𝑇$ reaches.

Definition 2.6.5 (Kernel). The kernel of $𝑇 \in L (𝑉, 𝑊)$ is the set of vectors that $𝑇$ sends to $𝟎$ ,

ker 𝑇 = {𝐯 \in 𝑉 | 𝑇 (𝐯) = 𝟎_{𝑊}} \subseteq 𝑉 .

Definition 2.6.6 (Image). The image of $𝑇 \in L (𝑉, 𝑊)$ is the set of values of $𝑇$ ,

im 𝑇 = {𝐰 \in 𝑊 | 𝐰 = 𝑇 (𝐯) for some 𝐯 \in 𝑉} = 𝑇 (𝑉) \subseteq 𝑊 .

The kernel is also called the null space of $𝑇$ , and the image its range. Note where each one lives: $ker 𝑇$ is a subset of the domain $𝑉$ , and $im 𝑇$ is a subset of the target $𝑊$ . Directly from Definition 2.6.2, $𝑇$ is surjective if and only if $im 𝑇 = 𝑊$ . The kernel always contains $𝟎_{𝑉}$ , since $𝑇 (𝟎_{𝑉}) = 𝟎_{𝑊}$ (Proposition 2.1.4).

Example 2.6.7 (First kernels and images).

(a)The zero map $𝑉 \to 𝑊$ has kernel $𝑉$ and image ${𝟎}$ . The identity ${id}_{𝑉}$ has kernel ${𝟎}$ and image $𝑉$ .
(b)For $𝑇 (𝑥) = (𝑥, 0)$ from Example 2.6.4(a), $ker 𝑇 = {0}$ and $im 𝑇 =$ ${(𝑥, 0) | 𝑥 \in 𝐹} =$ $span (𝐞_{1})$ . For $𝑃 (𝑥, 𝑦) = 𝑥$ from part (b) of that example, $ker 𝑃 = span (𝐞_{2})$ and $im 𝑃 = 𝐹$ .
(c)Let $𝑄 : 𝐹^{3} \to 𝐹^{3}$ , $𝑄 (𝑥, 𝑦, 𝑧) =$ $(𝑥, 𝑦, 0)$ . Then $𝑄 (𝑥, 𝑦, 𝑧) = 𝟎$ if and only if $𝑥 = 𝑦 = 0$ , so $ker 𝑄 = span (𝐞_{3})$ , the $𝑧$ -axis. Every value of $𝑄$ has third entry $0$ , and every such vector is a value, since $𝑄 (𝑥, 𝑦, 0) =$ $(𝑥, 𝑦, 0)$ . So $im 𝑄 =$ $span (𝐞_{1}, 𝐞_{2})$ , the $𝑥 𝑦$ -plane.
(d)Let $ℓ = span (𝐮)$ with $𝐮 =$ $(𝑐, 𝑠) =$ $(cos 𝜃, sin 𝜃)$ . By Example 2.1.11, ${Pr}_{ℓ} (𝑥, 𝑦) =$ $(𝑥 𝑐 + 𝑦 𝑠) 𝐮$ . Since $𝐮 \neq 𝟎$ , ${Pr}_{ℓ} (𝑥, 𝑦) = 𝟎$ if and only if $𝑥 𝑐 + 𝑦 𝑠 = 0$ (Proposition 1.2.35). Using $𝑐^{2} + 𝑠^{2} = 1$ , every $(𝑥, 𝑦) \in ℝ^{2}$ satisfies
$(𝑥, 𝑦) = (𝑥 𝑐 + 𝑦 𝑠) (𝑐, 𝑠) + (𝑦 𝑐 - 𝑥 𝑠) (- 𝑠, 𝑐),$
so the solutions of $𝑥 𝑐 + 𝑦 𝑠 = 0$ are exactly the multiples of $(- 𝑠, 𝑐)$ . Hence $ker {Pr}_{ℓ} =$ $span ((- 𝑠, 𝑐))$ , the line through $𝟎$ perpendicular to $ℓ$ . Every value of ${Pr}_{ℓ}$ lies on $ℓ$ , and every $𝑡 𝐮 \in ℓ$ is a value, since ${Pr}_{ℓ} (𝑡 𝐮) =$ $𝑡 (𝑐^{2} + 𝑠^{2}) 𝐮 =$ $𝑡 𝐮$ . So $im {Pr}_{ℓ} = ℓ$ .

Intuition (What the kernel and the image measure). The kernel measures how far $𝑇$ is from being injective, and the image measures how close it comes to being surjective. Two vectors with the same value differ by an element of the kernel, so a large kernel means that $𝑇$ glues many vectors together and loses information. Theorem 2.6.11 below shows that $𝑇$ loses nothing exactly when the kernel is as small as it can be, namely ${𝟎}$ . The image is what $𝑇$ reaches, and $𝑇$ is surjective exactly when the image is as large as it can be, namely $𝑊$ . In Example 2.6.7(d), ${Pr}_{ℓ}$ forgets the component perpendicular to $ℓ$ and reaches only $ℓ$ .

Both sets are subspaces (Proposition 2.6.8), so in finite dimensions their sizes can be measured by dimension. Rank–nullity (Section 4.2) shows that the two measurements are tied together: the more $𝑇$ loses, the less it reaches.

Proposition 2.6.8 (Kernels and images are subspaces). For every

𝑇 \in L (𝑉, 𝑊)

ker 𝑇

is a subspace of

𝑉

and

im 𝑇

is a subspace of

𝑊

Proof.routine

Uses. Theorem 1.4.3 (Subspace criterion), Proposition 2.1.4 (Linear maps preserve $𝟎$ and negatives), Definition 2.1.1 (Linear map), Proposition 1.2.33 (The zero vector)

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

Hint 1. Use part (III) of the subspace criterion (Theorem 1.4.3): check that each set contains the zero vector and is closed under addition and scalar multiplication.

Hint 2. For the kernel, apply

𝑇

𝐮 + 𝐯

and to

𝑎 𝐮

, where

𝑇 (𝐮) = 𝑇 (𝐯) = 𝟎

. For the image, write

𝑇 (𝐮) + 𝑇 (𝐯)

and

𝑎 𝑇 (𝐮)

as single values of

𝑇

Sketch. Both sets contain

𝟎

because

𝑇 (𝟎) = 𝟎

. For

𝐮, 𝐯 \in ker 𝑇

, linearity gives

𝑇 (𝐮 + 𝐯) =

𝟎 +

𝟎 =

𝟎

and

𝑇 (𝑎 𝐮) = 𝑎 𝟎 = 𝟎

. For values,

𝑇 (𝐮) +

𝑇 (𝐯) =

𝑇 (𝐮 + 𝐯)

and

𝑎 𝑇 (𝐮) = 𝑇 (𝑎 𝐮)

The kernel. By Proposition 2.1.4, $𝑇 (𝟎_{𝑉}) = 𝟎_{𝑊}$ , so $𝟎_{𝑉} \in ker 𝑇$ . Let $𝐮, 𝐯 \in ker 𝑇$ and $𝑎 \in 𝐹$ . Since $𝑇$ is linear,

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

where the last step is Proposition 1.2.33. So $𝐮 + 𝐯$ and $𝑎 𝐮$ lie in $ker 𝑇$ , and $ker 𝑇$ is a subspace of $𝑉$ by Theorem 1.4.3.

The image. Again $𝟎_{𝑊} =$ $𝑇 (𝟎_{𝑉}) \in$ $im 𝑇$ . Let $𝐰, 𝐰^{'} \in im 𝑇$ and $𝑎 \in 𝐹$ , say $𝐰 = 𝑇 (𝐮)$ and $𝐰^{'} = 𝑇 (𝐯)$ with $𝐮, 𝐯 \in 𝑉$ . Since $𝑇$ is linear,

𝐰 + 𝐰^{'} = 𝑇 (𝐮 + 𝐯), 𝑎 𝐰 = 𝑇 (𝑎 𝐮),

so both vectors lie in $im 𝑇$ , and $im 𝑇$ is a subspace of $𝑊$ by Theorem 1.4.3. □

The proof used only that ${𝟎}$ is a subspace of $𝑊$ and that $𝑉$ is a subspace of itself. The same argument shows that the image $im (𝑇 ↾_{𝑋})$ of a subspace $𝑋$ of $𝑉$ is a subspace of $𝑊$ , and that the set of $𝐯$ with $𝑇 (𝐯) \in 𝑈$ is a subspace of $𝑉$ for every subspace $𝑈$ of $𝑊$ (Exercise 2.6.51). Linearity cannot be dropped.

Non-example 2.6.9 (Without linearity).

(a)The function $𝑓 : ℝ^{2} \to ℝ$ , $𝑓 (𝑥, 𝑦) = 𝑥 𝑦$ , sends exactly the points of the two coordinate axes to $0$ . This set contains $(1, 0)$ and $(0, 1)$ but not their sum.
(b)The values of $𝑔 : ℝ \to ℝ^{2}$ , $𝑔 (𝑡) = (𝑡, 𝑡^{2})$ , form the parabola $𝑦 = 𝑥^{2}$ . It contains $(1, 1)$ but not $2 (1, 1) = (2, 2)$ , since $2 \neq 2^{2}$ .
(c)An affine map $𝐯 \mapsto 𝑇 (𝐯) + 𝐛$ with $𝐛 \neq 𝟎$ (Non-example 2.1.12) sends $𝟎$ to $𝐛$ , so the set of vectors it sends to $𝟎$ does not contain $𝟎$ .

Exercise 2.6.10.routine Find the kernel and the image of each map, and decide whether it is injective and whether it is surjective.

(a) $𝑇 : ℝ^{2} \to ℝ^{2}$ , $𝑇 (𝑥, 𝑦) =$ $(𝑥 - 𝑦, 2 𝑦 - 2 𝑥)$ .
(b) $𝑆 : 𝐹^{2} \to 𝐹^{3}$ , $𝑆 (𝑥, 𝑦) =$ $(𝑥, 𝑥 + 𝑦, 𝑦)$ .
(c) ${ev}_{1} : {𝐹 [𝑥]}_{\leq 1} \to 𝐹$ , ${ev}_{1} (𝑝) = 𝑝 (1)$ .

Solution.

Uses. Definition 2.6.5 (Kernel), Definition 2.6.6 (Image), Definition 1.3.10 (Span)

Hint 1. For the kernel, set the value equal to

𝟎

and solve. For the image, write a general value as a linear combination of fixed vectors.

Hint 2. (a) Every value is a multiple of

(1, - 2)

. (b) A value

(𝑎, 𝑏, 𝑐)

𝑆

satisfies

𝑏 = 𝑎 + 𝑐

. (c)

{ev}_{1} (𝑎 + 𝑏 𝑥) =

𝑎 +

𝑏

Sketch. (a)

𝑇 (𝑥, 𝑦) =

(𝑥 - 𝑦) (1, - 2)

: kernel

span ((1, 1))

, image

span ((1, - 2))

, neither injective nor surjective. (b)

𝑆 (𝑥, 𝑦) = 𝟎

forces

𝑥 = 𝑦 = 0

; the image is

{(𝑎, 𝑏, 𝑐) | 𝑏 = 𝑎 + 𝑐}

, which misses

(0, 1, 0)

; injective, not surjective. (c) Kernel

span (1 - 𝑥)

, image

𝐹

; surjective, not injective.

(a) $𝑇 (𝑥, 𝑦) =$ $(𝑥 - 𝑦, - 2 (𝑥 - 𝑦)) =$ $(𝑥 - 𝑦) (1, - 2)$ . This is $𝟎$ if and only if $𝑥 = 𝑦$ , so $ker 𝑇 = span ((1, 1))$ , and $im 𝑇 =$ ${𝑡 (1, - 2) | 𝑡 \in ℝ} =$ $span ((1, - 2))$ , since $𝑥 - 𝑦$ takes every real value. As $(1, 1) \neq 𝟎$ lies in the kernel, $𝑇$ is not injective, and as $(1, 0)$ is not a multiple of $(1, - 2)$ , $𝑇$ is not surjective.

(b) If $𝑆 (𝑥, 𝑦) = 𝟎$ , then $𝑥 = 0$ and $𝑦 = 0$ , so $ker 𝑆 = {𝟎}$ and $𝑆$ is injective (by definition: $𝑆 (𝑥, 𝑦) =$ $𝑆 (𝑥^{'}, 𝑦^{'})$ forces $𝑥 = 𝑥^{'}$ and $𝑦 = 𝑦^{'}$ from the first and last entries). Every value $(𝑎, 𝑏, 𝑐) =$ $(𝑥, 𝑥 + 𝑦, 𝑦)$ satisfies $𝑏 = 𝑎 + 𝑐$ , and conversely each $(𝑎, 𝑎 + 𝑐, 𝑐)$ equals $𝑆 (𝑎, 𝑐)$ . So $im 𝑆 =$ ${(𝑎, 𝑏, 𝑐) \in 𝐹^{3} | 𝑏 = 𝑎 + 𝑐} =$ $span ((1, 1, 0), (0, 1, 1))$ . It does not contain $(0, 1, 0)$ , so $𝑆$ is not surjective.

(c) ${ev}_{1} (𝑎 + 𝑏 𝑥) =$ $𝑎 +$ $𝑏$ . This is $0$ if and only if $𝑏 = - 𝑎$ , so $ker {ev}_{1} =$ ${𝑎 - 𝑎 𝑥 | 𝑎 \in 𝐹} =$ $span (1 - 𝑥)$ , and ${ev}_{1}$ is not injective. Every $𝑐 \in 𝐹$ equals ${ev}_{1} (𝑐)$ , so $im {ev}_{1} = 𝐹$ and ${ev}_{1}$ is surjective.

2.6.3Injectivity and the kernel

Intuition. Suppose

𝑇 (𝐯) = 𝑇 (𝐯^{'})

. Then

𝑇 (𝐯 - 𝐯^{'}) = 𝟎

, so the difference

𝐯 - 𝐯^{'}

lies in

ker 𝑇

. Conversely, adding a vector

𝐳 \in ker 𝑇

does not change values:

𝑇 (𝐯 + 𝐳) =

𝑇 (𝐯) +

𝟎 =

𝑇 (𝐯)

. So

𝑇

gives two vectors the same value exactly when they differ by an element of the kernel. A coincidence anywhere in

𝑉

is a coincidence with

𝟎

, moved over, and the coincidences with

𝟎

are recorded by the nonzero vectors of

ker 𝑇

Theorem 2.6.11 (Injectivity is decided at

𝟎

). A linear map

𝑇 \in L (𝑉, 𝑊)

is injective if and only if

ker 𝑇 = {𝟎}

Proof.routine

Uses. Definition 2.6.1 (Injective), Definition 2.6.5 (Kernel), Proposition 2.1.4 (Linear maps preserve $𝟎$ and negatives)

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

Hint 1. For “only if”, recall that

𝟎 \in ker 𝑇

always. If

𝐯 \in ker 𝑇

, compare

𝑇 (𝐯)

with

𝑇 (𝟎)

Hint 2. For “if”, suppose

𝑇 (𝐯) = 𝑇 (𝐯^{'})

and apply

𝑇

𝐯 - 𝐯^{'}

Sketch. If

𝑇

is injective and

𝑇 (𝐯) = 𝟎 = 𝑇 (𝟎)

, then

𝐯 = 𝟎

. If

ker 𝑇 = {𝟎}

and

𝑇 (𝐯) = 𝑇 (𝐯^{'})

, then

𝑇 (𝐯 - 𝐯^{'}) = 𝟎

by linearity, so

𝐯 - 𝐯^{'} = 𝟎

Only if. Suppose $𝑇$ is injective. We always have $𝟎 \in ker 𝑇$ (Proposition 2.1.4). If $𝐯 \in ker 𝑇$ , then $𝑇 (𝐯) = 𝟎 = 𝑇 (𝟎)$ , and injectivity gives $𝐯 = 𝟎$ . So $ker 𝑇 = {𝟎}$ .

If. Suppose $ker 𝑇 = {𝟎}$ , and let $𝐯, 𝐯^{'} \in 𝑉$ with $𝑇 (𝐯) = 𝑇 (𝐯^{'})$ . By Proposition 2.1.4,

𝑇 (𝐯 - 𝐯^{'}) = 𝑇 (𝐯) - 𝑇 (𝐯^{'}) = 𝟎,

so $𝐯 - 𝐯^{'} \in ker 𝑇 = {𝟎}$ . Hence $𝐯 = 𝐯^{'}$ , and $𝑇$ is injective. □

Linearity is essential here. The function $𝑓 : ℝ \to ℝ$ , $𝑓 (𝑥) = 𝑥^{2}$ , sends only $0$ to $0$ , but $𝑓 (1) = 𝑓 (- 1)$ .

Technique 2.6.12 (Test injectivity at $𝟎$ ). To prove that a linear map $𝑇$ is injective, take a vector $𝐯$ with $𝑇 (𝐯) = 𝟎$ and show that $𝐯 = 𝟎$ (Theorem 2.6.11). This replaces a statement about all pairs of vectors by a statement about one equation, and $𝑇 (𝐯) = 𝟎$ is usually the easiest equation to analyze. To prove that $𝑇$ is not injective, exhibit a single nonzero vector of $ker 𝑇$ .

The same move is how injectivity gets used. A relation $𝑎_{1} 𝑇 (𝐯_{1}) +$ $\dots +$ $𝑎_{𝑘} 𝑇 (𝐯_{𝑘}) =$ $𝟎$ can be pulled back through $𝑇$ : by linearity it says $𝑇 (𝑎_{1} 𝐯_{1} + \dots + 𝑎_{𝑘} 𝐯_{𝑘}) =$ $𝟎$ , and if $𝑇$ is injective, then $𝑎_{1} 𝐯_{1} +$ $\dots +$ $𝑎_{𝑘} 𝐯_{𝑘} =$ $𝟎$ .

Reach for it whenever injectivity of a linear map is to be proved or used. Check linearity first: for other functions the test proves nothing, as $𝑥 \mapsto 𝑥^{2}$ shows. The canonical example is Example 2.6.13.

Used in 14 places

Proposition 2.6.14, Exercise 2.6.16, Proposition 2.6.22, Exercise 2.6.37, Exercise 2.8.7, Exercise 2.8.10, Problem 2.1, Problem 2.15, Problem 2.18, Problem 2.19, Exercise 4.4.35, Proposition 6.5.13, Proposition 6.5.21, Corollary 8.4.6.

Example 2.6.13 (Testing at

𝟎

(a)Let $𝑇 : 𝐹 [𝑥] \to 𝐹 [𝑥]$ , $𝑇 (𝑝) = 𝑝 + 𝐷 (𝑝)$ , where $𝐷$ is differentiation. We claim that $𝑇$ is injective over every field. Differentiation lowers the degree of every nonzero polynomial: if $deg 𝑝 = 𝑑 \geq 1$ , then $𝐷 (𝑝)$ has degree at most $𝑑 - 1$ by the formula in Example 2.1.9, and if $𝑝$ is a nonzero constant, then $𝐷 (𝑝) = 𝟎$ , whose degree is $- \infty$ . So for $𝑝 \neq 𝟎$ the degrees of $𝑝$ and $𝐷 (𝑝)$ differ, and Proposition 1.2.18(a) gives $deg (𝑝 + 𝐷 (𝑝)) =$ $deg 𝑝 \geq$ $0$ . In particular $𝑇 (𝑝) \neq 𝟎$ . Hence $ker 𝑇 = {𝟎}$ , and $𝑇$ is injective by Theorem 2.6.11. Differentiation alone is not injective; adding the identity repairs this.
(b)Let $𝑆 : 𝐹^{2} \to 𝐹^{2}$ , $𝑆 (𝑥, 𝑦) =$ $(𝑥 + 𝑦, 𝑥 - 𝑦)$ . If $𝑆 (𝑥, 𝑦) = 𝟎$ , then $𝑥 + 𝑦 = 0$ and $𝑥 - 𝑦 = 0$ . Adding and subtracting the two equations gives $2 𝑥 = 0$ and $2 𝑦 = 0$ , where $2 = 1 + 1$ . If $2 \neq 0$ in $𝐹$ , then $𝑥 = 𝑦 = 0$ , so $𝑆$ is injective. Over $𝔽_{2}$ , $𝑆 (1, 1) =$ $(1 + 1, 1 - 1) =$ $𝟎$ , so $(1, 1)$ is a nonzero vector of $ker 𝑆$ , and $𝑆$ is not injective.

Proposition 2.1.16(b) says that every linear map sends linearly dependent lists to linearly dependent lists. It may also send independent lists to dependent ones, as ${Pr}_{ℓ}$ does with $𝐞_{1}, 𝐞_{2}$ . The injective maps are exactly the ones that never do.

Proposition 2.6.14 (Injective maps preserve independence). For

𝑇 \in L (𝑉, 𝑊)

, the following are equivalent.

(I) $𝑇$ is injective.
(II)For every linearly independent list $ℒ︀ = 𝐯_{1}, \dots, 𝐯_{𝑘}$ in $𝑉$ , the list $𝑇 (ℒ︀) =$ $𝑇 (𝐯_{1}),$ $\dots,$ $𝑇 (𝐯_{𝑘})$ is linearly independent.

Proof.routine

Uses. Theorem 2.6.11 (Injectivity is decided at $𝟎$ ), Theorem 1.5.5 (Independence is decided at zero), Proposition 1.5.15 (Short lists, zeros, and repeats), Definition 2.1.1 (Linear map)

Techniques. Test injectivity at $𝟎$ (Technique 2.6.12), Contrapositive (Technique 1.5.4)

Hint 1. For (I)

\Rightarrow

(II), start from a relation

\sum_{𝑖} 𝑎_{𝑖} 𝑇 (𝐯_{𝑖}) =

𝟎

and pull it back through

𝑇

Hint 2. For (II)

\Rightarrow

(I), work with the kernel. Which list could you feed to (II) to learn something about a vector

𝐯 \in ker 𝑇

Hint 3. A single nonzero vector forms a linearly independent list, and the list

𝟎

is linearly dependent.

Sketch. (I)

\Rightarrow

(II):

\sum_{𝑖} 𝑎_{𝑖} 𝑇 (𝐯_{𝑖}) =

𝟎

gives

𝑇 (\sum_{𝑖} 𝑎_{𝑖} 𝐯_{𝑖}) =

𝟎

, so

\sum_{𝑖} 𝑎_{𝑖} 𝐯_{𝑖} = 𝟎

, and all

𝑎_{𝑖} = 0

. (II)

\Rightarrow

(I): if

ker 𝑇

contained some

𝐯 \neq 𝟎

, the list

𝐯

would be independent while its image, the list

𝟎

, is not.

(I) $\Rightarrow$ (II). Suppose $𝑇$ is injective, let $ℒ︀$ be linearly independent, and suppose that $𝑎_{1} 𝑇 (𝐯_{1}) +$ $\dots +$ $𝑎_{𝑘} 𝑇 (𝐯_{𝑘}) =$ $𝟎$ . By linearity, $𝑇 (𝑎_{1} 𝐯_{1} + \dots + 𝑎_{𝑘} 𝐯_{𝑘}) =$ $𝟎$ , so $𝑎_{1} 𝐯_{1} +$ $\dots +$ $𝑎_{𝑘} 𝐯_{𝑘} \in$ $ker 𝑇 =$ ${𝟎}$ by Theorem 2.6.11. As $ℒ︀$ is linearly independent, $𝑎_{1} = \dots = 𝑎_{𝑘} = 0$ (Theorem 1.5.5). So $𝟎$ has only the trivial representation by $𝑇 (ℒ︀)$ , and $𝑇 (ℒ︀)$ is linearly independent by Theorem 1.5.5.

(II) $\Rightarrow$ (I). We prove the contrapositive. Suppose $𝑇$ is not injective. By Theorem 2.6.11 there is a vector $𝐯 \neq 𝟎$ in $ker 𝑇$ . The list $𝐯$ is linearly independent, but its image is the list $𝑇 (𝐯) = 𝟎$ , which is linearly dependent (Proposition 1.5.15). So (II) fails. □

Composites

Under composition, kernels can only grow and images can only shrink.

Proposition 2.6.15 (Kernels and images of composites). Let

𝑆 \in L (𝑈, 𝑉)

and

𝑇 \in L (𝑉, 𝑊)

(a) $ker 𝑆 \subseteq ker (𝑇 𝑆)$ , with equality if $𝑇$ is injective.
(b) $im (𝑇 𝑆) =$ $im (𝑇 ↾_{im 𝑆}) \subseteq$ $im 𝑇$ , with equality if $𝑆$ is surjective.
(c)If $𝑇 𝑆$ is injective, then $𝑆$ is injective. If $𝑇 𝑆$ is surjective, then $𝑇$ is surjective.

Proof.routine

Uses. Definition 2.6.5 (Kernel), Definition 2.6.6 (Image), Proposition 2.1.4 (Linear maps preserve $𝟎$ and negatives), Theorem 2.6.11 (Injectivity is decided at $𝟎$ )

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

Hint 1. For (a): if

𝑆 (𝐮) = 𝟎

, what is

𝑇 (𝑆 (𝐮))

? For the equality, if

𝑇 (𝑆 (𝐮)) = 𝟎

, which vector lies in

ker 𝑇

Hint 2. For (b), unwind the definitions: a value of

𝑇 𝑆

𝑇

applied to a value of

𝑆

Hint 3. (c) follows from (a) and (b):

ker 𝑆 \subseteq ker (𝑇 𝑆)

and

im (𝑇 𝑆) \subseteq im 𝑇

Sketch. (a)

𝑇 (𝑆 (𝐮)) =

𝑇 (𝟎) =

𝟎

for

𝐮 \in ker 𝑆

; if

𝑇

is injective and

𝑇 (𝑆 (𝐮)) = 𝟎

, then

𝑆 (𝐮) \in ker 𝑇 = {𝟎}

. (b) Both

im (𝑇 𝑆)

and

im (𝑇 ↾_{im 𝑆})

are the set of all

𝑇 (𝑆 (𝐮))

, which lie in

im 𝑇

; if

im 𝑆 = 𝑉

, then

𝑇 ↾_{im 𝑆} = 𝑇

. (c)

ker 𝑆 \subseteq

ker (𝑇 𝑆) =

{𝟎}

, and

𝑊 = im (𝑇 𝑆) \subseteq im 𝑇

(a) If $𝐮 \in ker 𝑆$ , then $(𝑇 𝑆) (𝐮) =$ $𝑇 (𝟎) =$ $𝟎$ by Proposition 2.1.4, so $𝐮 \in ker (𝑇 𝑆)$ . If $𝑇$ is injective and $𝐮 \in ker (𝑇 𝑆)$ , then $𝑆 (𝐮) \in ker 𝑇 = {𝟎}$ by Theorem 2.6.11, so $𝐮 \in ker 𝑆$ .

(b) By definition, $im (𝑇 𝑆) =$ ${𝑇 (𝑆 (𝐮)) | 𝐮 \in 𝑈}$ , and

im (𝑇 ↾_{im 𝑆}) = {𝑇 (𝐯) | 𝐯 \in im 𝑆} = {𝑇 (𝑆 (𝐮)) | 𝐮 \in 𝑈}

as well. Each $𝑇 (𝑆 (𝐮))$ is a value of $𝑇$ , so this set is contained in $im 𝑇$ . If $𝑆$ is surjective, then $im 𝑆 = 𝑉$ , so $𝑇 ↾_{im 𝑆} = 𝑇$ and the two images agree.

(c) If $𝑇 𝑆$ is injective, then by (a) and Theorem 2.6.11, $ker 𝑆 \subseteq$ $ker (𝑇 𝑆) =$ ${𝟎}$ . So $ker 𝑆 = {𝟎}$ , and $𝑆$ is injective. If $𝑇 𝑆$ is surjective, then by (b), $𝑊 =$ $im (𝑇 𝑆) \subseteq$ $im 𝑇 \subseteq$ $𝑊$ , so $im 𝑇 = 𝑊$ . □

Part (c) is the counterpart of Proposition 2.5.26(b) for injectivity and surjectivity, and it cannot be improved. With $𝐴 = (\begin{matrix} 1 & 0 \end{matrix})$ and $𝐵 = (\begin{matrix} 1 \\ 0 \end{matrix})$ , the composite $𝑇_{𝐴} 𝑇_{𝐵} =$ $𝑇_{𝐴 𝐵} =$ ${id}_{𝐹}$ is bijective, but $𝑇_{𝐴} (𝑥, 𝑦) = 𝑥$ is not injective and $𝑇_{𝐵} (𝑥) = (𝑥, 0)$ is not surjective (Example 2.5.27).

Exercise 2.6.16.routine

(a)Show that $𝑇 : 𝐹^{2} \to 𝐹^{3}$ , $𝑇 (𝑥, 𝑦) =$ $(𝑥, 𝑥 + 𝑦, 𝑥 + 2 𝑦)$ , is injective over every field.
(b)Show that $𝑇 : 𝐹 [𝑥] \to 𝐹 [𝑥]$ , $𝑇 (𝑝) = 𝑥 𝑝 + 𝑝$ , is injective. (Look at the coefficient of the highest power of $𝑥$ in $𝑇 (𝑝)$ .)
(c)Let $𝑇 \in L (𝑉)$ with $𝑇^{2} = 0$ . Show that $𝑇$ is not injective unless $𝑉 = {𝟎}$ .

Solution.

Uses. Theorem 2.6.11 (Injectivity is decided at $𝟎$ ), Definition 2.6.5 (Kernel)

Techniques. Test injectivity at $𝟎$ (Technique 2.6.12)

Hint 1. In each part, use the kernel: suppose the value is

𝟎

and deduce that the input is

𝟎

, or find a nonzero input with value

𝟎

Hint 2. (b) If

𝑝 = 𝑎_{0} + \dots + 𝑎_{𝑑} 𝑥^{𝑑}

with

𝑎_{𝑑} \neq 0

, the coefficient of

𝑥^{𝑑 + 1}

𝑥 𝑝 + 𝑝

𝑎_{𝑑}

. (c) If

𝑇 \neq 0

, pick

𝐯

with

𝑇 (𝐯) \neq 𝟎

and look at

𝑇 (𝐯)

Sketch. (a)

𝑇 (𝑥, 𝑦) = 𝟎

gives

𝑥 = 0

and then

𝑦 = 0

. (b) For

𝑝 \neq 0

of degree

𝑑

𝑇 (𝑝)

has coefficient

𝑎_{𝑑} \neq 0

𝑥^{𝑑 + 1}

. (c) If

𝑇 = 0

, any nonzero vector is in the kernel; otherwise

𝑇 (𝐯) \neq 𝟎

for some

𝐯

, and

𝑇 (𝑇 (𝐯)) = 𝟎

(a) If $𝑇 (𝑥, 𝑦) = 𝟎$ , the first entry gives $𝑥 = 0$ , and then the second gives $𝑦 = 0$ . So $ker 𝑇 = {𝟎}$ , and $𝑇$ is injective by Theorem 2.6.11. No division was needed, so this holds over every field.

(b) Let $𝑝 =$ $𝑎_{0} +$ $𝑎_{1} 𝑥 +$ $\dots +$ $𝑎_{𝑑} 𝑥^{𝑑}$ with $𝑎_{𝑑} \neq 0$ . Then $𝑥 𝑝 +$ $𝑝 =$ $𝑎_{0} +$ $(𝑎_{0} + 𝑎_{1}) 𝑥 +$ $\dots +$ $(𝑎_{𝑑 - 1} + 𝑎_{𝑑}) 𝑥^{𝑑} +$ $𝑎_{𝑑} 𝑥^{𝑑 + 1}$ , whose coefficient of $𝑥^{𝑑 + 1}$ is $𝑎_{𝑑} \neq 0$ . So $𝑇 (𝑝) \neq 𝟎$ for every $𝑝 \neq 𝟎$ , that is, $ker 𝑇 = {𝟎}$ , and $𝑇$ is injective.

(c) Suppose $𝑉 \neq {𝟎}$ . If $𝑇 = 0$ , then every nonzero vector of $𝑉$ lies in $ker 𝑇$ . Otherwise there is $𝐯$ with $𝐰 = 𝑇 (𝐯) \neq 𝟎$ , and $𝑇 (𝐰) = 𝑇^{2} (𝐯) = 𝟎$ , so $𝐰$ is a nonzero vector of $ker 𝑇$ . In either case $ker 𝑇 \neq {𝟎}$ , and $𝑇$ is not injective.

2.6.4Null space and column space

A matrix $𝐴 \in 𝐹^{𝑚 \times 𝑛}$ gives the linear map $𝑇_{𝐴} : 𝐹^{𝑛} \to 𝐹^{𝑚}$ , $𝐱 \mapsto 𝐴 𝐱$ (Definition 2.3.15), and the kernel and image of $𝑇_{𝐴}$ have names of their own.

Definition 2.6.17 (Null space and column space). Let $𝐴 \in 𝐹^{𝑚 \times 𝑛}$ . The null space of $𝐴$ is

ker 𝑇_{𝐴} = {𝐱 \in 𝐹^{𝑛} | 𝐴 𝐱 = 𝟎},

a subspace of $𝐹^{𝑛}$ , and the column space of $𝐴$ is

im 𝑇_{𝐴} = {𝐴 𝐱 | 𝐱 \in 𝐹^{𝑛}},

a subspace of $𝐹^{𝑚}$ . We sometimes abbreviate them to $ker 𝐴$ and $im 𝐴$ .

By Proposition 3.1.8, the null space of $𝐴$ is the solution set of the homogeneous system $𝐴 𝐱 = 𝟎$ , and the system $𝐴 𝐱 = 𝐲$ has a solution exactly when $𝐲$ lies in the column space. The name “column space” comes from the following description.

Proposition 2.6.18 (The column space is spanned by the columns). Let $𝐴 \in 𝐹^{𝑚 \times 𝑛}$ have columns $𝐚_{1}, \dots, 𝐚_{𝑛}$ . Then

im 𝑇_{𝐴} = span (𝐚_{1}, \dots, 𝐚_{𝑛}) .

Proof (from the definition of $𝐴 𝐱$ ).routine

Uses. Definition 2.3.8 (Matrix–vector product), Definition 1.3.10 (Span), Definition 2.6.6 (Image)

Hint 1. What is

𝐴 𝐱

, by definition, in terms of the columns of

𝐴

Hint 2. As

𝐱

runs over

𝐹^{𝑛}

, its entries

𝑥_{1}, \dots, 𝑥_{𝑛}

run over all lists of

𝑛

scalars.

Sketch.

𝐴 𝐱 =

𝑥_{1} 𝐚_{1} +

\dots +

𝑥_{𝑛} 𝐚_{𝑛}

, and as

𝐱

runs over

𝐹^{𝑛}

these are all the linear combinations of the columns.

By Definition 2.3.8,

𝐴 𝐱 =

𝑥_{1} 𝐚_{1} +

\dots +

𝑥_{𝑛} 𝐚_{𝑛}

for

𝐱 = (𝑥_{1}, \dots, 𝑥_{𝑛})

. Every list of scalars

𝑥_{1}, \dots, 𝑥_{𝑛}

is the list of entries of some

𝐱 \in 𝐹^{𝑛}

, so the values of

𝑇_{𝐴}

are exactly the linear combinations of

𝐚_{1}, \dots, 𝐚_{𝑛}

. By Definition 1.3.10, the set of these is

span (𝐚_{1}, \dots, 𝐚_{𝑛})

. □

Proof (from a spanning list).routine

Uses. Proposition 2.1.16 (Linear maps preserve spans and linear relations), Proposition 2.3.13 (Linearity of the product), Proposition 1.6.7 (Standard bases)

Techniques. Check it on a basis (Technique 2.2.7)

Hint 1. The standard basis

𝐞_{1}, \dots, 𝐞_{𝑛}

spans

𝐹^{𝑛}

. What does Proposition 2.1.16(a) say about the image of

𝑇_{𝐴}

Hint 2.

𝐴 𝐞_{𝑗} = 𝐚_{𝑗}

by Proposition 2.3.13(c).

Sketch. Since

𝐞_{1}, \dots, 𝐞_{𝑛}

spans

𝐹^{𝑛}

im 𝑇_{𝐴} =

span (𝐴 𝐞_{1}, \dots, 𝐴 𝐞_{𝑛}) =

span (𝐚_{1}, \dots, 𝐚_{𝑛})

The standard basis $𝐞_{1}, \dots, 𝐞_{𝑛}$ spans $𝐹^{𝑛}$ (Proposition 1.6.7). By Proposition 2.1.16(a) and Proposition 2.3.13(c),

im 𝑇_{𝐴} = span (𝐴 𝐞_{1}, \dots, 𝐴 𝐞_{𝑛}) = span (𝐚_{1}, \dots, 𝐚_{𝑛}) .

□

The first proof is a reading of the definition of the product. The second uses nothing about matrices except $𝐴 𝐞_{𝑗} = 𝐚_{𝑗}$ , so it proves more: for every $𝑇 \in L (𝑉, 𝑊)$ and every list $ℒ︀$ that spans $𝑉$ ,

im 𝑇 = span 𝑇 (ℒ︀) .

(2.6.1)

That is, the image of a linear map is spanned by the values on any spanning list. Section 2.6.5 builds on this.

Example 2.6.19 (Computing a null space and a column space).

(a)Let $𝐴 = (\begin{matrix} 1 & 2 & 3 \\ 2 & 4 & 7 \end{matrix})$ , over any field $𝐹$ .
Null space. A vector $𝐱$ lies in $ker 𝑇_{𝐴}$ if and only if $𝑥_{1} + 2 𝑥_{2} + 3 𝑥_{3} = 0$ and $2 𝑥_{1} + 4 𝑥_{2} + 7 𝑥_{3} = 0$ . If $𝐱$ satisfies both, then subtracting twice the first equation from the second gives $𝑥_{3} = 0$ , and then the first gives $𝑥_{1} = - 2 𝑥_{2}$ . Conversely, $(- 2 𝑡, 𝑡, 0)$ satisfies both equations for every $𝑡 \in 𝐹$ . So $ker 𝑇_{𝐴} =$ $span ((- 2, 1, 0))$ .
Column space. By Proposition 2.6.18, $im 𝑇_{𝐴} =$ $span ((1, 2), (2, 4), (3, 7))$ . The second column is twice the first, so it can be dropped without changing the span (Corollary 1.3.20(c)). The remaining two columns span all of $𝐹^{2}$ , since
$7 (1, 2) - 2 (3, 7) = (1, 0), - 3 (1, 2) + (3, 7) = (0, 1),$
and a subspace containing $𝐞_{1}$ and $𝐞_{2}$ contains $span (𝐞_{1}, 𝐞_{2}) = 𝐹^{2}$ (Proposition 1.4.14). So $𝑇_{𝐴}$ is surjective but not injective: every system $𝐴 𝐱 = 𝐲$ has solutions, and Section 2.6.7 shows that the solutions of each one form a translate of the line $ker 𝑇_{𝐴}$ .
(b)Let $𝐴 =$ $(\begin{matrix} 1 & 2 𝑖 \\ 𝑖 & - 2 \end{matrix}) \in$ $ℂ^{2 \times 2}$ . Its columns are $𝐚_{1} = (1, 𝑖)$ and $𝐚_{2} =$ $(2 𝑖, - 2) =$ $2 𝑖 𝐚_{1}$ , so $im 𝑇_{𝐴} =$ $span (𝐚_{1}, 2 𝑖 𝐚_{1}) =$ $span ((1, 𝑖))$ , a line in $ℂ^{2}$ . Also, $𝐴 𝐱 =$ $𝑥_{1} 𝐚_{1} +$ $𝑥_{2} 𝐚_{2} =$ $(𝑥_{1} + 2 𝑖 𝑥_{2}) 𝐚_{1}$ , which is $𝟎$ if and only if $𝑥_{1} + 2 𝑖 𝑥_{2} = 0$ , since $𝐚_{1} \neq 𝟎$ (Proposition 1.2.35). So $ker 𝑇_{𝐴} =$ ${(- 2 𝑖 𝑡, 𝑡) | 𝑡 \in ℂ} =$ $span ((- 2 𝑖, 1))$ , and $𝑇_{𝐴}$ is neither injective nor surjective.

For larger matrices both computations are done by Gauss–Jordan elimination. Theorem 3.4.9 writes every vector of the null space as a linear combination of the basic solutions of $𝐴$ , one for each free variable, and Section 4.2 reads off from the reduced matrix which columns of $𝐴$ suffice to span the column space.

In the span explorer below, the vectors $𝐯_{1}, 𝐯_{2}, 𝐯_{3}$ are the columns of

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

so their span is the column space of $𝐴$ , the plane ${(𝑎, 𝑏, 𝑐) | 𝑐 = 𝑎 + 𝑏}$ in $ℝ^{3}$ . The explorer finds the linear relation $𝐯_{1} + 𝐯_{2} - 𝐯_{3} = 𝟎$ , which says that $(1, 1, - 1) \in ker 𝑇_{𝐴}$ . For a target $𝐰$ in the plane it shows coefficients $𝑥_{1}, 𝑥_{2}, 𝑥_{3}$ with $𝑥_{1} 𝐯_{1} +$ $𝑥_{2} 𝐯_{2} +$ $𝑥_{3} 𝐯_{3} =$ $𝐰$ , and $(𝑥_{1}, 𝑥_{2}, 𝑥_{3})$ is a solution of $𝐴 𝐱 = 𝐰$ . Move the target off the plane and the system becomes inconsistent. Drag $𝐯_{3}$ off the plane and the columns span $ℝ^{3}$ : then $𝑇_{𝐴}$ is surjective, and the explorer finds no relation, so $𝑇_{𝐴}$ is also injective (Corollary 2.6.24).

\begin{matrix} 𝐯_{1} + 𝐯_{2} - 𝐯_{3} = 𝟎, (1, - 1, 0) = 𝐯_{1} - 𝐯_{2}, \\ im 𝑇_{𝐴} = {(𝑎, 𝑏, 𝑐) \in ℝ^{3} | 𝑐 = 𝑎 + 𝑏} . \end{matrix}

The columns

𝐯_{1}, 𝐯_{2}, 𝐯_{3}

𝐴

span the plane

𝑐 = 𝑎 + 𝑏

, the column space of

𝐴

. The relation

𝐯_{1} + 𝐯_{2} - 𝐯_{3} = 𝟎

gives the vector

(1, 1, - 1)

ker 𝑇_{𝐴}

, and the target

𝐰 =

(1, - 1, 0) =

𝐯_{1} -

𝐯_{2}

lies in the column space, so

𝐴 𝐱 = 𝐰

is consistent.

Kernels and images through a matrix

By Proposition 2.3.16, ${[𝑇_{𝐴}]}_{ℰ︀ ℰ︀} = 𝐴$ , so the null space and the column space of $𝐴$ are the kernel and the image of the linear map whose matrix in the standard bases is $𝐴$ . In the other direction, the kernel and the image of any linear map between spaces with bases can be computed from any matrix of the map.

Proposition 2.6.20 (Kernel and image in coordinates). Let

ℬ︀

be a basis of

𝑉

of length

𝑛

and

𝒞︀

a basis of

𝑊

of length

𝑚

, let

𝑇 \in L (𝑉, 𝑊)

, and let

𝐴 = {[𝑇]}_{𝒞︀ ℬ︀}

(a)For $𝐯 \in 𝑉$ : $𝐯 \in ker 𝑇$ if and only if ${[𝐯]}_{ℬ︀} \in ker 𝑇_{𝐴}$ .
(b)For $𝐰 \in 𝑊$ : $𝐰 \in im 𝑇$ if and only if ${[𝐰]}_{𝒞︀} \in im 𝑇_{𝐴}$ .
(c) $𝑇$ is injective if and only if $𝑇_{𝐴}$ is, and $𝑇$ is surjective if and only if $𝑇_{𝐴}$ is.

In particular,

𝐯 \mapsto {[𝐯]}_{ℬ︀}

maps

ker 𝑇

bijectively onto

ker 𝑇_{𝐴}

, and

𝐰 \mapsto {[𝐰]}_{𝒞︀}

maps

im 𝑇

bijectively onto

im 𝑇_{𝐴}

Proof (by chasing coordinates).standard

Uses. Theorem 2.3.9 (The matrix of a map computes the map), Proposition 1.6.16 (The coordinate map is a bijection), Proposition 1.6.19 (Coordinates respect linear combinations), Theorem 2.6.11 (Injectivity is decided at $𝟎$ )

Techniques. Translate between maps and matrices (Technique 2.3.27)

Hint 1. The bridge between

𝑇

and

𝑇_{𝐴}

𝐴 {[𝐯]}_{ℬ︀} =

{[𝑇 (𝐯)]}_{𝒞︀}

(Theorem 2.3.9).

Hint 2. The coordinate maps are bijections that send

𝟎

𝟎

. So

{[𝐰]}_{𝒞︀} = 𝟎

if and only if

𝐰 = 𝟎

, and every

𝐱 \in 𝐹^{𝑛}

{[𝐯]}_{ℬ︀}

for some

𝐯

Sketch. (a)

𝐴 {[𝐯]}_{ℬ︀} =

{[𝑇 (𝐯)]}_{𝒞︀}

, which is

𝟎

iff

𝑇 (𝐯) = 𝟎

. (b) If

𝐰 = 𝑇 (𝐯)

, then

{[𝐰]}_{𝒞︀} = 𝐴 {[𝐯]}_{ℬ︀}

. If

{[𝐰]}_{𝒞︀} = 𝐴 𝐱

, write

𝐱 = {[𝐯]}_{ℬ︀}

; then

{[𝐰]}_{𝒞︀} =

{[𝑇 (𝐯)]}_{𝒞︀}

, so

𝐰 = 𝑇 (𝐯)

. The bijections follow because the coordinate maps are bijections, and (c) follows from the bijections, which match

{𝟎}

with

{𝟎}

and

𝑊

with

𝐹^{𝑚}

By Theorem 2.3.9, $𝐴 {[𝐯]}_{ℬ︀} =$ ${[𝑇 (𝐯)]}_{𝒞︀}$ for every $𝐯 \in 𝑉$ . The coordinate maps $𝑉 \to 𝐹^{𝑛}$ and $𝑊 \to 𝐹^{𝑚}$ are bijections (Proposition 1.6.16) that send $𝟎$ to $𝟎$ (Proposition 1.6.19).

(a) ${[𝐯]}_{ℬ︀} \in ker 𝑇_{𝐴}$ means ${[𝑇 (𝐯)]}_{𝒞︀} =$ $𝐴 {[𝐯]}_{ℬ︀} =$ $𝟎$ . Since the coordinate map of $𝑊$ is injective and ${[𝟎]}_{𝒞︀} = 𝟎$ , this holds if and only if $𝑇 (𝐯) = 𝟎$ .

(b) If $𝐰 = 𝑇 (𝐯)$ , then ${[𝐰]}_{𝒞︀} =$ $𝐴 {[𝐯]}_{ℬ︀} \in$ $im 𝑇_{𝐴}$ . Conversely, suppose that ${[𝐰]}_{𝒞︀} = 𝐴 𝐱$ for some $𝐱 \in 𝐹^{𝑛}$ . Since the coordinate map of $𝑉$ is surjective, $𝐱 = {[𝐯]}_{ℬ︀}$ for some $𝐯 \in 𝑉$ , and then ${[𝐰]}_{𝒞︀} =$ $𝐴 {[𝐯]}_{ℬ︀} =$ ${[𝑇 (𝐯)]}_{𝒞︀}$ . Since the coordinate map of $𝑊$ is injective, $𝐰 = 𝑇 (𝐯) \in im 𝑇$ .

The bijections. By (a), the coordinate map of $𝑉$ , which is injective, sends $ker 𝑇$ into $ker 𝑇_{𝐴}$ . Every $𝐱 \in ker 𝑇_{𝐴}$ is ${[𝐯]}_{ℬ︀}$ for some $𝐯 \in 𝑉$ , and this $𝐯$ lies in $ker 𝑇$ by (a). So $ker 𝑇 \to ker 𝑇_{𝐴}$ , $𝐯 \mapsto {[𝐯]}_{ℬ︀}$ , is a bijection. The same argument with (b) and the coordinate map of $𝑊$ shows that $im 𝑇 \to im 𝑇_{𝐴}$ , $𝐰 \mapsto {[𝐰]}_{𝒞︀}$ , is a bijection.

(c) The bijection $ker 𝑇 \to ker 𝑇_{𝐴}$ sends $𝟎$ to $𝟎$ , so $ker 𝑇 = {𝟎}$ if and only if $ker 𝑇_{𝐴} = {𝟎}$ . By Theorem 2.6.11, $𝑇$ is injective if and only if $𝑇_{𝐴}$ is. If $im 𝑇 = 𝑊$ , then $im 𝑇_{𝐴}$ is the image of all of $𝑊$ under the coordinate map of $𝑊$ , which is $𝐹^{𝑚}$ . Conversely, if $im 𝑇_{𝐴} = 𝐹^{𝑚}$ , then ${[𝐰]}_{𝒞︀} \in im 𝑇_{𝐴}$ for every $𝐰 \in 𝑊$ , so every $𝐰$ lies in $im 𝑇$ by (b). □

Proof (through the square of maps).standard

Uses. Theorem 2.3.9 (The matrix of a map computes the map), Example 2.5.14 (Invertible maps and matrices), Proposition 2.6.15 (Kernels and images of composites), Proposition 2.1.17 (Compositions of linear maps are linear)

Techniques. Translate between maps and matrices (Technique 2.3.27)

Hint 1. Write

𝐶_{ℬ︀} : 𝑉 \to 𝐹^{𝑛}

and

𝐶_{𝒞︀} : 𝑊 \to 𝐹^{𝑚}

for the coordinate maps. Rewrite Theorem 2.3.9 as an equation between composites.

Hint 2.

𝐶_{𝒞︀} 𝑇 = 𝑇_{𝐴} 𝐶_{ℬ︀}

, and both coordinate maps are invertible. Apply Proposition 2.6.15 to each side, using that

𝐶_{𝒞︀}

is injective and

𝐶_{ℬ︀}

is surjective.

Sketch. With

𝐶_{𝒞︀} 𝑇 = 𝑇_{𝐴} 𝐶_{ℬ︀}

: since

𝐶_{𝒞︀}

is injective,

ker 𝑇 =

ker (𝐶_{𝒞︀} 𝑇) =

ker (𝑇_{𝐴} 𝐶_{ℬ︀})

, which is (a). Since

𝐶_{ℬ︀}

is surjective,

im (𝐶_{𝒞︀} ↾_{im 𝑇}) =

im (𝐶_{𝒞︀} 𝑇) =

im (𝑇_{𝐴} 𝐶_{ℬ︀}) =

im 𝑇_{𝐴}

, which with injectivity of

𝐶_{𝒞︀}

is (b). For (c),

𝑇_{𝐴} =

𝐶_{𝒞︀} 𝑇 𝐶_{ℬ︀}^{- 1}

and

𝑇 =

𝐶_{𝒞︀}^{- 1} 𝑇_{𝐴} 𝐶_{ℬ︀}

, and composites of injective (surjective) maps are injective (surjective).

Let $𝐶_{ℬ︀} : 𝑉 \to 𝐹^{𝑛}$ and $𝐶_{𝒞︀} : 𝑊 \to 𝐹^{𝑚}$ be the coordinate maps. They are linear and invertible (Example 2.5.14(c)), hence bijective, and Theorem 2.3.9 says that

𝐶_{𝒞︀} \circ 𝑇 = 𝑇_{𝐴} \circ 𝐶_{ℬ︀},

that is, the two routes around the square of Figure 2.3.1 agree.

(a) Since $𝐶_{𝒞︀}$ is injective, Proposition 2.6.15(a) gives $ker 𝑇 =$ $ker (𝐶_{𝒞︀} 𝑇) =$ $ker (𝑇_{𝐴} 𝐶_{ℬ︀})$ . A vector $𝐯$ lies in $ker (𝑇_{𝐴} 𝐶_{ℬ︀})$ exactly when $𝐶_{ℬ︀} (𝐯) = {[𝐯]}_{ℬ︀}$ lies in $ker 𝑇_{𝐴}$ .

(b) Since $𝐶_{ℬ︀}$ is surjective, Proposition 2.6.15(b) gives

im (𝐶_{𝒞︀} ↾_{im 𝑇}) = im (𝐶_{𝒞︀} 𝑇) = im (𝑇_{𝐴} 𝐶_{ℬ︀}) = im 𝑇_{𝐴} .

So $𝐶_{𝒞︀}$ maps $im 𝑇$ onto $im 𝑇_{𝐴}$ , and it is injective. In particular, if ${[𝐰]}_{𝒞︀} \in im 𝑇_{𝐴}$ , then ${[𝐰]}_{𝒞︀} = {[𝐰^{'}]}_{𝒞︀}$ for some $𝐰^{'} \in im 𝑇$ , and $𝐰 = 𝐰^{'} \in im 𝑇$ . The converse is immediate.

(c) By Proposition 2.6.15(a) and (b), a composite of injective linear maps is injective (its kernel is the kernel of the first map) and a composite of surjective linear maps is surjective (its image is the image of the last map). Now $𝑇_{𝐴} =$ $𝐶_{𝒞︀} 𝑇 𝐶_{ℬ︀}^{- 1}$ and $𝑇 =$ $𝐶_{𝒞︀}^{- 1} 𝑇_{𝐴} 𝐶_{ℬ︀}$ , and the coordinate maps and their inverses are bijective. So if $𝑇$ is injective or surjective, so is $𝑇_{𝐴}$ , and conversely. □

The first proof keeps track of individual vectors. The second works with whole maps: it records ${[𝑇 (𝐯)]}_{𝒞︀} =$ $𝐴 {[𝐯]}_{ℬ︀}$ as the equation $𝐶_{𝒞︀} 𝑇 = 𝑇_{𝐴} 𝐶_{ℬ︀}$ between linear maps and lets Proposition 2.6.15 do the rest. Any property that is preserved by composing with invertible maps on both sides transfers between $𝑇$ and $𝑇_{𝐴}$ in this way, and Section 4.5 uses the same idea for change of basis.

Example 2.6.21 (Differentiation in coordinates). Let $𝐷 : {𝐹 [𝑥]}_{\leq 3} \to$ ${𝐹 [𝑥]}_{\leq 2}$ be differentiation. By Example 2.3.4, its matrix in the bases $1, 𝑥, 𝑥^{2}, 𝑥^{3}$ and $1, 𝑥, 𝑥^{2}$ is

𝐴 = (\begin{matrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \end{matrix}), so 𝐴 𝐱 = (𝑥_{2}, 2 𝑥_{3}, 3 𝑥_{4}) .

If $2 \neq 0$ and $3 \neq 0$ in $𝐹$ (for instance if $𝐹$ has characteristic $0$ ), then $𝐴 𝐱 = 𝟎$ if and only if $𝑥_{2} = 𝑥_{3} = 𝑥_{4} = 0$ , so $ker 𝑇_{𝐴} = span (𝐞_{1})$ . The columns $𝟎, 𝐞_{1}, 2 𝐞_{2}, 3 𝐞_{3}$ span $𝐹^{3}$ . By Proposition 2.6.20, $ker 𝐷$ consists of the polynomials with coordinates in $span (𝐞_{1})$ , the constants, and $𝐷$ is surjective.

Over $𝔽_{3}$ , the entry $3$ is $0$ , so $𝐴 𝐱 = (𝑥_{2}, 2 𝑥_{3}, 0)$ . Now $ker 𝑇_{𝐴} =$ $span (𝐞_{1}, 𝐞_{4})$ and $im 𝑇_{𝐴} =$ $span (𝐞_{1}, 𝐞_{2})$ . Translating back, $ker 𝐷 = span (1, 𝑥^{3})$ and $im 𝐷 =$ $span (1, 𝑥) =$ $𝔽_{3} {[𝑥]}_{\leq 1}$ . Indeed $𝐷 (𝑥^{3}) = 3 𝑥^{2} = 0$ , and no polynomial of degree at most $3$ over $𝔽_{3}$ has derivative $𝑥^{2}$ .

2.6.5Testing on a basis

Intuition. By Theorem 2.2.1, a linear map is determined by its values on a basis

ℬ︀

𝑉

, and those values can be chosen freely. So every property of

𝑇

must be visible in the list

𝑇 (ℬ︀)

. Injectivity and surjectivity are visible in the simplest possible way. They become linear independence and spanning, the two halves of “basis” (Proposition 1.6.2), carried over to

𝑊

Proposition 2.6.22 (Testing injectivity and surjectivity on a basis). Let

ℬ︀ = 𝐯_{1}, \dots, 𝐯_{𝑛}

be a basis of

𝑉

, let

𝑇 \in L (𝑉, 𝑊)

, and write

𝑇 (ℬ︀) =

𝑇 (𝐯_{1}),

\dots,

𝑇 (𝐯_{𝑛})

(a) $𝑇$ is injective if and only if $𝑇 (ℬ︀)$ is linearly independent.
(b) $𝑇$ is surjective if and only if $𝑇 (ℬ︀)$ spans $𝑊$ .
(c) $𝑇$ is bijective if and only if $𝑇 (ℬ︀)$ is a basis of $𝑊$ .
(d)In every case $𝑇 (ℬ︀)$ spans $im 𝑇$ , and it is a basis of $im 𝑇$ if and only if $𝑇$ is injective.

Proof.routine

Uses. Proposition 2.6.14 (Injective maps preserve independence), Theorem 2.6.11 (Injectivity is decided at $𝟎$ ), Theorem 1.5.5 (Independence is decided at zero), Proposition 2.1.16 (Linear maps preserve spans and linear relations), Proposition 1.6.2 (Bases are independent spanning lists), Definition 2.6.3 (Bijective)

Techniques. Test injectivity at $𝟎$ (Technique 2.6.12), Check it on a basis (Technique 2.2.7)

Hint 1. For (a), one direction is Proposition 2.6.14. For the other, take

𝐯 \in ker 𝑇

, expand it in the basis

ℬ︀

, and apply

𝑇

Hint 2. For (b) and (d), express

im 𝑇

in terms of

𝑇 (ℬ︀)

with Proposition 2.1.16(a).

Hint 3. (c) is (a) and (b) together.

Sketch. (a) An injective map preserves independence. Conversely, if

𝐯 =

\sum_{𝑖} 𝑎_{𝑖} 𝐯_{𝑖} \in

ker 𝑇

, then

\sum_{𝑖} 𝑎_{𝑖} 𝑇 (𝐯_{𝑖}) =

𝟎

, so all

𝑎_{𝑖} = 0

and

𝐯 = 𝟎

. (b)

im 𝑇 = span 𝑇 (ℬ︀)

since

ℬ︀

spans

𝑉

, and this equals

𝑊

iff

𝑇 (ℬ︀)

spans

𝑊

. (c) Combine (a) and (b). (d) A list spanning

im 𝑇

is a basis of it iff it is independent, iff

𝑇

is injective by (a).

(a) If $𝑇$ is injective, then $𝑇 (ℬ︀)$ is linearly independent by Proposition 2.6.14, since $ℬ︀$ is. Conversely, suppose that $𝑇 (ℬ︀)$ is linearly independent, and let $𝐯 \in ker 𝑇$ . Since $ℬ︀$ spans $𝑉$ , $𝐯 =$ $𝑎_{1} 𝐯_{1} +$ $\dots +$ $𝑎_{𝑛} 𝐯_{𝑛}$ for some scalars $𝑎_{𝑖}$ , and by linearity

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

As $𝑇 (ℬ︀)$ is linearly independent, $𝑎_{1} = \dots = 𝑎_{𝑛} = 0$ (Theorem 1.5.5), so $𝐯 = 𝟎$ . Hence $ker 𝑇 = {𝟎}$ , and $𝑇$ is injective by Theorem 2.6.11.

(b) Since $ℬ︀$ spans $𝑉$ , Proposition 2.1.16(a) gives

im 𝑇 = span 𝑇 (ℬ︀) .

So $im 𝑇 = 𝑊$ , which is surjectivity, holds if and only if $span 𝑇 (ℬ︀) = 𝑊$ .

(c) By Definition 2.6.3 and Proposition 1.6.2, (c) is (a) and (b) together.

(d) The first claim was shown in (b). The list $𝑇 (ℬ︀)$ lies in $im 𝑇$ and spans it, so it is a basis of $im 𝑇$ if and only if it is linearly independent (Proposition 1.6.2), which by (a) happens if and only if $𝑇$ is injective. □

So whether $𝑇$ is injective, and whether it is surjective, can be decided on one basis of $𝑉$ : a list of $𝑛$ vectors instead of all of $𝑉$ . This is Technique 2.2.7 in a new form. There, the property to be checked was an equation between linear maps; here it is a property of the list of values.

Remark 2.6.23 (Which half of “basis” is used where). The proof of (a) used both halves of “basis”: independence of

ℬ︀

in the “only if” direction (through Proposition 2.6.14), and spanning in the “if” direction, to expand

𝐯

. Part (b) used only that

ℬ︀

spans

𝑉

, so it holds for every spanning list: if

ℒ︀

spans

𝑉

, then

𝑇

is surjective if and only if

𝑇 (ℒ︀)

spans

𝑊

. The “if” direction of (a) fails for lists that do not span. The map

𝑃 (𝑥, 𝑦) = 𝑥

sends the independent list

𝐞_{1}

to the independent list

1

, yet

𝑃

is not injective. Compare Remark 2.2.2, where the two halves played similar roles in constructing maps.

For a matrix, the values on the standard basis are the columns.

Corollary 2.6.24 (The columns decide). Let

𝐴 \in 𝐹^{𝑚 \times 𝑛}

have columns

𝐚_{1}, \dots, 𝐚_{𝑛}

(a) $𝑇_{𝐴}$ is injective if and only if $𝐚_{1}, \dots, 𝐚_{𝑛}$ is linearly independent.
(b) $𝑇_{𝐴}$ is surjective if and only if $𝐚_{1}, \dots, 𝐚_{𝑛}$ spans $𝐹^{𝑚}$ .
(c) $𝑇_{𝐴}$ is bijective if and only if $𝐚_{1}, \dots, 𝐚_{𝑛}$ is a basis of $𝐹^{𝑚}$ .

Proof.routine

Uses. Proposition 2.6.22 (Testing injectivity and surjectivity on a basis), Proposition 1.6.7 (Standard bases), Proposition 2.3.13 (Linearity of the product)

Hint 1. Apply Proposition 2.6.22 to

𝑇_{𝐴}

and a basis of

𝐹^{𝑛}

whose values you know.

Hint 2. The standard basis:

𝑇_{𝐴} (𝐞_{𝑗}) =

𝐴 𝐞_{𝑗} =

𝐚_{𝑗}

Sketch.

𝐞_{1}, \dots, 𝐞_{𝑛}

is a basis of

𝐹^{𝑛}

and

𝑇_{𝐴} (𝐞_{𝑗}) = 𝐚_{𝑗}

, so

𝑇_{𝐴} (ℰ︀)

is the list of columns. Apply Proposition 2.6.22(a)–(c).

The standard basis

ℰ︀ = 𝐞_{1}, \dots, 𝐞_{𝑛}

is a basis of

𝐹^{𝑛}

(Proposition 1.6.7), and

𝑇_{𝐴} (𝐞_{𝑗}) =

𝐴 𝐞_{𝑗} =

𝐚_{𝑗}

by Proposition 2.3.13(c). So

𝑇_{𝐴} (ℰ︀)

is the list of columns of

𝐴

, and (a)–(c) are parts (a)–(c) of Proposition 2.6.22. □

Part (a) can also be read off from the definition of the product. Since $𝐴 𝐱 =$ $𝑥_{1} 𝐚_{1} +$ $\dots +$ $𝑥_{𝑛} 𝐚_{𝑛}$ , a nonzero vector of $ker 𝑇_{𝐴}$ is the same thing as a nontrivial linear relation among the columns, so $ker 𝑇_{𝐴} = {𝟎}$ says exactly that the columns are linearly independent (Theorem 1.5.5). The three parts of the corollary are the equivalences (I) $\Leftrightarrow$ (III) in the Fundamental Theorems of Section 2.7. Combined with Proposition 2.6.20, they decide injectivity and surjectivity of any map between spaces with bases from the columns of one of its matrices.

Example 2.6.25 (Evaluating at three points). Let $𝐸 : {𝐹 [𝑥]}_{\leq 2} \to 𝐹^{3}$ , $𝐸 (𝑝) =$ $(𝑝 (0), 𝑝 (1), 𝑝 (2))$ , where $2 = 1 + 1$ . Each entry of $𝐸$ is an evaluation map, so $𝐸$ is linear (Example 2.1.9). On the basis $1, 𝑥, 𝑥^{2}$ ,

𝐸 (1) = (1, 1, 1), 𝐸 (𝑥) = (0, 1, 2), 𝐸 (𝑥^{2}) = (0, 1, 4) .

Suppose $𝑎 (1, 1, 1) +$ $𝑏 (0, 1, 2) +$ $𝑐 (0, 1, 4) =$ $𝟎$ . The first entry gives $𝑎 = 0$ , then $𝑏 + 𝑐 = 0$ and $2 𝑏 + 4 𝑐 = 0$ , so $2 𝑐 =$ $(2 𝑏 + 4 𝑐) -$ $2 (𝑏 + 𝑐) =$ $0$ .

If $2 \neq 0$ in $𝐹$ , then $𝑐 = 0$ and $𝑏 = 0$ , so the three values are linearly independent and $𝐸$ is injective by Proposition 2.6.22(a). They also span $𝐹^{3}$ , because

𝐞_{3} = 2^{- 1} (𝐸 (𝑥^{2}) - 𝐸 (𝑥)), 𝐞_{2} = 𝐸 (𝑥) - 2 𝐞_{3}, 𝐞_{1} = 𝐸 (1) - 𝐞_{2} - 𝐞_{3},

so $𝐸$ is bijective by Proposition 2.6.22(c): for every $(𝑦_{0}, 𝑦_{1}, 𝑦_{2}) \in 𝐹^{3}$ there is exactly one $𝑝 \in {𝐹 [𝑥]}_{\leq 2}$ with $𝑝 (0) = 𝑦_{0}$ , $𝑝 (1) = 𝑦_{1}$ , and $𝑝 (2) = 𝑦_{2}$ . This is a case of polynomial interpolation (Theorem 3.4.24).

Over $𝔽_{2}$ , $2 = 4 = 0$ , so $𝐸 (𝑥) =$ $𝐸 (𝑥^{2}) =$ $(0, 1, 0)$ , and the values are linearly dependent. So $𝐸$ is not injective; indeed $𝐸 (𝑥^{2} + 𝑥) =$ $(0, 1 + 1, 0) =$ $𝟎$ . Nor is it surjective: every value of $𝐸$ has equal first and third entries, because over $𝔽_{2}$ the points $0$ and $2$ coincide, so $(1, 0, 0) \notin im 𝐸$ .

Exercise 2.6.26.routine For $𝑐 \in 𝐹$ , let $𝑇_{𝑐} : {𝐹 [𝑥]}_{\leq 2} \to$ ${𝐹 [𝑥]}_{\leq 2}$ , $𝑇_{𝑐} (𝑝) =$ $𝐷 (𝑥 𝑝) -$ $𝑐 𝑝$ , where $𝐷$ is differentiation.

(a)Compute $𝑇_{𝑐} (1)$ , $𝑇_{𝑐} (𝑥)$ , and $𝑇_{𝑐} (𝑥^{2})$ .
(b)Over $ℚ$ , find all $𝑐$ for which $𝑇_{𝑐}$ is injective, and all $𝑐$ for which it is surjective.
(c)Answer the same questions over $𝔽_{2}$ , over $𝔽_{3}$ , and over $𝔽_{5}$ .

Solution.

Uses. Proposition 2.6.22 (Testing injectivity and surjectivity on a basis), Proposition 1.5.15 (Short lists, zeros, and repeats), Proposition 1.5.24 (Distinct degrees), Example 2.1.9 (Maps of polynomials)

Techniques. Check it on a basis (Technique 2.2.7)

Hint 1.

𝑥 \cdot 𝑥^{𝑘} = 𝑥^{𝑘 + 1}

, whose derivative is

(𝑘 + 1) 𝑥^{𝑘}

Hint 2. So

𝑇_{𝑐} (𝑥^{𝑘}) =

(𝑘 + 1 - 𝑐) 𝑥^{𝑘}

. When is the list

(1 - 𝑐) 1,

(2 - 𝑐) 𝑥,

(3 - 𝑐) 𝑥^{2}

independent, and when does it span?

Sketch.

𝑇_{𝑐} (𝑥^{𝑘}) =

(𝑘 + 1 - 𝑐) 𝑥^{𝑘}

. If

1 - 𝑐

2 - 𝑐

3 - 𝑐

are all nonzero, the values are nonzero multiples of

1, 𝑥, 𝑥^{2}

, so they form a basis and

𝑇_{𝑐}

is bijective. If one of them is

0

, the list of values contains

0

(not injective) and its span misses the corresponding monomial (not surjective). Over

ℚ

: bijective iff

𝑐 \notin {1, 2, 3}

, otherwise neither. Over

𝔽_{2}

and

𝔽_{3}

the set

{1, 2, 3}

is all of

𝐹

, so

𝑇_{𝑐}

is never injective or surjective; over

𝔽_{5}

it is bijective iff

𝑐 \in {0, 4}

(a) For $𝑘 = 0, 1, 2$ , $𝐷 (𝑥 \cdot 𝑥^{𝑘}) =$ $𝐷 (𝑥^{𝑘 + 1}) =$ $(𝑘 + 1) 𝑥^{𝑘}$ , so $𝑇_{𝑐} (𝑥^{𝑘}) =$ $(𝑘 + 1 - 𝑐) 𝑥^{𝑘}$ :

𝑇_{𝑐} (1) = (1 - 𝑐) 1, 𝑇_{𝑐} (𝑥) = (2 - 𝑐) 𝑥, 𝑇_{𝑐} (𝑥^{2}) = (3 - 𝑐) 𝑥^{2} .

(b) and (c). Let $ℬ︀ = 1, 𝑥, 𝑥^{2}$ . Suppose first that $1 - 𝑐$ , $2 - 𝑐$ , and $3 - 𝑐$ are all nonzero. Then $𝑇_{𝑐} (ℬ︀)$ consists of nonzero multiples of $1$ , $𝑥$ , and $𝑥^{2}$ , which have distinct degrees, so it is linearly independent (Proposition 1.5.24). It also spans ${𝐹 [𝑥]}_{\leq 2}$ , since $𝑥^{𝑘} =$ ${(𝑘 + 1 - 𝑐)}^{- 1} 𝑇_{𝑐} (𝑥^{𝑘})$ for each $𝑘$ . By Proposition 2.6.22, $𝑇_{𝑐}$ is injective and surjective.

Now suppose $𝑘 + 1 - 𝑐 = 0$ for some $𝑘 \in {0, 1, 2}$ . Then $𝑇_{𝑐} (𝑥^{𝑘}) = 𝟎$ , so $𝑇_{𝑐} (ℬ︀)$ contains $𝟎$ and is linearly dependent (Proposition 1.5.15), and $𝑇_{𝑐}$ is not injective. Every value of $𝑇_{𝑐}$ is a combination of the other two monomials, so its coefficient of $𝑥^{𝑘}$ is $0$ ; hence $𝑥^{𝑘} \notin im 𝑇_{𝑐}$ and $𝑇_{𝑐}$ is not surjective.

So $𝑇_{𝑐}$ is bijective if $𝑐 \notin {1, 2, 3}$ and neither injective nor surjective if $𝑐 \in {1, 2, 3}$ , where $1, 2, 3$ are read in $𝐹$ . Over $ℚ$ these are three distinct values. Over $𝔽_{2}$ , ${1, 2, 3} =$ ${1, 0, 1} =$ $𝔽_{2}$ , and over $𝔽_{3}$ , ${1, 2, 3} =$ ${1, 2, 0} =$ $𝔽_{3}$ , so in both cases $𝑇_{𝑐}$ is neither injective nor surjective for every $𝑐$ . Over $𝔽_{5}$ , ${1, 2, 3}$ leaves $𝑐 \in {0, 4}$ , and these are the values for which $𝑇_{𝑐}$ is bijective.

2.6.6Computing kernels and images

Here are the kernels and images of the main examples of Section 2.1.

Example 2.6.27 (Differentiation on $𝐹 [𝑥]$ ). Let $𝐷 : 𝐹 [𝑥] \to 𝐹 [𝑥]$ be differentiation. For $𝑓 =$ $\sum_{𝑘 = 0}^{𝑛} 𝑎_{𝑘} 𝑥^{𝑘}$ , Example 2.1.9 gives $𝐷 (𝑓) =$ $\sum_{𝑘 = 1}^{𝑛} (𝑘 𝑎_{𝑘}) 𝑥^{𝑘 - 1}$ , where $𝑘 𝑎_{𝑘} = (𝑘 * 1) 𝑎_{𝑘}$ , so the coefficient of $𝑥^{𝑘 - 1}$ in $𝐷 (𝑓)$ is $(𝑘 * 1) 𝑎_{𝑘}$ .

Kernel. $𝐷 (𝑓) = 𝟎$ if and only if $(𝑘 * 1) 𝑎_{𝑘} = 0$ for every $𝑘 \geq 1$ . If $𝐹$ has characteristic $0$ (Definition 1.1.28), then $𝑘 * 1 \neq 0$ for $𝑘 \geq 1$ , so this forces $𝑎_{𝑘} = 0$ for all $𝑘 \geq 1$ (Proposition 1.1.17). Hence $ker 𝐷 = span (1)$ , the constant polynomials. Over $𝔽_{𝑝}$ , $𝑘 * 1 = 0$ exactly when $𝑝$ divides $𝑘$ , so $𝐷 (𝑓) = 𝟎$ if and only if $𝑎_{𝑘} = 0$ for every $𝑘$ not divisible by $𝑝$ . Hence $ker 𝐷$ consists of the polynomials $𝑏_{0} +$ $𝑏_{1} 𝑥^{𝑝} +$ $𝑏_{2} 𝑥^{2 𝑝} +$ $\dots$ , those in which only powers of $𝑥^{𝑝}$ occur.

Image. In characteristic $0$ , $𝐷 𝐽 = id$ for the antiderivative $𝐽$ of Example 2.5.24, so $𝐷$ is surjective by Proposition 2.5.5. Over $𝔽_{𝑝}$ it is not. The coefficient of $𝑥^{𝑘 - 1}$ in $𝐷 (𝑓)$ is $(𝑘 * 1) 𝑎_{𝑘}$ , which is $0$ whenever $𝑝$ divides $𝑘$ , so every value of $𝐷$ has coefficient $0$ at $𝑥^{𝑝 - 1},$ $𝑥^{2 𝑝 - 1},$ $𝑥^{3 𝑝 - 1},$ $\dots$ . Conversely, if $𝑞 = \sum_{𝑗} 𝑐_{𝑗} 𝑥^{𝑗}$ has $𝑐_{𝑗} = 0$ whenever $𝑝$ divides $𝑗 + 1$ , then

𝑞 = 𝐷 (\sum_{𝑝 ∤ 𝑗 + 1} {((𝑗 + 1) * 1)}^{- 1} 𝑐_{𝑗} 𝑥^{𝑗 + 1}) .

So $im 𝐷$ consists of the polynomials whose coefficients at $𝑥^{𝑝 - 1}, 𝑥^{2 𝑝 - 1}, \dots$ vanish. In particular $𝑥^{𝑝} \in ker 𝐷$ and $𝑥^{𝑝 - 1} \notin im 𝐷$ , as in Exercise 2.5.42.

Example 2.6.28 (The trace). For $𝑛 \geq 1$ , the trace $tr : 𝐹^{𝑛 \times 𝑛} \to 𝐹$ is linear (Proposition 2.4.27). Write $𝑈_{𝑖 𝑗} = 𝐞_{𝑖} 𝐞_{𝑗}^{𝖳}$ for the matrix units, which form a basis of $𝐹^{𝑛 \times 𝑛}$ (Lemma 2.4.24). Then $tr 𝑈_{𝑖 𝑗}$ is $1$ if $𝑖 = 𝑗$ and $0$ otherwise.

Image. The values $tr 𝑈_{𝑖 𝑗}$ include $1$ , which spans $𝐹$ , so $tr$ is surjective by Proposition 2.6.22(b). Explicitly, $𝑐 = tr (𝑐 𝑈_{11})$ .

Kernel. $ker tr$ is the subspace of trace-zero matrices, ${𝐴 \in 𝐹^{𝑛 \times 𝑛} | 𝑎_{11} + \dots + 𝑎_{𝑛 𝑛} = 0}$ . For $𝑛 \geq 2$ it contains $𝑈_{12} \neq 0$ , so $tr$ is not injective. To find a spanning list, let $tr 𝐴 = 0$ . Then $𝑎_{𝑛 𝑛} =$ $- (𝑎_{11} + \dots + 𝑎_{𝑛 - 1, 𝑛 - 1})$ , and expanding $𝐴$ in matrix units gives

𝐴 = \sum_{𝑖 \neq 𝑗} 𝑎_{𝑖 𝑗} 𝑈_{𝑖 𝑗} + \sum_{𝑖 = 1}^{𝑛 - 1} 𝑎_{𝑖 𝑖} (𝑈_{𝑖 𝑖} - 𝑈_{𝑛 𝑛}) .

So $ker tr$ is spanned by the $𝑛^{2} - 𝑛$ matrices $𝑈_{𝑖 𝑗}$ with $𝑖 \neq 𝑗$ and the $𝑛 - 1$ matrices $𝑈_{𝑖 𝑖} - 𝑈_{𝑛 𝑛}$ , all of which have trace $0$ . By Proposition 2.4.28, every commutator $𝐴 𝐵 - 𝐵 𝐴$ lies in $ker tr$ , which is why $𝐴 𝐵 - 𝐵 𝐴 = 𝐼_{𝑛}$ is impossible when $𝑛 * 1 \neq 0$ (Proposition 2.4.30).

Example 2.6.29 (Evaluation at $0$ , and multiplication by $𝑥$ ). Let ${ev}_{0} : 𝐹 [𝑥] \to 𝐹$ , ${ev}_{0} (𝑝) = 𝑝 (0)$ , and let $𝑀 : 𝐹 [𝑥] \to 𝐹 [𝑥]$ , $𝑀 (𝑝) = 𝑥 𝑝$ . For $𝑝 =$ $𝑎_{0} +$ $𝑎_{1} 𝑥 +$ $\dots +$ $𝑎_{𝑛} 𝑥^{𝑛}$ , ${ev}_{0} (𝑝) = 𝑎_{0}$ . So ${ev}_{0}$ is surjective, since $𝑐 = {ev}_{0} (𝑐)$ for each constant $𝑐$ , and its kernel is the set of polynomials with constant term $0$ .

That kernel is exactly $im 𝑀$ . Every $𝑥 𝑞$ has constant term $0$ , and conversely $𝑎_{1} 𝑥 +$ $𝑎_{2} 𝑥^{2} +$ $\dots +$ $𝑎_{𝑛} 𝑥^{𝑛} =$ $𝑀 (𝑎_{1} + 𝑎_{2} 𝑥 + \dots + 𝑎_{𝑛} 𝑥^{𝑛 - 1})$ . So in

𝐹 [𝑥] \overset{𝑀}{⟶} 𝐹 [𝑥] \overset{{ev}_{0}}{⟶} 𝐹,

$𝑀$ is injective (Example 2.6.4(e)), ${ev}_{0}$ is surjective, and the vectors that the second map sends to $0$ are precisely the values of the first.

For $𝑡 \in 𝐹$ , the kernel of ${ev}_{𝑡} (𝑝) = 𝑝 (𝑡)$ consists of the polynomials with root $𝑡$ . It contains every value of $𝑀 - 𝑡 id$ : if $𝑞 = \sum_{𝑘} 𝑏_{𝑘} 𝑥^{𝑘}$ , then $(𝑥 𝑞) (𝑡) =$ $\sum_{𝑘} 𝑏_{𝑘} 𝑡^{𝑘 + 1} =$ $𝑡 𝑞 (𝑡)$ , so $(𝑥 𝑞 - 𝑡 𝑞) (𝑡) =$ $𝑡 𝑞 (𝑡) -$ $𝑡 𝑞 (𝑡) =$ $0$ by Proposition 1.2.18(c). In fact $ker {ev}_{𝑡} =$ $im (𝑀 - 𝑡 id)$ : this is the factor theorem, a consequence of division with remainder (Theorem 7.1.14).

2.6.7Preimages and the equation $𝑇 (𝐯) = 𝐰$

The kernel describes the solutions of $𝑇 (𝐯) = 𝟎$ , and the image records which equations $𝑇 (𝐯) = 𝐰$ have solutions at all. Together they describe the solutions of every such equation.

Notation 2.6.30 (Preimages). For $𝑌 \subseteq 𝑊$ , the preimage of $𝑌$ under $𝑇$ is

𝑇^{- 1} (𝑌) = {𝐯 \in 𝑉 | 𝑇 (𝐯) \in 𝑌} .

For $𝐰 \in 𝑊$ , we write $𝑇^{- 1} (𝐰)$ for $𝑇^{- 1} ({𝐰})$ , the set of solutions of $𝑇 (𝐯) = 𝐰$ , also called the fiber of $𝑇$ over $𝐰$ .

The notation does not presuppose that $𝑇$ is invertible: $𝑇^{- 1} (𝐰)$ is a set, possibly empty. If $𝑇$ happens to be invertible, it is the one-element set ${𝑇^{- 1} (𝐰)}$ , and we will never need to tell the two apart. In this notation $ker 𝑇 = 𝑇^{- 1} (𝟎)$ , and $𝑇$ is injective if and only if every fiber has at most one element, surjective if and only if every fiber is nonempty.

Corollary 2.6.31 (Fibers are translates of the kernel). Let

𝐰 \in 𝑊

(a)If $𝐰 \notin im 𝑇$ , then $𝑇^{- 1} (𝐰) = \emptyset$ .
(b)If $𝐰 \in im 𝑇$ and $𝐯_{0} \in 𝑇^{- 1} (𝐰)$ , then $𝑇^{- 1} (𝐰) =$ $𝐯_{0} +$ $ker 𝑇$ , and $𝐳 \mapsto 𝐯_{0} + 𝐳$ is a bijection from $ker 𝑇$ onto $𝑇^{- 1} (𝐰)$ .
(c)Every $𝐯 \in 𝑉$ lies in exactly one fiber, namely $𝑇^{- 1} (𝑇 (𝐯))$ . So $𝑉$ is the disjoint union of the fibers $𝑇^{- 1} (𝐰)$ with $𝐰 \in im 𝑇$ , each of which is a translate of $ker 𝑇$ .

Proof.routine

Uses. Definition 2.6.6 (Image), Definition 2.6.5 (Kernel), Notation 3.1.17 (Translates), Proposition 3.1.19 (Solution sets are translates of the kernel)

Techniques. Particular plus homogeneous (Technique 3.1.18), Double containment (Technique 1.3.14)

Hint 1. (a) is the definition of the image. For (b), prove the two containments: apply

𝑇

𝐯_{0} + 𝐳

with

𝐳 \in ker 𝑇

, and for a solution

𝐯

, look at

𝐯 - 𝐯_{0}

Hint 2. For (c),

𝐯 \in 𝑇^{- 1} (𝐰)

means

𝑇 (𝐯) = 𝐰

. How many

𝐰

satisfy this for a fixed

𝐯

Sketch. (a) No

𝐯

has

𝑇 (𝐯) = 𝐰

. (b)

𝑇 (𝐯_{0} + 𝐳) = 𝐰 + 𝟎

for

𝐳 \in ker 𝑇

, and if

𝑇 (𝐯) = 𝐰

, then

𝐯 - 𝐯_{0} \in ker 𝑇

; this is Proposition 3.1.19. The map

𝐳 \mapsto 𝐯_{0} + 𝐳

is injective by cancellation. (c)

𝐯

lies in

𝑇^{- 1} (𝐰)

exactly when

𝐰 = 𝑇 (𝐯)

(a) If $𝐰 \notin im 𝑇$ , then no $𝐯 \in 𝑉$ satisfies $𝑇 (𝐯) = 𝐰$ (Definition 2.6.6).

(b) This is Proposition 3.1.19; we repeat the argument. If $𝐳 \in ker 𝑇$ , then $𝑇 (𝐯_{0} + 𝐳) =$ $𝐰 +$ $𝟎 =$ $𝐰$ , so $𝐯_{0} +$ $ker 𝑇 \subseteq$ $𝑇^{- 1} (𝐰)$ . If $𝑇 (𝐯) = 𝐰$ , then $𝑇 (𝐯 - 𝐯_{0}) =$ $𝐰 -$ $𝐰 =$ $𝟎$ , so $𝐯 =$ $𝐯_{0} +$ $(𝐯 - 𝐯_{0}) \in$ $𝐯_{0} +$ $ker 𝑇$ . Hence $𝑇^{- 1} (𝐰) =$ $𝐯_{0} +$ $ker 𝑇$ , which says that $𝐳 \mapsto 𝐯_{0} + 𝐳$ maps $ker 𝑇$ onto the fiber. It is injective, because $𝐯_{0} + 𝐳 = 𝐯_{0} + 𝐳^{'}$ implies $𝐳 = 𝐳^{'}$ .

(c) $𝐯 \in 𝑇^{- 1} (𝐰)$ means $𝑇 (𝐯) = 𝐰$ , which holds for exactly one $𝐰$ , namely $𝐰 = 𝑇 (𝐯) \in im 𝑇$ . So the fibers over the vectors of $im 𝑇$ are disjoint and cover $𝑉$ . By (b), each of them is a translate of $ker 𝑇$ . □

So the domain is sliced into parallel copies of $ker 𝑇$ , one for each vector of $im 𝑇$ . The map $𝑇$ collapses each slice to a single vector, and different slices go to different vectors. In this picture, Theorem 2.6.11 says that $𝑇$ is injective exactly when the slices are single points. Figure 2.6.1 shows the slices for

𝐴 = (\begin{matrix} 1 & 2 \\ 1 & 2 \end{matrix}), 𝑇_{𝐴} (𝐱) = (𝑥_{1} + 2 𝑥_{2}) (\begin{matrix} 1 \\ 1 \end{matrix}) .

Here $ker 𝑇_{𝐴}$ is the line $𝑥_{1} + 2 𝑥_{2} = 0$ , the column space is $im 𝑇_{𝐴} =$ $span ((1, 1))$ , and the fiber over $𝑡 (1, 1)$ is the line $𝑥_{1} + 2 𝑥_{2} = 𝑡$ , parallel to the kernel.

Figure 2.6.1. The fibers of

𝑇_{𝐴}

for

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

over

ℝ

. Left:

ker 𝑇_{𝐴}

(red) and the fibers

𝑥_{1} + 2 𝑥_{2} = 𝑡

for

𝑡 = - 4, - 2, 2, 4

, parallel translates of the kernel. The blue one is the fiber over

𝐲 = (2, 2)

, the solution set of

𝐴 𝐱 = 𝐲

. Right: the column space

im 𝑇_{𝐴} =

span ((1, 1))

(grey), with the vectors

𝑡 (1, 1)

onto which the fibers collapse. The vector

𝐲^{'} = (2, - 2)

is not in

im 𝑇_{𝐴}

, so its fiber is empty.

The row and column pictures in Section 3.4.4 show the same phenomenon interactively for a single fiber: there $𝐲$ can be dragged along the column space, where the solutions form a line parallel to the kernel, and off it, where there are none.

Remark 2.6.32 (What the slicing leads to). The slicing suggests a count:

𝑉

consists of as many copies of

ker 𝑇

as there are vectors in

im 𝑇

. Over a finite field this is literally true,

| 𝑉 | =

| ker 𝑇 | \cdot

| im 𝑇 |

(Exercise 2.6.44). For finite-dimensional spaces over any field, the count becomes the dimension formula

dim 𝑉 =

dim ker 𝑇 +

dim im 𝑇

, which is rank–nullity (Theorem 4.2.11). Treating each slice as a single object leads to the quotient space

𝑉 / ker 𝑇

, and the bijection between the slices and

im 𝑇

becomes the first isomorphism theorem (Theorem 4.6.34).

Exercise 2.6.33.routine Let

𝐴 = (\begin{matrix} 1 & 2 & 3 \\ 2 & 4 & 7 \end{matrix})

as in Example 2.6.19(a). Find one solution of

𝐴 𝐱 = (1, 3)

by inspection, and then describe all of them.

Solution.

Uses. Corollary 2.6.31 (Fibers are translates of the kernel), Example 2.6.19 (Computing a null space and a column space)

Techniques. Particular plus homogeneous (Technique 3.1.18)

Hint 1. The third column minus the first is

(2, 5)

. Is there a simpler combination of columns that gives

(1, 3)

Hint 2. Once you have one solution

𝐱_{0}

, every solution is

𝐱_{0} + 𝐳

with

𝐳 \in

ker 𝑇_{𝐴} =

span ((- 2, 1, 0))

Sketch.

(1, 3) =

(3, 7) -

2 (1, 2)

, so

𝐱_{0} = (- 2, 0, 1)

is a solution. All solutions:

(- 2 - 2 𝑡, 𝑡, 1)

for

𝑡 \in 𝐹

The columns are $(1, 2)$ , $(2, 4)$ , and $(3, 7)$ , and $(3, 7) -$ $2 (1, 2) =$ $(1, 3)$ . So $𝐱_{0} = (- 2, 0, 1)$ satisfies $𝐴 𝐱_{0} = (1, 3)$ . By Example 2.6.19(a), $ker 𝑇_{𝐴} =$ $span ((- 2, 1, 0))$ , so by Corollary 2.6.31(b) the solutions are exactly

(- 2, 0, 1) + 𝑡 (- 2, 1, 0) = (- 2 - 2 𝑡, 𝑡, 1), 𝑡 \in 𝐹 .

As a check, $(- 2 - 2 𝑡) + 2 𝑡 + 3 = 1$ and $2 (- 2 - 2 𝑡) + 4 𝑡 + 7 = 3$ .

Exercises

Exercise 2.6.34.routine Let $𝑇 : ℝ^{3} \to ℝ^{3}$ , $𝑇 (𝑥, 𝑦, 𝑧) =$ $(𝑥 + 𝑦, 𝑦 + 𝑧, 𝑥 - 𝑧)$ .

(a)Find $ker 𝑇$ , and describe $im 𝑇$ by a single linear equation.
(b)Decide whether $(1, 2, 3)$ and $(2, 1, 1)$ lie in $im 𝑇$ . For the one that does, find all of its preimages.

Solution.

Uses. Definition 2.6.5 (Kernel), Proposition 2.6.18 (The column space is spanned by the columns), Corollary 1.3.20 (Operations that preserve the span), Corollary 2.6.31 (Fibers are translates of the kernel)

Techniques. Particular plus homogeneous (Technique 3.1.18)

Hint 1.

𝑇 (𝑥, 𝑦, 𝑧) =

𝑥 (1, 0, 1) +

𝑦 (1, 1, 0) +

𝑧 (0, 1, - 1)

. Is one of these three vectors a combination of the other two?

Hint 2. Every value

(𝑎, 𝑏, 𝑐)

𝑇

satisfies one linear equation. Find it by eliminating

𝑥

𝑦

, and

𝑧

from

𝑎 = 𝑥 + 𝑦

𝑏 = 𝑦 + 𝑧

𝑐 = 𝑥 - 𝑧

Sketch. The kernel is

span ((1, - 1, 1))

. Since

(0, 1, - 1) =

(1, 1, 0) -

(1, 0, 1)

im 𝑇 =

span ((1, 0, 1), (1, 1, 0)) =

{(𝑎, 𝑏, 𝑐) | 𝑎 - 𝑏 - 𝑐 = 0}

. So

(1, 2, 3) \notin im 𝑇

and

(2, 1, 1) =

𝑇 (1, 1, 0) \in

im 𝑇

, with preimages

(1 + 𝑡, 1 - 𝑡, 𝑡)

(a) $𝑇 (𝑥, 𝑦, 𝑧) = 𝟎$ means $𝑥 + 𝑦 = 0$ , $𝑦 + 𝑧 = 0$ , and $𝑥 - 𝑧 = 0$ . The first two give $𝑦 = - 𝑥$ and $𝑧 = 𝑥$ , and then the third holds. So $ker 𝑇 =$ ${(𝑡, - 𝑡, 𝑡) | 𝑡 \in ℝ} =$ $span ((1, - 1, 1))$ .

$𝑇$ is $𝑇_{𝐴}$ for the matrix with columns $𝐚_{1} = (1, 0, 1)$ , $𝐚_{2} = (1, 1, 0)$ , and $𝐚_{3} =$ $(0, 1, - 1) =$ $𝐚_{2} -$ $𝐚_{1}$ , so by Proposition 2.6.18 and Corollary 1.3.20(c), $im 𝑇 =$ $span (𝐚_{1}, 𝐚_{2})$ . Every value $(𝑎, 𝑏, 𝑐) =$ $(𝑥 + 𝑦, 𝑦 + 𝑧, 𝑥 - 𝑧)$ satisfies $𝑎 - 𝑏 - 𝑐 = 0$ . Conversely, if $𝑎 - 𝑏 - 𝑐 = 0$ , then

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

So $im 𝑇 =$ ${(𝑎, 𝑏, 𝑐) \in ℝ^{3} | 𝑎 - 𝑏 - 𝑐 = 0}$ , a plane.

(b) Since $1 - 2 - 3 = - 4 \neq 0$ , $(1, 2, 3) \notin im 𝑇$ . Since $2 - 1 - 1 = 0$ , $(2, 1, 1) \in im 𝑇$ , and by the computation above $(2, 1, 1) =$ $𝐚_{1} +$ $𝐚_{2} =$ $𝑇 (1, 1, 0)$ . By Corollary 2.6.31, its preimages are

(1, 1, 0) + 𝑡 (1, - 1, 1) = (1 + 𝑡, 1 - 𝑡, 𝑡), 𝑡 \in ℝ .

As a check, $𝑇 (1 + 𝑡, 1 - 𝑡, 𝑡) =$ $(2, 1, 1)$ .

Exercise 2.6.35.routine Let $𝐴 =$ $(\begin{matrix} 1 & 𝑖 & 1 + 𝑖 \\ 𝑖 & - 1 & 𝑖 - 1 \end{matrix}) \in$ $ℂ^{2 \times 3}$ .

(a)Show that every column of $𝐴$ is a multiple of $(1, 𝑖)$ , and find $im 𝑇_{𝐴}$ .
(b)Find $ker 𝑇_{𝐴}$ as the span of two vectors.
(c)Is $𝑇_{𝐴}$ injective? Is it surjective? Find all solutions of $𝐴 𝐱 = (1 + 𝑖, 𝑖 - 1)$ .

Solution.

Uses. Proposition 2.6.18 (The column space is spanned by the columns), Definition 2.3.8 (Matrix–vector product), Proposition 1.2.35 (Zero products), Corollary 2.6.31 (Fibers are translates of the kernel)

Hint 1.

𝑖 (1, 𝑖) = (𝑖, - 1)

. What is

(1 + 𝑖) (1, 𝑖)

Hint 2. Then

𝐴 𝐱 =

(𝑥_{1} + 𝑖 𝑥_{2} + (1 + 𝑖) 𝑥_{3}) (1, 𝑖)

, which is

𝟎

exactly when the scalar factor is

0

Sketch. The columns are

(1, 𝑖)

𝑖 (1, 𝑖)

, and

(1 + 𝑖) (1, 𝑖)

, so

im 𝑇_{𝐴} =

span ((1, 𝑖))

𝐴 𝐱 = 𝟎

iff

𝑥_{1} =

- 𝑖 𝑥_{2} -

(1 + 𝑖) 𝑥_{3}

, so

ker 𝑇_{𝐴} =

span ((- 𝑖, 1, 0), (- 1 - 𝑖, 0, 1))

. Neither injective nor surjective.

(1 + 𝑖, 𝑖 - 1)

is the third column, so the solutions are

𝐞_{3} + ker 𝑇_{𝐴}

(a) The columns are $𝐚_{1} = (1, 𝑖)$ , $𝐚_{2} =$ $(𝑖, - 1) =$ $𝑖 (1, 𝑖)$ , and $𝐚_{3} =$ $(1 + 𝑖, 𝑖 - 1) =$ $(1 + 𝑖) (1, 𝑖)$ , since $(1 + 𝑖) 𝑖 =$ $𝑖 +$ $𝑖^{2} =$ $𝑖 -$ $1$ . By Proposition 2.6.18, $im 𝑇_{𝐴} =$ $span ((1, 𝑖), 𝑖 (1, 𝑖), (1 + 𝑖) (1, 𝑖)) =$ $span ((1, 𝑖))$ .

(b) By Definition 2.3.8, $𝐴 𝐱 =$ $𝑥_{1} 𝐚_{1} +$ $𝑥_{2} 𝐚_{2} +$ $𝑥_{3} 𝐚_{3} =$ $(𝑥_{1} + 𝑖 𝑥_{2} + (1 + 𝑖) 𝑥_{3}) (1, 𝑖)$ . Since $(1, 𝑖) \neq 𝟎$ , this is $𝟎$ if and only if $𝑥_{1} =$ $- 𝑖 𝑥_{2} -$ $(1 + 𝑖) 𝑥_{3}$ (Proposition 1.2.35). So

ker 𝑇_{𝐴} = {(- 𝑖 𝑠 - (1 + 𝑖) 𝑡, 𝑠, 𝑡) | 𝑠, 𝑡 \in ℂ} = span ((- 𝑖, 1, 0), (- 1 - 𝑖, 0, 1)) .

(c) $𝑇_{𝐴}$ is not injective, since $(- 𝑖, 1, 0)$ is a nonzero vector of its kernel. It is not surjective: $(1, 0) \notin$ $span ((1, 𝑖))$ , since $𝑐 (1, 𝑖) = (1, 0)$ would need $𝑐 = 1$ and $𝑐 𝑖 = 0$ . Finally, $(1 + 𝑖, 𝑖 - 1) =$ $𝐚_{3} =$ $𝐴 𝐞_{3}$ , so by Corollary 2.6.31 the solutions of $𝐴 𝐱 = (1 + 𝑖, 𝑖 - 1)$ are

𝐞_{3} + 𝑠 (- 𝑖, 1, 0) + 𝑡 (- 1 - 𝑖, 0, 1) = (- 𝑖 𝑠 - (1 + 𝑖) 𝑡, 𝑠, 1 + 𝑡), 𝑠, 𝑡 \in ℂ .

Exercise 2.6.36.routine Let $𝑆, 𝑇 \in L (𝑉, 𝑊)$ .

(a)Show that $ker 𝑆 \cap$ $ker 𝑇 \subseteq$ $ker (𝑆 + 𝑇)$ .
(b)Show that $im (𝑆 + 𝑇) \subseteq$ $im 𝑆 +$ $im 𝑇$ , the sum of subspaces from Proposition 1.4.24.
(c)Show that both inclusions can be strict.

Solution.

Uses. Definition 2.1.21 (Pointwise operations), Definition 2.6.5 (Kernel), Definition 2.6.6 (Image), Proposition 1.4.24 (The sum of two subspaces)

Techniques. Fix an arbitrary element and chase it (Technique 1.3.13), Finding counterexamples and minimal examples (Technique 1.4.18)

Hint 1.

(𝑆 + 𝑇) (𝐯) =

𝑆 (𝐯) +

𝑇 (𝐯)

by definition of the sum of maps.

Hint 2. For (c), try

𝑇 = - 𝑆

Sketch. (a) If

𝑆 (𝐯) = 𝑇 (𝐯) = 𝟎

, then

(𝑆 + 𝑇) (𝐯) = 𝟎

. (b)

(𝑆 + 𝑇) (𝐯) =

𝑆 (𝐯) +

𝑇 (𝐯) \in

im 𝑆 +

im 𝑇

. (c) With

𝑉 = 𝑊 \neq {𝟎}

𝑆 = id

, and

𝑇 = - id

ker 𝑆 \cap ker 𝑇 = {𝟎}

but

ker (𝑆 + 𝑇) = 𝑉

, and

im (𝑆 + 𝑇) = {𝟎}

but

im 𝑆 + im 𝑇 = 𝑉

Recall that $(𝑆 + 𝑇) (𝐯) =$ $𝑆 (𝐯) +$ $𝑇 (𝐯)$ (Definition 2.1.21).

(a) If $𝐯 \in ker 𝑆 \cap ker 𝑇$ , then $(𝑆 + 𝑇) (𝐯) =$ $𝟎 +$ $𝟎 =$ $𝟎$ .

(b) For every $𝐯 \in 𝑉$ , $(𝑆 + 𝑇) (𝐯) =$ $𝑆 (𝐯) +$ $𝑇 (𝐯)$ with $𝑆 (𝐯) \in im 𝑆$ and $𝑇 (𝐯) \in im 𝑇$ , so it lies in $im 𝑆 + im 𝑇$ .

(c) Let $𝑉 = 𝑊$ be any nonzero space, $𝑆 = {id}_{𝑉}$ , and $𝑇 = - {id}_{𝑉}$ . Then $ker 𝑆 \cap ker 𝑇 = {𝟎}$ , while $𝑆 + 𝑇 = 0$ has kernel $𝑉 \neq {𝟎}$ . And $im (𝑆 + 𝑇) = {𝟎}$ , while $im 𝑆 +$ $im 𝑇 =$ $𝑉 +$ $𝑉 =$ $𝑉$ .

Exercise 2.6.37.standard Prove or disprove each statement.

(a)If $𝑆, 𝑇 \in L (𝑉, 𝑊)$ satisfy $ker 𝑆 = ker 𝑇$ and $im 𝑆 = im 𝑇$ , then $𝑆 = 𝑇$ .
(b)For every $𝑇 \in L (𝑉)$ , $ker 𝑇 \cap im 𝑇 = {𝟎}$ .
(c)If $𝑃 \in L (𝑉)$ satisfies $𝑃^{2} = 𝑃$ , then $ker 𝑃 \cap im 𝑃 = {𝟎}$ .
(d)If $𝑇 \in L (𝑉, 𝑊)$ is injective and $𝑈$ is a subspace of $𝑉$ , then the restriction $𝑇 ↾_{𝑈} : 𝑈 \to 𝑊$ is injective.
(e)If $𝑇 \in L (𝑉, 𝑊)$ is not injective, then $𝑇$ is not surjective.

Solution.

Uses. Theorem 2.6.11 (Injectivity is decided at $𝟎$ ), Definition 2.6.5 (Kernel), Definition 2.6.6 (Image)

Techniques. Finding counterexamples and minimal examples (Technique 1.4.18), Test injectivity at $𝟎$ (Technique 2.6.12)

Hint 1. (a) Look for two different bijections of

𝐹^{2}

. (b) Try

𝑁 (𝑥, 𝑦) = (𝑦, 0)

Hint 2. (c) If

𝐰 = 𝑃 (𝐯)

, then

𝑃 (𝐰) = 𝑃^{2} (𝐯) = 𝐰

. (d) What is the kernel of the restriction? (e) Look at Example 2.6.4.

Sketch. (a) False:

id

and the swap

(𝑥, 𝑦) \mapsto (𝑦, 𝑥)

𝐹^{2}

. (b) False: for

𝑁 (𝑥, 𝑦) = (𝑦, 0)

ker 𝑁 =

im 𝑁 =

span (𝐞_{1})

. (c) True:

𝐰 \in im 𝑃

gives

𝑃 (𝐰) = 𝐰

, and

𝐰 \in ker 𝑃

gives

𝑃 (𝐰) = 𝟎

. (d) True:

ker (𝑇 ↾_{𝑈}) =

𝑈 \cap

ker 𝑇 =

{𝟎}

. (e) False:

𝑃 (𝑥, 𝑦) = 𝑥

(a) False. On $𝐹^{2}$ , let $𝑆 = id$ and $𝑇 (𝑥, 𝑦) = (𝑦, 𝑥)$ . Both have kernel ${𝟎}$ and image $𝐹^{2}$ , since each is its own inverse ( $𝑇^{2} = id$ ), but $𝑆 (𝐞_{1}) =$ $𝐞_{1} \neq$ $𝐞_{2} =$ $𝑇 (𝐞_{1})$ .

(b) False. Let $𝑁 : 𝐹^{2} \to 𝐹^{2}$ , $𝑁 (𝑥, 𝑦) = (𝑦, 0)$ . Then $𝑁 (𝑥, 𝑦) = 𝟎$ if and only if $𝑦 = 0$ , so $ker 𝑁 = span (𝐞_{1})$ , and the values of $𝑁$ are the vectors $(𝑦, 0)$ , so $im 𝑁 = span (𝐞_{1})$ as well. Their intersection is $span (𝐞_{1}) \neq {𝟎}$ .

(c) True. Let $𝐰 \in ker 𝑃 \cap im 𝑃$ , say $𝐰 = 𝑃 (𝐯)$ . Then $𝐰 =$ $𝑃 (𝐯) =$ $𝑃^{2} (𝐯) =$ $𝑃 (𝐰) =$ $𝟎$ , the last step because $𝐰 \in ker 𝑃$ .

(d) True. The restriction is linear, and its kernel is ${𝐮 \in 𝑈 | 𝑇 (𝐮) = 𝟎} =$ $𝑈 \cap$ $ker 𝑇 =$ $𝑈 \cap$ ${𝟎} =$ ${𝟎}$ , by Theorem 2.6.11 applied to $𝑇$ . Applying Theorem 2.6.11 to $𝑇 ↾_{𝑈}$ , it is injective. (Directly: if $𝐮, 𝐮^{'} \in 𝑈$ and $𝑇 (𝐮) = 𝑇 (𝐮^{'})$ , then $𝐮 = 𝐮^{'}$ .)

(e) False. $𝑃 : 𝐹^{2} \to 𝐹$ , $𝑃 (𝑥, 𝑦) = 𝑥$ , is surjective but not injective (Example 2.6.4(b)).

Exercise 2.6.38.standard

(a)Find $𝑇 \in L (𝐹^{2})$ with $ker 𝑇 = im 𝑇$ .
(b)Find $𝑇 \in L (𝐹^{4})$ with $ker 𝑇 = im 𝑇$ .
(c)Show that every $𝑇 \in L (𝑉)$ with $ker 𝑇 = im 𝑇$ satisfies $𝑇^{2} = 0$ , and that $𝑇^{2} = 0$ does not imply $ker 𝑇 = im 𝑇$ .

Solution.

Uses. Definition 2.6.5 (Kernel), Definition 2.6.6 (Image)

Techniques. Finding counterexamples and minimal examples (Technique 1.4.18)

Hint 1. Decide what

𝑇

does to the standard basis. Which basis vectors should be sent to

𝟎

, and where should the others go so that their values fill up exactly the kernel?

Hint 2. In

𝐹^{2}

: kill

𝐞_{1}

and send

𝐞_{2}

𝐞_{1}

. In

𝐹^{4}

: kill

𝐞_{1}, 𝐞_{2}

and send

𝐞_{3} \mapsto 𝐞_{1}

𝐞_{4} \mapsto 𝐞_{2}

Hint 3. For (c),

𝑇 (𝐯) \in im 𝑇

for every

𝐯

. For the converse, try

𝑇 = 0

Sketch. (a)

𝑇 (𝑥, 𝑦) = (𝑦, 0)

: kernel and image are

span (𝐞_{1})

. (b)

𝑇 (𝑥_{1}, 𝑥_{2}, 𝑥_{3}, 𝑥_{4}) =

(𝑥_{3}, 𝑥_{4}, 0, 0)

: kernel and image are

span (𝐞_{1}, 𝐞_{2})

. (c)

𝑇 (𝐯) \in im 𝑇 = ker 𝑇

gives

𝑇^{2} (𝐯) = 𝟎

; the zero map on

𝐹

has

𝑇^{2} = 0

, kernel

𝐹

, and image

{0}

(a) Let $𝑇 (𝑥, 𝑦) = (𝑦, 0)$ . As in Exercise 2.6.37(b), $ker 𝑇 =$ $im 𝑇 =$ $span (𝐞_{1})$ .

(b) Let $𝑇 (𝑥_{1}, 𝑥_{2}, 𝑥_{3}, 𝑥_{4}) =$ $(𝑥_{3}, 𝑥_{4}, 0, 0)$ . Then $𝑇 (𝐱) = 𝟎$ if and only if $𝑥_{3} = 𝑥_{4} = 0$ , so $ker 𝑇 =$ $span (𝐞_{1}, 𝐞_{2})$ . The values of $𝑇$ are the vectors $(𝑎, 𝑏, 0, 0)$ with $𝑎, 𝑏 \in 𝐹$ , each of which occurs (take $𝑥_{3} = 𝑎$ , $𝑥_{4} = 𝑏$ ), so $im 𝑇 =$ $span (𝐞_{1}, 𝐞_{2})$ as well.

(c) Let $ker 𝑇 = im 𝑇$ and $𝐯 \in 𝑉$ . Then $𝑇 (𝐯) \in im 𝑇 = ker 𝑇$ , so $𝑇^{2} (𝐯) =$ $𝑇 (𝑇 (𝐯)) =$ $𝟎$ . Hence $𝑇^{2} = 0$ . The converse fails: the zero map $𝑇$ on $𝐹$ satisfies $𝑇^{2} = 0$ , but $ker 𝑇 = 𝐹$ and $im 𝑇 = {0}$ .

No $𝑇 \in L (𝐹^{3})$ has $ker 𝑇 = im 𝑇$ . Rank–nullity (Theorem 4.2.11) gives $dim ker 𝑇 +$ $dim im 𝑇 =$ $3$ , which is impossible if the two dimensions are equal.

Exercise 2.6.39.standard Let $𝑆 \in L (𝑈, 𝑉)$ and $𝑇 \in L (𝑉, 𝑊)$ .

(a)Show that $ker (𝑇 𝑆) = ker 𝑆$ if and only if $ker 𝑇 \cap im 𝑆 = {𝟎}$ .
(b)Show that $im (𝑇 𝑆) = im 𝑇$ if and only if $im 𝑆 + ker 𝑇 = 𝑉$ .
(c)Deduce that for $𝑇 \in L (𝑉)$ , $ker 𝑇^{2} = ker 𝑇$ if and only if $ker 𝑇 \cap im 𝑇 = {𝟎}$ , and $im 𝑇^{2} = im 𝑇$ if and only if $ker 𝑇 + im 𝑇 = 𝑉$ .

Solution.

Uses. Proposition 2.6.15 (Kernels and images of composites), Proposition 1.4.24 (The sum of two subspaces), Proposition 2.6.8 (Kernels and images are subspaces)

Techniques. Double containment (Technique 1.3.14), Fix an arbitrary element and chase it (Technique 1.3.13)

Hint 1. By Proposition 2.6.15,

ker 𝑆 \subseteq ker (𝑇 𝑆)

and

im (𝑇 𝑆) \subseteq im 𝑇

always hold, so only one containment is at stake in each part.

Hint 2. (a) If

𝑇 (𝑆 (𝐮)) = 𝟎

, then

𝑆 (𝐮)

lies in

ker 𝑇 \cap im 𝑆

. (b) If

𝑇 (𝐯) = 𝑇 (𝑆 (𝐮))

, then

𝐯 - 𝑆 (𝐮) \in ker 𝑇

Sketch. (a) “If”:

𝐮 \in ker (𝑇 𝑆)

puts

𝑆 (𝐮)

ker 𝑇 \cap im 𝑆 = {𝟎}

. “Only if”: for

𝐯 = 𝑆 (𝐮) \in ker 𝑇

𝐮 \in

ker (𝑇 𝑆) =

ker 𝑆

, so

𝐯 = 𝟎

. (b) “If”: for

𝐯 = 𝑆 (𝐮) + 𝐳

with

𝐳 \in ker 𝑇

𝑇 (𝐯) = 𝑇 𝑆 (𝐮)

. “Only if”:

𝑇 (𝐯) = 𝑇 𝑆 (𝐮)

for some

𝐮

, so

𝐯 =

𝑆 (𝐮) +

(𝐯 - 𝑆 (𝐮))

. (c) Take

𝑈 = 𝑉

and

𝑆 = 𝑇

(a) By Proposition 2.6.15(a), $ker 𝑆 \subseteq ker (𝑇 𝑆)$ always. Suppose $ker 𝑇 \cap im 𝑆 = {𝟎}$ , and let $𝐮 \in ker (𝑇 𝑆)$ . Then $𝑆 (𝐮) \in im 𝑆$ and $𝑇 (𝑆 (𝐮)) = 𝟎$ , so $𝑆 (𝐮) \in$ $ker 𝑇 \cap$ $im 𝑆 =$ ${𝟎}$ , and $𝐮 \in ker 𝑆$ . Conversely, suppose $ker (𝑇 𝑆) = ker 𝑆$ , and let $𝐯 \in ker 𝑇 \cap im 𝑆$ , say $𝐯 = 𝑆 (𝐮)$ . Then $𝑇 (𝑆 (𝐮)) =$ $𝑇 (𝐯) =$ $𝟎$ , so $𝐮 \in$ $ker (𝑇 𝑆) =$ $ker 𝑆$ , and $𝐯 = 𝑆 (𝐮) = 𝟎$ .

(b) By Proposition 2.6.15(b), $im (𝑇 𝑆) \subseteq im 𝑇$ always. Suppose $im 𝑆 + ker 𝑇 = 𝑉$ , and let $𝐯 \in 𝑉$ . Write $𝐯 = 𝑆 (𝐮) + 𝐳$ with $𝐮 \in 𝑈$ and $𝐳 \in ker 𝑇$ . Then $𝑇 (𝐯) =$ $𝑇 (𝑆 (𝐮)) +$ $𝑇 (𝐳) =$ $𝑇 (𝑆 (𝐮)) \in$ $im (𝑇 𝑆)$ . So $im 𝑇 \subseteq im (𝑇 𝑆)$ . Conversely, suppose $im (𝑇 𝑆) = im 𝑇$ , and let $𝐯 \in 𝑉$ . Then $𝑇 (𝐯) \in$ $im 𝑇 =$ $im (𝑇 𝑆)$ , so $𝑇 (𝐯) = 𝑇 (𝑆 (𝐮))$ for some $𝐮 \in 𝑈$ . The vector $𝐳 = 𝐯 - 𝑆 (𝐮)$ satisfies $𝑇 (𝐳) = 𝟎$ , and $𝐯 =$ $𝑆 (𝐮) +$ $𝐳 \in$ $im 𝑆 +$ $ker 𝑇$ . So $𝑉 \subseteq im 𝑆 + ker 𝑇$ , and the other containment holds because $im 𝑆 + ker 𝑇$ is a subspace of $𝑉$ (Proposition 2.6.8, Proposition 1.4.24).

(c) Apply (a) and (b) with $𝑈 = 𝑉$ and $𝑆 = 𝑇$ , so that $𝑇 𝑆 = 𝑇^{2}$ .

Exercise 2.6.40.routine Find the error in the following argument, show that its claim is false, and state a correct version.

Claim. Let $ℬ︀ = 𝐯_{1}, \dots, 𝐯_{𝑛}$ be a basis of $𝑉$ and $𝑇 \in L (𝑉, 𝑊)$ . If $𝑇 (𝐯_{𝑖}) \neq 𝟎$ for every $𝑖$ , then $𝑇$ is injective.

Proof. Let $𝐯 \in ker 𝑇$ and write $𝐯 =$ $𝑎_{1} 𝐯_{1} +$ $\dots +$ $𝑎_{𝑛} 𝐯_{𝑛}$ . Then $𝟎 =$ $𝑇 (𝐯) =$ $𝑎_{1} 𝑇 (𝐯_{1}) +$ $\dots +$ $𝑎_{𝑛} 𝑇 (𝐯_{𝑛})$ . None of the vectors $𝑇 (𝐯_{𝑖})$ is $𝟎$ , so every coefficient $𝑎_{𝑖}$ must be $0$ . Hence $𝐯 = 𝟎$ , and $𝑇$ is injective by Theorem 2.6.11.

Solution.

Uses. Proposition 2.6.22 (Testing injectivity and surjectivity on a basis), Theorem 1.5.5 (Independence is decided at zero)

Techniques. Finding counterexamples and minimal examples (Technique 1.4.18)

Hint 1. Can a linear combination of nonzero vectors with nonzero coefficients be

𝟎

Hint 2. Try

𝑇 : 𝐹^{2} \to 𝐹

𝑇 (𝑥, 𝑦) = 𝑥 + 𝑦

, with the standard basis.

Sketch. The step “none of the

𝑇 (𝐯_{𝑖})

𝟎

, so every

𝑎_{𝑖} = 0

” confuses “nonzero entries” with “linearly independent”.

𝑇 (𝑥, 𝑦) = 𝑥 + 𝑦

has

𝑇 (𝐞_{1}) =

𝑇 (𝐞_{2}) =

1 \neq

0

but

𝑇 (1, - 1) = 0

. Correct version:

𝑇

is injective iff

𝑇 (ℬ︀)

is linearly independent (Proposition 2.6.22(a)).

The error is the step “none of the vectors $𝑇 (𝐯_{𝑖})$ is $𝟎$ , so every coefficient $𝑎_{𝑖}$ must be $0$ ”. A linear combination of nonzero vectors can vanish with nonzero coefficients; what the step needs is that $𝑇 (𝐯_{1}),$ $\dots,$ $𝑇 (𝐯_{𝑛})$ is linearly independent (Theorem 1.5.5), and nonzero entries do not guarantee that.

The claim is false. Let $𝑇 : 𝐹^{2} \to 𝐹$ , $𝑇 (𝑥, 𝑦) = 𝑥 + 𝑦$ , with the standard basis. Then $𝑇 (𝐞_{1}) =$ $𝑇 (𝐞_{2}) =$ $1 \neq$ $0$ , but $𝑇 (1, - 1) = 0$ with $(1, - 1) \neq 𝟎$ , so $𝑇$ is not injective.

The correct statement is Proposition 2.6.22(a): $𝑇$ is injective if and only if $𝑇 (ℬ︀)$ is linearly independent. The hypothesis of the claim only says that each one-entry list $𝑇 (𝐯_{𝑖})$ is linearly independent.

Exercise 2.6.41.standard

(a)Let $𝐷 \in L (𝔽_{2} {[𝑥]}_{\leq 3})$ be differentiation. Find $ker 𝐷$ and $im 𝐷$ , and show that they are equal.
(b)Find $ker 𝐷$ and $im 𝐷$ for differentiation $𝐷 \in L (𝔽_{3} {[𝑥]}_{\leq 3})$ . Are they equal?
(c)Show that $𝐷^{2} = 0$ on all of $𝔽_{2} [𝑥]$ .

Solution.

Uses. Example 2.1.9 (Maps of polynomials), Definition 2.6.5 (Kernel), Definition 2.6.6 (Image)

Hint 1. Write

𝑝 =

𝑎 +

𝑏 𝑥 +

𝑐 𝑥^{2} +

𝑑 𝑥^{3}

, so

𝐷 (𝑝) =

𝑏 +

2 𝑐 𝑥 +

3 𝑑 𝑥^{2}

, and reduce the coefficients

2

and

3

in each field.

Hint 2. For (c),

𝐷^{2} (𝑥^{𝑘}) =

𝑘 (𝑘 - 1) 𝑥^{𝑘 - 2}

, and

𝑘 (𝑘 - 1)

is always even.

Sketch. (a) Over

𝔽_{2}

𝐷 (𝑝) = 𝑏 + 𝑑 𝑥^{2}

, so

ker 𝐷 =

span (1, 𝑥^{2}) =

im 𝐷

. (b) Over

𝔽_{3}

𝐷 (𝑝) = 𝑏 + 2 𝑐 𝑥

, so

ker 𝐷 = span (1, 𝑥^{3})

and

im 𝐷 = span (1, 𝑥)

; they are not equal. (c)

𝐷^{2} (\sum 𝑎_{𝑘} 𝑥^{𝑘}) =

\sum 𝑘 (𝑘 - 1) 𝑎_{𝑘} 𝑥^{𝑘 - 2}

, and

𝑘 (𝑘 - 1) * 1 = 0

𝔽_{2}

Write $𝑝 =$ $𝑎 +$ $𝑏 𝑥 +$ $𝑐 𝑥^{2} +$ $𝑑 𝑥^{3}$ , so that $𝐷 (𝑝) =$ $𝑏 +$ $2 𝑐 𝑥 +$ $3 𝑑 𝑥^{2}$ (Example 2.1.9).

(a) In $𝔽_{2}$ , $2 = 0$ and $3 = 1$ , so $𝐷 (𝑝) = 𝑏 + 𝑑 𝑥^{2}$ . This is $𝟎$ if and only if $𝑏 = 𝑑 = 0$ , so $ker 𝐷 =$ ${𝑎 + 𝑐 𝑥^{2} | 𝑎, 𝑐 \in 𝔽_{2}} =$ $span (1, 𝑥^{2})$ . The values $𝑏 + 𝑑 𝑥^{2}$ , with $𝑏, 𝑑$ arbitrary, form $span (1, 𝑥^{2})$ as well. So $ker 𝐷 = im 𝐷$ .

(b) In $𝔽_{3}$ , $3 = 0$ , so $𝐷 (𝑝) = 𝑏 + 2 𝑐 𝑥$ . Since $2 \neq 0$ in $𝔽_{3}$ , this is $𝟎$ if and only if $𝑏 = 𝑐 = 0$ , so $ker 𝐷 = span (1, 𝑥^{3})$ . Every $𝑒 + 𝑓 𝑥$ is a value, namely $𝐷 (𝑒 𝑥 + 2 𝑓 𝑥^{2})$ , because $2 \cdot 2 𝑓 = 4 𝑓 = 𝑓$ in $𝔽_{3}$ . So $im 𝐷 = span (1, 𝑥)$ . These are not equal, since $𝑥^{3} \in ker 𝐷$ but $𝑥^{3} \notin im 𝐷$ . They share the constants.

(c) For $𝑓 =$ $\sum_{𝑘} 𝑎_{𝑘} 𝑥^{𝑘} \in$ $𝔽_{2} [𝑥]$ , applying the formula for $𝐷$ twice gives $𝐷^{2} (𝑓) =$ $\sum_{𝑘 \geq 2} (𝑘 (𝑘 - 1) * 1) 𝑎_{𝑘} 𝑥^{𝑘 - 2}$ . One of $𝑘$ and $𝑘 - 1$ is even, so $𝑘 (𝑘 - 1) * 1 = 0$ in $𝔽_{2}$ , and $𝐷^{2} (𝑓) = 0$ . In particular $im 𝐷 \subseteq ker 𝐷$ on $𝔽_{2} [𝑥]$ , which is what made (a) possible.

Exercise 2.6.42.standard Let $𝑇 \in L (𝑉, 𝑊)$ be injective, and let $𝑆 \in L (𝑈, 𝑊)$ .

(a)Suppose $im 𝑆 \subseteq im 𝑇$ . Show that there is exactly one linear map $𝑅 : 𝑈 \to 𝑉$ with $𝑇 𝑅 = 𝑆$ .
(b)Show that the hypothesis $im 𝑆 \subseteq im 𝑇$ in (a) is necessary.

Solution.

Uses. Definition 2.6.1 (Injective), Definition 2.6.6 (Image), Proposition 2.6.15 (Kernels and images of composites), Definition 2.1.1 (Linear map)

Techniques. Uniqueness via two candidates (Technique 1.1.8)

Hint 1. For each

𝐮 \in 𝑈

, the vector

𝑆 (𝐮)

lies in

im 𝑇

. How many

𝐯 \in 𝑉

satisfy

𝑇 (𝐯) = 𝑆 (𝐮)

Hint 2. Define

𝑅 (𝐮)

to be that vector. To prove that

𝑅

is linear, apply

𝑇

𝑅 (𝑎 𝐮 + 𝐮^{'})

and to

𝑎 𝑅 (𝐮) + 𝑅 (𝐮^{'})

, and use injectivity.

Hint 3. For (b), compute

im (𝑇 𝑅)

with Proposition 2.6.15(b).

Sketch. Since

𝑇

is injective and

𝑆 (𝐮) \in im 𝑇

, there is exactly one

𝐯

with

𝑇 (𝐯) = 𝑆 (𝐮)

; call it

𝑅 (𝐮)

. Then

𝑇 𝑅 = 𝑆

, and

𝑇 (𝑅 (𝑎 𝐮 + 𝐮^{'})) =

𝑆 (𝑎 𝐮 + 𝐮^{'}) =

𝑇 (𝑎 𝑅 (𝐮) + 𝑅 (𝐮^{'}))

, so

𝑅

is linear by injectivity. If

𝑇 𝑅^{'} = 𝑆

, then

𝑇 (𝑅^{'} (𝐮)) =

𝑆 (𝐮)

forces

𝑅^{'} (𝐮) = 𝑅 (𝐮)

. (b)

𝑇 𝑅 = 𝑆

gives

im 𝑆 =

im (𝑇 𝑅) \subseteq

im 𝑇

(a) Existence. Let $𝐮 \in 𝑈$ . Since $𝑆 (𝐮) \in im 𝑆 \subseteq im 𝑇$ , there is some $𝐯 \in 𝑉$ with $𝑇 (𝐯) = 𝑆 (𝐮)$ , and since $𝑇$ is injective there is only one. Let $𝑅 (𝐮)$ be this vector. This defines a function $𝑅 : 𝑈 \to 𝑉$ with $𝑇 (𝑅 (𝐮)) = 𝑆 (𝐮)$ for every $𝐮$ , that is, $𝑇 𝑅 = 𝑆$ . To see that $𝑅$ is linear, let $𝐮, 𝐮^{'} \in 𝑈$ and $𝑎 \in 𝐹$ . Since $𝑇$ and $𝑆$ are linear,

𝑇 (𝑎 𝑅 (𝐮) + 𝑅 (𝐮^{'})) = 𝑎 𝑆 (𝐮) + 𝑆 (𝐮^{'}) = 𝑆 (𝑎 𝐮 + 𝐮^{'}) = 𝑇 (𝑅 (𝑎 𝐮 + 𝐮^{'})) .

As $𝑇$ is injective, $𝑅 (𝑎 𝐮 + 𝐮^{'}) =$ $𝑎 𝑅 (𝐮) +$ $𝑅 (𝐮^{'})$ .

Uniqueness. If $𝑅^{'} : 𝑈 \to 𝑉$ also satisfies $𝑇 𝑅^{'} = 𝑆$ , then $𝑇 (𝑅^{'} (𝐮)) =$ $𝑆 (𝐮) =$ $𝑇 (𝑅 (𝐮))$ for every $𝐮$ , so $𝑅^{'} (𝐮) = 𝑅 (𝐮)$ by injectivity. Hence $𝑅^{'} = 𝑅$ .

(b) If $𝑇 𝑅 = 𝑆$ for some linear $𝑅$ , then $im 𝑆 =$ $im (𝑇 𝑅) \subseteq$ $im 𝑇$ by Proposition 2.6.15(b). So without $im 𝑆 \subseteq im 𝑇$ there is no such $𝑅$ at all.

Exercise 2.6.43.hard Let $𝑇 \in L (𝑉, 𝑊)$ be surjective, and let $𝑆 \in L (𝑉, 𝑈)$ .

(a)Suppose $ker 𝑇 \subseteq ker 𝑆$ . Show that there is exactly one linear map $𝑅 : 𝑊 \to 𝑈$ with $𝑅 𝑇 = 𝑆$ .
(b)Show that the hypothesis $ker 𝑇 \subseteq ker 𝑆$ in (a) is necessary.

Solution.

Uses. Definition 2.6.2 (Surjective), Definition 2.6.5 (Kernel), Proposition 2.6.15 (Kernels and images of composites), Definition 2.1.1 (Linear map)

Techniques. Uniqueness via two candidates (Technique 1.1.8)

Hint 1. If

𝑅 𝑇 = 𝑆

, then

𝑅 (𝑇 (𝐯)) = 𝑆 (𝐯)

, so there is only one candidate: for

𝐰 \in 𝑊

, pick any

𝐯

with

𝑇 (𝐯) = 𝐰

and set

𝑅 (𝐰) = 𝑆 (𝐯)

Hint 2. The danger is that

𝑅 (𝐰)

might depend on the choice of

𝐯

. If

𝑇 (𝐯) = 𝑇 (𝐯^{'}) = 𝐰

, where does

𝐯 - 𝐯^{'}

lie?

Hint 3. For linearity, if

𝑇 (𝐯) = 𝐰

and

𝑇 (𝐯^{'}) = 𝐰^{'}

, then

𝑎 𝐯 + 𝐯^{'}

is an allowed choice for

𝑎 𝐰 + 𝐰^{'}

Sketch. For

𝐰 \in 𝑊

choose

𝐯

with

𝑇 (𝐯) = 𝐰

and put

𝑅 (𝐰) = 𝑆 (𝐯)

. Another choice

𝐯^{'}

has

𝐯 -

𝐯^{'} \in

ker 𝑇 \subseteq

ker 𝑆

, so

𝑆 (𝐯) = 𝑆 (𝐯^{'})

𝑅

is well defined. Choosing

𝑎 𝐯 + 𝐯^{'}

for

𝑎 𝐰 + 𝐰^{'}

shows that

𝑅

is linear, and

𝑅 (𝑇 (𝐯)) = 𝑆 (𝐯)

by construction. Uniqueness:

𝑅^{'} 𝑇 = 𝑅 𝑇

and

𝑇

surjective give

𝑅^{'} = 𝑅

. (b) If

𝑅 𝑇 = 𝑆

, then

ker 𝑇 \subseteq

ker (𝑅 𝑇) =

ker 𝑆

(a) Definition of $𝑅$ . Let $𝐰 \in 𝑊$ . Since $𝑇$ is surjective, there is some $𝐯 \in 𝑉$ with $𝑇 (𝐯) = 𝐰$ , and we would like to set $𝑅 (𝐰) = 𝑆 (𝐯)$ . For this to make sense, the value must not depend on which preimage $𝐯$ we pick. If $𝑇 (𝐯) = 𝑇 (𝐯^{'}) = 𝐰$ , then $𝑇 (𝐯 - 𝐯^{'}) = 𝟎$ , so $𝐯 -$ $𝐯^{'} \in$ $ker 𝑇 \subseteq$ $ker 𝑆$ , and $𝑆 (𝐯) -$ $𝑆 (𝐯^{'}) =$ $𝑆 (𝐯 - 𝐯^{'}) =$ $𝟎$ . So $𝑆 (𝐯)$ is the same for every preimage $𝐯$ of $𝐰$ , and $𝑅 (𝐰) = 𝑆 (𝐯)$ , for any such $𝐯$ , defines a function $𝑅 : 𝑊 \to 𝑈$ .

$𝑅$ is linear. Let $𝐰, 𝐰^{'} \in 𝑊$ and $𝑎 \in 𝐹$ , and choose $𝐯, 𝐯^{'}$ with $𝑇 (𝐯) = 𝐰$ and $𝑇 (𝐯^{'}) = 𝐰^{'}$ . Then $𝑇 (𝑎 𝐯 + 𝐯^{'}) =$ $𝑎 𝐰 +$ $𝐰^{'}$ , so $𝑎 𝐯 + 𝐯^{'}$ is a preimage of $𝑎 𝐰 + 𝐰^{'}$ , and

𝑅 (𝑎 𝐰 + 𝐰^{'}) = 𝑆 (𝑎 𝐯 + 𝐯^{'}) = 𝑎 𝑆 (𝐯) + 𝑆 (𝐯^{'}) = 𝑎 𝑅 (𝐰) + 𝑅 (𝐰^{'}) .

$𝑅 𝑇 = 𝑆$ . For $𝐯 \in 𝑉$ , the vector $𝐯$ is a preimage of $𝑇 (𝐯)$ , so $𝑅 (𝑇 (𝐯)) = 𝑆 (𝐯)$ .

Uniqueness. If $𝑅^{'} 𝑇 = 𝑆$ as well, then $𝑅^{'} (𝑇 (𝐯)) =$ $𝑅 (𝑇 (𝐯))$ for every $𝐯 \in 𝑉$ . Every $𝐰 \in 𝑊$ is $𝑇 (𝐯)$ for some $𝐯$ , so $𝑅^{'} = 𝑅$ .

(b) If $𝑅 𝑇 = 𝑆$ for some linear $𝑅$ , then $ker 𝑇 \subseteq$ $ker (𝑅 𝑇) =$ $ker 𝑆$ by Proposition 2.6.15(a).

The construction is the general pattern for defining a map on $𝑊$ through a surjection $𝑇$ : define it on preimages, and check that the result does not depend on the choice. Since the preimages of $𝐰$ form the fiber $𝑇^{- 1} (𝐰)$ , a translate of $ker 𝑇$ (Corollary 2.6.31), the check is exactly that $𝑆$ is constant on translates of $ker 𝑇$ , which is what $ker 𝑇 \subseteq ker 𝑆$ says. The same pattern defines the maps out of quotient spaces in Section 4.6.

Exercise 2.6.44.standard Let $𝑝$ be a prime, let $𝑉$ and $𝑊$ be vector spaces over $𝔽_{𝑝}$ with $𝑉$ finite, and let $𝑇 \in L (𝑉, 𝑊)$ .

(a)Show that $| 𝑉 | =$ $| ker 𝑇 | \cdot$ $| im 𝑇 |$ .
(b)Deduce that if $𝑇 : 𝔽_{𝑝}^{𝑛} \to 𝔽_{𝑝}^{𝑚}$ is linear and injective, then $𝑛 \leq 𝑚$ , and if it is linear and surjective, then $𝑛 \geq 𝑚$ .

Solution.

Uses. Corollary 2.6.31 (Fibers are translates of the kernel)

Hint 1. By Corollary 2.6.31,

𝑉

is the disjoint union of the fibers

𝑇^{- 1} (𝐰)

over

𝐰 \in im 𝑇

. How many elements does each fiber have?

Hint 2. For (b),

| 𝔽_{𝑝}^{𝑛} | = 𝑝^{𝑛}

. What are

| ker 𝑇 |

and

| im 𝑇 |

when

𝑇

is injective? When it is surjective?

Sketch. (a) The fibers over the

| im 𝑇 |

vectors of

im 𝑇

partition

𝑉

, and each is in bijection with

ker 𝑇

, so

| 𝑉 | =

| im 𝑇 | \cdot

| ker 𝑇 |

. (b) Injective:

| ker 𝑇 | = 1

, so

𝑝^{𝑛} = | im 𝑇 | \leq 𝑝^{𝑚}

. Surjective:

𝑝^{𝑛} =

| ker 𝑇 | 𝑝^{𝑚} \geq

𝑝^{𝑚}

(a) By Corollary 2.6.31(c), $𝑉$ is the disjoint union of the fibers $𝑇^{- 1} (𝐰)$ with $𝐰 \in im 𝑇$ , and by part (b) of that corollary each of these fibers is in bijection with $ker 𝑇$ , so it has $| ker 𝑇 |$ elements. Since $𝑉$ is finite, so are $ker 𝑇$ and $im 𝑇$ , and adding up the sizes of the fibers gives

| 𝑉 | = \sum_{𝐰 \in im 𝑇} | 𝑇^{- 1} (𝐰) | = | im 𝑇 | \cdot | ker 𝑇 | .

(b) $𝔽_{𝑝}^{𝑛}$ has $𝑝^{𝑛}$ elements. If $𝑇$ is injective, then $ker 𝑇 = {𝟎}$ , so by (a) $| im 𝑇 | = 𝑝^{𝑛}$ . Since $im 𝑇 \subseteq 𝔽_{𝑝}^{𝑚}$ , $𝑝^{𝑛} \leq 𝑝^{𝑚}$ , and $𝑛 \leq 𝑚$ . If $𝑇$ is surjective, then $im 𝑇 = 𝔽_{𝑝}^{𝑚}$ , so $𝑝^{𝑛} =$ $| ker 𝑇 | \cdot$ $𝑝^{𝑚} \geq$ $𝑝^{𝑚}$ , and $𝑛 \geq 𝑚$ .

Part (b) is the “moreover” clause of the Fundamental Theorems over $𝔽_{𝑝}$ . Over other fields it needs a different argument (Section 3.5). Section 4.2 proves $dim 𝑉 =$ $dim ker 𝑇 +$ $dim im 𝑇$ over every field; over $𝔽_{𝑝}$ this is equivalent to (a), because a space of dimension $𝑑$ over $𝔽_{𝑝}$ has $𝑝^{𝑑}$ elements.

Exercise 2.6.45.hard Let

𝑝

be a prime. Show that exactly

(𝑝^{3} - 1) (𝑝^{3} - 𝑝)

linear maps

𝔽_{𝑝}^{3} \to 𝔽_{𝑝}^{2}

are surjective. (Every such map is

𝑇_{𝐴}

for exactly one

𝐴 \in 𝔽_{𝑝}^{2 \times 3}

, by Theorem 2.3.20 and Proposition 2.3.16. Use Corollary 2.6.24 and Theorem 1.4.25.)

Solution.

Uses. Corollary 2.6.24 (The columns decide), Theorem 1.4.25 (The subspaces of $𝐹^{2}$ ), Theorem 2.3.20 (Linear maps correspond to matrices), Proposition 2.3.16 (The matrix of $𝑇_{𝐴}$ ), Proposition 2.6.18 (The column space is spanned by the columns)

Hint 1. By Corollary 2.6.24, count the matrices

𝐴 \in 𝔽_{𝑝}^{2 \times 3}

whose three columns span

𝔽_{𝑝}^{2}

. It is easier to count the complement.

Hint 2. By Theorem 1.4.25, if the columns do not span

𝔽_{𝑝}^{2}

, their span is

{𝟎}

or a line. How many lines through

𝟎

does

𝔽_{𝑝}^{2}

have, and how many column triples lie in a given line?

Hint 3. Each line has

𝑝 - 1

nonzero vectors, distinct lines share only

𝟎

, and

𝔽_{𝑝}^{2}

has

𝑝^{2} - 1

nonzero vectors.

Sketch. Count triples of columns that do not span: their span is

{𝟎}

or one of the

𝑝 + 1

lines. All three columns lie in a given line in

𝑝^{3}

ways, and a triple other than the zero triple lies in at most one line. So

1 + (𝑝 + 1) (𝑝^{3} - 1)

triples fail to span, and

𝑝^{6} -

1 -

(𝑝 + 1) (𝑝^{3} - 1) =

𝑝^{6} -

𝑝^{4} -

𝑝^{3} +

𝑝 =

(𝑝^{3} - 1) (𝑝^{3} - 𝑝)

succeed.

The maps $𝑇_{𝐴}$ with $𝐴 \in 𝔽_{𝑝}^{2 \times 3}$ are all the linear maps $𝔽_{𝑝}^{3} \to 𝔽_{𝑝}^{2}$ , each exactly once. By Corollary 2.6.24(b), $𝑇_{𝐴}$ is surjective if and only if the columns $𝐚_{1}, 𝐚_{2}, 𝐚_{3}$ span $𝔽_{𝑝}^{2}$ . There are $𝑝^{6}$ matrices in all, so we count those whose columns do not span $𝔽_{𝑝}^{2}$ .

Lines. By Theorem 1.4.25, the span of the columns is ${𝟎}$ , a line $span (𝐯)$ with $𝐯 \neq 𝟎$ , or $𝔽_{𝑝}^{2}$ . A line $span (𝐯)$ consists of the $𝑝$ vectors $𝑐 𝐯$ , $𝑐 \in 𝔽_{𝑝}$ , which are distinct since $𝐯 \neq 𝟎$ ; so it has $𝑝 - 1$ nonzero vectors. Two different lines share only $𝟎$ : if $𝐮 \neq 𝟎$ lies on $span (𝐯)$ , then $𝐮 = 𝑐 𝐯$ with $𝑐 \neq 0$ , and $span (𝐮) = span (𝐯)$ by Theorem 1.4.25. So the $𝑝^{2} - 1$ nonzero vectors of $𝔽_{𝑝}^{2}$ are split among the lines, $𝑝 - 1$ to a line, and there are $(𝑝^{2} - 1) / (𝑝 - 1) =$ $𝑝 +$ $1$ lines.

Counting. The columns fail to span $𝔽_{𝑝}^{2}$ exactly when all three lie in a common line (this includes the zero matrix, whose columns lie in every line). For a fixed line there are $𝑝^{3}$ choices of three columns in it, one of which is the zero triple. A triple with a nonzero column $𝐚_{𝑗}$ lies only in the line $span (𝐚_{𝑗})$ , so it is counted once. Hence the number of matrices whose columns do not span is

1 + (𝑝 + 1) (𝑝^{3} - 1) = 𝑝^{4} + 𝑝^{3} - 𝑝,

and the number of surjective maps is

𝑝^{6} - 𝑝^{4} - 𝑝^{3} + 𝑝 = (𝑝^{3} - 1) (𝑝^{3} - 𝑝) .

For $𝑝 = 2$ this is $7 \cdot 6 = 42$ of the $64$ linear maps $𝔽_{2}^{3} \to 𝔽_{2}^{2}$ .

The factored form has its own meaning: $(𝑝^{3} - 1) (𝑝^{3} - 𝑝)$ is the number of ways to choose a nonzero first row and then a second row that is not a multiple of it, so it counts the $2 \times 3$ matrices with linearly independent rows. That the two counts agree is a first sign that row rank equals column rank (Section 4.3).

Exercise 2.6.46.standard Let $Φ : 𝐹^{𝑛 \times 𝑛} \to 𝐹^{𝑛 \times 𝑛}$ , $Φ (𝐴) = 𝐴 - 𝐴^{𝖳}$ .

(a)Show that $Φ$ is linear and that $ker Φ$ is the subspace of symmetric matrices.
(b)Suppose $1 + 1 \neq 0$ in $𝐹$ . Show that $im Φ$ is the subspace of skew-symmetric matrices.
(c)Over $𝔽_{2}$ , show that $im Φ$ consists of the symmetric matrices with all diagonal entries $0$ . Deduce that $im Φ$ is strictly smaller than the set of skew-symmetric matrices when $𝑛 \geq 1$ .

Solution.

Uses. Proposition 2.4.22 (Transpose is a linear involution), Definition 2.4.21 (Transpose), Definition 2.6.5 (Kernel), Definition 2.6.6 (Image)

Techniques. Compute entry by entry (Technique 2.4.15)

Hint 1. Transposition is linear and

{(𝐴^{𝖳})}^{𝖳} = 𝐴

(Proposition 2.4.22). For (b), compute

{Φ (𝐴)}^{𝖳}

, and for the converse try

𝐴 = 2^{- 1} 𝐵

Hint 2. For (c), the

(𝑖, 𝑗)

entry of

Φ (𝐴)

over

𝔽_{2}

𝑎_{𝑖 𝑗} + 𝑎_{𝑗 𝑖}

. Given a symmetric

𝐵

with zero diagonal, try

𝐴

= the part of

𝐵

strictly below the diagonal.

Sketch. (a)

Φ (𝐴) = 0

iff

𝐴^{𝖳} = 𝐴

. (b)

{Φ (𝐴)}^{𝖳} = - Φ (𝐴)

, and

Φ (𝐵 / 2) = 𝐵

for skew-symmetric

𝐵

. (c) Over

𝔽_{2}

Φ (𝐴) = 𝐴 + 𝐴^{𝖳}

is symmetric with diagonal entries

2 𝑎_{𝑖 𝑖} = 0

; conversely

𝐵 = 𝐿 + 𝐿^{𝖳} = Φ (𝐿)

for the strictly lower part

𝐿

𝐵

. Over

𝔽_{2}

skew-symmetric means symmetric, and

𝐼_{𝑛}

is symmetric but not in

im Φ

(a) By Proposition 2.4.22, ${(𝑐 𝐴 + 𝐵)}^{𝖳} =$ $𝑐 𝐴^{𝖳} +$ $𝐵^{𝖳}$ , so

Φ (𝑐 𝐴 + 𝐵) = 𝑐 𝐴 + 𝐵 - 𝑐 𝐴^{𝖳} - 𝐵^{𝖳} = 𝑐 Φ (𝐴) + Φ (𝐵) .

And $Φ (𝐴) = 0$ if and only if $𝐴^{𝖳} = 𝐴$ , that is, $𝐴$ is symmetric.

(b) For every $𝐴$ , ${Φ (𝐴)}^{𝖳} =$ $𝐴^{𝖳} -$ $𝐴 =$ $- Φ (𝐴)$ , so $im Φ$ consists of skew-symmetric matrices. Conversely, let $𝐵^{𝖳} = - 𝐵$ . Since $2 = 1 + 1 \neq 0$ ,

Φ (2^{- 1} 𝐵) = 2^{- 1} 𝐵 - 2^{- 1} 𝐵^{𝖳} = 2^{- 1} 𝐵 + 2^{- 1} 𝐵 = 𝐵 .

So $im Φ$ is the set of skew-symmetric matrices.

(c) Over $𝔽_{2}$ , $- 1 = 1$ , so $Φ (𝐴) = 𝐴 + 𝐴^{𝖳}$ , whose $(𝑖, 𝑗)$ entry is $𝑎_{𝑖 𝑗} + 𝑎_{𝑗 𝑖}$ . This is symmetric in $𝑖$ and $𝑗$ , and on the diagonal it is $𝑎_{𝑖 𝑖} + 𝑎_{𝑖 𝑖} = 0$ . Conversely, let $𝐵$ be symmetric with zero diagonal, and let $𝐿$ have entries $𝑙_{𝑖 𝑗} = 𝑏_{𝑖 𝑗}$ for $𝑖 > 𝑗$ and $𝑙_{𝑖 𝑗} = 0$ for $𝑖 \leq 𝑗$ . The $(𝑖, 𝑗)$ entry of $Φ (𝐿) = 𝐿 + 𝐿^{𝖳}$ is $𝑙_{𝑖 𝑗} + 𝑙_{𝑗 𝑖}$ , which is $𝑏_{𝑖 𝑗}$ if $𝑖 > 𝑗$ , $𝑏_{𝑗 𝑖} = 𝑏_{𝑖 𝑗}$ if $𝑖 < 𝑗$ , and $0 = 𝑏_{𝑖 𝑖}$ if $𝑖 = 𝑗$ . So $Φ (𝐿) = 𝐵$ , and $im Φ$ is exactly the set of symmetric matrices with zero diagonal.

Over $𝔽_{2}$ , $𝐵^{𝖳} = - 𝐵$ says the same thing as $𝐵^{𝖳} = 𝐵$ , so every symmetric matrix is skew-symmetric. In particular $𝐼_{𝑛}$ is skew-symmetric, but its diagonal entries are $1$ , so $𝐼_{𝑛} \notin im Φ$ . So the conclusion of (b) fails over $𝔽_{2}$ , and the image is a proper subset of $ker Φ$ .

Exercise 2.6.47.hard Let $𝐴 \in 𝐹^{𝑚 \times 𝑛}$ and $𝐵 \in 𝐹^{𝑛 \times 𝑚}$ .

(a)Show that if $𝐱 \in ker 𝑇_{𝐼_{𝑛} - 𝐵 𝐴}$ , then $𝐴 𝐱 \in$ $ker 𝑇_{𝐼_{𝑚} - 𝐴 𝐵}$ , and that if $𝐲 \in ker 𝑇_{𝐼_{𝑚} - 𝐴 𝐵}$ , then $𝐵 𝐲 \in$ $ker 𝑇_{𝐼_{𝑛} - 𝐵 𝐴}$ .
(b)Show that $𝐱 \mapsto 𝐴 𝐱$ is a bijection from $ker 𝑇_{𝐼_{𝑛} - 𝐵 𝐴}$ onto $ker 𝑇_{𝐼_{𝑚} - 𝐴 𝐵}$ , with inverse $𝐲 \mapsto 𝐵 𝐲$ .
(c)Deduce that $𝑇_{𝐼_{𝑛} - 𝐵 𝐴}$ is injective if and only if $𝑇_{𝐼_{𝑚} - 𝐴 𝐵}$ is.

Solution.

Uses. Proposition 2.4.14 (Algebra of matrices), Definition 2.6.5 (Kernel), Theorem 2.6.11 (Injectivity is decided at $𝟎$ )

Hint 1.

𝐱 \in ker 𝑇_{𝐼_{𝑛} - 𝐵 𝐴}

means

𝐵 𝐴 𝐱 = 𝐱

. Compute

𝐴 𝐵 (𝐴 𝐱)

by regrouping.

Hint 2. For (b), compose the two maps from (a) in both orders:

𝐵 (𝐴 𝐱) = 𝐵 𝐴 𝐱

and

𝐴 (𝐵 𝐲) = 𝐴 𝐵 𝐲

Hint 3. For (c), a bijection between the two kernels matches

{𝟎}

with

{𝟎}

Sketch. (a) If

𝐵 𝐴 𝐱 = 𝐱

, then

𝐴 𝐵 (𝐴 𝐱) =

𝐴 (𝐵 𝐴 𝐱) =

𝐴 𝐱

; symmetrically for

𝐲

. (b) On the kernels,

𝐵 (𝐴 𝐱) =

𝐵 𝐴 𝐱 =

𝐱

and

𝐴 (𝐵 𝐲) =

𝐴 𝐵 𝐲 =

𝐲

, so the maps are mutually inverse. (c) The bijection sends

𝟎

𝟎

, so one kernel is

{𝟎}

iff the other is.

(a) Let $𝐱 \in ker 𝑇_{𝐼_{𝑛} - 𝐵 𝐴}$ , that is, $𝐱 - 𝐵 𝐴 𝐱 = 𝟎$ , or $𝐵 𝐴 𝐱 = 𝐱$ . By associativity (Proposition 2.4.14),

𝐴 𝐵 (𝐴 𝐱) = 𝐴 (𝐵 𝐴 𝐱) = 𝐴 𝐱,

so $(𝐼_{𝑚} - 𝐴 𝐵) (𝐴 𝐱) =$ $𝟎$ , and $𝐴 𝐱 \in$ $ker 𝑇_{𝐼_{𝑚} - 𝐴 𝐵}$ . In the same way, if $𝐴 𝐵 𝐲 = 𝐲$ , then $𝐵 𝐴 (𝐵 𝐲) =$ $𝐵 (𝐴 𝐵 𝐲) =$ $𝐵 𝐲$ , so $𝐵 𝐲 \in$ $ker 𝑇_{𝐼_{𝑛} - 𝐵 𝐴}$ .

(b) By (a), $𝑓 (𝐱) = 𝐴 𝐱$ maps $𝐾_{1} =$ $ker 𝑇_{𝐼_{𝑛} - 𝐵 𝐴}$ into $𝐾_{2} =$ $ker 𝑇_{𝐼_{𝑚} - 𝐴 𝐵}$ , and $𝑔 (𝐲) = 𝐵 𝐲$ maps $𝐾_{2}$ into $𝐾_{1}$ . For $𝐱 \in 𝐾_{1}$ , $𝑔 (𝑓 (𝐱)) =$ $𝐵 𝐴 𝐱 =$ $𝐱$ , and for $𝐲 \in 𝐾_{2}$ , $𝑓 (𝑔 (𝐲)) =$ $𝐴 𝐵 𝐲 =$ $𝐲$ . So $𝑓$ and $𝑔$ are inverse bijections between $𝐾_{1}$ and $𝐾_{2}$ .

(c) The bijection $𝑓$ sends $𝟎$ to $𝟎$ , so $𝐾_{1} = {𝟎}$ if and only if $𝐾_{2} = {𝟎}$ . By Theorem 2.6.11, $𝑇_{𝐼_{𝑛} - 𝐵 𝐴}$ is injective if and only if $𝑇_{𝐼_{𝑚} - 𝐴 𝐵}$ is.

Compare Exercise 2.5.47, which produces an explicit inverse. In the language of Section 7.2, (b) says that $𝐴 𝐵$ and $𝐵 𝐴$ have “the same” eigenvectors for the eigenvalue $1$ ; the same argument works for every nonzero eigenvalue.

Exercise 2.6.48.routine Let $ℓ = span (𝐮)$ be a line in $ℝ^{2}$ , with $𝐮 = (cos 𝜃, sin 𝜃)$ , and let $ℓ^{⟂} =$ $span ((- sin 𝜃, cos 𝜃))$ . Let ${Pr}_{ℓ}$ be the projection onto $ℓ$ and $𝑆_{ℓ} = 2 {Pr}_{ℓ} - id$ the reflection across $ℓ$ (Example 2.1.11). Find the kernel and the image of each map.

(a) ${Pr}_{ℓ} - id$ .
(b) $𝑆_{ℓ} - id$ .
(c) $𝑆_{ℓ} + id$ .
(d) $𝑅_{𝜑} - id$ , for a rotation angle $𝜑$ that is not a multiple of $2 𝜋$ .

Solution.

Uses. Example 2.6.7 (First kernels and images), Example 2.1.11 (Rotations, projections, and reflections), Proposition 2.5.28 (Inverse of a $2 \times 2$ matrix), Proposition 2.5.20 (Inverses and matrices), Definition 2.5.12 (Invertible)

Hint 1. By Example 2.6.7(d), every

𝐯 = (𝑥, 𝑦)

𝐯 = 𝛼 𝐮 + 𝛽 𝐮^{⟂}

with

𝛼 =

𝑥 cos 𝜃 +

𝑦 sin 𝜃

𝛽 =

𝑦 cos 𝜃 -

𝑥 sin 𝜃

, and

𝐮^{⟂} =

(- sin 𝜃, cos 𝜃)

, and

{Pr}_{ℓ} (𝐯) = 𝛼 𝐮

Hint 2. So

({Pr}_{ℓ} - id) (𝐯) =

- 𝛽 𝐮^{⟂}

. Express (b) and (c) through

{Pr}_{ℓ}

. For (d), use Proposition 2.5.28.

Sketch. With

𝐯 = 𝛼 𝐮 + 𝛽 𝐮^{⟂}

({Pr}_{ℓ} - id) (𝐯) =

- 𝛽 𝐮^{⟂}

, so kernel

ℓ

and image

ℓ^{⟂}

. (b)

𝑆_{ℓ} - id = 2 ({Pr}_{ℓ} - id)

: same. (c)

𝑆_{ℓ} + id = 2 {Pr}_{ℓ}

: kernel

ℓ^{⟂}

, image

ℓ

. (d)

𝑎 𝑑 -

𝑏 𝑐 =

{(cos 𝜑 - 1)}^{2} +

{sin}^{2} 𝜑 =

2 -

2 cos 𝜑 \neq

0

, so

𝑅_{𝜑} - id

is invertible: kernel

{𝟎}

, image

ℝ^{2}

Write $𝑐 = cos 𝜃$ , $𝑠 = sin 𝜃$ , and $𝐮^{⟂} = (- 𝑠, 𝑐)$ . By Example 2.6.7(d), every $𝐯 = (𝑥, 𝑦)$ satisfies $𝐯 = 𝛼 𝐮 + 𝛽 𝐮^{⟂}$ with $𝛼 = 𝑥 𝑐 + 𝑦 𝑠$ and $𝛽 = 𝑦 𝑐 - 𝑥 𝑠$ , and ${Pr}_{ℓ} (𝐯) = 𝛼 𝐮$ . Note that $𝐯 = 𝑡 𝐮$ gives $𝛽 =$ $𝑡 𝑠 𝑐 -$ $𝑡 𝑐 𝑠 =$ $0$ , and $𝐯 = 𝑡 𝐮^{⟂}$ gives $𝛽 = 𝑡 𝑐^{2} + 𝑡 𝑠^{2} = 𝑡$ .

(a) $({Pr}_{ℓ} - id) (𝐯) =$ $𝛼 𝐮 -$ $𝐯 =$ $- 𝛽 𝐮^{⟂}$ . This is $𝟎$ if and only if $𝛽 = 0$ , that is, $𝐯 = 𝛼 𝐮 \in ℓ$ ; and every vector of $ℓ$ has $𝛽 = 0$ . So $ker ({Pr}_{ℓ} - id) = ℓ$ . Every value is a multiple of $𝐮^{⟂}$ , and $({Pr}_{ℓ} - id) (- 𝑡 𝐮^{⟂}) =$ $𝑡 𝐮^{⟂}$ , so $im ({Pr}_{ℓ} - id) = ℓ^{⟂}$ .

(b) $𝑆_{ℓ} -$ $id =$ $2 {Pr}_{ℓ} -$ $2 id =$ $2 ({Pr}_{ℓ} - id)$ . Multiplying a map by the nonzero scalar $2$ changes neither its kernel nor its image, so the kernel is $ℓ$ and the image is $ℓ^{⟂}$ : the reflection fixes $ℓ$ and moves every other vector by a multiple of $𝐮^{⟂}$ .

(c) $𝑆_{ℓ} + id = 2 {Pr}_{ℓ}$ , so by Example 2.6.7(d) its kernel is $ker {Pr}_{ℓ} = ℓ^{⟂}$ and its image is $im {Pr}_{ℓ} = ℓ$ .

(d) By Example 2.1.11, $𝑅_{𝜑} - id$ has matrix $(\begin{matrix} cos 𝜑 - 1 & - sin 𝜑 \\ sin 𝜑 & cos 𝜑 - 1 \end{matrix})$ , with $𝑎 𝑑 -$ $𝑏 𝑐 =$ ${(cos 𝜑 - 1)}^{2} +$ ${sin}^{2} 𝜑 =$ $2 -$ $2 cos 𝜑$ . Since $𝜑$ is not a multiple of $2 𝜋$ , $cos 𝜑 \neq 1$ , so this is nonzero, and the matrix is invertible by Proposition 2.5.28, so $𝑅_{𝜑} - id$ is invertible (Proposition 2.5.20(c)). An invertible map is bijective, so $ker (𝑅_{𝜑} - id) = {𝟎}$ and $im (𝑅_{𝜑} - id) = ℝ^{2}$ . Geometrically, a nontrivial rotation fixes only the origin.

Exercise 2.6.49.hard Let $𝑇 \in L (𝑉)$ satisfy $𝑇^{3} = 𝑇$ , and suppose $1 + 1 \neq 0$ in $𝐹$ .

(a)Show that every $𝐯 \in 𝑉$ can be written as $𝐯 = 𝐯_{0} + 𝐯_{+} + 𝐯_{-}$ with $𝐯_{0} \in ker 𝑇$ , $𝐯_{+} \in ker (𝑇 - id)$ , and $𝐯_{-} \in ker (𝑇 + id)$ .
(b)Show that the decomposition in (a) is unique.
(c)Show that (a) fails over $𝔽_{2}$ for $𝑇 = 𝑇_{𝐴}$ , $𝐴 = (\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix})$ .

Solution.

Uses. Definition 2.6.5 (Kernel), Proposition 2.1.24 (Algebra of composition), Proposition 1.2.35 (Zero products)

Techniques. Uniqueness via two candidates (Technique 1.1.8)

Hint 1. Look for the three pieces as combinations of

𝐯

𝑇 (𝐯)

, and

𝑇^{2} (𝐯)

. The piece in

ker 𝑇

must be killed by

𝑇

, and

𝑇 (𝐯 - 𝑇^{2} (𝐯)) =

𝑇 (𝐯) -

𝑇^{3} (𝐯) =

𝟎

Hint 2. Try

𝐯_{\pm} =

2^{- 1} (𝑇^{2} (𝐯) \pm 𝑇 (𝐯))

. Where do these come from? When

2 \neq 0

, the identity

1 =

(1 - 𝑥^{2}) +

(𝑥^{2} + 𝑥) / 2 +

(𝑥^{2} - 𝑥) / 2

splits

1

along the roots

0, 1, - 1

𝑥^{3} - 𝑥

: each term is

1

at one root and

0

at the other two.

Hint 3. For (b), suppose

𝐯_{0} + 𝐯_{+} + 𝐯_{-} = 𝟎

and apply

𝑇

, then

𝑇

again.

Sketch. (a)

𝐯_{0} = 𝐯 - 𝑇^{2} 𝐯

𝐯_{\pm} =

2^{- 1} (𝑇^{2} 𝐯 \pm 𝑇 𝐯)

; these add up to

𝐯

, and

𝑇^{3} = 𝑇

shows

𝑇 𝐯_{0} = 𝟎

and

𝑇 𝐯_{\pm} = \pm 𝐯_{\pm}

. (b) If the three pieces add to

𝟎

, applying

𝑇

gives

𝐯_{+} - 𝐯_{-} = 𝟎

, and applying

𝑇

twice gives

𝐯_{+} + 𝐯_{-} = 𝟎

; as

2 \neq 0

, all pieces are

𝟎

. (c)

𝐴^{2} = 𝐼

over

𝔽_{2}

, so

𝐴^{3} = 𝐴

, but

ker 𝑇_{𝐴} = {𝟎}

and

ker 𝑇_{𝐴 - 𝐼} =

ker 𝑇_{𝐴 + 𝐼} =

span (𝐞_{1})

, which do not add up to

𝔽_{2}^{2}

(a) Write $𝑇 𝐯$ for $𝑇 (𝐯)$ , and let

𝐯_{0} = 𝐯 - 𝑇^{2} 𝐯, 𝐯_{+} = 2^{- 1} (𝑇^{2} 𝐯 + 𝑇 𝐯), 𝐯_{-} = 2^{- 1} (𝑇^{2} 𝐯 - 𝑇 𝐯) .

Then $𝐯_{0} +$ $𝐯_{+} +$ $𝐯_{-} =$ $𝐯 -$ $𝑇^{2} 𝐯 +$ $𝑇^{2} 𝐯 =$ $𝐯$ . Using $𝑇^{3} = 𝑇$ ,

\begin{aligned} 𝑇 𝐯_{0} & = 𝑇 𝐯 - 𝑇^{3} 𝐯 = 𝟎, \\ 𝑇 𝐯_{+} & = 2^{- 1} (𝑇^{3} 𝐯 + 𝑇^{2} 𝐯) = 2^{- 1} (𝑇 𝐯 + 𝑇^{2} 𝐯) = 𝐯_{+}, \\ 𝑇 𝐯_{-} & = 2^{- 1} (𝑇^{3} 𝐯 - 𝑇^{2} 𝐯) = 2^{- 1} (𝑇 𝐯 - 𝑇^{2} 𝐯) = - 𝐯_{-} . \end{aligned}

So $𝐯_{0} \in ker 𝑇$ , $𝐯_{+} \in ker (𝑇 - id)$ , and $𝐯_{-} \in ker (𝑇 + id)$ .

(b) It suffices to show that if $𝐳_{0} + 𝐳_{+} + 𝐳_{-} = 𝟎$ with $𝐳_{0} \in ker 𝑇$ and $𝑇 𝐳_{\pm} = \pm 𝐳_{\pm}$ , then all three are $𝟎$ : the difference of two decompositions of $𝐯$ is such a decomposition of $𝟎$ , since each of the three kernels is a subspace. Applying $𝑇$ gives $𝐳_{+} - 𝐳_{-} = 𝟎$ , and applying $𝑇$ once more gives $𝐳_{+} + 𝐳_{-} = 𝟎$ . Adding, $2 𝐳_{+} = 𝟎$ , so $𝐳_{+} = 𝟎$ because $2 \neq 0$ (Proposition 1.2.35). Then $𝐳_{-} = 𝟎$ , and $𝐳_{0} = 𝟎$ .

(c) Over $𝔽_{2}$ , $𝐴^{2} = (\begin{matrix} 1 & 2 \\ 0 & 1 \end{matrix}) = 𝐼_{2}$ , so $𝐴^{3} = 𝐴$ . Since $𝐴$ is invertible, $ker 𝑇_{𝐴} = {𝟎}$ . Also $- 1 = 1$ , so $𝐴 -$ $𝐼_{2} =$ $𝐴 +$ $𝐼_{2} =$ $(\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix})$ , whose kernel is $span (𝐞_{1})$ . Every sum $𝐯_{0} + 𝐯_{+} + 𝐯_{-}$ as in (a) therefore lies in $span (𝐞_{1})$ , and $𝐞_{2}$ has no such decomposition.

The polynomial identity in the second hint is Lagrange interpolation (Proposition 3.4.27) at the points $0, 1, - 1$ . Decompositions of this kind, along the roots of a polynomial that $𝑇$ satisfies, are the subject of Section 7.4.

Exercise 2.6.50.challenge

(a)Let $𝑆, 𝑇 \in L (𝑉, 𝑊)$ both be nonzero. Show that some $𝐯 \in 𝑉$ has $𝑆 (𝐯) \neq 𝟎$ and $𝑇 (𝐯) \neq 𝟎$ .
(b)Find three nonzero linear maps $𝑇_{1},$ $𝑇_{2},$ $𝑇_{3} : 𝔽_{2}^{2} \to$ $𝔽_{2}$ such that every $𝐯 \in 𝔽_{2}^{2}$ lies in some $ker 𝑇_{𝑖}$ .
(c)Let $𝐹$ be infinite, and let $𝑇_{1},$ $\dots,$ $𝑇_{𝑘} \in$ $L (𝑉, 𝑊)$ all be nonzero. Show that some $𝐯 \in 𝑉$ has $𝑇_{𝑖} (𝐯) \neq 𝟎$ for every $𝑖$ .

Solution.

Uses. Proposition 1.4.20 (Unions of two subspaces), Proposition 2.6.8 (Kernels and images are subspaces), Proposition 1.2.35 (Zero products)

Techniques. Finding counterexamples and minimal examples (Technique 1.4.18), Induction on the length of a list (Technique 1.6.9)

Hint 1. (a) asks for a vector outside

ker 𝑆 \cup ker 𝑇

. When can a union of two subspaces be all of

𝑉

Hint 2. (b)

𝔽_{2}^{2}

has three lines through

𝟎

Hint 3. (c) Induct on

𝑘

. Given

𝐯

that works for

𝑇_{1}, \dots, 𝑇_{𝑘 - 1}

and

𝐮

with

𝑇_{𝑘} (𝐮) \neq 𝟎

, look at the line

𝐯 + 𝑡 𝐮

𝑡 \in 𝐹

. For how many

𝑡

can

𝑇_{𝑖} (𝐯 + 𝑡 𝐮) = 𝟎

Sketch. (a)

ker 𝑆

and

ker 𝑇

are proper subspaces, and if their union were

𝑉

it would be a subspace, so one contains the other (Proposition 1.4.20) and equals

𝑉

. (b)

𝑥

𝑦

𝑥 + 𝑦

. (c) Induct: for each

𝑖 \leq 𝑘

𝑇_{𝑖} (𝐯 + 𝑡 𝐮) =

𝑇_{𝑖} (𝐯) +

𝑡 𝑇_{𝑖} (𝐮)

vanishes for at most one

𝑡

, since two such

𝑡

would give

𝑇_{𝑖} (𝐮) = 𝟎

and then

𝑇_{𝑖} (𝐯) = 𝟎

. So at most

𝑘

values of

𝑡

are bad, and

𝐹

is infinite.

(a) Since $𝑆 \neq 0$ and $𝑇 \neq 0$ , the kernels $ker 𝑆$ and $ker 𝑇$ are subspaces of $𝑉$ (Proposition 2.6.8) different from $𝑉$ . Suppose $ker 𝑆 \cup ker 𝑇 = 𝑉$ . Then the union is a subspace, so by Proposition 1.4.20 one kernel contains the other, and the union is the larger kernel. That kernel would be $𝑉$ , a contradiction. So some $𝐯$ lies in neither kernel.

(b) Let $𝑇_{1} (𝑥, 𝑦) = 𝑥$ , $𝑇_{2} (𝑥, 𝑦) = 𝑦$ , and $𝑇_{3} (𝑥, 𝑦) = 𝑥 + 𝑦$ . Then $(0, 0)$ lies in every kernel, $(1, 0) \in ker 𝑇_{2}$ , $(0, 1) \in ker 𝑇_{1}$ , and $(1, 1) \in ker 𝑇_{3}$ , since $1 + 1 = 0$ . These are all the vectors of $𝔽_{2}^{2}$ .

(c) We induct on $𝑘$ . For $𝑘 = 1$ , $𝑇_{1} \neq 0$ means that $𝑇_{1} (𝐯) \neq 𝟎$ for some $𝐯$ . Let $𝑘 \geq 2$ , and suppose by induction that $𝐯$ satisfies $𝑇_{𝑖} (𝐯) \neq 𝟎$ for $𝑖 = 1, \dots, 𝑘 - 1$ . Choose $𝐮$ with $𝑇_{𝑘} (𝐮) \neq 𝟎$ , and consider the vectors $𝐯 + 𝑡 𝐮$ for $𝑡 \in 𝐹$ .

Claim: for each $𝑖 \in {1, \dots, 𝑘}$ , there is at most one $𝑡 \in 𝐹$ with $𝑇_{𝑖} (𝐯 + 𝑡 𝐮) = 𝟎$ . Suppose $𝑡 \neq 𝑡^{'}$ both work. Then

𝟎 = 𝑇_{𝑖} (𝐯 + 𝑡 𝐮) - 𝑇_{𝑖} (𝐯 + 𝑡^{'} 𝐮) = (𝑡 - 𝑡^{'}) 𝑇_{𝑖} (𝐮),

so $𝑇_{𝑖} (𝐮) = 𝟎$ by Proposition 1.2.35, and then $𝑇_{𝑖} (𝐯) =$ $𝑇_{𝑖} (𝐯 + 𝑡 𝐮) =$ $𝟎$ . For $𝑖 = 𝑘$ this contradicts $𝑇_{𝑘} (𝐮) \neq 𝟎$ , and for $𝑖 < 𝑘$ it contradicts $𝑇_{𝑖} (𝐯) \neq 𝟎$ .

So at most $𝑘$ values of $𝑡$ are excluded. Since $𝐹$ is infinite, some $𝑡$ is not excluded, and $𝐯 + 𝑡 𝐮$ is the required vector.

The proof of (c) only needs $𝐹$ to have more than $𝑘$ elements at the $𝑘$ th step, so the statement holds for $𝑘$ maps over any field with more than $𝑘$ elements; (b) shows that some such bound is needed. Every proper subspace $𝑈$ of a finite-dimensional space $𝑉$ is the kernel of a nonzero operator on $𝑉$ (Exercise 2.2.35). Applied to such operators, (c) says that over an infinite field a finite-dimensional vector space is never the union of finitely many proper subspaces.

Exercise 2.6.51.standard Let $𝑇 \in L (𝑉, 𝑊)$ .

(a)Show that $im (𝑇 ↾_{𝑋})$ is a subspace of $𝑊$ for every subspace $𝑋$ of $𝑉$ , and that $𝑇^{- 1} (𝑈)$ is a subspace of $𝑉$ containing $ker 𝑇$ for every subspace $𝑈$ of $𝑊$ .
(b)Show that $im (𝑇 ↾_{𝑇^{- 1} (𝑈)}) =$ $𝑈 \cap$ $im 𝑇$ for every subspace $𝑈$ of $𝑊$ .
(c)Show that $𝑇^{- 1} (im (𝑇 ↾_{𝑋})) =$ $𝑋 +$ $ker 𝑇$ for every subspace $𝑋$ of $𝑉$ .

Solution.

Uses. Theorem 1.4.3 (Subspace criterion), Notation 2.6.30 (Preimages), Proposition 1.4.24 (The sum of two subspaces), Proposition 2.1.4 (Linear maps preserve $𝟎$ and negatives)

Techniques. Double containment (Technique 1.3.14), Fix an arbitrary element and chase it (Technique 1.3.13)

Hint 1. (a) is the proof of Proposition 2.6.8 with

{𝟎}

replaced by

𝑈

and

𝑉

replaced by

𝑋

Hint 2. (c) If

𝑇 (𝐯) = 𝑇 (𝐱)

with

𝐱 \in 𝑋

, then

𝐯 - 𝐱 \in ker 𝑇

Sketch. (a)

𝑇 (𝟎) = 𝟎

, and

𝑇 (𝐱 + 𝐱^{'}) =

𝑇 (𝐱) +

𝑇 (𝐱^{'})

𝑇 (𝑎 𝐱) = 𝑎 𝑇 (𝐱)

give closure for both sets;

ker 𝑇 =

𝑇^{- 1} (𝟎) \subseteq

𝑇^{- 1} (𝑈)

. (b) A value

𝑇 (𝐯)

with

𝑇 (𝐯) \in 𝑈

lies in

𝑈 \cap im 𝑇

, and

𝐰 = 𝑇 (𝐯) \in 𝑈

comes from

𝐯 \in 𝑇^{- 1} (𝑈)

. (c)

𝑇 (𝐱 + 𝐳) = 𝑇 (𝐱)

for

𝐳 \in ker 𝑇

, and

𝑇 (𝐯) = 𝑇 (𝐱)

gives

𝐯 = 𝐱 + (𝐯 - 𝐱)

(a) $im (𝑇 ↾_{𝑋})$ contains $𝑇 (𝟎) = 𝟎$ (Proposition 2.1.4), and for $𝐱, 𝐱^{'} \in 𝑋$ and $𝑎 \in 𝐹$ , $𝑇 (𝐱) +$ $𝑇 (𝐱^{'}) =$ $𝑇 (𝐱 + 𝐱^{'})$ and $𝑎 𝑇 (𝐱) = 𝑇 (𝑎 𝐱)$ lie in $im (𝑇 ↾_{𝑋})$ because $𝐱 + 𝐱^{'}, 𝑎 𝐱 \in 𝑋$ . So $im (𝑇 ↾_{𝑋})$ is a subspace by Theorem 1.4.3. Similarly, $𝑇 (𝟎) = 𝟎 \in 𝑈$ , so $𝟎 \in 𝑇^{- 1} (𝑈)$ , and if $𝑇 (𝐯), 𝑇 (𝐯^{'}) \in 𝑈$ , then $𝑇 (𝐯 + 𝐯^{'}) =$ $𝑇 (𝐯) +$ $𝑇 (𝐯^{'}) \in$ $𝑈$ and $𝑇 (𝑎 𝐯) =$ $𝑎 𝑇 (𝐯) \in$ $𝑈$ . So $𝑇^{- 1} (𝑈)$ is a subspace. Since $𝟎 \in 𝑈$ , $ker 𝑇 =$ $𝑇^{- 1} (𝟎) \subseteq$ $𝑇^{- 1} (𝑈)$ .

(b) An element of $im (𝑇 ↾_{𝑇^{- 1} (𝑈)})$ has the form $𝑇 (𝐯)$ with $𝑇 (𝐯) \in 𝑈$ , so it lies in $𝑈$ and in $im 𝑇$ . Conversely, if $𝐰 \in 𝑈 \cap im 𝑇$ , then $𝐰 = 𝑇 (𝐯)$ for some $𝐯$ , and $𝑇 (𝐯) = 𝐰 \in 𝑈$ means $𝐯 \in 𝑇^{- 1} (𝑈)$ , so $𝐰 \in$ $im (𝑇 ↾_{𝑇^{- 1} (𝑈)})$ .

(c) If $𝐱 \in 𝑋$ and $𝐳 \in ker 𝑇$ , then $𝑇 (𝐱 + 𝐳) =$ $𝑇 (𝐱) \in$ $im (𝑇 ↾_{𝑋})$ , so $𝐱 +$ $𝐳 \in$ $𝑇^{- 1} (im (𝑇 ↾_{𝑋}))$ . Conversely, if $𝑇 (𝐯) \in im (𝑇 ↾_{𝑋})$ , then $𝑇 (𝐯) = 𝑇 (𝐱)$ for some $𝐱 \in 𝑋$ , so $𝐳 = 𝐯 - 𝐱 \in ker 𝑇$ , and $𝐯 = 𝐱 + 𝐳 \in 𝑋 + ker 𝑇$ .

By (b), $im (𝑇 ↾_{𝑇^{- 1} (𝑈)}) =$ $𝑈$ exactly when $𝑈 \subseteq im 𝑇$ , and by (c), $𝑇^{- 1} (im (𝑇 ↾_{𝑋})) =$ $𝑋$ exactly when $ker 𝑇 \subseteq 𝑋$ . So taking images and preimages are inverse operations between the subspaces of $𝑉$ containing $ker 𝑇$ and the subspaces of $im 𝑇$ .

Exercise 2.6.52.standard Let $𝑆, 𝑇 \in L (𝑉)$ with $𝑆 𝑇 = 𝑇 𝑆$ .

(a)Show that $𝑆$ maps $ker 𝑇$ into $ker 𝑇$ and $im 𝑇$ into $im 𝑇$ .
(b)Deduce that for every $𝜆 \in 𝐹$ , $𝑇$ maps $ker (𝑇 - 𝜆 id)$ into itself and $im (𝑇 - 𝜆 id)$ into itself.
(c)Show by an example that (a) can fail if $𝑆 𝑇 \neq 𝑇 𝑆$ .

Solution.

Uses. Definition 2.6.5 (Kernel), Definition 2.6.6 (Image), Proposition 2.1.24 (Algebra of composition)

Techniques. Fix an arbitrary element and chase it (Technique 1.3.13), Finding counterexamples and minimal examples (Technique 1.4.18)

Hint 1. For

𝐯 \in ker 𝑇

, compute

𝑇 (𝑆 (𝐯))

using

𝑇 𝑆 = 𝑆 𝑇

. For

𝐰 = 𝑇 (𝐯)

, rewrite

𝑆 (𝐰)

as a value of

𝑇

Hint 2. For (b),

𝑇

commutes with

𝑇 - 𝜆 id

. For (c), use a swap and a projection on

𝐹^{2}

Sketch. (a)

𝑇 (𝑆 (𝐯)) =

𝑆 (𝑇 (𝐯)) =

𝑆 (𝟎) =

𝟎

, and

𝑆 (𝑇 (𝐯)) =

𝑇 (𝑆 (𝐯)) \in

im 𝑇

. (b)

𝑇 (𝑇 - 𝜆 id) =

𝑇^{2} -

𝜆 𝑇 =

(𝑇 - 𝜆 id) 𝑇

. (c)

𝑆 (𝑥, 𝑦) = (𝑦, 𝑥)

and

𝑇 (𝑥, 𝑦) = (𝑥, 0)

ker 𝑇 = span (𝐞_{2})

, but

𝑆 (𝐞_{2}) = 𝐞_{1} \notin ker 𝑇

(a) If $𝐯 \in ker 𝑇$ , then $𝑇 (𝑆 (𝐯)) =$ $𝑆 (𝑇 (𝐯)) =$ $𝑆 (𝟎) =$ $𝟎$ , so $𝑆 (𝐯) \in ker 𝑇$ . If $𝐰 = 𝑇 (𝐯) \in im 𝑇$ , then $𝑆 (𝐰) =$ $𝑆 (𝑇 (𝐯)) =$ $𝑇 (𝑆 (𝐯)) \in$ $im 𝑇$ .

(b) By Proposition 2.1.24, $𝑇 (𝑇 - 𝜆 id) =$ $𝑇^{2} -$ $𝜆 𝑇 =$ $(𝑇 - 𝜆 id) 𝑇$ . Apply (a) with $𝑇 - 𝜆 id$ in place of $𝑇$ and $𝑇$ in place of $𝑆$ .

(c) On $𝐹^{2}$ , let $𝑆 (𝑥, 𝑦) = (𝑦, 𝑥)$ and $𝑇 (𝑥, 𝑦) = (𝑥, 0)$ . Then $ker 𝑇 = span (𝐞_{2})$ and $im 𝑇 = span (𝐞_{1})$ , but $𝑆 (𝐞_{2}) = 𝐞_{1} \notin ker 𝑇$ and $𝑆 (𝐞_{1}) = 𝐞_{2} \notin im 𝑇$ . Indeed $𝑆 𝑇 (𝑥, 𝑦) =$ $(0, 𝑥)$ while $𝑇 𝑆 (𝑥, 𝑦) =$ $(𝑦, 0)$ .

In the language of Section 7.2, (b) says that eigenspaces of $𝑇$ are invariant under $𝑇$ , and (a) that they are invariant under everything that commutes with $𝑇$ .

More practice

The practice bank has 55 exercises on this section’s topics.