Chapter 3 Linear Maps

Section 3A Vector Space of Linear Maps

3A.1

Suppose $𝑏, 𝑐 ∈ 𝐑$. Define $𝑇 ∶ 𝐑^3 → 𝐑^2$ by

$$𝑇(𝑥, 𝑦, 𝑧) = (2𝑥 − 4𝑦 + 3𝑧 + 𝑏, 6𝑥 + 𝑐𝑥𝑦𝑧).$$

Show that $𝑇$ is linear if and only if $𝑏 = 𝑐 = 0$.

Proof:

$\Rightarrow$

$T(0) = (b, 0) = (0,0)$, so $b = 0$.

$T(2,2,2) = (2, 12 + 8c) = 2T(1,1,1) = 2(1, 6 + c)$

So $8c = 2c$, then $c = 0$.

$\square$

3A.2

Suppose $𝑏, 𝑐 ∈ 𝐑$. Define $𝑇 ∶ 𝒫(𝐑) → 𝐑^2$ by

$$𝑇𝑝 = (3𝑝(4) + 5𝑝'(6) + 𝑏𝑝(1)𝑝(2), \int_{-1}^{2} 𝑥^3 𝑝(𝑥) 𝑑𝑥 + 𝑐 \sin 𝑝(0)).$$

Show that $𝑇$ is linear if and only if $𝑏 = 𝑐 = 0$.

Proof:

Consider $p(x) = 1$, then

$$3𝑝(4) + 5𝑝'(6) + 𝑏𝑝(1)𝑝(2) = 3 + 0 + b \\ 3(2𝑝)(4) + 5(2𝑝)'(6) + 𝑏(2𝑝)(1)(2𝑝)(2) = 6 + 0 + 4b \\ $$

So $b = 0$.

Consider $p(x) = \pi / 4$, then

$$T(2p)[1] = \int_{-1}^{2} 𝑥^3 2 (\frac{\pi}{4}) dx + c \cdot 1$$

On the other hand

$$T(2p)[1] = 2 T(p) [1] = 2 \int_{-1}^{2} 𝑥^3 (\frac{\pi}{4}) dx + c \cdot \frac{\sqrt[]{2}}{2}$$

So $c = 0$.

$\square$

3A.3

Suppose that $𝑇 ∈ ℒ(𝐅^𝑛, 𝐅^𝑚)$. Show that there exist scalars $𝐴_{𝑗,𝑘} ∈ 𝐅$ for $𝑗 = 1, …, 𝑚$ and $𝑘 = 1, …, 𝑛$ such that

$$𝑇(𝑥_1, …, 𝑥_𝑛) = (𝐴_{1, 1} 𝑥_1 +⋯ + 𝐴_{1, 𝑛} 𝑥_𝑛, …, 𝐴_{𝑚, 1} 𝑥_1 +⋯ + 𝐴_{𝑚, 𝑛} 𝑥_𝑛)$$

for every $(𝑥_1, …, 𝑥_𝑛) ∈ 𝐅_𝑛$.

Proof:

Let

$$T(e_1) = (A_{1,1}, A_{2,1}, \cdots, A_{m,1}) \\ T(e_2) = (A_{1,2}, A_{2,2}, \cdots, A_{m,2}) \\ \cdots \\ T(e_n) = (A_{1,n}, A_{2,n}, \cdots, A_{m,n}) \\ $$

$\square$

3A.4

Suppose $𝑇 ∈ ℒ(𝑉, 𝑊)$ and $𝑣_1, …, 𝑣_𝑚$ is a list of vectors in $𝑉$ such that $𝑇𝑣_1, …, 𝑇𝑣_𝑚$ is a linearly independent list in $𝑊$. Prove that $𝑣_1, …, 𝑣_𝑚$ is linearly independent.

Proof:

Assume

$$a_1 v_1 + \cdots + a_m v_m = 0$$

Then

$$T(a_1 v_1 + \cdots + a_m v_m) = a_1 Tv_1 + \cdots + a_m Tv_m = 0$$

So $a_1 = \cdots = a_m = 0$.

$\square$

3A.5

3A.6

3A.7

Show that every linear map from a one-dimensional vector space to itself is multiplication by some scalar. More precisely, prove that if $\dim 𝑉 = 1$ and $𝑇 ∈ ℒ(𝑉)$, then there exists $𝜆 ∈ 𝐅$ such that $𝑇𝑣 = 𝜆𝑣$ for all $𝑣 ∈ 𝑉$.

Proof:

Assume $v_0$ is a basis of $V$. Let $w_0 = Tv_0 \in V$. We can assume $w_0 = 𝜆 v_0$ Then assume $v = a v_0$.

$$Tv = T a v_0 = a T v_0 = a 𝜆 v_0 = 𝜆 a v_0 = 𝜆 v$$

$\square$

3A.8

Give an example of a function $𝜑 ∶ 𝐑^2 → 𝐑$ such that

$$𝜑(𝑎𝑣) = 𝑎𝜑(𝑣)$$

for all $𝑎 ∈ 𝐑$ and all $𝑣 ∈ 𝐑^2$ but $𝜑$ is not linear.

Solution:

Consider the norm function:

$$𝜑(𝑣) = \left\| v \right\| = \sqrt[]{v_1^2 + v_2^2}$$

$\square$

3A.9

Give an example of a function $𝜑 ∶ 𝐂 → 𝐂$ such that

$$𝜑(𝑤 + 𝑧) = 𝜑(𝑤) + 𝜑(𝑧)$$

for all $𝑤, 𝑧 ∈ 𝐂$ but $𝜑$ is not linear. (Here $𝐂$ is thought of as a complex vector space.)

Solution:

Consider $𝜑(𝑤) = \overline{w}$.

Then let $w = i, a = i$.

$$𝜑(a𝑤) = 𝜑(-1) = -1 \\ a 𝜑(𝑤) = i 𝜑(i) = i (-i) = 1 \\ $$

$\square$

3A.10

Prove or give a counterexample: If $𝑞 ∈ 𝒫(𝐑)$ and $𝑇 ∶ 𝒫(𝐑) → 𝒫(𝐑)$ is defined by 𝑇𝑝 = 𝑞 ∘ 𝑝, then $𝑇$ is a linear map.

Solution: We provide a counterexample.

Let $q = x + 1, p = 1$.

$$T(p)(0) = q(p(0)) = q(1) = 2 \\ T(2p)(0) = q(2p(0)) = q(2) = 3 \\ $$

$\square$

3A.11

Suppose $𝑉$ is finite-dimensional and $𝑇 ∈ ℒ(𝑉)$. Prove that 𝑇 is a scalar multiple of the identity if and only if $𝑆𝑇 = 𝑇𝑆$ for every $𝑆 ∈ ℒ(𝑉)$.

Proof:

$\Rightarrow$ is obvious.

Given $v_1, \cdots, v_m$ is a basis with $m \geq 2$. $T$ is not a zero map. Without loss of generality, assume

$$T(v_1) = w_1 = a_1 v_1 + \cdots + a_m v_m \neq 0.$$

Let $S(v_1) = v_1, S(v_i) = 0$ for $i \neq 1$. Then

$$TS(v_1) = T(v_1) = a_1 v_1 + \cdots + a_m v_m \\ ST(v_1) = S(a_1 v_1 + \cdots + a_m v_m) = a_1 v_1$$

So $T(v_1) = a_1 v_1$. For the same reason, $T(v_i) = a_i v_i$.

Now consider $S(v_1) = v_2, S(v_2) = v_1, S(v_i) = 0$ for all other $i$.

$$TS(v_1) = T(v2) = a_2 v_2 \\ ST(v_1) = S(a_1 v_1) = a_1 v_2 \\ $$

So $a_1 = a_2 = \cdots = a_n$.

$\square$

3A.12

Suppose $𝑈$ is a subspace of $𝑉$ with $𝑈 ≠ 𝑉$. Suppose $𝑆 ∈ ℒ(𝑈, 𝑊)$ and $𝑆 ≠ 0$ (which means that $𝑆𝑢 ≠ 0$ for some $𝑢 ∈ 𝑈$). Define $𝑇 ∶ 𝑉 → 𝑊$ by

$$T v = \begin{cases} S v &\text{if } v \in U \\ 0 &\text{if } v \in V \text{ and } v \not\in U. \\ \end{cases}$$

Prove that 𝑇 is not a linear map on 𝑉 .

Proof:

Consider $v \in U, S(v) \neq 0$, $w \not\in U$, then $v + w \not\in U$.

$$0 = S(v + w) = S(v) + S(w) = S(v)$$

So we have a contradition.

$\square$

3A.13

Suppose $𝑉$ is finite-dimensional. Prove that every linear map on a subspace of $𝑉$ can be extended to a linear map on $𝑉$ . In other words, show that if $𝑈$ is a subspace of $𝑉$ and $𝑆 ∈ ℒ(𝑈, 𝑊)$, then there exists $𝑇 ∈ ℒ(𝑉, 𝑊)$ such that $𝑇𝑢 = 𝑆𝑢$ for all $𝑢 ∈ 𝑈$.

Proof:

Consider $U$ is a subspace, $u_1, \cdots, u_m$ is a basis of $U$. Let $w_i = S(u_i)$. We can extend $u_1, \cdots, u_m$ to $u_1, \cdots, u_m, v_1, \cdots, v_n$.

Let $T(u_i) = w_i, T(v_j) = 0$, then from 3.4 linear map lemma, $T$ satisfy $𝑇𝑢 = 𝑆𝑢$ for all $𝑢 ∈ 𝑈$.

$\square$

3A.14

Suppose $𝑉$ is finite-dimensional with $\dim 𝑉 > 0$, and suppose $𝑊$ is infinite-dimensional. Prove that $ℒ(𝑉, 𝑊)$ is infinite-dimensional.

Proof:

Assume $\dim ℒ(𝑉, 𝑊) = n$, and $v_1, \cdots, v_m$ is a basis of of $V$.

Also assume $w_1, \cdots, w_{n+1}$ is linear independent in $W$.

Let $T_1, \cdots, T_{n+1}$ are linear mappings such that $T_i(v_1) = w_i$.

Since $\dim ℒ(𝑉, 𝑊) = n$, then $T_1, \cdots, T_{n+1}$ are linear dependent, i.e.

$$a_1 T_1 + \cdots + a_{n+1} T_{n+1} = 0, a_i \text{ are not all 0}$$

Then

$$(a_1 T_1 + \cdots + a_{n+1} T_{n+1})(v_1) = \\ a_1 w_1 + \cdots + a_{n+1} w_{n+1} = 0$$

So $w_1, \cdots, w_{n+1}$ is linear dependent. We have a contradition.

3A.15

Suppose $𝑣_1, …, 𝑣_𝑚$ is a linearly dependent list of vectors in $𝑉$ . Suppose also that $𝑊 ≠ \{0\}$. Prove that there exist $𝑤_1, …, 𝑤_𝑚 ∈ 𝑊$ such that no $𝑇 ∈ ℒ(𝑉, 𝑊)$ satisfies $𝑇𝑣_𝑘 = 𝑤_𝑘$ for each $𝑘 = 1, …, 𝑚$.

Proof:

We can find

$$a_1 v_1 + \cdots + a_m v_m = 0, a_i \text{ are not all 0}$$

Without loss of generality, assume $a_1 \neq 0$.

Then let $w_1 \neq 0, w_2 = \cdots = w_m = 0$.

If $𝑇𝑣_𝑘 = 𝑤_𝑘$ for each $𝑘 = 1, …, 𝑚$

Then

$$0 = T(0) = T(a_1 v_1 + \cdots + a_m v_m) = a_1 T(v_1) + \cdots + a_m T(v_m) = a_1 w_1 \neq 0$$

So $T$ does not exist.

3A.16

Suppose $𝑉$ is finite-dimensional with $\dim 𝑉 > 1$. Prove that there exist $𝑆, 𝑇 ∈ ℒ(𝑉)$ such that $𝑆𝑇 ≠ 𝑇𝑆$.

Proof:

Consider $v_1, v_2, \cdots, v_m$ is a basis of $V$.

Let $T(v_1) = v_1, T(v_2) = 2 v_2$, $S(v_1) = v_2, S(v_2) = v_1$. And $T(v_i) = S(v_i) = 0$ for all other $i \neq 1, 2$.

Then

$$ST(v_1) = v_2 \\ TS(v_1) = T(v_2) = 2 v_2 \\ $$

Since $v_2 \neq 0$, then $ST(v_1) \neq TS(v_2)$.

$\square$

3A.17

Suppose $𝑉$ is finite-dimensional. Show that the only two-sided ideals of $ℒ(𝑉)$ are $\{0\}$ and $ℒ(𝑉)$.

A subspace $ℰ$ of $ℒ(𝑉)$ is called a two-sided ideal of $ℒ(𝑉)$ if $𝑇𝐸 ∈ ℰ$ and $𝐸𝑇 ∈ ℰ$ for all $𝐸 ∈ ℰ$ and all $𝑇 ∈ ℒ(𝑉)$.

Proof: I cannot figure it out now.

$\square$

Section 3B Null Spaces and Ranges

3B.1

Give an example of a linear map $𝑇$ with dim null $𝑇 = 3$ and dim range $𝑇 = 2$.

Solution:

Consider $T : \mathbb{R}^5 \longrightarrow \mathbb{R}^2$

$$T(r_1, r_2, r_3, r_4, r_5) = (r_1, r_2)$$

It is surjective, so dim range $𝑇 = 2$.

Note $(0, 0, 1, 0, 0), (0,0,0,1,0), (0,0,0,0,1)$ is a basis of $\text{null } T$, so dim null $𝑇 = 3$.

$\square$

3B.2

Suppose $𝑆, 𝑇 ∈ ℒ(𝑉)$ are such that range $𝑆 ⊆ \text{null } T$. Prove that $(𝑆𝑇)^2 = 0$.

Proof:

Given any $v \in S$

$$((𝑆𝑇)^2)(v) = (ST)((ST)(v)) = S(T(S(T(v))))$$

Note $S(T(v)) \in \text{range } S \subseteq \text{null } T$.

So $S(T(S(T(v)))) = S(0) = 0$.

$\square$

3B.3

Suppose $𝑣_1, …, 𝑣_𝑚$ is a list of vectors in $𝑉$. Define $𝑇 ∈ ℒ(𝐅^𝑚 , 𝑉)$ by

$$T (z_1, \cdots, z_m) = z_1 v_1 + \cdots + z_m v_m$$

Anwser:

(a) What property of $𝑇$ corresponds to $v_1, \cdots, v_m$ spanning $𝑉$?

Solution:

When $T$ is surjective.

(b) What property of $𝑇$ corresponds to the list $v_1, \cdots, v_m$ being linearly independent?

Solution:

When $T$ is injective.

3B.4

Show that $\{𝑇 ∈ ℒ(𝐑^5 , 𝐑^4) ∶ \text{dim null } T > 2\}$ is not a subspace of $ℒ(\mathbb{R}^5, \mathbb{R}^4)$.

Solution:

Consider the following 2 maps:

$$S(r_1, r_2, r_3, r_4, r_5) = (r_1, r_2, 0, 0) \\ T(r_1, r_2, r_3, r_4, r_5) = (0, 0, r_3, r_4) \\ $$

Then we have

$$(S+T)(r_1, r_2, r_3, r_4, r_5) = (r_1, r_2, r_3, r_4)$$

So $\text{dim null } (S+T) = 1$.

This set is not closed under addition.

$\square$

3B.5

Give an example of $𝑇 ∈ ℒ(𝐑^4)$ such that $\text{range } T = \text{null } T$.

Solution:

Consider

$$T(r_1, r_2, r_3, r_4) = (0, 0, r_1, r_2)$$

Then

$$\text{null } T = \{(0, 0, r_3, r_4)\} = \text{range } T$$

$\square$

3B.6

Prove that there does not exist $𝑇 ∈ ℒ(𝐑^5)$ such that $\text{range } T = \text{null } T$.

Proof:

From 3.21 fundamental theorem of linear maps,

$$\text{dim null } T + \text{dim range } T = 5$$

So

$$\text{dim null } T \neq \text{dim range } T$$

Then $\text{null } T \neq \text{range } T$.

$\square$

3B.7

Suppose $𝑉$ and $𝑊$ are finite-dimensional with $2 ≤ \dim 𝑉 ≤ \dim 𝑊$. Show that $\{𝑇 ∈ ℒ(𝑉, 𝑊) ∶ 𝑇 \text{ is not injective}\}$ is not a subspace of $ℒ(𝑉, 𝑊)$.

Proof:

Let $v_1, \cdots, v_m$ is a basis of $V$ and $w_1, \cdots, w_m$ is linear independent vector.

Let $T(v_1) = w_1, T(v_i) = 0$, $S(v_1) = 0, S(v_i) = w_i$.

So $(S+T)(v_i) = w_i$. $S+T$ is injective.

This set is not closed under addition.

$\square$

3B.8

Suppose $𝑉$ and $𝑊$ are finite-dimensional with $2 ≤ \dim W ≤ \dim V$. Show that $\{𝑇 ∈ ℒ(𝑉, 𝑊) ∶ 𝑇 \text{ is not surjective}\}$ is not a subspace of $ℒ(𝑉, 𝑊)$.

Proof:

Let $w_1, \cdots, w_m$ is a basis of $W$ and $v_1, \cdots, v_m$ is linear independent vector.

Let $T(v_1) = w_1, T(v_i) = 0$, $S(v_1) = 0, S(v_i) = w_i$.

So $(S+T)(v_i) = w_i$. $S+T$ is surjective.

This set is not closed under addition.

$\square$

3B.9

Suppose $𝑇 ∈ ℒ(𝑉, 𝑊)$ is injective and $v_1, \cdots, v_n$ is linearly independent in $𝑉$. Prove that $𝑇𝑣_1, …, 𝑇𝑣_𝑛$ is linearly independent in 𝑊.

Proof:

If

$$a_1 Tv_1 + \cdots + a_m Tv_m = 0$$

Then

$$T(a_1 v_1 + \cdots + a_m v_m) = 0$$

Since $T$ is injective, so

$$a_1 v_1 + \cdots + a_m v_m = 0$$

Then $a_1 = \cdots = a_m = 0$. Thus $𝑇𝑣_1, …, 𝑇𝑣_𝑛$ is linearly independent in 𝑊.

$\square$

3B.10

Suppose $𝑣_1, …, 𝑣_𝑛$ spans $𝑉$ and $𝑇 ∈ ℒ(𝑉, 𝑊)$. Show that $𝑇𝑣_1, …, 𝑇𝑣_𝑛$ spans $\text{range } T$.

Proof:

Let $w \in \text{range } T$, then $T(v) = w$.

Let $v = a_1 v_1 + \cdots + a_n v_n$, so

$$T (a_1 v_1 + \cdots + a_n v_n) = a_1 Tv_1 + \cdots + a_n Tv_n = w$$

$\square$

3B.11

Suppose that $𝑉$ is finite-dimensional and that 𝑇 ∈ ℒ(𝑉, 𝑊). Prove that there exists a subspace 𝑈 of 𝑉 such that

$$U \cap \text{null } T = \{0\} \quad \text{and} \quad \text{range } T = \{Tu : u \in U\}$$

Proof:

Let $v_1, \cdots, v_m$ be a basis of $\text{null } T$, we can extend a basis of $V$ it to $v_1, \cdots, v_m, w_1, \cdots, w_n$.

Let $U = \text{span}(w_1, \cdots, w_n)$.

If $u \in \text{range } T$, and $T(v) = u$

$$v = a_1 v_1 + \cdots + a_m v_m + b_1 w_1 + \cdots + b_n w_n$$

So

$$u = T(v) = T (a_1 v_1 + \cdots + a_m v_m + b_1 w_1 + \cdots + b_n w_n) \\ = T(a_1 v_1 + \cdots + a_m v_m) + T(b_1 w_1 + \cdots + b_n w_n) \\ = T(b_1 w_1 + \cdots + b_n w_n)$$

$\square$

3B.12

Suppose $𝑇$ is a linear map from $𝐅^4$ to $𝐅^2$ such that

$$\text{null } T = \{ (𝑥_1, 𝑥_2, 𝑥_3, 𝑥_4) ∈ 𝐅^4 ∶ 𝑥_1 = 5𝑥_2 \text{ and } 𝑥_3 = 7𝑥_4 \}$$

Prove that $𝑇$ is surjective.

Proof:

Note

$$v_1 = (5, 1, 0, 0) \\ v_2 = (0, 0, 7, 1) \\ $$

is a basis of $\text{null } T$, so $\dim \text{range } T = 2$.

Then $𝑇$ is surjective.

3B.13

Suppose $𝑈$ is a three-dimensional subspace of $𝐑^8$ and that $𝑇$ is a linear map from $𝐑^8$ to $𝐑^5$ such that $\text{null } T = 𝑈$. Prove that $𝑇$ is surjective.

Proof:

Note $\text{dim range } T = 5$, and $\dim R^5 = 5$. So $𝑇$ is surjective.

3B.14

Prove that there does not exist a linear map from $𝐅_5$ to $𝐅_2$ whose null space equals

$$\{ (𝑥_1, 𝑥_2, 𝑥_3, 𝑥_4, 𝑥_5) ∈ 𝐅^5 ∶ 𝑥_1 = 3𝑥_2 \text{ and } 𝑥_3 = 𝑥_4 = 𝑥_5 \}$$

Proof:

Assume $T$ exists. Note

$$v_1 = (3, 1, 0, 0, 0) \\ v_2 = (0, 0, 1, 1, 1) \\ $$

is a basis of $\text{null } T$, so $\text{dim range } T$ has to be $3$.

But $\mathbb{R}^2$ is only 2 dimensional. So $T$ does not exist.

$\square$

3B.15

Suppose there exists a linear map on $𝑉$ whose null space and range are both finite-dimensional. Prove that $𝑉$ is finite-dimensional.

Proof:

Assume otherwise, $u_1, \cdots, u_m$ is a basis of $\text{null } T$ and $v_1, \cdots, v_n$ such that $Tv_1, \cdots, Tv_n$ is a basis of $\text{range } T$.

We can see $V = \text{null } T + \text{span}(𝑣_1, \cdots, 𝑣_n)$.

From 2.43 dimension of a sum, we can see

$$\dim V \leq \text{dim null } T + \dim \text{span}(𝑣_1, \cdots, 𝑣_n)$$

$\square$

3B.16

Suppose 𝑉 and 𝑊 are both finite-dimensional. Prove that there exists an injective linear map from 𝑉 to 𝑊 if and only if dim 𝑉 ≤ dim 𝑊.

Proof:

$\Rightarrow$

It's the contrapositive of 3.22, i.e. linear map to a lower-dimensional space is not injective.

$\Leftarrow$

Assume $\dim V = m$, and $v_1, \cdots, v_m$ is a basis of $V$, and $w_1, \cdots, w_m$ is linearly independent in $W$.

Let $T$ be a linear map such that $T(v_i) = w_i$. Then if

$$T(a_1 v_1 + \cdots + a_m v_m) = 0 \\ \Rightarrow \\ a_1 Tv_1 + \cdots + a_m Tv_m \\ \Rightarrow \\ a_1 w_1 + \cdots + a_m w_m = 0 \\ \Rightarrow \\ a_1 = \cdots = a_m = 0$$

So $\text{null } T = 0$, so $T$ is injective.

3B.17

Suppose 𝑉 and 𝑊 are both finite-dimensional. Prove that there exists a surjective linear map from 𝑉 onto 𝑊 if and only if dim 𝑉 ≥ dim 𝑊.

Proof:

$\Rightarrow$

It is the contrapositive of 3.24: linear map to a higher-dimensional space is not surjective.

$\Leftarrow$

Assume $w_1, \cdots, w_m$ is a basis of $W$. $v_1, \cdots, v_m$ is linear independent in $V$. Extend it to a basis of $V$ by adding $u_1, \cdots, u_n$.

Let $T v_i = w_i, T u_j = 0$. This map is surjective.

$\square$

3B.18

Suppose 𝑉 and 𝑊 are finite-dimensional and that 𝑈 is a subspace of 𝑉 . Prove that there exists 𝑇 ∈ ℒ(𝑉, 𝑊) such that null 𝑇 = 𝑈 if and only if dim 𝑈 ≥ dim 𝑉 − dim 𝑊.

Proof:

$\Rightarrow$

$$\dim V = \text{dim null } T + \text{dim range } T \\ \Rightarrow \\ \dim U = \dim V - \text{dim range } T \\ \Rightarrow \\ \dim U \geq \dim V - \dim W \\ $$

$\Leftarrow$

Let $u_1, \cdots, u_m$ be a basis of $U$, extend it to a basis of $V$ by adding $v_1, \cdots, v_n$.

Since $\dim U \geq \dim V - \dim W$, then $\dim W \geq n$. We can find $w_1, \cdots, w_n \in W$ such that they are independent in $W$.

Let $T u_i = 0, T v_j = w_j$.

$\square$

3B.19

Suppose 𝑊 is finite-dimensional and 𝑇 ∈ ℒ(𝑉, 𝑊). Prove that 𝑇 is injective if and only if there exists 𝑆 ∈ ℒ(𝑊, 𝑉) such that 𝑆𝑇 is the identity operator on 𝑉 .

Proof:

$\Rightarrow$

First, since $T$ is injective, $V$ has to be finite-dimensional because of 3.22.

Assume $v_1, \cdots, v_m$ is a basis of $V$.

Also assume

$$a_1 Tv_1 + \cdots + a_m Tv_m = 0 \\ \Rightarrow \\ T(a_1 v_1 + \cdots + a_m v_m) = 0 \\ \Rightarrow \\ a_1 v_1 + \cdots + a_m v_m \in \text{null } T \\ \Rightarrow \text{ T is injective } \\ a_1 v_1 + \cdots + a_m v_m = 0 \\ \Rightarrow \\ a_1 = \cdots = a_m = 0 \\ \Rightarrow \\ Tv_1, \cdots, Tv_m \text{ is independent }$$

Then extend $Tv_1, \cdots, Tv_m$ to a basis of $W$ by adding $w_1, \cdots, w_n$. Consider $S$ as $S(T v_i) = v_i, S(w_j) = 0$.

Then we have $ST$ is the identity operator on 𝑉.

$\Leftarrow$

If $T(v) = 0$, then $v = ST(v) = S(0) = 0$. So $T$ is injective.

$\square$

3B.20

Suppose 𝑊 is finite-dimensional and 𝑇 ∈ ℒ(𝑉, 𝑊). Prove that 𝑇 is surjective if and only if there exists 𝑆 ∈ ℒ(𝑊, 𝑉) such that 𝑇𝑆 is the identity operator on 𝑊.

Proof:

$\Rightarrow$

$w_1, \cdots, w_m$ is a basis of $W$, and $T(v_i) = w_i$.

$$a_1 v_1 + \cdots + a_m v_m = 0 \\ \Rightarrow \\ T(a_1 v_1 + \cdots + a_m v_m) = 0 \\ \Rightarrow \\ a_1 Tv_1 + \cdots + a_m Tv_m = 0 \\ \Rightarrow \\ a_1 w_1 + \cdots + a_m w_m = 0 \\ \Rightarrow \\ a_1 = \cdots = a_m = 0 \\ \Rightarrow \\ v_1, \cdots, v_m \text{ is linear independent.}$$

Let $S(w_i) = v_i$, then if $w = a_1 w_1 + \cdots + a_m w_m$, then

$$TS(w) = T(a_1 Sw_1 + \cdots + a_m Sw_m) \\ = T(a_1 v_1 + \cdots + a_m v_m) \\ = a_1 Tv_1 + \cdots + a_m Tv_m \\ = a_1 w_1 + \cdots + a_m w_m \\ = w$$

$\Leftarrow$

If $TS$ is the identity operator. For any given $w$, $TS(w) = w$, $S(w) \in V$. So $T$ is surjective.

$\square$

3B.21

Suppose $𝑉$ is finite-dimensional, $𝑇 ∈ ℒ(𝑉, 𝑊)$, and $𝑈$ is a subspace of $𝑊$. Prove that $\{𝑣 ∈ 𝑉 ∶ 𝑇𝑣 ∈ 𝑈\}$ is a subspace of $𝑉$ and

$$\dim \{ v \in 𝑉 ∶ 𝑇𝑣 ∈ 𝑈 \} = \text{dim null } T + \dim (𝑈 ∩ \text{range } T)$$

Proof:

We first show $\{𝑣 ∈ 𝑉 ∶ 𝑇𝑣 ∈ 𝑈\}$ is a subspace of $V$.

Let $A = \{𝑣 ∈ 𝑉 ∶ 𝑇𝑣 ∈ 𝑈\}$, $v_1, v_2 \in A$, then we can find $Tv_1 = a_1 \in U, Tv_2 = a_2 \in U$, so $T(v_1 + v_2) = a_1 + a_2 \in U$.

$T(k v_1) = k T(v_1) = k a_1 \in U$.

So $A$ is a subspace of $V$.

Consider the linear mapping that limits $T$ on $A$ and call it $T|_A$.

$$\dim A = \text{dim null } T|_A + \text{dim range } T|_A$$

Note that $\text{null } T \subseteq A$, so $\text{dim null } T = \text{dim null } T|_A$.

Also $\text{range } T|_A = 𝑈 ∩ \text{range } T$, so

$$\text{dim range } T|_A = \dim (𝑈 ∩ \text{range } T)$$

$\square$

3B.22

Suppose 𝑈 and 𝑉 are finite-dimensional vector spaces and 𝑆 ∈ ℒ(𝑉, 𝑊) and 𝑇 ∈ ℒ(𝑈, 𝑉). Prove that

$$\text{dim null } ST \leq \text{dim null } S + \text{dim null } T$$

Proof:

Use the exercise 3B.21, Let $B = \text{null } S$, then

$$\text{null } ST = \{u: Tu \in B\} = C$$

So

$$\dim C = \text{dim null } T + \dim (B \cap \text{range } T)$$

And $\dim (B \cap \text{range } T) \leq \dim B = \text{dim null } S$.

$\square$

3B.23

Suppose 𝑈 and 𝑉 are finite-dimensional vector spaces and 𝑆 ∈ ℒ(𝑉, 𝑊) and 𝑇 ∈ ℒ(𝑈, 𝑉). Prove that

$$\text{dim range } ST \leq \min \{\text{dim range } S, \text{dim range } T\}$$

Proof:

If $w \in \text{range } ST$, then we can find $u$ such that $ST(u) = w$, i.e. $S(T(u)) = w$, $w \in \text{range } S$. So $\text{range } ST \subseteq \text{range } S$.

Limit $S$ on $\text{range } T$.

$$\text{dim range } T = \text{dim null } S|_{\text{range } T} + \text{dim range } S|_{\text{range } T} \geq \text{dim range } S|_{\text{range } T}$$

Also $\text{range } S|_{\text{range } T} = \text{range } ST$.

So $\text{dim range } T \geq \text{dim range } ST$.

$\square$

3B.24

(a) Suppose dim 𝑉 = 5 and 𝑆, 𝑇 ∈ ℒ(𝑉) are such that 𝑆𝑇 = 0. Prove that dim range 𝑇𝑆 ≤ 2.

Proof:

$\text{dim null } ST = 5$. So $\text{dim null } T + \text{dim null } S \geq 5$.

So $\max \{\text{dim null } T , \text{dim null } S\} \geq 3$, $\min \{\text{dim range } T , \text{dim range } S\} \leq 2$.

So $\text{dim range } TS \leq 2$.

$\square$

(b) Give an example of 𝑆, $𝑇 ∈ ℒ(𝐅^5)$ with 𝑆𝑇 = 0 and dim range 𝑇𝑆 = 2.

Solution:

Consider $T(x_1, x_2, x_3, x_4, x_5) = (0, 0, 0, x_1, x_2)$, $S(x_1, x_2, x_3, x_4, x_5) = (x_1, x_2, 0, 0, 0)$

$TS(x_1, x_2, x_3, x_4, x_5) = T(x_1, x_2, 0, 0, 0) = (0, 0, 0, x_1, x_2)$

$\square$

3B.25

Suppose that 𝑊 is finite-dimensional and 𝑆, 𝑇 ∈ ℒ(𝑉, 𝑊). Prove that null 𝑆 ⊆ null 𝑇 if and only if there exists 𝐸 ∈ ℒ(𝑊) such that 𝑇 = 𝐸𝑆.

Proof:

$\Rightarrow$

Assume $Tv_1, \cdots, Tv_m$ is a basis of $\text{range } T$. Then $v_1, \cdots, v_m$ is independent in $V$.

Then $Sv_1, \cdots, Sv_m$ is independent in $W$. This is because assume

$$a_1 Sv_1 + \cdots + a_m Sv_m = 0$$

Then $a_1 v_1 + \cdots + a_m v_m \in \text{null } S \subseteq \text{null } T$

Then

$$a_1 Tv_1 + \cdots + a_m Tv_m = 0$$

So $a_1 = \cdots = a_m = 0$.

Then extend $Sv_1, \cdots, Sv_m$ to a basis of $\text{range } S$ by adding $Sr_1, \cdots, Sr_n$.

Consider

$$A = \text{span}(𝑣_1, \cdots, 𝑣_𝑚, r_1, \cdots, r_n)$$

First $𝑣_1, \cdots, 𝑣_𝑚, r_1, \cdots, r_n$ are independent:

$$a_1 v_1 + \cdots + a_m v_m + b_1 r_1 + \cdots + b_n r_n = 0 \\ \Rightarrow \\ a_1 Sv_1 + \cdots + a_m Sv_m + b_1 Sr_1 + \cdots + b_n Sr_n = 0 \\ \Rightarrow \\ a_1 = \cdots = a_m = b_1 = \cdots = b_n = 0$$

Note that we have $S|_A$ is injective. Because

$$\dim A = m + n = \text{dim range } S = \text{dim range } S|_A \\ \Rightarrow \\ \text{dim null } S|_A = 0$$

Then we have

$$V = A \oplus \text{null } S$$

Now, consider $T|_A$ We also have

$$\dim A = \text{dim range } T|_A + \text{dim null } T|_A \\ = \text{dim range } T + \text{dim null } T \cap A$$

So we can find a basis of $\text{null } T \cap A$, i.e., $q_1, \cdots, q_n$, such that

$$A = \text{span}(𝑣_1, \cdots, 𝑣_𝑚, q_1, \cdots, q_n)$$

Note that $Sv_1, \cdots, Sv_m, Sq_1, \cdots, Sq_n$ are also independent in $W$ because $S|_A$ is injective.

Now, we can construct $E$ this way. We extend $Sv_1, \cdots, Sv_m, Sq_1, \cdots, Sq_n$ to a basis of $W$ by adding $w_1, \cdots, w_k$. Let $E(Sv_i) = T v_i, E(Sq_j) = E(w_j) = 0$.

Given $v \in V$, we can right it as

$$v = (a_1 v_1 + \cdots + a_m v_m) + (b_1 q_1 + \cdots + b_n q_n) + c$$

Where $c \in \text{null } S$.

Then

$$ES(v) = (a_1 ESv_1 + \cdots + a_m ESv_m) + (b_1 ESr_1 + \cdots + b_n ESr_n) + ESc \\ = a_1 Tv_1 + \cdots + a_m Tv_m$$

On the other hand

$$T(v) = (a_1 Tv_1 + \cdots + a_m Tv_m) + (b_1 Tq_1 + \cdots + b_n Tq_n) + Tc \\ = a_1 Tv_1 + \cdots + a_m Tv_m$$

So $T = ES$.

$\Leftarrow$

If $v \in \text{null } S$, then $T(v) = ES(v) = E(0) = 0$. So $v \in \text{null } T$. So $\text{null } S \subseteq \text{null } T$

$\square$

3B.26

Suppose that 𝑉 is finite-dimensional and 𝑆, 𝑇 ∈ ℒ(𝑉, 𝑊). Prove that range 𝑆 ⊆ range 𝑇 if and only if there exists 𝐸 ∈ ℒ(𝑉) such that 𝑆 = 𝑇𝐸.

Proof:

$\Rightarrow$

Consider $w_1, \cdots, w_m$ is a basis of $\text{range } S$. We can find

$$Tv_1 = w_1, \cdots, Tv_m = w_m \\ Su_1 = w_1, \cdots, Su_m = w_m \\ $$

It's easy to prove $V = \text{span}(u_1, \cdots, u_𝑚) \oplus \text{null } S$.

Then let $p_1, \cdots, p_n$ be a basis of $\text{null } S$. Together $u_1, \cdots, u_m, p_1, \cdots, p_n$ is a basis of $V$.

Define $E(u_i) = v_i, E(p_j) = 0$.

Given $v \in V$, then

$$v = a_1 u_1 + \cdots + a_m u_m + b_1 p_1 + \cdots + b_n p_n \\ \Rightarrow \\ S(v) = a_1 Su_1 + \cdots + a_m Su_m + b_1 Sp_1 + \cdots + b_n Sp_n \\ = a_1 Su_1 + \cdots + a_m Su_m \\ = a_1 w_1 + \cdots + a_m w_m$$

Also

$$TE(v) = a_1 TEu_1 + \cdots + a_m TEu_m + b_1 TEp_1 + \cdots + b_n TEp_n \\ = a_1 Tv_1 + \cdots + a_m Tv_m \\ = a_1 w_1 + \cdots + a_m w_m$$

So $S = TE$.

$\Leftarrow$

Assume $w \in \text{range } S$, and assume $S(v) = w$, then $TE(v) = w$, so $w \in \text{range } T$. Then $\text{range } S ⊆ \text{range } T$.

$\square$

3B.27

Suppose 𝑃 ∈ ℒ(𝑉) and $𝑃^2 = 𝑃$. Prove that $𝑉 = \text{null } P ⊕ \text{range } P$.

Proof:

Consider $v \in P$, then $v - P(v) \in \text{null } P$.

So $V = \text{null } P + \text{range } P$.

Assume $v \in \text{null } P \cap \text{range } P$.

Let $P(u) = v$. So $P^2(u) = P(P(u)) = P(v) = 0$.

On the other hand, $P^2(u) = P(u) = v$, so $v = 0$.

$\square$

3B.28

Suppose 𝐷 ∈ ℒ(𝒫(𝐑)) is such that deg 𝐷𝑝 = (deg 𝑝) − 1 for every non-constant polynomial 𝑝 ∈ 𝒫(𝐑). Prove that 𝐷 is surjective.

The notation 𝐷 is used above to remind you of the differentiation map that sends a polynomial 𝑝 to 𝑝′

Proof:

Given a $p$ such that $\deg Dp = 0$, so $Dp = c \neq 0$. Then $D \frac{1}{c} p = 1$.

Then we can also find $p$ such that $Dp = x$ and so on.

So $D$ is surjective.

$\square$

3B.29

Suppose 𝑝 ∈ 𝒫(𝐑). Prove that there exists a polynomial 𝑞 ∈ 𝒫(𝐑) such that 5𝑞″ + 3𝑞′ = 𝑝. This exercise can be done without linear algebra, but it’s more fun to do it using linear algebra.

Proof:

Note that $Dq = 5𝑞″ + 3𝑞'$ satisfies 3B.28. So $D$ is surjective.

$\square$

3B.30

Suppose 𝜑 ∈ ℒ(𝑉, 𝐅) and 𝜑 ≠ 0. Suppose 𝑢 ∈ 𝑉 is not in null 𝜑. Prove that 𝑉 = null 𝜑 ⊕ {𝑎𝑢 ∶ 𝑎 ∈ 𝐅}.

Proof:

Assume $\varphi u = c \neq 0$.

Give $v \in V$ and $\varphi (v) = b$.

Then

$$𝜑(v - \frac{b}{c}u) = 𝜑(v) - \frac{b}{c} 𝜑(u) = b - b = 0$$

So $v - \frac{b}{c}u \in \text{null } 𝜑$.

Thus $𝑉 = \text{null } 𝜑 + \{𝑎𝑢 ∶ 𝑎 ∈ 𝐅\}.$

Assume $v \in \text{null } 𝜑 \cap \{𝑎𝑢 ∶ 𝑎 ∈ 𝐅\}$,

Then $0 = 𝜑(v) = 𝜑(au) = a 𝜑(u) = ac = 0$, so $a = 0 \cdot c^{-1} = 0$.

Then $v = 0$.

So $𝑉 = \text{null } 𝜑 ⊕ \{𝑎𝑢 ∶ 𝑎 ∈ 𝐅\}.$

$\square$

3B.31

Suppose 𝑉 is finite-dimensional, 𝑋 is a subspace of 𝑉 , and 𝑌 is a finite-dimensional subspace of 𝑊. Prove that there exists 𝑇 ∈ ℒ(𝑉, 𝑊) such that null 𝑇 = 𝑋 and range 𝑇 = 𝑌 if and only if dim 𝑋 + dim 𝑌 = dim 𝑉 .

Proof:

$\Rightarrow$

From 3.21 fundamental theorem of linear maps

$$\dim V = \dim \text{null } T + \text{dim range } T$$

Since $\text{null } T = X$ and $\text{range } T = Y$, then

$$\dim 𝑋 + \dim 𝑌 = \dim 𝑉$$

$\Leftarrow$

Let $w_1, \cdots, w_m$ be a basis of $Y$. $v_1, \cdots, v_n$ is a basis of $X$, and we can extend it to a basis of $V$ by adding $m$ elements $u_1, \cdots, u_m$.

Define $T(v_i) = 0, T(u_j) = w_j$.

$\square$

3B.32

Suppose 𝑉 is finite-dimensional with dim 𝑉 > 1. Show that if 𝜑 ∶ ℒ(𝑉) → 𝐅 is a linear map such that 𝜑(𝑆𝑇) = 𝜑(𝑆)𝜑(𝑇) for all 𝑆, 𝑇 ∈ ℒ(𝑉), then 𝜑 = 0.

Hint: The description of the two-sided ideals of ℒ(𝑉) given by Exercise 17 in Section 3A might be useful.

Proof:

Consider the $\text{null } 𝜑$, for all $T \in ℒ(𝑉)$ and all $E \in \text{null } \varphi$,

$$𝜑(E𝑇) = 𝜑(E) 𝜑(T) = 0 𝜑(T) = 0 = 𝜑(T) 0 = 𝜑(T) 𝜑(E) = 𝜑(ET)$$

So $ET \in \text{null } \varphi$ and $TE \in \text{null } \varphi$.

$\text{null } \varphi$ is a two-sided ideal.

Since 𝑉 is finite-dimensional with $\dim 𝑉 > 1$, then $\dim ℒ(V) > 1$. Since the following 2 linear maps are linearly independent.

$$T_1(v_1) = v_1, T_1(v_2) = v_2 \\ T_2(v_1) = v_2, T_2(v_2) = v_1 \\ $$

Also note $\dim F = 1$. So from 3.22 linear map to a lower-dimensional space is not injective and 3.15 injectivity ⟺ null space equals {0}.

Then $\text{null } \varphi \neq \{0\}$, so $\text{null } \varphi = ℒ(V)$, then 𝜑 = 0.

$\square$

3B.33

Suppose that 𝑉 and 𝑊 are real vector spaces and 𝑇 ∈ ℒ(𝑉, 𝑊). Define $𝑇_𝐂 ∶ 𝑉_𝐂 → 𝑊_𝐂$ by

$$𝑇_𝐂 (𝑢 + 𝑖𝑣) = 𝑇𝑢 + 𝑖𝑇𝑣$$

for all 𝑢, 𝑣 ∈ 𝑉.

(a) Show that $𝑇_𝐂$ is a (complex) linear map from $𝑉_𝐂$ to $𝑊_𝐂$.

Proof:

$$T_C((a + 𝑖b) + (c + di)) = T_C ((a+c)+i(b+d)) \\ = T(a+c) + iT(b+d) \\ = (T(a) + iT(b)) + (T(c) + iT(d)) \\ = T_C(a + 𝑖b) + T_C(c+id)$$

Also

$$T_C((x+iy)(a+ib)) = \\ T_C((ax-by) + i(xb + ya)) \\ = T(xa - yb) + i T(ya + xb) \\ = xTa - yT(b) + i yT(a) + i x T(b) \\ = (x+iy) (T(a) + iT(b)) \\ = (x+iy) T(a + ib) \\ $$

$\square$

(b) Show that $𝑇_𝐂$ is injective if and only if 𝑇 is injective.

Proof:

$\Rightarrow$

$𝑇_𝐂$ is injective

Assume $T(u) = 0$, then $T_C(u + ui) = 0$.

So $u + iu = 0$, so $u = 0$.

$\Leftarrow$

$T$ is injective.

Assume $T_C(u + iv) = 0$, then $T(u) + i T(v) = 0$, so $T(u) = T(v) = 0$, then $u = v = 0$.

$\square$

(c) Show that range $𝑇_𝐂 = 𝑊_𝐂$ if and only if range 𝑇 = 𝑊.

Proof:

$\Rightarrow$

Given $w \in W$, since $T_C$ is surjective, then we can find $T_C(a + ib) = w + 0i$, i.e. $T(a) = w$.

$\Leftarrow$

Given $w = a + ib$, we can fine $T(u) = a, T(v) = b$, then $T(u+iv) = a + ib$.

$\square$

Section 3C Matrices

3C.1

Suppose 𝑇 ∈ ℒ(𝑉, 𝑊). Show that with respect to each choice of bases of 𝑉 and 𝑊, the matrix of 𝑇 has at least dim range 𝑇 nonzero entries.

Proof:

Consider the matrix for $T$ is $A$.

Then let $k = \text{dim null } T, n = \text{dim range } T$, then $k + n = \dim V$. If only $m < n$ entries are nonzero, then it means the matrix has more than $k$ columns are $0$.

That means $\text{dim null } T > k$, we reach a contradition.

$\square$

3C.5

Suppose 𝑉 and 𝑊 are finite-dimensional and 𝑇 ∈ ℒ(𝑉, 𝑊). Prove that there exist a basis of 𝑉 and a basis of 𝑊 such that with respect to these bases, all entries of ℳ(𝑇) are 0 except that the entries in row 𝑘, column 𝑘, equal 1 if 1 ≤ 𝑘 ≤ dim range 𝑇.

Proof:

Let $m = \text{dim range } T$, and $w_1, \cdots, w_m$ is a basis of $\text{range } T$.

Assume $v_1, \cdots, v_m \in V$ such that $Tv_1 = w_1, \cdots, Tv_m = w_m$.

Assume $p = n - m = \text{dim null } T$, then expand $v_1, \cdots, v_m$ with $u_1, \cdots, u_p \in \text{null } T$.

With these 2 basis of $V$ and $W$, $ℳ(𝑇)$ is what we want.

$\square$

3C.6

Suppose $v_1, \cdots, v_m$ is a basis of 𝑉 and 𝑊 is finite-dimensional. Suppose 𝑇 ∈ ℒ(𝑉, 𝑊). Prove that there exists a basis $w_1, \cdots, w_n$ of 𝑊 such that all entries in the first column of ℳ(𝑇) [with respect to the bases $v_1, \cdots, v_m$ and $w_1, \cdots, w_n$] are $0$ except for possibly a $1$ in the first row, first column.

In this exercise, unlike Exercise 5, you are given the basis of 𝑉 instead of being able to choose a basis of 𝑉 .

Proof:

Let $w_1 = T v_1$ and extend $w_1$ to a basis of $W$ by adding $w_2, \cdots, w_m$.

Then $M(T)$ is desired matrix.

$\square$

3C.7

Suppose $w_1, \cdots, w_m$ is a basis of 𝑊 and 𝑉 is finite-dimensional. Suppose 𝑇 ∈ ℒ(𝑉, 𝑊). Prove that there exists a basis $v_1, \cdots, v_n$ of 𝑉 such that all entries in the first row of ℳ(𝑇) [with respect to the bases $v_1, \cdots, v_n$ and $w_1, \cdots, w_m$] are 0 except for possibly a $1$ in the first row, first column.

In this exercise, unlike Exercise 5, you are given the basis of 𝑊 instead of being able to choose a basis of 𝑊.

Proof:

Let $v_1, \cdots, v_n \in V$ is a basis of $V$. Assume the matrix $\mathcal{M}(T) = [A_{i, j}]$. If $A_{1, .} = 0$, then then $v_1, \cdots, v_m$ satisfies the requirement.

If $A_{1, .} \neq 0$, then $A_{1, j} \neq 0$ for some $j$. We can swap $v_1$ and $v_j$. Then $A_{1, 1} \neq 0$. Let $v_1' = \frac{1}{A_{1, 1}}v_1$, then $A_{1, 1} = 1$.

For any $v_j$ such that $A_{1, j} \neq 0$, let

$$v_j' = v_j - A_{1, j}v_1'$$

With $v'_1, \cdots, v'_n$, the first row of $\mathcal{M}(T)$, $A_{1, .} = [1, 0, \cdot, 0]$, as required.

$\square$

3C.16

Suppose $𝐴$ is an $𝑚$-by-$𝑛$ matrix with $𝐴 ≠ 0$. Prove that the rank of $𝐴$ is 1 if and only if there exist $(𝑐_1, …, 𝑐_𝑚) ∈ 𝐅^𝑚$ and $(𝑑_1, …, 𝑑_𝑛) ∈ 𝐅^𝑛$ such that $A_{j, k} = 𝑐_𝑗 𝑑_𝑘$ for every $𝑗 = 1, …, 𝑚$ and every $𝑘 = 1, …, 𝑛$.

Proof:

$\Rightarrow$

From 3.56 column–row factorization, since the rank of $A$ is $1$, we can find a $m$-by-$1$ matrix $C$ and $1$-by-$n$ matrix $R$ such that $A = CR$.

Then $C$ and $R$ are what we need.

$\Leftarrow$

We have

$$A = \begin{pmatrix} c_1 \\ \vdots \\ c_m \end{pmatrix} (𝑑_1, …, 𝑑_𝑛)$$

So from 3.50 and 3.51, we have

$$A_{., j} = C d_j$$

That means the columns of $A$ are the scalar of $C$. So $A$'s rank is 1.

$\square$

3C.17

Suppose $𝑇 ∈ ℒ(𝑉)$, and $u_1, \cdots, u_n$ and $v_1, \cdots, v_n$ are bases of $𝑉$. Prove that the following are equivalent.

(a) $𝑇$ is injective.

(b) The columns of $ℳ(𝑇)$ are linearly independent in $𝐅^{𝑛, 1}$.

(c) The columns of $ℳ(𝑇)$ span $𝐅^{𝑛, 1}$.

(d) The rows of $ℳ(𝑇)$ span $𝐅^{𝑛, 1}$.

(e) The rows of $ℳ(𝑇)$ are linearly independent in $𝐅^{𝑛, 1}$.

Proof:

Our strategy is to prove

$$(a) \Leftrightarrow (b) \\ (b) \Rightarrow (c) \Rightarrow (d) \Rightarrow (e) \\ (e) \Rightarrow (d) \Rightarrow (c) \Rightarrow (b)$$

$(a) \Leftrightarrow (b)$.

If $T$ is injective, then $\text{dim null } T = 0$.

Assume $A = \mathcal{M}(T)$, and there are $A b = 0$, where $b = F^{n,1}$, i.e.

$$0 = A_{., 1} b_1 + \cdots + A_{., n} b_n$$

Then consider

$$T(b_1 v_1 + \cdots + b_n v_n) = \\ b_1 Tv_1 + \cdots + b_n Tv_n = \\ (Tv_1, \cdots, Tv_m) \begin{pmatrix} b_1 \\ \vdots \\ b_n \end{pmatrix} \\ = ((w_1, \cdots, w_n)A)b \\ = (w_1, \cdots, w_n) (Ab) \\ = (w_1, \cdots, w_n) \cdot 0 \\ = 0$$

Then $b_1 v_1 + \cdots + b_n v_n \in \text{null } T$, so $b_1 v_1 + \cdots + b_n v_n = 0$, i.e. $b_1 = \cdots = b_n = 0$.

On the other hand, if

$$b_1 v_1 + \cdots + b_n v_n \in \text{null } T$$

Then

$$0 = T(b_1 v_1 + \cdots + b_n v_n) = \\ b_1 Tv_1 + \cdots + b_n Tv_n = \\ (Tv_1, \cdots, Tv_m) \begin{pmatrix} b_1 \\ \vdots \\ b_n \end{pmatrix} \\ = ((w_1, \cdots, w_n)A)b \\ = (w_1, \cdots, w_n) (Ab)$$

Since $w_1, \cdots, w_n$ is linearly independent, we have $Ab = 0$.

Since

$$Ab = A_{., 1} b_1 + \cdots + A_{., n} b_n$$

And $A_{., 1}, \cdots, A_{., n}$ are linearly independent, we must have $b_1 = \cdots = b_n = 0$.

Then we have $\text{null } T = \{0\}$.

This completes the proof of $(a) \Leftrightarrow (b)$.

$(b) \Rightarrow (c)$

Since $A_{., 1}, \cdots, A_{., n}$ are linearly independent, and $\dim F^{n,1} = n$, then $A_{., 1}, \cdots, A_{., n}$ is a basis of $F^{n,1}$, so it spans $F^{n,1}$.

$(c) \Rightarrow (d)$

Since $A_{., 1}, \cdots, A_{., n}$ spans $F^{n,1}$, the column rank of $A$ is $n$. Since the column rank of $A$ is $n$, the row rank is also $n$. Then $A_{1, .}, \cdots, A_{n, .}$ spans $F^{1,n}$.

$(d) \Rightarrow (e)$

$A_{1, .}, \cdots, A_{n, .}$ spans $F^{1,n}$, and the dimension of $F^{1,n}$ is $n$, then $A_{1, .}, \cdots, A_{n, .}$ is a basis of $F^{1,n}$, then $A_{1, .}, \cdots, A_{n, .}$ are linearly independent.

The reverse direction is similar.

$\square$

Section 3D Invertibility and Isomorphisms

3D.1

Suppose 𝑇 ∈ ℒ(𝑉, 𝑊) is invertible. Show that $𝑇^{−1}$ is invertible and

$$(𝑇^{−1})^{-1} = T$$

Proof:

This is because

$$𝑇^{−1} T = I \\ T 𝑇^{−1} = I \\ $$

So

$$(𝑇^{−1})^{-1} = T$$

$\square$

3D.2

Suppose 𝑇 ∈ ℒ(𝑈, 𝑉) and 𝑆 ∈ ℒ(𝑉, 𝑊) are both invertible linear maps. Prove that 𝑆𝑇 ∈ ℒ(𝑈, 𝑊) is invertible and that $(𝑆𝑇)^{−1} = 𝑇^{−1}𝑆^{−1}$.

Proof:

This is because

$$𝑇^{−1}𝑆^{−1} (ST) = I \\ (ST) 𝑇^{−1}𝑆^{−1} = I \\ $$

$\square$

3D.3

Suppose 𝑉 is finite-dimensional and 𝑇 ∈ ℒ(𝑉). Prove that the following are equivalent.

(a) 𝑇 is invertible.

(b) $Tv_1, \cdots, Tv_n$ is a basis of 𝑉 for every basis $v_1, \cdots, v_n$ of 𝑉 .

(c) $Tv_1, \cdots, Tv_n$ is a basis of 𝑉 for some basis $v_1, \cdots, v_n$ of 𝑉 .

Proof:

Our order is

$$(a) \Rightarrow (b) \Rightarrow (c) \Rightarrow (a)$$

$(a) \Rightarrow (b)$

If 𝑇 is invertible. Then $T$ is surjective.

Given any basis $v_1, \cdots, v_n$, $Tv_1, \cdots, Tv_n$ spans $V$. So $Tv_1, \cdots, Tv_n$ is a basis of $V$.

$(b) \Rightarrow (c)$ is obvious.

$(c) \Rightarrow (a)$

If $Tv_1, \cdots, Tv_n$ is a basis of 𝑉, then $T$ is surjective, so $𝑇$ is invertible.

$\square$

3D.4

Suppose 𝑉 is finite-dimensional and dim 𝑉 > 1. Prove that the set of noninvertible linear maps from 𝑉 to itself is not a subspace of ℒ(𝑉).

Proof:

Assume $\dim V = 2$. $v_1, v_2$ is a basis of $V$.

Let $T(v_1) = v_1, T(v_2) = 0, S(v_1) = 0, S(v_2) = v_2$.

Since $T, S$ are not injective, they are not invertible.

But $T_1 + T_2 = I$.

$\square$

3D.5

Suppose 𝑉 is finite-dimensional, 𝑈 is a subspace of 𝑉 , and 𝑆 ∈ ℒ(𝑈, 𝑉). Prove that there exists an invertible linear map 𝑇 from 𝑉 to itself such that 𝑇𝑢 = 𝑆𝑢 for every 𝑢 ∈ 𝑈 if and only if 𝑆 is injective.

Proof:

$\Rightarrow$

If $Su = Tu = 0$, since $T$ is injective, then $u = 0$, so $S$ is injective.

$\Leftarrow$

If $S$ is injective, and $u_1, \cdots, u_m$ is a basis of $U$. We can extend it to a basis of $V$ by adding $v_1, \cdots, v_n$.

Let $Tu_i = Su_i$. $Tu_i$ is linear independent since $S$ is injective. Extend $Tu_i$ to a basis of $V$ by adding $w_1, \cdots, w_n$, then let $Tv_i = w_i$.

Now assume $Tu = 0$, let

$$u = a_1 u_1 + \cdots + a_m u_m + b_1 v_1 + \cdots + b_n v_n$$

Then

$$0 = a_1 Tu_1 + \cdots + a_m Tu_m + b_1 Tv_1 + \cdots + b_n Tv_n \\ \Rightarrow \\ a_1 = \cdots = a_m = b_1 = \cdots = b_n = 0$$

So $u = 0$ and $T$ is invertible.

$\square$

3D.6

Suppose that 𝑊 is finite-dimensional and 𝑆, 𝑇 ∈ ℒ(𝑉, 𝑊). Prove that null 𝑆 = null 𝑇 if and only if there exists an invertible 𝐸 ∈ ℒ(𝑊) such that 𝑆 = 𝐸𝑇.

Proof:

$\Rightarrow$

Assume null 𝑆 = null 𝑇, let $u_1, \cdots, u_m$ be a basis of $\text{null } T$.

Extend $u_1, \cdots, u_m$ with $v_1, \cdots, v_n$ so they become a basis of $V$.

So $Tv_1, \cdots, Tv_n$ is basis of $\text{range } T$ since $\text{dim range } T = (n+m)-m = n$.

For the same reason, $Sv_1, \cdots, Sv_n$ is basis of $\text{range } S$

Let $ETv_i = Sv_i$.

$\square$

$\Leftarrow$

Assume $v \in \text{null } T$, then $S(v) = ET(v) = E(0) = 0$. So $\text{null } T \subseteq \text{null } S$.

Assume $w in \text{null } S$, then $ET(w) = 0$, since $E$ is injective, then $T(w) = 0$. So $\text{null } S \subseteq \text{null } T$.

$\square$

3D.7

Suppose that 𝑉 is finite-dimensional and 𝑆, 𝑇 ∈ ℒ(𝑉, 𝑊). Prove that range 𝑆 = range 𝑇 if and only if there exists an invertible 𝐸 ∈ ℒ(𝑉) such that 𝑆 = 𝑇𝐸.

Proof:

$\Rightarrow$

Assume $w_1, \cdots, w_m$ is a basis of $\text{range } S$.

$Sv_1 = w_1, \cdots, Sv_m = w_m$ and $Tu_1 = w_1, \cdots, Tu_m = w_m$.

Let $x_1, \cdots, x_n$ be a basis of $\text{null } S$. Let $y_1, \cdots, y_n$ be a basis of $\text{null } T$.

$$a_1 v_1 + \cdots + a_m v_m + b_1 x_1 + \cdots + b_n x_n = 0 \\ \Rightarrow \\ S( a_1 v_1 + \cdots + a_m v_m + b_1 x_1 + \cdots + b_n x_n) = 0 \\ \Rightarrow \\ a_1 Sv_1 + \cdots + a_m Sv_m + b_1 Sx_1 + \cdots + b_n Sx_n = 0 \\ \Rightarrow \\ a_1 = \cdots = a_m = b_1 = \cdots = b_n = 0$$

So $v_1, \cdots, v_m, x_1, \cdots, x_n$ is a basis of $V$. Also $u_1, \cdots, u_m, y_1, \cdots, y_n$ is a basis of $U$.

Let $Eu_i = v_i, Ey_j = x_j$.

$E$ is an isomorphism and $S = TE$.

$\Leftarrow$

$S = TE$, assume $w \in \text{range } T$ then $T(v) = w$. Since $E$ is surjective, we can find $u$ such that $E(u) = v$, then $S(u) = TE(u) = Tv = w$.

On the other hand, assume $w in \text{range } S$, then $w = S(u) = TE(u) = T(E(u))$, so $w \in \text{range } T$.

$\square$

3D.8

Suppose 𝑉 and 𝑊 are finite-dimensional and 𝑆, 𝑇 ∈ ℒ(𝑉, 𝑊). Prove that there exist invertible $𝐸_1 ∈ ℒ(𝑉)$ and $𝐸_2 ∈ ℒ(𝑊)$ such that $𝑆 = 𝐸_2𝑇𝐸_1$ if and only if $\text{dim null } T = \text{dim null } S$.

Proof:

$\Leftarrow$

If $\text{dim null } T = \text{dim null } S$, then $\text{null } T$ and $\text{null } S$ are isomorphic.

Assume $E_1: \text{null } S \longrightarrow \text{null } T$ is an isomorphism and we can extend it to $E_1: S \longrightarrow T$.

If $u \in \text{null } S \Rightarrow E_1(u) \in \text{null } T \Rightarrow T E_1(u) = 0$.

If $u \in \text{null } T E_1 \Rightarrow E_1 u \in \text{null } T \Rightarrow u \in \text{null } S$

Therefore $\text{null } TE_1 = \text{null } S$

Apply exercise 3D.6, we can have $E_2 \in \mathcal{L}(W)$, such that $S = E_2 T E_1$.

$\Rightarrow$

Consider the subspace $E_1(\text{null } S)$.

Assume $u \in \text{null } S$, then $E_2 T E_1 (u) = 0$, since $E_2$ is injective, $E_1u \in \text{null } T$.

So $E_1(\text{null } S) \subseteq \text{null } T$.

Assume $v \in \text{null } T$, $E_1$ is injective, then we can find $u$, $E_1(u) = v$, then $S(u) = E_2 T E_1 (u) = E_2 T v = 0$, i.e. $\text{null } T \subseteq E_1(\text{null } S)$.

So $\text{null } T = \subseteq E_1(\text{null } S)$. We have $\text{dim null } T = \text{dim null } S$ as required.

$\square$

3D.9

Suppose 𝑉 is finite-dimensional and 𝑇 ∶ 𝑉 → 𝑊 is a surjective linear map of 𝑉 onto 𝑊. Prove that there is a subspace 𝑈 of 𝑉 such that $𝑇|_𝑈$ is an isomorphism of 𝑈 onto 𝑊.

Proof:

Let $w_1, \cdots, w_m$ be a basis of $W$. And $v_1, \cdots, v_m$ such that $Tv_1 = w_1, \cdots, Tv_m = w_m$.

$v_1, \cdots, v_m$ is independent.

Let $U = \text{span}(𝑣_1, \cdots, 𝑣_𝑚)$, $𝑇|_𝑈$ is an isomorphism.

$\square$

3D.10

Suppose 𝑉 and 𝑊 are finite-dimensional and 𝑈 is a subspace of 𝑉. Let

$$ℰ = \{ 𝑇 ∈ ℒ(𝑉, 𝑊) ∶ 𝑈 ⊆ null 𝑇 \}.$$

Prove the following

(a) Show that ℰ is a subspace of ℒ(𝑉, 𝑊).

Proof:

If $T_1, T_2 \in ℰ$ and $u \in U$, then

$$(T_1 + T_2)(u) = T_1(u) + T_2(u) = 0 + 0 = 0$$

So $T_1 + T_2 \in ℰ$.

Same for $kT_1$.

$\square$

(b) Find a formula for dim ℰ in terms of dim 𝑉 , dim 𝑊, and dim 𝑈.

Hint: Define $Φ ∶ ℒ(𝑉, 𝑊) → ℒ(𝑈, 𝑊)$ by $Φ(𝑇) = 𝑇|_𝑈$. What is null $Φ$? What is range $Φ$?

Proof:

If $T \in \text{null } Φ$, then $Φ(T) = T|_U = 0$, i.e. for $u \in U, T(u) = 0$, then $U \subseteq \text{null } T$. So $\text{null } Φ \subseteq ℰ$.

On the other hand, if $T \in ℰ$, then $Φ(T) = T|_U = 0$, so $ℰ \subseteq \text{null } Φ$.

So $ℰ = \text{null } Φ$

$Φ$ is surjective because for any $S \in \mathcal{L}(U, W)$, we can find $T$ such that $Φ(T) = T|_U = S$.

Then

$$\dim \mathcal{L}(V, W) = \dim V \times \dim W \\ \text{dim range } Φ = \dim \mathcal{L}(U, W) = \dim U \times \dim W \\ \dim ℰ = \text{dim null } Φ = (\dim V - \dim U) \times \dim W$$

3D.11

Suppose 𝑉 is finite-dimensional and 𝑆, 𝑇 ∈ ℒ(𝑉). Prove that

$$ST \text{ is invertible } \Longleftrightarrow \text{𝑆 and 𝑇 are invertible.}$$

Proof:

$\Rightarrow$

$ST$ is invertible, then $ST$ is injective. So if $v_1 \neq v_2$ then $STv_1 \neq STv_2$. That means $Tv_1 \neq Tv_2$, so $T$ is injective, so $T$ is invertible.

$ST$ is invertible, then $ST$ is surjective. Given any $v$ we can find $u$, such that $STu - v$, so $S(Tu) = v$. So $S$ is surjective, then it's invertible.

$\Leftarrow$

𝑆 and 𝑇 are invertible, then

$$T^{-1}S^{-1} ST = 1 \\ ST T^{-1}S^{-1} = 1$$

So $T^{-1}S^{-1}$ is the inverse of $ST$.

$\square$

3D.12

Suppose 𝑉 is finite-dimensional and 𝑆, 𝑇, 𝑈 ∈ ℒ(𝑉) and 𝑆𝑇𝑈 = 𝐼. Show that 𝑇 is invertible and that $𝑇^{−1} = 𝑈𝑆$.

Proof:

From 3.68, $STU = I$ then $TUS = I$. Then use 3.68 again $(US)T = I$.

$\square$

3D.13

Show that the result in Exercise 12 can fail without the hypothesis that 𝑉 is finite-dimensional.

Solution:

Consider $V = \mathbf{F}^{\infty }$, $S$ is the identity mapping. And

$$T: (x_1, x_2, x_3, \cdots) \longrightarrow (x_2, x_3, \cdots) \\ U: (x_1, x_2, x_3, \cdots) \longrightarrow (0, x_1, x_2, x_3, \cdots) \\ $$

We know from the second example of 3.64 that $T$ is not invertible, but $TUS = I$.

$\square$

3D.14

Prove or give a counterexample: If 𝑉 is a finite-dimensional vector space and 𝑅, 𝑆, 𝑇 ∈ ℒ(𝑉) are such that 𝑅𝑆𝑇 is surjective, then 𝑆 is injective.

Proof:

$RST$ is surjective, then it's invertible. From exercise 3D.11, $ST$ is invertible. Again use 3D.11, $S$ is invertible, then $S$ is injective.

$\square$

3D.15

Suppose 𝑇 ∈ ℒ(𝑉) and $v_1, \cdots, v_m$ is a list in 𝑉 such that $𝑇𝑣_1, …, 𝑇𝑣_𝑚$ spans 𝑉 . Prove that $𝑣_1, …, 𝑣_𝑚$ spans 𝑉 .

Proof:

Since $𝑇𝑣_1, …, 𝑇𝑣_𝑚$ spans $V$, then $V$ is finite dimension. So $T$ is surjective means $T$ is invertible. Let $v \in V, S = T^{-1}$, and assume

$$ Tv = a_1 Tv_1 + \cdots + a_m Tv_m \\ \Rightarrow \ STv = S(a_1 Tv_1 + \cdots + a_m Tv_m) \ \Rightarrow \ v = a_1 STv_1 + \cdots + a_m STv_m \ \Rightarrow (\text{$ST$ is identity operator and $S$ is linear}) \ v = a_1 v_1 + \cdots + a_m v_m $$

$\square$

3D.16

Prove that every linear map from $\mathbf{F}^{n, 1}$ to $\mathbf{F}^{m, 1}$ is given by a matrix multiplication. In other words, prove that if $𝑇 ∈ ℒ(\mathbf{F}^{n, 1}, \mathbf{F}^{m, 1})$, then there exists an 𝑚-by-𝑛 matrix 𝐴 such that $𝑇𝑥 = 𝐴𝑥$ for every $𝑥 ∈ \mathbf{F}^{n, 1}$.

Proof:

Given $𝑇 ∈ ℒ(\mathbf{F}^{n, 1}, \mathbf{F}^{m, 1})$, consider

$$A = \mathcal{M}(T, (e_1, \cdots, e_n), (e_1, \cdots, e_m))$$

Then

$$T(x) = T(x_1 e_1 + \cdots + x_n e_n) \\ = x_1 Te_1 + \cdots + x_n Te_n \\ = x_1 A_{.,1} + \cdots + x_n A_{.,n} \\ = Ax$$

The 2nd $=$ is because $T$ is linear. The 3rd $=$ is from the definitions of $ℳ(𝑇)$. The last equality is from 3.50.

$\square$

3D.17

Suppose 𝑉 is finite-dimensional and 𝑆 ∈ ℒ(𝑉). Define 𝒜 ∈ ℒ(ℒ(𝑉)) by

$$𝒜(𝑇) = 𝑆𝑇$$

for 𝑇 ∈ ℒ(𝑉).

(a) Show that $\text{dim null } 𝒜 = (\dim 𝑉)(\text{dim null } 𝑆)$.

Proof:

Consider this subspace of $ℒ(ℒ(𝑉))$:

$$T : V \rightarrow \text{null } S$$

It's the $\text{null } 𝒜$. And its dimension is $(\dim 𝑉)(\text{dim null } 𝑆)$.

(b) Show that $\text{dim range } 𝒜 = (\dim 𝑉)(\text{dim range } 𝑆)$.

Proof:

Note $𝒜$ is a linear map from $ℒ(𝑉) \rightarrow ℒ(𝑉)$, so

$$\text{dim range } 𝒜 + \text{dim null } 𝒜 = \dim ℒ(𝑉) = \dim 𝑉 \times \dim 𝑉$$

Then

$$\text{dim range } 𝒜 = \dim 𝑉 \times \dim 𝑉 - (\dim 𝑉)(\text{dim null } 𝑆) = (\dim 𝑉)(\text{dim range } 𝑆)$$

as required.

$\square$

3D.18

Show that 𝑉 and ℒ(𝐅, 𝑉) are isomorphic vector spaces.

Proof:

Given $v \in V$, define this map from $T: 𝑉 \longrightarrow ℒ(𝐅, 𝑉), T(v) = S_v, S_v(1) = v$.

We first show $T$ is injective.

If $T(v) = 0$, then $v = S_v(1) = 0(1) = 0$. So $T$ is surjective.

Next we show $T$ is surjective.

Assume $S \in ℒ(𝐅, 𝑉)$ and $S(1) = v$, then given any $a \in 𝐅$,

$$S(a) = S (a \cdot 1) = a S(1) = a v = a S_v(1) = S_v(a) \\ \Rightarrow \\ S = S_v = T(v)$$

So $T$ is surjective.

In summary, $T$ is an isomorphism.

$\square$

3D.19

Suppose 𝑉 is finite-dimensional and 𝑇 ∈ ℒ(𝑉). Prove that 𝑇 has the same matrix with respect to every basis of 𝑉 if and only if 𝑇 is a scalar multiple of the identity operator.

Proof:

$\Rightarrow$

We first consider when $\dim V = 2$ and consider 2 basis: $v_1, v_2$ and $v_2, v_1$. Assume this same matrix is

$$A = \begin{bmatrix} a & b \\ c & d \\ \end{bmatrix}$$

From 3.84 change-of-basis formula

$$𝐶 = ℳ(𝐼, (v_1, v_2), (v_2, v_1)) \\ = \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix}$$

Then we have

$$C A = A C \\ \begin{bmatrix} c & d \\ a & b \\ \end{bmatrix} = \begin{bmatrix} b & a \\ d & c \\ \end{bmatrix}$$

So we must have $a = d, b = c$. Then

$$A = \begin{bmatrix} a & b \\ b & a \\ \end{bmatrix}$$

Then we consider

$$𝐶 = ℳ(𝐼, (v_1 + v_2, v_2), (v_1, v_2)) \\ = \begin{bmatrix} 1 & 0 \\ 1 & 1 \\ \end{bmatrix}$$

Then we have

$$C A = A C \\ \begin{bmatrix} a & b \\ a+b & a+b \\ \end{bmatrix} = \begin{bmatrix} a+b & b \\ a+b & a \\ \end{bmatrix}$$

So we must have $b = 0$, then

$$A = \begin{bmatrix} a & 0 \\ 0 & a \\ \end{bmatrix}$$

We next consider when $\dim V = 3$ Assume this same matrix is:

$$A = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \\ \end{bmatrix}$$

and consider 2 basis: $v_1, v_2, v_3$ and $v_2, v_1, v_3$.

$$C = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}$$

Then

$$CA = AC \\ \begin{bmatrix} d & e & f \\ a & b & c \\ g & h & i \\ \end{bmatrix} = \begin{bmatrix} b & a & c \\ e & d & f \\ h & g & i \\ \end{bmatrix}$$

So we have $a = e, d = b, c = f, g = h$

$$A = \begin{bmatrix} a & b & c \\ b & a & c \\ g & g & i \\ \end{bmatrix}$$

Next consider $v_1, v_2, v_3$ and $v_3, v_2, v_1$.

$$C = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \\ \end{bmatrix}$$

$$CA = AC \\ \begin{bmatrix} g & g & i \\ b & a & c \\ a & b & c \\ \end{bmatrix} = \begin{bmatrix} c & b & a \\ c & a & b \\ i & g & g \\ \end{bmatrix}$$

Then we have $a = i, b = c = g$.

$$A = \begin{bmatrix} a & b & b \\ b & a & b \\ b & b & a \\ \end{bmatrix}$$

Next consider $v_1+v_2, v_2, v_3$ and $v_1, v_2, v_3$.

$$C = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}$$

$$CA = AC \\ \begin{bmatrix} a & b & b \\ b+a & a+b & b+b \\ b & b & a \\ \end{bmatrix} = \begin{bmatrix} a+b & b & b \\ b+a & a & b \\ b+b & b & a \\ \end{bmatrix}$$

Then $b = 0$.

We have

$$A = \begin{bmatrix} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a \\ \end{bmatrix}$$

For other dimensions, it's similar. So, we finished one direction.

$\Leftarrow$

Now assume if 𝑇 is a scalar multiple of the identity operator. Then $T(v) = av$, so $\mathcal{M}(T) = a I$. as required.

$\square$

3D.20

Suppose 𝑞 ∈ 𝒫(𝐑). Prove that there exists a polynomial 𝑝 ∈ 𝒫(𝐑) such that

$$𝑞(𝑥) = (𝑥^2 + 𝑥)𝑝''(𝑥) + 2𝑥𝑝'(𝑥) + 𝑝(3)$$

for all 𝑥 ∈ 𝐑.

Proof:

First note if $p(x) = ax+b$, then $q(x) = 2ax + 3a + b \neq 0$. Then $ax+b \not\in \text{null } T$

Assume $p(x) = a x^n + b x^{n-1} + r(x), r(x) \in 𝒫_{n-2}(𝐑), p(x) \in \text{null } T$, then $n > 1$.

Then we have

$$p'(x) = an x^{n-1} + b(n-1) x^{n-2} + r'(x)\\ p''(x) = an(n-1) x^{n-2} + b(n-1)(n-2) x^{n-3} + r''(x) \\ 2x p'(x) = 2 an x^n + 2b (n-1) x^{n-1} \\ (x^2 + x) p''(x) = an(n-1) x^n + b(n-1)(n-2)x^{n-1} + an(n-1) x^{n-1} + b(n-1)(n-2)x^{n-2}$$

So we have

$$2an + an(n-1) = 0 \Rightarrow \\ an(n+1) = 0 \\ \Rightarrow \\ a = 0 2b (n-1) + b(n-1)(n-2) + an(n-1) = 0 \\ \Rightarrow \\ bn(n-1) = 0 \\ \Rightarrow \\ b = 0$$

Then we have a contradition. So $\text{null } T = \{0\}$.

Then $T$ is surjective, we can find $p$ for $q$.

$\square$

3D.21

Suppose 𝑛 is a positive integer and $A_{j, k} ∈ 𝐅$ for all $𝑗, 𝑘 = 1, …, 𝑛$. Prove that the following are equivalent (note that in both parts below, the number of equations equals the number of variables).

(a) The trivial solution $𝑥_1 = ⋯ = 𝑥_𝑛 = 0$ is the only solution to the homogeneous system of equations

$$\sum_{k=1}^{n} A_{1, k} x_k = 0 \\ \vdots \\ \sum_{k=1}^{n} A_{n, k} x_k = 0. \\ $$

And

(b) For every $𝑐_1, …, 𝑐_𝑛 ∈ 𝐅$, there exists a solution to the system of equations

$$\sum_{k=1}^{n} A_{1, k} x_k = c_1 \\ \vdots \\ \sum_{k=1}^{n} A_{n, k} x_k = c_n. \\ $$

Proof:

We can have 2 perspectives.

(i) Let $A = [A_{j, k}]$ then view $A$ as a liner map from $\mathbb{F}^{n,1}$ to $\mathbb{F}^{n,1}$ by mapping $x$ to $Ax$.

Then (a) says $A$ is injective and (b) says $A$ is surjective. So they are equivalent.

(ii) We can treat $A_{., k}$ as a vector in $\mathbb{F}^{n,1}$.

Then (a) says $A_{., k}$ are independent and (b) says $A_{., k}$ spans the $\mathbb{F}^{n,1}$. From the fundamental theorem of linear mapping, they are equivalent.

$\square$

3D.22

Suppose 𝑇 ∈ ℒ(𝑉) and $𝑣_1, …, 𝑣_𝑛$ is a basis of 𝑉. Prove that

$$ℳ(𝑇, (𝑣_1, …, 𝑣_𝑛)) \text{ is invertible} \Longleftrightarrow T \text{ is invertible}$$

Proof:

$\Rightarrow$

Let $A = ℳ(𝑇, (𝑣_1, …, 𝑣_𝑛))$ and $B$ be its inverse matrix. Let $S$ be the linear map that $B = ℳ(S, (𝑣_1, …, 𝑣_𝑛))$.

Then

$$ℳ(𝑇S, (𝑣_1, …, 𝑣_𝑛)) = ℳ(𝑇, (𝑣_1, …, 𝑣_𝑛))ℳ(S, (𝑣_1, …, 𝑣_𝑛)) = AB = I$$

Then $TS = I$, from 3.68, we have $ST = I$.

$\Leftarrow$

Assume $TS = ST = I$, then

$$ℳ(𝑇, (𝑣_1, …, 𝑣_𝑛))ℳ(S, (𝑣_1, …, 𝑣_𝑛)) =\\ ℳ(𝑇S, (𝑣_1, …, 𝑣_𝑛)) = \\ ℳ(I, (𝑣_1, …, 𝑣_𝑛)) = I$$

$\square$

3D.23

Suppose that $u_1, \cdots, u_n$ and $𝑣_1, …, 𝑣_𝑛$ are bases of $𝑉$. Let 𝑇 ∈ ℒ(𝑉) be such that $𝑇𝑣_𝑘 = 𝑢_𝑘$ for each $𝑘 = 1, …, 𝑛$. Prove that

$$ℳ(𝑇, (𝑣_1, …, 𝑣_𝑛)) = ℳ(𝐼, (𝑢_1, …, 𝑢_𝑛), (𝑣_1, …, 𝑣_𝑛))$$

Proof:

Since $𝑇𝑣_𝑘 = 𝑢_𝑘$, so

$$ℳ(T, (v_1, …, v_𝑛), (u_1, …, u_𝑛)) = I$$

Then from

$$ℳ(T, (v_1, …, v_𝑛), (u_1, …, u_𝑛)) ℳ(I, (u_1, …, u_𝑛), (v_1, …, v_𝑛)) =\\ ℳ(T, (𝑣_1, …, 𝑣_𝑛))$$

So we proved.

$\square$

3D.24

Suppose 𝐴 and 𝐵 are square matrices of the same size and 𝐴𝐵 = 𝐼. Prove that 𝐵𝐴 = 𝐼.

Proof:

Let $A, B$ are $n$ by $n$ square matrices. Each of them induces a linear map from $\mathbb{F}^{n,1}$ to $\mathbb{F}^{n,1}$, say they are $T, S$. And their matrices w.r.t the standard basis $(e_1, …, e_n)$ are $A, B$.

Since $AB = I$, then $TS = I$. From 3.68, we have $ST = I$.

From 3.81,

$$ℳ(𝑆𝑇, (e_1, …, e_n), (e_1, …, e_n)) = \\ ℳ(𝑆, (e_1, …, e_𝑛), (e_1, …, e_n))ℳ(𝑇, (e_1, …, e_n), (e_1, …, e_𝑛)).$$

i.e.

$$ℳ(I, (e_1, …, e_n)) = \\ ℳ(𝑆, (e_1, …, e_𝑛))ℳ(𝑇, (e_1, …, e_n)).$$

Then we have $I = BA$ as required.

$\square$

Section 3E Products and Quotients of Vector Spaces

3E.2

Suppose that $V_1, \cdots, V_m$ are vector spaces such that $𝑉_1 × ⋯ × 𝑉_𝑚$ is finite-dimensional. Prove that $𝑉_𝑘$ is finite-dimensional for each $𝑘 = 1, …, 𝑚$.

Proof:

Since $𝑉_1 × ⋯ × 𝑉_𝑚$ is finite-dimensional, then assume it can be spanned by $n$ vectors.

Take the 1st component from these $n$ vectors and they can span $V_1$. So $V_1$ is finite-dimensional.

$\square$

3E.3

Suppose $𝑉_1, …, 𝑉_𝑚$ are vector spaces. Prove that $ℒ(𝑉_1 × ⋯ × 𝑉_𝑚, 𝑊)$ and $ℒ(𝑉_1, 𝑊) × ⋯ × ℒ(𝑉_𝑚, 𝑊)$ are isomorphic vector spaces.

Proof:

Assume

$$T \in ℒ(𝑉_1 × ⋯ × 𝑉_𝑚, 𝑊)$$

Let

$$\tau_1 : T \longrightarrow T_1$$

Let $T_1$ be a mapping, such that

$$T_1(v) = T((v, 0, \cdots, 0))$$

It's easy to see, $T_1 \in ℒ(𝑉_1, 𝑊)$.

Then we can have this mapping

$$\tau: T \longrightarrow (\tau_1(T), \cdots, \tau_n(T)) = (T_1, \cdots, T_m)$$

To see it's injective, assume

$T_1 = \cdots = T_m = 0$, then

$$T((v_1, \cdots, v_m)) = T((v_1, \cdots, 0)) + \cdots + T((0, \cdots, v_m)) = T_1 v_1 + \cdots + T_m v_m = 0 \\ \Rightarrow \\ T = 0$$

To see it's surjective, given $(T_1, \cdots, T_m)$, and let

$$T((v_1, \cdots, v_m)) = T_1 v_1 + \cdots + T_m v_m$$

It's easy to see $T$ is a linear map.

We also have

$$\tau_1(T)(v_1) = T(v_1, \cdots, 0) = T_1 v_1 \\ \Rightarrow \\ \tau_1(T) = T_1$$

so $\tau$ is surjective.

Then $T$ is an isomorphism.

$\square$

3E.4

Suppose $𝑊_1, …, 𝑊_𝑚$ are vector spaces. Prove that $ℒ(𝑉, 𝑊_1 × ⋯ × 𝑊_𝑚)$ and $ℒ(𝑉, 𝑊_1) × ⋯ × ℒ(𝑉, 𝑊_𝑚)$ are isomorphic vector spaces.

Proof:

Given $T \in ℒ(𝑉, 𝑊_1 × ⋯ × 𝑊_𝑚)$, let $\tau_1 : T \longrightarrow T_1 \in ℒ(𝑉, 𝑊_1)$.

$$T(v) = (\tau_1(T)(v), \cdots, \tau_m(T)(v))$$

$$\tau(T) = (\tau_1(T)(v), \cdots, \tau_m(T)(v))$$

$\square$

3E.5

Suppose that $𝑣, 𝑥$ are vectors in $𝑉$ and that $𝑈, 𝑊$ are subspaces of $𝑉$ such that $𝑣 + 𝑈 = 𝑥 + 𝑊$. Prove that $𝑈 = 𝑊$.

Proof:

Since $U, W$ are subspaces, then $0 \in U, W$. So $v \in x + W, x \in v + U$. Then we can assume

$$v = x + w_0 \\ x = v + u_0$$

Then given $u \in U$, we have $v + u = x + w_1$, i.e.

$$(x + w_0) + u = x + w_1 \\ \Rightarrow \\ u = w_1 - w_0 \in W \\ \Rightarrow \\ U \subseteq W$$

For the same reason, $W \subseteq U$, i.e. $U = W$.

$\square$

3E.9

Prove that a nonempty subset 𝐴 of 𝑉 is a translate of some subspace of 𝑉 if and only if 𝜆𝑣 + (1 − 𝜆)𝑤 ∈ 𝐴 for all 𝑣, 𝑤 ∈ 𝐴 and all 𝜆 ∈ 𝐅.

Proof:

$\Rightarrow$

Assume $A = u + U$ where $U$ is a subspace of $V$. And if $𝑣, 𝑤 ∈ 𝐴$, then

$$v = u + u_0 \\ w = u + u_1 \\ $$

Then we have

$$\lambda 𝑣 + (1 − \lambda )𝑤 = \lambda (u + u_0) + (1 - \lambda ) (u + u_1) = u + (\lambda u_0 + (1 - \lambda) u_1) \in A$$

$\Leftarrow$

We will prove $-v + A$ is a subspace.

If $a, b \in -v + A$.

$$\begin{align*} a &= -v + 𝑣_a \\ b &= -v + 𝑣_b \\ a + b &= -v + (v_a + v_b - v) \\ &= -v + 2(\frac{1}{2}v_a + \frac{1}{2}v_b) - v \end{align*}$$

Note $w = \frac{1}{2}v_a + \frac{1}{2}v_b \in A$, then $2w - v \in A$. So $a + b \in -v + A$.

Similarly,

$$ka = k (-v + v_a) = -v + k v_a - (k-1)v \in -v + A$$

$\square$

3E.10

Suppose $𝐴_1 = 𝑣 + 𝑈_1$ and $𝐴_2 = 𝑤 + 𝑈_2$ for some $𝑣, 𝑤 ∈ 𝑉$ and some subspaces $𝑈_1, 𝑈_2$ of $𝑉$. Prove that the intersection $𝐴_1 ∩ 𝐴_2$ is either a translate of some subspace of 𝑉 or is the empty set.

Proof:

Assume $𝐴_1 ∩ 𝐴_2 \neq \emptyset, a \in 𝐴_1 ∩ 𝐴_2$.

Let $A = 𝐴_1 ∩ 𝐴_2, U = -a + A$.

Assume $v_1, v_2 \in U$, then $v_1 = -a + a_1, v_2 = -a + a_2, a_1 \in A, a_2 \in A$.

We will show $v_1 + v_2 \in U$.

$$v_1 = -a + v + u_1 = -(v + u_0) + v + u_1 \\ v_2 = -a + v + u_2 = -(v + u_0) + v + u_2 \\ \Rightarrow \\ v_1 + v_2 = u_1 + u_2 - 2 u_0 = - (v + u_0) + (v + u_0) + (u_1 + u_2 - 2 u_0) \\ = -a + v + (u_1 + u_2 - u_0) \\ \Rightarrow \\ v_1 + v_2 + a = v + (u_1 + u_2 - u_0) \in A_1$$

For the same reason, $v_1 + v_2 + a \in A_2$. Then $v_1 + v_2 + a \in 𝐴_1 ∩ 𝐴_2$, i.e. $v_1 + v_2 \in -a + A$.

Now consider $k v_1 = -k a + k a_1 = k (-(v + u_0) + v + u_1) = k (u_1 - u0)$.

$$a + k (u_1 - u0) = v + u_0 + k(u_1 - u0) \in A_1$$

So $k v_1 \in U$.

Then $U$ is a subspace.

$\square$

3E.11

Suppose $𝑈 = \{(𝑥_1, 𝑥_2, … ) ∈ 𝐅^∞ ∶ 𝑥_𝑘 ≠ 0 \text{ for only finitely many } 𝑘\}$.

(a) Show that $𝑈$ is a subspace of $𝐅^∞$

Proof:

Assume $a, b \in U$, then assume for all $n > N_a, a_n = 0$ and $n > N_b, b_n = 0$.

Then $a+b$ has only finite non-zero items. So $a+b \in U$.

Similarly, $ka \in U$.

$\square$

(b) Prove that $𝐅^∞ /𝑈$ is infinite-dimensional.

Proof:

We prove by contradition. Assume it's finite-dimensional, and the dimension is $N$, all we need to do is to find $N+1$ independent vectors in $𝐅^∞ /𝑈$.

Consider the following group of vectors

$$v_{1,i} =\begin{cases} 1 &\text{ if } i = k(N+1)+1, k = 0, 1, 2, \cdots \\ 0 &\text{ otherwise}\\ \end{cases} \\ v_{2,i} =\begin{cases} 1 &\text{ if } i = k(N+1)+2, k = 0, 1, 2, \cdots \\ 0 &\text{ otherwise}\\ \end{cases} \\ v_{N+1,i} =\begin{cases} 1 &\text{ if } i = k(N+1)+(N+1), k = 0, 1, 2, \cdots \\ 0 &\text{ otherwise}\\ \end{cases} \\ $$

Any linear combination of them cannot have finite non-zero items.

Anything example is

$$v_{1,i} =\begin{cases} 1 &\text{ if } p | 2 \\ 0 &\text{ otherwise}\\ \end{cases} \\ v_{2,i} =\begin{cases} 1 &\text{ if } p | 3 \\ 0 &\text{ otherwise}\\ \end{cases} \\ \cdots$$

$\square$

3E.12

Suppose $𝑣_1, …, 𝑣_𝑚 ∈ 𝑉$ . Let

$$A = \{ \lambda_1 v_1 + \cdots + \lambda_m v_m : \lambda_1, \cdots, \lambda_m \in F \text{ and } \lambda_1 + \cdots + \lambda_m = 1 \}$$

(a) Prove that $𝐴$ is a translate of some subspace of $𝑉$.

Proof:

We will prove $-v_1 + A$ is a subspace.

$$(-v_1 + a) + (-v_1 + a') = \\ (-v_1 + \lambda_1 v_1 + \cdots + \lambda_m v_m) + (-v_1 + \lambda'_1 v_1 + \cdots + \lambda'_m v_m) \\ = -v_1 + (\lambda_1 + \lambda'_1 - 1) v_1 + \cdots + (\lambda_m + \lambda'_m) v_m$$

Note that

$$(\lambda_1 + \lambda'_1 - 1) + \cdots + (\lambda_m + \lambda'_m) = 1$$

So $(-v_1 + a) + (-v_1 + a') \in -v_1 + A$.

Next

$$k(-v_1 + a) = k (-v_1 + \lambda_1 v_1 + \cdots + \lambda_m v_m) \\ = -v_1 + ((1-k)v_1 + k\lambda_1 v_1 + \cdots + k\lambda_m v_m)$$

Note that

$$(1-k) + k\lambda_1 + \cdots + k\lambda_m = (1-k) + k = 1$$

So $k(-v_1 + a) \in -v_1 + A$, then $-v_1 + A$ is a subspace.

$\square$

(b) Prove that if $𝐵$ is a translate of some subspace of $𝑉$ and $\{𝑣_1, …, 𝑣_𝑚\} ⊆ 𝐵$, then $𝐴 ⊆ 𝐵$.

Proof:

Assume $B = u + U$ and $U$ is a subspace. Then $v_i - u \in U$ and

$$\lambda_1 (v_1-u) + \cdots + \lambda_m (v_m-u) \\ = (\lambda_1 v_1 + \cdots + \lambda_m v_m) - (\lambda_1 + \cdots + \lambda_m) u \\ = (\lambda_1 v_1 + \cdots + \lambda_m v_m) - u \in U$$

Then $\lambda_1 v_1 + \cdots + \lambda_m v_m \in B$. So $𝐴 ⊆ 𝐵$.

$\square$

(c) Prove that $𝐴$ is a translate of some subspace of $𝑉$ of dimension less than $𝑚$.

Proof:

Now consider the space spanned by $v_2-v_1, \cdots , v_m - v_1$ and call it $U$.

For any $u \in U, u = \lambda_2 (v_2-v_1) + \cdots + \lambda_m (v_m-v_1)$,

Consider $v_1 + u = v_1 + \lambda_2 (v_2-v_1) + \cdots + \lambda_m (v_m-v_1)$, then

$$v_1 + u = (1 - \lambda_2 - \cdots - \lambda_m) v_1 + \lambda_2 v_2 + \cdots + \lambda_m v_m \in A$$

So $v_1 + U \subseteq A$.

On the other hand, for any $\lambda_1 v_1 + \cdots + \lambda_m v_m \in A$, we have

$$\lambda_1 v_1 + \cdots + \lambda_m v_m \\ = v_1 + \lambda_2 (v_2-v_1) + \cdots + \lambda_m (v_m-v_1)$$

So $A \subseteq v_1+ U$, i.e. $A = v_1 + U$.

Since $U = \text{span}(𝑣_2-v_1, \cdots, 𝑣_𝑚-v_1)$, $\dim U < m$.

$\square$

3E.13

Suppose $𝑈$ is a subspace of $𝑉$ such that $𝑉/𝑈$ is finite-dimensional. Prove that $𝑉$ is isomorphic to $𝑈 × (𝑉/𝑈)$.

Proof:

Given $v \in V$, and $w_1+U, \cdots, w_m+U$ is a basis of $V/U$. Assume $v+U = a_1 (w_1+U) + \cdots + a_m (w_m+U)$, and let $u = v - (a_1 w_1 + \cdots + a_m w_m) \in U$. We have a map:

$$T: v \rightarrow (u, v + U)$$

It's easy to verify it's linear.

injective: if $T(v) = 0 = (0, 0 + U)$, then $u = 0 + 0 = 0$.

surjective: Given $(u, v+U)$, Then $T(v+u) = (u, v+U)$.

So $T$ is an isomorphism.

$\square$

3E.14

Suppose $𝑈$ and $𝑊$ are subspaces of $𝑉$ and $𝑉 = 𝑈 ⊕ 𝑊$. Suppose $𝑤_1, …, 𝑤_𝑚$ is a basis of $𝑊$. Prove that $𝑤_1 + 𝑈, …, 𝑤_𝑚 + 𝑈$ is a basis of $𝑉/𝑈$.

Proof:

Assume $v \in V$, and $v = u + w, u \in U, w \in W$. Then $v + U = w+U$, because $v-w \in U$. Then $v+U = a_1 (w_1+U) + \cdots + a_m (w_m+U)$. So $𝑤_1 + 𝑈, …, 𝑤_𝑚 + 𝑈$ spans $V/U$.

Now if $a_1 (w_1+U) + \cdots + a_m (w_m+U) = 0 + U$, then $$a_1 w_1 + \cdots + a_m w_m \in U \\ a_1 w_1 + \cdots + a_m w_m \in W \\ $$

Since $U \cap W = 0$, then $a_1 w_1 + \cdots + a_m w_m = 0$, then $a_1 = \cdots = a_m = 0$.

So $𝑤_1 + 𝑈, …, 𝑤_𝑚 + 𝑈$ is a basis of $𝑉/𝑈$.

$\square$

3E.15

Suppose $𝑈$ is a subspace of $𝑉$ and $𝑣_1 + 𝑈, …, 𝑣_𝑚 + 𝑈$ is a basis of $𝑉/𝑈$ and $𝑢_1, …, 𝑢_𝑛$ is a basis of $𝑈$. Prove that $𝑣_1, …, 𝑣_𝑚, 𝑢_1, …, 𝑢_𝑛$ is a basis of $𝑉$.

Proof:

Given $w \in V$, then $w+U \in V/U$, then

$$w+U = a_1 (v_1+U) + \cdots + a_m (v_m+U)$$

So

$$w - (a_1 v_1 + \cdots + a_m v_m) \in U \\ \Rightarrow \\ w - (a_1 v_1 + \cdots + a_m v_m) = b_1 u_1 + \cdots + b_n u_n$$

Then $𝑣_1, …, 𝑣_𝑚, 𝑢_1, …, 𝑢_𝑛$ spans $V$.

Assume

$$a_1 v_1 + \cdots + a_m v_m + b_1 u_1 + \cdots + b_n u_n = 0$$

Then $a_1 v_1 + \cdots + a_m v_m \in U$, then

$$(a_1 v_1 + \cdots + a_m v_m) + U = \\ a_1 (v_1+U) + \cdots + a_m (v_m+U) = 0 + U \\ \Rightarrow \\ a_1 = \cdots = a_m = 0$$

Then we have $b_1 = \cdots = b_n = 0$.

Then $𝑣_1, …, 𝑣_𝑚, 𝑢_1, …, 𝑢_𝑛$ are independent.

$\square$

3E.16

Suppose $𝜑 ∈ ℒ(𝑉, 𝐅)$ and $𝜑 ≠ 0$. Prove that

$$\dim V / \text{null } \varphi = 1$$

Proof:

Assume $u, v \not\in \text{null } \varphi$.

Then $\varphi (u) \neq 0, \varphi (v) \neq 0$.

Then we can find $k \neq 0$ such that $\varphi (u) = k \varphi (v)$, so $\varphi (u - kv) = 0$.

Then $u - kv \in \text{null } \varphi$.

Then we have $u + \text{null } \varphi = kv + \text{null } \varphi = k (v + \text{null } \varphi)$.

Since $u$ is arbitary, then any $u + \text{null } \varphi$ can be represented by a scalar of $v + \text{null } \varphi$.

So

$$\dim V / \text{null } \varphi = 1$$

$\square$

3E.17

Suppose $𝑈$ is a subspace of $𝑉$ such that $\dim 𝑉/𝑈 = 1$. Prove that there exists $𝜑 ∈ ℒ(𝑉, 𝐅)$ such that $\text{null } \varphi = 𝑈$.

Proof:

Since $\dim V/U = 1$, let $v + U$ be a basis of $V/U$.

For any $w \in V$, we have $w + U = k(v+U)$. Then $\mathcal{M}(w) = k$.

Then we can consider $\varphi = \mathcal{M} \circ \pi$.

If $w \in \text{null } \varphi$, then $\mathcal{M}(\pi(w)) = 0$, so $\pi(w) = 0 + U$, i.e. $w \in U$. So $\text{null } \varphi = U$.

$\square$

3E.18

Suppose that $𝑈$ is a subspace of $𝑉$ such that $𝑉/𝑈$ is finite-dimensional.

(a) Show that if $𝑊$ is a finite-dimensional subspace of $𝑉$ and $𝑉 = 𝑈 + 𝑊$, then $\dim 𝑊 ≥ \dim 𝑉/𝑈$.

Proof:

All we need to prove is that the basis $w_1+U, \cdots, w_m+U$ of $W$ spans $V/U$.

Assume $v \in V$, we can write $v = u + w = u + (a_1 w_1 + \cdots + a_m w_m)$. Then $v - w \in U$, so $v + U = w + U$.

So $w_1+U, \cdots, w_m+U$ of $W$ spans $V/U$. Then $\dim 𝑊 ≥ \dim 𝑉/𝑈$.

$\square$

(b) Prove that there exists a finite-dimensional subspace $𝑊$ of $𝑉$ such that $\dim 𝑊 = \dim 𝑉/𝑈$ and $𝑉 = 𝑈 ⊕ 𝑊$.

Proof:

Let $w_1+U, \cdots, w_m+U$ be a basis of $V/U$, then consider the $W = \text{span}(w_1, \cdots, w_𝑚)$.

If $u \in U$ and $u = a_1 w_1 + \cdots + a_m w_m$, then $a_1 w_1 + \cdots + a_m w_m \in U$, then

$$a_1 (w_1+U) + \cdots + a_m (w_m+U) = 0 + U$$

So $a_1 = \cdots = a_m = 0$, then $u = 0$.

So $U + W = 𝑈 ⊕ 𝑊$.

If $v \in V$, then $v + U = a_1 (w_1+U) + \cdots + a_m (w_m+U)$. Then $v - (a_1 w_1 + \cdots + a_m w_m) \in U$.

So $V = U + W$.

Combine these 2 parts, we have $V = 𝑈 ⊕ 𝑊$.

$\square$

3E.19

Suppose $𝑇 ∈ ℒ(𝑉, 𝑊)$ and $𝑈$ is a subspace of $𝑉$. Let $𝜋$ denote the quotient map from $𝑉$ onto $𝑉/𝑈$. Prove that there exists $𝑆 ∈ ℒ(𝑉/𝑈, 𝑊)$ such that $𝑇 = 𝑆 ∘ 𝜋$ if and only if $𝑈 ⊆ \text{null } 𝑇$.

Proof:

$\Rightarrow$

Assume $u \in U$, then $\pi (u) = 0 + U$, so $S(\pi (u)) = 0$.

So $𝑈 ⊆ \text{null } 𝑇$.

$\Leftarrow$

If $v \in V$, then let $S(v + U) = T(v)$.

It's well defined, because if $v_1 + U = v_2 + U$, then

$$S((v_1+U)-(v_2+U)) = S((v_1-v_2)+U) = T(v_1-v_2) = 0 \\ \Rightarrow (\because v_1-v_2 \in U \And U \subseteq \text{null } T) \\ S(v_1 + U) = S(v_2 + U)$$

Next, we show it's linear.

$$S((v_1+U)+(v_2+U)) = S((v_1+v_2) + U) = T(v_1+v_2) = T(v_1) + T(v_2) \\ = S(v_1+U) + S(v_2+U) \\ S(k(v_1+U)) = S(k v_1 + U) = T(k v_1) = k T(v_1) = k S(v_1 + U)$$

$\square$

Section 3F Duality

3F.1

Explain why each linear functional is surjective or is the zero map.

Proof:

Assume $\varphi \in V'$, if $\varphi(v) = c \neq 0$.

Then for any $d \in F$

$$\varphi ((d/c) v) = (d/c) \varphi(v) = d/c \cdot c = d$$

So $\varphi$ is surjective.

$\square$

3F.2

Give three distinct examples of linear functionals on $𝐑^{[0, 1]}$.

Proof:

$f(0), f(1), f(1/2)$

$\square$

3F.3

Suppose $𝑉$ is finite-dimensional and $𝑣 ∈ 𝑉$ with $𝑣 ≠ 0$. Prove that there exists $\varphi ∈ 𝑉'$ such that $\varphi(𝑣) = 1$.

Proof:

Extend $v$ to a basis of $V$, then its dual basis satisfies $\varphi(𝑣) = 1$.

$\square$

3F.4

Suppose $𝑉$ is finite-dimensional and $𝑈$ is a subspace of $𝑉$ such that $𝑈 ≠ 𝑉$. Prove that there exists $\varphi ∈ 𝑉'$ such that $\varphi(𝑢) = 0$ for every $𝑢 ∈ 𝑈$ but $\varphi ≠ 0$.

Proof:

$$\dim U^0 = \dim V - \dim U > 0$$

So $U^0 \neq 0$, then pick any $\varphi \in U^0$.

$\square$

3F.5

Suppose 𝑇 ∈ ℒ(𝑉, 𝑊) and $𝑤_1, …, 𝑤_𝑚$ is a basis of range $𝑇$. Hence for each $𝑣 ∈ 𝑉$ , there exist unique numbers $\varphi_1(𝑣), …, \varphi_𝑚(𝑣)$ such that

$$Tv = \varphi_1(v) w_1 + \cdots + \varphi_m(v) w_m$$

thus defining functions $\varphi_1, …, \varphi_𝑚$ from 𝑉 to 𝐅. Show that each of the functions $\varphi_1, …, \varphi_𝑚$ is a linear functional on $𝑉$.

Proof:

Let $\psi_1, \cdots, \psi_m$ be the dual basis of $𝑤_1, …, 𝑤_𝑚$. Then

$$T'(\psi_i)(v) = \psi_i(T(v)) = \varphi_i(v) \psi_i(w_i) = \varphi_i(v)$$

So $T'(\psi_i) = \varphi_i$, which is a linear functional on $V$.

$\square$

3F.6

Suppose 𝜑, 𝛽 ∈ 𝑉′ . Prove that null 𝜑 ⊆ null 𝛽 if and only if there exists 𝑐 ∈ 𝐅 such that 𝛽 = 𝑐𝜑.

Proof:

$\Leftarrow$

Assume $v \in \text{null } \varphi$, then

$$\beta (v) = c \varphi (v) = c 0 = 0$$

So $\text{null } \varphi \subseteq \text{null } \beta$

$\Rightarrow$

If $\beta = 0$, then $c = 0$ can satisfy $\beta = c \varphi$.

If $\beta \neq 0$, then $\text{null } \varphi \subseteq \text{null } \beta \subset V$,

we can find $u \in V$ such that $\varphi (u) \neq 0$.

Then $V = \text{null } \varphi \oplus \text{span}(u)$.

If $\beta (u) = 0$, then $\text{null } \beta = V$, we have a contradition.

So $\beta (u) \neq 0$. Then let $c = \beta (u) / \varphi (u)$.

$\square$

3F.7

Suppose that $V_1, \cdots, V_m$ are vector spaces. Prove that $(𝑉_1 × ⋯ × 𝑉_𝑚)'$ and $𝑉_1' × ⋯ × 𝑉_𝑚'$ are isomorphic vector spaces.

Proof:

Note $V_i' = \mathcal{L}(V_i, R)$, then we just need to apply exercise 3E.3 by replacing $W$ with $R$.

$\square$

3F.8

Suppose $𝑣_1, …, 𝑣_𝑛$ is a basis of 𝑉 and $\varphi_1, …, \varphi_𝑛$ is the dual basis of $𝑉'$. Define $\Gamma ∶ 𝑉 → 𝐅^𝑛$ and $Λ ∶ 𝐅^𝑛 → 𝑉$ by

$$\Gamma(𝑣) = (\varphi_1(𝑣), …, \varphi_𝑛(𝑣)), \Delta(a_1, \cdots, a_n) = a_1 v_1 + \cdots + a_n v_n.$$

Explain why $\Gamma$ and $\Delta$ are inverses of each other.

Proof:

First, we need to prove $\Gamma (\Delta (a_1, \cdots, a_n)) = (a_1, \cdots, a_n)$.

$$\Gamma (\Delta (a_1, \cdots, a_n)) =\\ \Gamma (a_1 v_1 + \cdots + a_n v_n) = \\ (\varphi_1(a_1 v_1 + \cdots + a_n v_n), \cdots, \varphi (a_1 v_1 + \cdots + a_n v_n)) = \\ (a_1, \cdots, a_n)$$

Secondly, we need to prove $\Delta(\Gamma(v)) = v$

$$\Delta(\Gamma(v)) = \Delta(\varphi_1(𝑣), …, \varphi_𝑛(𝑣)) \\ = \varphi_1(v) v_1 + \cdots + \varphi_n(v) v_n \\ = v$$

The last equation is from 3.114.

$\square$

3F.9

Suppose $𝑚$ is a positive integer. Show that the dual basis of the basis $1, 𝑥, …, 𝑥_𝑚$ of $𝒫_𝑚(𝐑)$ is $\varphi_0, \varphi_1, …, \varphi_𝑚$, where

$$\varphi_k(p) = \frac{p^{(k)}(0)}{k!}$$

Proof:

$$\varphi_k(x^{j}) = \begin{cases} 0 &\text{ if } k > j \\ 1 &\text{ if } k = j \\ x^{j-k} = 0 &\text{ if } k < j \\ \end{cases}$$

$\square$

3F.10

Suppose $𝑚$ is a positive integer.

(a) Show that $1, 𝑥 − 5, …, (𝑥 − 5)^𝑚$ is a basis of $𝒫_{m}(\mathbf{R})$.

Proof:

$$a_0 \cdot 1 + a_1 \cdot (x-5) + \cdots + a_m \cdot (x-5)^m = 0 \\ \Rightarrow \\ a_m = 0 \\ \Rightarrow \\ a_{m-1} = 0 \\ \Rightarrow \\ \cdots \\ \Rightarrow \\ a_0 = 0$$

$\square$

(b) What is the dual basis of the basis in (a)?

Solution:

$$\varphi_k(p) = \frac{p^{(k)}(5)}{k!}$$

$\square$

3F.11

Suppose $𝑣_1, …, 𝑣_𝑛$ is a basis of $𝑉$ and $\varphi_1, …, \varphi_𝑛$ is the corresponding dual basis of $𝑉'$ . Suppose $\psi ∈ 𝑉'$. Prove that

$$\psi = \psi(𝑣_1)\varphi_1 +⋯ + \psi(𝑣_𝑛)\varphi_𝑛.$$

Proof:

For all $j$,

$$(\psi(𝑣_1)\varphi_1 +⋯ + \psi(𝑣_𝑛)\varphi_𝑛)(v_j) \\ \psi(𝑣_1)\varphi_1(v_j) +⋯ + \psi(𝑣_𝑛)\varphi_𝑛(v_j) \\ = \psi(v_j) \varphi (v_j) = \psi (v_j)$$

So

$$\psi = \psi(𝑣_1)\varphi_1 +⋯ + \psi(𝑣_𝑛)\varphi_𝑛.$$

$\square$

3F.12

Suppose $𝑆, 𝑇 ∈ ℒ(𝑉, 𝑊)$.

(a) Prove that $(𝑆 + 𝑇)' = 𝑆' + 𝑇'$

Proof:

Assume $\varphi \in W'$ and $v \in V$.

$$\begin{align*} (𝑆 + 𝑇)'(\varphi)(v) &= (\varphi \circ (S+T))(v) & \text{ (definition of dual mapping) } \\ &= \varphi((S+T)(v)) & \text{ (definition of composition) } \\ &= \varphi (S(v) + T(v)) & \text{ (the definition of addition of linear maps in } L(V, W)) \\ &= \varphi (S(v)) + \varphi (T(v)) & \varphi \text{ is linear map of } W'\\ &= (S'(\varphi)) (v) + (T'(\varphi)) (v) & \text{ (definition of dual mapping) } \\ &= (S'(\varphi) + T'(\varphi)) (v) & \text{ the definition of addition of linear maps in } V'\\ &= ((S' + T')(\varphi)) (v) & \text{ the definition of addition of linear maps in } ℒ(W', V') \\ \end{align*}$$

So $(𝑆 + 𝑇)' = 𝑆' + 𝑇'$.

$\square$

(b) Prove that $(\lambda 𝑇)' = \lambda 𝑇'$ for all $𝜆 ∈ 𝐅$.

Proof:

Assume $\varphi \in W'$ and $v \in V$.

$$\begin{align*} ((\lambda 𝑇)'(\varphi))(v) &= (\varphi \circ (\lambda 𝑇))(v) & \text{ (definition of dual mapping) } \\ &= \varphi ((\lambda 𝑇)(v)) & \text{ (definition of composition)} \\ &= \varphi (\lambda 𝑇(v)) & \text{ definition of scalar multiplication in } L(V, W)\\ &= \lambda (\varphi(T(v))) & \varphi \text{ is a linear map in } W' \\ &= \lambda ((T'(\varphi))(v)) & \text{ (definition of dual mapping) } \\ &= (\lambda T'(\varphi))(v) & \text{ definition of scalar multiplication in V'} \\ &= ((\lambda T')(\varphi))(v) & \text{ definition of scalar multiplication in } ℒ(W', V') \\ \end{align*}$$

$$\begin{align*} \end{align*}$$

So $(\lambda 𝑇)' = \lambda 𝑇'$

$\square$

3F.13

Show that the dual map of the identity operator on $𝑉$ is the identity operator on $𝑉'$.

Proof:

Let $i$ be the identity operator, and $\varphi$ be a linear map in $V'$.

$i'(\varphi) = \varphi \circ i = \varphi$.

So $i'$ is the identity operator on $𝑉'$.

$\square$

3F.14

Define $𝑇 ∶ 𝐑^3 → 𝐑^2$ by

$$𝑇(𝑥, 𝑦, 𝑧) = (4𝑥 + 5𝑦 + 6𝑧, 7𝑥 + 8𝑦 + 9𝑧).$$

Suppose $\varphi_1, \varphi_2$ denotes the dual basis of the standard basis of $𝐑^2$ and $\psi_1, \psi_2, \psi_3$ denotes the dual basis of the standard basis of $𝐑^3$

(a) Describe the linear functionals $𝑇'(\varphi_1)$ and $𝑇'(\varphi_2)$.

Solution:

$$T'(\varphi_1) = \varphi_1 \circ T = 4x+5y+6z \\ T'(\varphi_2) = \varphi_2 \circ T = 7x+8y+9z \\ $$

$\square$

(b) Write $𝑇'(\varphi_1)$ and $𝑇'(\varphi_2)$ as linear combinations $\psi_1, \psi_2, \psi_3$.

Solution:

$$T'(\varphi_1) = \varphi_1 \circ T = 4x+5y+6z = 4\psi_1+5\psi_2+6\psi_3 \\ T'(\varphi_2) = \varphi_2 \circ T = 7x+8y+9z = 7\psi_1+8\psi_2+9\psi_3 \\ $$

$\square$

3F.15

Define $𝑇 ∶ 𝒫(𝐑) → 𝒫(𝐑)$ by

$$(𝑇𝑝)(𝑥) = 𝑥^2𝑝(𝑥) + 𝑝''(𝑥)$$

for each $𝑥 ∈ 𝐑$.

(a) Suppose $\varphi ∈ 𝒫(𝐑)'$ is defined by $\varphi(𝑝) = 𝑝'(4)$. Describe the linear functional $𝑇' (\varphi)$ on $𝒫(𝐑)$.

Solution:

$$(𝑇' (\varphi))(p(x)) = (\varphi \circ T)(p(x)) \\ = \varphi (T(p(x))) \\ = \varphi (𝑥^2p(𝑥) + p''(𝑥)) \\ = x^2 p'(x) + 2x p(x) + p'''(x) |_4 \\ = 16 p'(4) + 8 p(4) + p'''(4)$$

$\square$

(b) Suppose $\varphi ∈ 𝒫(𝐑)'$ is defined by $\varphi(𝑝) = \int_{0}^{1} 𝑝$. Evaluate $(𝑇'(\varphi))(𝑥^3)$.

Solution:

$$(𝑇' (\varphi))(x^3) = (\varphi \circ T)(x^3) \\ = \varphi (T(x^3)) \\ = \varphi (x^5 + 6x) \\ = \frac{1}{6} x^6 + 3x^2 \bigg|_0^1 \\ = 3 \frac{1}{6}$$

$\square$

3F.16

Suppose $𝑊$ is finite-dimensional and $𝑇 ∈ ℒ(𝑉, 𝑊)$. Prove that

$$T'=0 \Longleftrightarrow T = 0$$

Proof:

Assume $w_1, \cdots, w_m$ is a basis of $W$, $\varphi_1, \cdots, \varphi_m$ is the dual basis in $W'$, then

$\Rightarrow$

$$\begin{align*} T(v) &= \varphi_1(T(v)) w_1 + \cdots + \varphi_m(T(v)) w_m &\text{ (3.114) } \\ &= T'(\varphi_1)(v) w_1 + \cdots + T'(\varphi_m)(v) w_m &\text{ (3.118 definition: dual map) } \\ &= 0 &(T' = 0) \\ \end{align*}$$

$\Leftarrow$

$$\begin{align*} T'(\varphi ) (v) &= \varphi (T(0)) &\text{ (3.118 definition: dual map) } \\ &=0 &(T=0) \\ &\Rightarrow T'(\varphi ) = 0 \text{ for any } \varphi \\ &\Rightarrow T' = 0 \end{align*}$$

Another way to prove

$$\begin{align*} T' = 0 & \Longleftrightarrow \text{null } T' = W' \\ & \Longleftrightarrow \text{dim range } T' = 0 &\text{ (3.21 fundamental theorem of linear maps) } \\ & \Longleftrightarrow \text{dim range } T = 0 &\text{ (3.130 (a)) } \\ & \Longleftrightarrow \text{dim null } T = \dim V &\text{ (3.21 fundamental theorem of linear maps) } \\ & \Longleftrightarrow \text{null } T = V \\ & \Longleftrightarrow T = 0 \\ \end{align*}$$

$\square$

3F.17

Suppose 𝑉 and 𝑊 are finite-dimensional and 𝑇 ∈ ℒ(𝑉, 𝑊). Prove that 𝑇 is invertible if and only if 𝑇′ ∈ ℒ(𝑊′, 𝑉′) is invertible.

Proof:

$\Rightarrow$

Since $T$ is invertible, we can find $S \in \mathcal{L}(W, V)$ such that $ST = I, TS = I$.

$$\begin{align*} I &= I' &\text{ (ex. 3F.13) } \\ &= (ST)' \\ &= T' S' &\text{ (3.120 algebraic properties of dual maps) } \end{align*}$$

For the same reason $S'T' = I$. So $T'$ is invertible.

$\Leftarrow$

We can use a different way: $T'$ is invertible then, $W', V'$ are the same dimension, so are the $W, V$.

$$\begin{align*} T' \text{ is invertible } &\Rightarrow T' \text{ is injective } &\text{ (3.63) } \\ &\Rightarrow T \text{ is surjective } &\text{ (3.129) } \\ &\Rightarrow T \text{ is invertible } &(\text{ 3.65 and } \dim W = \dim V) \\ \end{align*}$$

$\square$

3F.18

Suppose 𝑉 and 𝑊 are finite-dimensional. Prove that the map that takes 𝑇 ∈ ℒ(𝑉, 𝑊) to 𝑇′ ∈ ℒ(𝑊′ , 𝑉′) is an isomorphism of ℒ(𝑉, 𝑊) onto ℒ(𝑊′ , 𝑉′).

Proof:

Since $𝑉$ and $𝑊$ are finite-dimensional, we just need to prove this map is injective is enough. From exercise 3F.16, it's indeed the case.

$\square$

3F.19

Suppose $𝑈 ⊆ 𝑉$. Explain why

$$U^0 = \{\varphi ∈ 𝑉' ∶ 𝑈 ⊆ \text{null } \varphi \}.$$

Proof:

Let $W = \{\varphi ∈ 𝑉' ∶ 𝑈 ⊆ \text{null } \varphi \}$.

If $\varphi \in W$, then for $u \in U$, $\varphi (u) = 0$, so $\varphi \in U^0$. So $W \subseteq U^0$.

If $\varphi \in U^0$, then for $u \in U$, $\varphi (u) = 0$, so $U \subseteq \text{null } \varphi$. So $\varphi \in W$. So $U^0 \subseteq W$.

In summary $U^0 = W$.

$\square$

3F.20

Suppose $𝑉$ is finite-dimensional and $𝑈$ is a subspace of $𝑉$. Show that

$$𝑈 = \{𝑣 ∈ 𝑉 ∶ \varphi(𝑣) = 0 \text{ for every } \varphi ∈ 𝑈^0\}.$$

Proof

Let $W = \{𝑣 ∈ 𝑉 ∶ \varphi(𝑣) = 0 \text{ for every } \varphi ∈ 𝑈^0\}$.

$\Rightarrow$

If $u \in U$, then for every $\varphi \in U^0$, $\varphi(u) = 0$. So $u \in W$, i.e. $U \subseteq W$.

$\Leftarrow$

Let $\dim U = m, \dim V = n$. $u_1, \cdots, u_m$ be a basis of $U$.

For $w \in W$, we consider $U' = \text{span}(u_1, \cdots, u_𝑚, w)$.

Given $\varphi \in U^0$, and any $u' \in U'$, $\varphi (u') = 0$, so $U^0 \subseteq U'^0$.

Then $\dim U' = \dim V - \dim U'^0 \leq \dim V - \dim U^0 = \dim U$.

So $U = U'$, then $w \in U$, so $W \subseteq U$.

In summary, $U = W$.

$\square$

3F.21

Suppose $𝑉$ is finite-dimensional and $𝑈$ and $𝑊$ are subspaces of $𝑉$.

(a) Prove that $𝑊^0 ⊆ 𝑈^0$ if and only if $𝑈 ⊆ 𝑊$.

Proof:

$\Rightarrow$

Assume $u \in U$, and give any $\varphi \in W^0$, since $\varphi \in U^0$, then $\varphi (u) = 0$. Then from ex 3F.20, we have $u \in W$.

$\Leftarrow$

Given $\varphi \in W^0$, and any $u \in U$, since $u \in W$, $\varphi (u) = 0$, so $\varphi \in U^0$. So $𝑊^0 ⊆ 𝑈^0$.

$\square$

3F.22

Suppose $𝑉$ is finite-dimensional and $𝑈$ and $𝑊$ are subspaces of $𝑉$.

(a) Show that $(𝑈 + 𝑊)^0 = 𝑈^0 ∩ 𝑊^0$.

Proof:

$\subseteq$

if $\psi \in (U+W)^0$, then given $u \in U$ then $u \in U+W$, so $\psi (u) = 0$. Then $\psi \in U^0$. Similarly, $\psi \in W^0$. So $\psi \in 𝑈^0 ∩ 𝑊^0$. Then $(𝑈 + 𝑊)^0 \subseteq 𝑈^0 ∩ 𝑊^0$.

$\supseteq$

If $\psi \in 𝑈^0 ∩ 𝑊^0$, then $\psi \in U^0$ and $\psi \in W^0$. Given $u+w \in U+W$

$$\begin{align*} \psi (u+w) &= \psi (u) + \psi (w) \\ &= 0+0 \\ &= 0 \\ &\Rightarrow \psi \in (U+W)^0 \\ &\Rightarrow (𝑈 + 𝑊)^0 \supseteq 𝑈^0 ∩ 𝑊^0 \end{align*}$$

$\square$

(b) Show that $(𝑈 ∩ 𝑊)^0 = 𝑈^0 + 𝑊^0$.

Proof:

If $\varphi \in U^0$ and $\psi \in W^0$, and $v \in 𝑈 ∩ 𝑊$.

Then $(\varphi + \psi)(v) = \varphi(v) + \psi (v) = 0 + 0 = 0$

So $𝑈^0 + 𝑊^0 \subseteq (𝑈 ∩ 𝑊)^0$.

On other other hand,

$$\begin{align*} \dim (𝑈 ∩ 𝑊)^0 &= \dim V - (\dim (𝑈 ∩ 𝑊)) \\ &= \dim (U+W) + \dim V - (\dim U + \dim W) \\ \dim 𝑈^0 + 𝑊^0 &= \dim 𝑈^0 + \dim W^0 - \dim (𝑈^0 ∩ 𝑊^0) \\ &= \dim V - \dim U + \dim V - \dim W - \dim (𝑈 + 𝑊)^0 \\ &= \dim (U+W) + \dim V - (\dim U + \dim W) \end{align*}$$

So $\dim (𝑈 ∩ 𝑊)^0 = \dim 𝑈^0 + 𝑊^0$

Then $(𝑈 ∩ 𝑊)^0 = 𝑈^0 + 𝑊^0$.

$\square$

3F.23

Suppose $𝑉$ is finite-dimensional and $\varphi_1, …, \varphi_𝑚 ∈ 𝑉'$ . Prove that the following three sets are equal to each other.

(a) $\text{span}(\varphi_1, \cdots, \varphi_𝑚)$

(b) $((\text{null } \varphi_1) \cap \cdots \cap (\text{null } \varphi_m))^0$

(c) $\{\varphi \in V' : (\text{null } \varphi_1) \cap \cdots \cap (\text{null } \varphi_m) \subseteq \text{null } \varphi\}$

Proof:

The equivalence of $(b)$ and $(c)$ is the result from ex 3F.19.

From ex 3F.22 (b)

$$\begin{align*} ((\text{null } \varphi_1) \cap \cdots \cap (\text{null } \varphi_m))^0 &= (\text{null } \varphi_1)^0 + \cdots + (\text{null } \varphi_m)^0 \end{align*}$$

$\varphi_i \in (\text{null } \varphi_i)^0$

$$\begin{align*} \text{dim} (\text{null } \varphi_i )^0 &= \dim V - \text{dim null } \varphi_i \\ &= \dim V - (\dim V - \text{dim range } \varphi_i) \\ &= \dim V - (\dim V - 1) \\ &= 1 \\ &\Rightarrow \text{span}(\varphi_i) = (\text{null } \varphi_i)^0 \\ &\Rightarrow (\text{null } \varphi_1)^0 + \cdots + (\text{null } \varphi_m)^0 = \text{span}(\varphi_1, \cdots, \varphi_𝑚) \end{align*}$$

So we proved (a) and (b) are equivalent.

$\square$

3F.24

Suppose $𝑉$ is finite-dimensional and $𝑣_1, …, 𝑣_𝑚 ∈ 𝑉$. Define a linear map

$$\Gamma ∶ 𝑉' → 𝐅^𝑚 \text{ by } \Gamma(\varphi) = (\varphi(𝑣_1), …, \varphi(𝑣_𝑚)).$$

(a) Prove that $𝑣_1, …, 𝑣_𝑚$ spans $𝑉$ if and only if $\Gamma$ is injective.

Proof:

$\Rightarrow$

If $\Gamma (\varphi) = 0$, then

$$\varphi(𝑣_1) = … = \varphi(𝑣_𝑚) = 0$$

Then $\varphi (v) = 0$, so $\varphi = 0$.

Then $\Gamma$ is injective.

$\Leftarrow$

If $\Gamma$ is injective, and let $U = \text{span}(𝑣_1, \cdots, 𝑣_𝑚)$. If $\varphi \in U^0$, then $\Gamma(\varphi)=(\varphi(𝑣_1), …, \varphi(𝑣_𝑚)) = 0$.

Then $\varphi = 0$, that means $U^0 = \{0\}$, i.e. $U = V$.

$\square$

(b) Prove that $𝑣_1, …, 𝑣_𝑚$ is linearly independent if and only if $\Gamma$ is surjective.

Proof:

$\Rightarrow$

Let $\varphi_1, \cdots, \varphi_m$ be the dual basis, then $\Gamma (\varphi_i) = e_i$ which is a standard basis of $𝐅^𝑚$. So $\Gamma$ is surjective.

$\Leftarrow$

Since $\Gamma$ is surjective, let $\varphi_i$ be the mapping such that $\Gamma (\varphi_i) = e_i$, so $\varphi_i(v_i) = 1, \varphi_i(v_j) = 0$.

Let

$$\begin{align*} 0 = a_1 v_1 + \cdots + a_m v_m & \Rightarrow \varphi_i (a_1 v_1 + \cdots + a_m v_m) = 0 \\ & \Rightarrow a_i \varphi_i (v_i) = 0 \\ & \Rightarrow a_i = 0 \end{align*}$$

So $𝑣_1, …, 𝑣_𝑚$ is linearly independent.

$\square$

Now we try a different approach. First, we consider the following isomorphism from $F^n \rightarrow (F^n)'$

$$i : (x_1, \cdots, x_m) \rightarrow x_1 \psi_1 + \cdots + x_m \psi_m$$

Where $\psi_i$ is the dual basis of the standard basis $e_i$.

Let $\Gamma' = i \circ \Gamma$, so $\Gamma' \in (F^n)'$

Also consider $\gamma$ from $F^n \mapsto V$

$$\gamma : (x_1, \cdots, x_m) \rightarrow x_1 v_1 + \cdots + x_m v_m$$

Let $\varphi \in V'$, then

$$\begin{align*} (\varphi \circ \gamma) (x_1, \cdots, x_m) &= x_1 \varphi(v_1) + \cdots + x_m \varphi(v_m) \end{align*}$$

On the other hand,

$$\begin{align*} \Gamma'(\varphi)(x_1, \cdots, x_m) &= (\varphi(v_1) \psi_1 + \cdots + \varphi(v_m) \psi_m) (x_1, \cdots, x_m) \\ &= x_1 \varphi(v_1) + \cdots + x_m \varphi(v_m) \end{align*}$$

This shows $\Gamma'$ is dual map of $\gamma$.

Now coming back the problem.

$$\begin{align*} \text{span}(𝑣_1, \cdots, 𝑣_𝑚) = V &\Longleftrightarrow \gamma \text{ is surjective} \\ &\Longleftrightarrow \Gamma' \text{ is injective} &\text{ (3.131) } \\ &\Longleftrightarrow \Gamma \text{ is injective} \end{align*}$$

Also

$$\begin{align*} 𝑣_1, …, 𝑣_𝑚 \text{ is linearly independent } &\Longleftrightarrow \gamma \text{ is injective} \\ &\Longleftrightarrow \Gamma' \text{ is surjective} &\text{ (3.129) } \\ &\Longleftrightarrow \Gamma \text{ is surjective} \end{align*}$$

$\square$

3F.25

Suppose $𝑉$ is finite-dimensional and $\varphi_1, …, \varphi_𝑚 ∈ 𝑉'$. Define a linear map

$\Gamma ∶ 𝑉 → 𝐅^𝑚$ by $\Gamma(𝑣) = (\varphi_1(𝑣), …, \varphi_𝑚(𝑣))$.

(a) Prove that $\varphi_1, …, \varphi_𝑚$ spans $𝑉'$ if and only if $Γ$ is injective.

Proof:

$\Rightarrow$

Consider $\text{null } \Gamma$, note $\varphi_i \in (\text{null } \Gamma )^0$, so $(\text{null } \Gamma )^0 = V'$, then from $\text{null } \Gamma = \{0\}$. $\Gamma$ is injective.

$\Leftarrow$

If $\Gamma$ is injective, then we consider $\text{span}(\varphi_1, \cdots, \varphi_𝑚)$. From 23.b we know

$$\text{span}(\varphi_1, \cdots, \varphi_𝑚) = ((\text{null } \varphi_1) \cap \cdots \cap (\text{null } \varphi_n))^0$$

Since $\Gamma$ is injective, then

$$(\text{null } \varphi_1) \cap \cdots \cap (\text{null } \varphi_n) = \{0\}$$

Then $\{0\}^0 = V'$.

$\square$

(b) Prove that $\varphi_1, …, \varphi_𝑚$ is linearly independent if and only if $Γ$ is surjective.

Proof

$\Leftarrow$

Since $Γ$ is surjective, we can find $\Gamma(v_i) = e_i$.

$$\begin{align*} a_1 v_1 + \cdots + a_m v_m = 0 &\Rightarrow a_1 \Gamma(v_1) + \cdots + a_m \Gamma(v_m) = 0 \\ &\Rightarrow a_1 e_1 + \cdots + a_m e_m = 0 \\ &\Rightarrow a_1 = \cdots = a_m = 0 \end{align*}$$

Then $\varphi_i$ is the dual basis $v_i$. Then $\varphi_1, …, \varphi_𝑚$ is linearly independent.

I cannot prove the other side easily.

So we consider the dual map of $\Gamma ' : (F^m)' \rightarrow V'$, and $\psi_1, \cdots, \psi_m$ is the dual basis of the standard basis of $e_i$.

$$\Gamma'(\psi_i)(v) = \psi_i (\Gamma(v)) = \psi_i (\varphi_1(𝑣), …, \varphi_𝑚(𝑣)) = \varphi_i(v)$$

So $\Gamma'(\psi_i) = \varphi_i$.

$$\begin{align*} \text{span}(\varphi_1, \cdots, \varphi_𝑚) = V' &\Longleftrightarrow \Gamma' \text{ is surjective } \\ &\Longleftrightarrow \Gamma \text{ is injective } \\ \end{align*}$$

Also

$$\begin{align*} \varphi_1, \cdots, \varphi_𝑚 \text{ is independent } &\Longleftrightarrow \Gamma' \text{ is injective } \\ &\Longleftrightarrow \Gamma \text{ is surjective } \\ \end{align*}$$

$\square$

3F.26

Suppose $𝑉$ is finite-dimensional and $Ω$ is a subspace of $𝑉'$ . Prove that

$$\Omega = \{ v \in V : \varphi (v) = 0 \text{ for every } \varphi \in \Omega \}^0$$

Proof:

Let

$$U = \{ v \in V : \varphi (v) = 0 \text{ for every } \varphi \in \Omega \}$$

And $\varphi_1, \cdots, \varphi_m$ be a basis of $\Omega$.

Then

$$U = \{ \text{null } \varphi_1 \cap \cdots \cap \text{null } \varphi_m \}$$

Then from ex 2.22

$$U^0 = \text{span}(\varphi_1, \cdots, \varphi_𝑚) = \Omega$$

$\square$

3F.27

Suppose 𝑇 ∈ ℒ(𝒫 5(𝐑)) and null 𝑇′ = span(𝜑), where 𝜑 is the linear functional on 𝒫 5(𝐑) defined by 𝜑(𝑝) = 𝑝(8). Prove that

$$\text{range } T = \{p \in 𝒫_{5}(𝐑) : p(8) = 0\}$$

Proof:

Let $U = \{p \in 𝒫_{5}(𝐑) : p(8) = 0\}$.

Assume $p \in \text{range } T$, since $\text{null } T' = (\text{range } T)^0$ We have $(\text{range } T)^0 = \text{span}(\varphi)$.

Then $\varphi (p) = 0$, i.e. $p(8) = 0$, so $\text{range } T \subseteq U$.

On the other hand

$$\begin{align*} U = \text{null } \varphi & \Rightarrow \dim U = \dim V - \text{dim range } \varphi \\ & \Rightarrow \dim U = \dim V - 1 \\ \text{dim range } T &= \dim V - \dim (\text{range } T)^0 \\ &= \dim V - \dim \text{span}(\varphi) \\ &= \dim V - 1 \end{align*}$$

So $U$ and $\text{range } T$ has the same dimension.

Then we conclude $U = \text{range } T$.

$\square$

3F.28

Suppose $𝑉$ is finite-dimensional and $\varphi_1, …, \varphi_𝑚$ is a linearly independent list in $𝑉'$. Prove that

$$\dim ( (\text{null } \varphi_1) \cap \cdots \cap (\text{null } \varphi_m) ) = (\dim V) - m$$

Proof:

Let

$$U = (\text{null } \varphi_1) \cap \cdots \cap (\text{null } \varphi_m)$$

From ex 3F.23, $U^0 = \text{span}(\varphi_1, \cdots, \varphi_𝑚)$. Since $\varphi_1, …, \varphi_𝑚$ is a linearly independent list, $\dim U^0 = m$.

Then $\dim U = \dim V - m$.

$\square$

3F.29

Suppose 𝑉 and 𝑊 are finite-dimensional and 𝑇 ∈ ℒ(𝑉, 𝑊).

(a) Prove that if $\varphi ∈ 𝑊'$ and $\text{null } 𝑇' = \text{span}(\varphi)$, then $\text{range } 𝑇 = \text{null } \varphi$.

Proof:

$$\begin{align*} (\text{range } T)^0 &= \text{null } T' \\ &= \text{span}(\varphi) \\ \end{align*}$$

If $w \in \text{range } T$, since $\varphi \in (\text{range } T)^0$ then $\varphi (w) = 0$, $w \in \text{null } \varphi$.

On the other hand,

$$\begin{align*} \dim (\text{range } T)^0 &= \dim \text{span}(\varphi) \\ &\Rightarrow \text{dim range } T = \dim W - 1 \\ &\Rightarrow \text{dim range } T = \text{dim null } \varphi \\ &\Rightarrow \text{range } T = \text{null } \varphi \\ \end{align*}$$

$\square$

(b) Prove that if $\psi ∈ 𝑉'$ and range $𝑇' = \text{span }(\psi)$, then $\text{null } 𝑇 = \text{null } \psi$.

Proof:

$$(\text{null } T)^0 = \text{range } 𝑇' = \text{span }(\psi)$$

If $v \in \text{null } T$, since $\psi \in (\text{null } T)^0$, $\psi(v) = 0$. Then $v \in \text{null } \psi$.

$$\begin{align*} \text{dim null } T &= \dim V - \dim (\text{null } T)^0 \\ &= \dim V - 1 \\ \text{dim null } \psi &= \dim V - 1 \end{align*}$$

So $\text{null } 𝑇 = \text{null } \psi$.

$\square$

3F.30

Suppose $𝑉$ is finite-dimensional and $\varphi_1, \cdots, \varphi_m$ is a basis of $𝑉'$ . Show that there exists a basis of $𝑉$ whose dual basis is $\varphi_1, \cdots, \varphi_m$.

Proof:

We will use ex 3F.26.

Let $\Omega_1 = \text{span}(\varphi_2, \cdots, \varphi_𝑚)$.

$$U_1 = \{ v \in V : \varphi (v) = 0 \text{ for every } \varphi \in \Omega_1 \}$$

$U_1^0 = \Omega_1$.

$$\dim U_1^0 = \dim \Omega_1 = n-1$$

So $\dim U_1 = 1$, assume $v \in U_1$. If $\varphi_1 (v_1) = 0$, then $\varphi_1 \in U_1^0 = \Omega_1$, we have a contradition.

So let $\varphi_1(v_1) \neq 0$. Assume $\varphi_1(v_1) = 1$.

The repeat this process, we can find the rest of $v_i$.

$\square$

3F.31

Suppose 𝑈 is a subspace of 𝑉 . Let 𝑖 ∶ 𝑈 → 𝑉 be the inclusion map defined by 𝑖(𝑢) = 𝑢. Thus 𝑖′ ∈ ℒ(𝑉′, 𝑈′).

(a) Show that null $𝑖' = 𝑈^0$.

Proof:

If $\varphi \in \text{null } i'$, then $(i'(\varphi ))(u) = 0$, i.e.

$$\begin{align*} \varphi (i(u)) = 0 & \Rightarrow \varphi (u) = 0 \\ & \Rightarrow \varphi \in U^0 \end{align*}$$

On the other hand,

$$\begin{align*} \varphi \in U^0 & \Rightarrow \varphi (u) = 0 \\ & \Rightarrow \varphi (i(u)) = 0 \\ & \Rightarrow (i'(\varphi )) (u) = 0 \\ & \Rightarrow i'(\varphi ) = 0 \\ & \Rightarrow \varphi \in \text{null } i' \end{align*}$$

$\square$

(b) Prove that if 𝑉 is finite-dimensional, then range 𝑖′ = 𝑈′.

Proof:

For any $\psi \in U'$, it's a linear mapping from $U$ to $F$. Since $U$ is a subspace of $V$, it's possible to extend $\psi$ from $U$ to $V$, such that for $u \in U$, $\psi (u) = \varphi (u)$. Then we have

$$\begin{align*} \varphi (u) = \psi (u) & \Rightarrow \varphi (i(u)) = \psi (u) \\ & \Rightarrow (i'(\varphi)) (u) = \psi (u) \\ & \Rightarrow i'(\varphi ) = \psi \end{align*}$$

So $\text{range } i' = U'$.

$\square$

(c) Prove that if $𝑉$ is finite-dimensional, then $\tilde{i'}$ is an isomorphism from $𝑉' /𝑈^0$ onto $𝑈'$.

Proof:

This is an immediate result from 3.107 (d) null space and range of $\tilde{T}$.

$\square$

3F.32

The double dual space of $𝑉$ , denoted by $𝑉″$ , is defined to be the dual space of $𝑉'$ . In other words, $𝑉″ = (𝑉')'$ . Define $Λ ∶ 𝑉 → 𝑉″$ by

$$(Λ𝑣)(\varphi) = \varphi(𝑣)$$

for each $𝑣 ∈ 𝑉$ and each $\varphi ∈ 𝑉'$.

(a) Show that $Λ$ is a linear map from $𝑉$ to $𝑉″$.

Proof:

$$\begin{align*} (Λ(𝑣_1 + 𝑣_2))(\varphi) &= \varphi(𝑣_1 + 𝑣_2) \\ &= \varphi(𝑣_1) + \varphi(𝑣_2) \\ &= (Λ𝑣_1)(\varphi) + (Λ𝑣_2)(\varphi) \\ &= (Λ(𝑣_1) + Λ(𝑣_2))(\varphi) \end{align*}$$

$$\begin{align*} (Λ(k𝑣))(\varphi) &= \varphi(k𝑣) \\ &= k\varphi(𝑣) \\ &= k(Λ𝑣)(\varphi) \\ &= (kΛ𝑣)(\varphi) \\ \end{align*}$$

$\square$

(b) Show that if 𝑇 ∈ ℒ(𝑉), then 𝑇″ ∘ Λ = Λ ∘ 𝑇, where 𝑇″ = (𝑇′)′.

Proof:

First note $T' : V' \rightarrow V'$ by $T'(\varphi) = \varphi \circ T$. $V' = \mathcal{L}(V, F)$

And $T'' : V'' \rightarrow V''$ by $T''(\varphi) = \varphi \circ T'$. $V'' = \mathcal{L}(V, W)' = \mathcal{L}(\mathcal{L}(V, F), F)$

Assume $v \in V$,

$$\begin{align*} (𝑇″ ∘ Λ(v))(\varphi) &= (𝑇″ (Λ(v))(\varphi)) \\ &= (Λ(v))(𝑇'(\varphi)) \\ &= T'(\varphi)(v) \\ &= \varphi(T(v)) \end{align*}$$

$$\begin{align*} ((Λ ∘ 𝑇)(v))(\varphi) &= (Λ(T(v)))(\varphi) \\ &= \varphi(T(v)) \end{align*}$$

$\square$

3F.33

Suppose 𝑈 is a subspace of 𝑉 . Let 𝜋 ∶ 𝑉 → 𝑉/𝑈 be the usual quotient map. Thus 𝜋′ ∈ ℒ((𝑉/𝑈)′, 𝑉′).

(a) Show that 𝜋′ is injective.

Proof:

This is because $𝜋$ is surjective, then from 3.129, $𝜋′$ is injective.

$\square$

(b) Show that $\text{range } \pi ' = 𝑈^0$.

Proof:

3.130 (b) states $\text{range } \pi ' = (\text{null } \pi)^0$.

Also $\text{null } \pi = U$, so $\text{range } \pi ' = 𝑈^0$.

$\square$

(c) Conclude that $𝜋′$ is an isomorphism from (𝑉/𝑈)′ onto $U^0$.

Proof:

From (a) and (b) $\pi '$ is both injective and surjective from $(V/U)'$ to $U^0$. So $𝜋′$ is an isomorphism.

$\square$