# Chapter 3 Exercise F

1. Solution: For any $\vp\in\ca L(V,\mb F)$, if $\dim \m{range} \vp=0$, then $\vp$ is the zero map. If $\dim \m{range} \vp=1$, then $\vp$ is surjective since $\dim\mb F=1$. Moreover, $\dim \m{range} \vp\leqslant \dim \mb F=1$. Hence, that is all the possible cases.

2. Solution: Let $\vp_1,\vp_2,\vp_3\in\ca L(\R^{[0,1]},\mb F)$ defined by $\vp_1(f)=f(0),\quad\vp_2(f)=f(0.5),\quad\vp_3(f)=f(1).$Please check that $\vp_1,\vp_2,\vp_3\in\ca L(\R^{[0,1]},\mb F)$ and they are different from each other.

3. Solution: Extend $v$ to a basis of $V$ and use 3.96.

4. Solution: Let $u_1$, $\cdots$, $u_m$ be a basis of $U$, since $U\ne V$ we can extend it to a basis of $V$ as $u_1$, $\cdots$, $u_m$, $u_{m+1}$, $\cdots$, $v_{m+n}$, where $n\geqslant 1$. Hence we can define $\vp\in V’$ by $\vp(u_i)=\left\{ \begin{array}{ll} 0, & \hbox{if i\ne m+1;} \\ 1, & \hbox{if i=m+1.} \end{array} \right.$Then $\vp\in V’$ and $\vp(u)=0$ for every $u\in U$ but $\vp\ne 0$.

5. Solution: Define $P_i\in\ca L(V_i,V_1\times\cdots\times V_m)$ by $P_i(x)=(0,\cdots,0,x,0,\cdots,0)$with $x$ in the $i$-th component. Define $\vp\in \ca L((V_1\times\cdots\times V_m)’,V’_1\times\cdots\times V’_m)$ by $\vp(f)=(P’_1f,\cdots,P’_mf).$Now let us check that $\vp$ is an isomorphism.

Injectivity: suppose $(P’_1f,\cdots,P’_mf)=0$, that is for any $(x_1,\cdots,x_m)\in V_1\times\cdots\times V_m$, we have $P’_if(x_i)=0\Longrightarrow f(0,\cdots,x_i,\cdots,0)=0$by the definition of $P_i$ and dual map. This implies $f(x_1,\cdots,x_m)=\sum_{i=1}^mf(0,\cdots,x_i,\cdots,0)=0,$namely $f=0$. Thus $\vp$ is injective. Here $(0,\cdots,x_i,\cdots,0)$ means the $i$-th component is $x_i$ and all other components are zero.

Surjectivity: for any $(f_1,\cdots,f_m)\in V’_1\times\cdots\times V’_m$, define $f\in (V_1\times\cdots\times V_m)’$ by $f(x_1,\cdots,x_m)=\sum_{i=1}^mf_i(x_i).$Then we can easily check that $\vp f=(f_1,\cdots,f_m)$.

By the arguments above, it follows that $(V_1\times\cdots\times V_m)’$ and $V’_1\times\cdots\times V’_m$ are isomorphic.

6. Solution: (a) If $v_1,\cdots,v_m$ spans $V$, then $\Gamma(\vp)=0$ implies $\vp(v_1)=\cdots=\vp(v_m)=0.$Hence $\vp=0$ since $v_1,\cdots,v_m$ spans $V$. Specifically, for any $v\in V$, we can write $v=\sum_{i=1}^mk_iv_i,\quad k_i\in\mb F.$Thus $\vp(v)=\vp\left(\sum_{i=1}^mk_iv_i\right)=\sum_{i=1}^mk_i\vp(v_i)=0.$This implies $\vp=0$. We conclude $\Gamma$ is injective.

If $\Gamma$ is injective and $\m{span}(v_1,\cdots,v_m)\ne V$, then by Problem 4, there exists a $\vp\in V’$ such that $\vp(\m{span}(v_1,\cdots,v_m))=0$and $\vp\ne 0$. This implies $\Gamma$ is not injective. We get a contradiction. Hence $v_1,\cdots,v_m$ spans $V$.

(b) If $v_1,\cdots,v_m$ is linearly independent, then for any $(f_1,\cdots,f_m)\in\mb F^m$, there exists a $\vp\in V’$ such that $\vp(v_i)=f_i,\quad i=1,\cdots,m.$This is easy to show by extending $v_1,\cdots,v_m$ to a basis of $V$ and using 3.5. Then by definition of $\Gamma$, we have$\Gamma(\vp)=(f_1,\cdots,f_m).$This implies $\Gamma$ is surjective.

If $\Gamma$ is surjective, suppose $v_1,\cdots,v_m$ is linearly dependent. Then there exist $k_1,\cdots,k_m\in\mb F$ such that $k_1v_1+\cdots+k_mv_m=0$and some $k_i$ is nonzero. Let $k_i\ne 0$, then $v_i$ can be written as a linear combination of $v_1,\cdots,v_{i-1}$,$v_{i+1},\cdots,v_n$. Hence, $(0,\cdots,0,1,0,\cdots,0)$ is not in $\m{range}\Gamma$, where $1$ is on the $i$-th component. Otherwise, we have $\vp\in V’$ such that $\Gamma(\vp)=(0,\cdots,0,1,0,\cdots,0)$. Then $\vp(v_j)=0,\vp(v_i)=1,j=1,\cdots,i-1,i+1,\cdots,m.$This implies $\vp(v)=0$ if $v$ is a linear combination of $v_1,\cdots,v_{i-1}$,$v_{i+1},\cdots,v_n$. Thus $\vp(v_i)=0$ by our previous argument. However, we also have $\vp(v_i)=1$. Therefore this can not happen, namely $\Gamma$ is not surjective. That means that the assumption that $v_1,\cdots,v_m$ is linearly dependent can never happen. Hence $v_1,\cdots,v_m$ is linearly independent.

7. Solution: By calculating them directly, we have $\vp_j(x^i)=\delta_{i,j},$where $\delta_{i,j}=1$ if $i=j$ and $\delta_{i,j}=0$ if $i\ne j$. Note that the dual basis of one given basis is unique(if exist). Hence we have the dual basis of the basis $1,x,\cdots,x_m$ of $\ca P_m(\R)$ is $\vp_0,\vp_1,\cdots,\vp_m$.

8. Solution: (a) This is easy, see Problem 10 of Exercise 2C.

(b) The dual basis of the basis $1,x-5,\cdots,(x-5)_m$ of $\ca P_m(\R)$ is $\vp_0,\vp_1,\cdots,\vp_m$, where $\vp_j(p)=\frac{p^{(j)}(5)}{j!}$. Here $p^{(j)}$ denotes the $j^{\m{th}}$ derivative of $p$, with the understanding that the $0^{\m{th}}$ derivative of $p$ is $p$. The proof is similar to Problem 7.

9. Solution: Note $v_1,\cdots,v_n$ is a basis of $V$ and $\vp_1,\cdots,\vp_n$ is the corresponding dual basis of $V’$, we have $(\psi(v_1)\vp_1+\cdots+\psi(v_n)\vp_n)(v_1)=\psi(v_1).$Similarly, we also have$(\psi(v_1)\vp_1+\cdots+\psi(v_n)\vp_n)(v_i)=\psi(v_i).$Hence$\psi=\psi(v_1)\vp_1+\cdots+\psi(v_n)\vp_n,$as they coincide at a basis of $V$.

10. Solution: (a) $(S+T)’=S’+T’$ for all $S,T\in\ca L(V,W)$. For each $\vp\in W’$, we have \begin{align*} (S+T)'(\vp)(x)&=\vp((S+T)x)=\vp(Sx+Tx)=\vp(Sx)+\vp(Tx)\\&=S'(\vp)(x)+T'(\vp)(x)=(S’+T’)(\vp)(x) \end{align*} for all $x\in W$. The first and forth equality hold by the definition of dual map (3.99). The other ones hold by 3.6. Hence $(S+T)'(\vp)=(S’+T’)(\vp)$ for each $\vp\in W’$, namely $(S+T)’=S’+T’$.

(b) $(\lambda T)’=\lambda T’$ for all $\lambda\in\mb F$ and all $T\in\ca L(V,W)$. For each $\vp\in W’$, we have \begin{align*} (\lambda T)'(\vp)(x)&=\vp((\lambda T)x)=\vp(\lambda Tx)=\lambda\vp( Tx)\\&=\lambda T'(\vp)(x)=(\lambda T’)(\vp)(x) \end{align*}for all $x\in W$. Here we also use 3.6 and 3.99. Similarly, we conclude $(\lambda T)’=\lambda T’$.

11. Solution: Suppose that the rank of $A$ is 1. We have that all the columns are multiples of each other. Then $A$ can be written in the following form

$$A = \begin{bmatrix} c_1\\ \vdots\\ c_m \end{bmatrix} \begin{bmatrix} d_1 & \dots & d_n \end{bmatrix} = \begin{bmatrix} c_1 d_1 & \dots & c_1 d_n\\ \vdots & \ddots & \vdots\\ c_m d_1 & \dots & c_m d_n \end{bmatrix}$$

Where the first vector is a non-zero scalar multiple of a column in $A$ and the $d$’s are the corresponding scalars of each column such that $d_j$ times the first vector equals the $j$-th column of $A$.

Conversely, suppose there are $(c_1, \dots, c_m) \in F^m$ and $(d_1, \dots, d_n) \in F^n$ such that $A_{j,k} = c_j d_k$. It is easy to see that $A$ takes the same previous form and thus each column is a scalar multiple of each other which implies that the rank of A is 1.

12. Solution: Suppose that $I$ is the identity map on $V$ and $\varphi$ is any linear functional in $V’$. We have

$$I'(\varphi) = \varphi \circ I = \varphi$$

As desired.

13. Solution:(a) We have

\begin{aligned} T'(\varphi_1)(x, y, z) &= \varphi_1 \circ T(x, y, z)\\ &= \varphi_1(4x + 5y + 6z, 7x + 8y + 9z)\\ &= 4x + 5y + 6z \end{aligned}

\begin{aligned} T'(\varphi_2)(x, y, z) &= \varphi_2 \circ T(x, y, z)\\ &= \varphi_2(4x + 5y + 6z, 7x + 8y + 9z)\\ &= 7x + 8y + 9z \end{aligned}

(b) Since $\psi_1(x, y, z) = x$, $\psi_2(x, y, z) = y$ and $\psi_3(x, y, z) = z$, substituting these in the results from item (a), we get

$$T'(\varphi_1)(x, y, z) = 4 \psi_1(x, y z) + 5 \psi_2(x, y, z) + 6 \psi_3(x, y, z)$$

Thus $T'(\varphi_1) = 4 \psi_1 + 5 \psi_2 + 6 \psi_3$. Similarly, we get $T'(\varphi_2) = 7 \psi_1 + 8 \psi_2 + 9 \psi_3$.

14. Solution: (a) Let $p \in \mathcal{P}(\mathbb{R})$. Then

\begin{aligned} T'(\varphi)(p) &= \varphi \circ Tp\\ &= \varphi(x^2 p + p”)\\ &= (2xp + x^2p’ + p”) \rvert_{x=4}\\ &= 8p(4) + 16p'(4) + p”'(4) \end{aligned}

(b) We have \begin{aligned} (T'(\varphi))(x^3) &= \varphi \circ T(x^3)\\ &= \varphi(x^5 + 6x)\\ &= \int_0^1 x^5 + 6x \,dx\\ &= (\frac{x^6}{6} + 3x^2) \biggr\rvert_{0}^{1} \end{aligned}

15. Solution: If $T=0$, then for any $f\in W’$ and any $v\in V$, we have $$(T’f)v=f(Tv)=f(0)=0.$$Therefore $T’f=0$ for all $f\in W’$ and hence $T’=0$.

Conversely, suppose $T’=0$, we are going to show that $T=0$ by contradiction. We assume that $T\ne 0$, then there exists $v\in V$ such that $Tv\ne 0$. Since $W$ is finite, it follows from Problem 3 that there exists $\vp\in W’$ such that $\vp(Tv)\ne 0$. Note that $(T’\vp)v=\vp(Tv)\ne 0$, which contradicts with the assumption that $T’=0$. Hence $T=0$.

16. Solution: Let $\Gamma:\ca L(V,W)\to \ca L(W’,V’)$ defined by $\Gamma(T)=T’.$By 3.60, we have $\dim \ca L(V,W)=\dim \ca L(W’,V’)$. Hence, by 3.69, it suffices to show $\Gamma$ is injective. Suppose $\Gamma(S)=0$ for some $S\in \ca L(V,W)$, that is $S’=0$. Hence for any $\vp\in W’$ and $v\in V$, we have $S'(\vp)(v)=\vp(Sv)=0.$By Problem 3, this can only happen when $Sv=0$. Hence $Sv=0$ for all $v\in V$. Thus $S=0$. We conclude $\Gamma$ is injective.

17. Solution: Note that$\vp(u)=0\text{ for all } u\in U\iff U\subset \m{null}\vp.$

18. Solution: By Problem 17, $U^0=V’$ if and only if $U\subset \m{null}\vp$ for all $\vp\in V’$. Note that by Problem 3, $v\in\m{null}\vp$ for all $\vp\in V’$ if and only if $v=0$. This implies $U^0=V’$ if and only if $U=\{0\}$.

Other solution: by 3.106, we have $\dim \mathrm{span}(U)+\dim U^0=\dim V.$Hence$\dim U^0=\dim V’\iff \dim \mathrm{span}(U)=0$ since $\dim V’=\dim V$.

19. Solution: By 3.106, we have $\dim U+\dim U^0=\dim V.$Hence $\dim U=\dim V\iff \dim U^0=0.$That is $U=V$ if and only if $U^0=\{0\}$.

20. Solution: If $\vp\in W^0$, then $\vp(w)=0$ for all $w\in W$. As $U\subset W$, we also have $\vp(u)=0$ for all $u\in W$, hence $\vp\in U^0$. Since $\vp$ is chosen arbitrarily, we deduce that $W^0\subset U^0$.

21. Solution: Since $W^0\subset U^0$, it follows from Problem
22 that $$(U+W)^0=U^0\cap W^0= W^0.$$Note that $V$ is finite-dimensional, by 3.106 we have $$\dim (U+W)^0=\dim V-\dim(U+W),\quad \dim W^0=\dim V-\dim W.$$Therefore, we have $\dim (U+W)=\dim W$. As $W\subset U+W$ and $\dim (U+W)=\dim W$, we conclude that $U+W=W$, which implies that $U\subset W$.

22. Solution: Note that $U\subset U+W$ and $W\subset U+W$, it follows from Problem 20 that $(U+W)^0\subset U^0$ and $(U+W)^0\subset W^0$. Therefore, $(U+W)^0\subset U^0 \cap W^0$.

On the other hand, for any given $f\in U^0\cap W^0$, we have $f(u)=0$ and $f(w)=0$ for any $u\in U$ and any $w\in W$. Therefore, $$f(u+w)=f(u)+f(w)=0$$for any $u\in U$ and any $w\in W$. Note that every vector $x\in U+W$ can be written in the form of $u+w$, where $u\in U$ and $w\in W$. Therefore, we prove that $f(x)=0$ for all $x\in U+W$. This implies that $f\in (U+W)^0$, hence we have $U^0 \cap W^0\subset (U+W)^0$.

Therefore, $(U+W)^0=U^0 \cap W^0$.

23. Solution: Note that $U\cap W\subset U$ and $U\cap W\subset W$, it follows from Problem 20 that $U^0\subset (U\cap W)^0$ and $W^0\subset (U\cap W)^0$. Hence $U^0+W^0\subset (U\cap W)^0$.

On the other hand, since $V$ is finite-dimensional, it follows from 3.106 that\begin{align*}\dim(U^0+W^0)=& \dim
U^0+\dim W^0-\dim (U^0\cap W^0)\\ \text{by Problem 22 and 3.106}\quad=&\dim V-\dim U+\dim V-\dim W-\dim((U+W)^0)\\ \text{by 3.106}\quad=&\dim V-\dim U+\dim V-\dim W-\dim V+\dim(U+W)\\ =&\dim V-\dim U-\dim W+(\dim U+\dim W-\dim (U\cap W))\\ \text{by 3.106}\quad=&\dim V-\dim(U\cap W)=\dim ((U\cap W)^0).\end{align*}Since $\dim(U^0+W^0)=\dim \dim ((U\cap W)^0)$ and $U^0+W^0\subset (U\cap W)^0$, they must equal. Therefore, $U^0+W^0= (U\cap W)^0$.

24. Solution:Let $u_1, \dots, u_m$ be a basis of $U$. It can be extended to a basis $u_1, \dots, u_m, v_1, \dots, v_n$ of V. Let $\psi_1, \dots, \psi_m, \varphi_1, \dots, \varphi_n$ be the dual basis.

Suppose $\varphi \in \operatorname{span}(\varphi_1, \dots, \varphi_n)$. There are $a_1, \dots, a_n \in \mathbb{F}$ such that

$$\varphi = a_1 \varphi_1 + \dots + a_n \varphi_n$$

Let $u \in U$. We have

$$\varphi(u) = (a_1 \varphi_1 + \dots + a_n \varphi_n)(u) = 0$$

Therefore $\varphi \in U^0$. Hence $\operatorname{span}(\varphi_1, \dots, \varphi_n) \subset U^0$.

Now suppose $\varphi \in U^0$. Because $\varphi \in V’$ there are $c_1, \dots, c_m, a_1, \dots, a_n \in \mathbb{F}$ such that

$$\varphi = c_1 \psi_1 + \dots + c_m \psi_m + a_1 \varphi_1 + \dots + a_n \varphi_n$$

For every $j \in \{ 1, \dots, m \}$, we have $\psi_j(u_j) = c_j$. But $\varphi \in U^0$, that implies $c_j = 0$ and, hence, $\varphi \in \operatorname{span}(\varphi_1, \dots, \varphi_m)$. Thus $U^0 \subset \operatorname{span}(\varphi_1, \dots, \varphi_m)$.

Since $\varphi_1, \dots, \varphi_m$ is linearly independent, $\operatorname{dim}(U^0) = m$. We get

\begin{aligned} \operatorname{dim} V &= m + n\\ &= \operatorname{dim} U^0 + \operatorname{dim} U .\end{aligned}

25. Solution: Let $B = \{v \in V: \varphi(v) = 0 \text{ for every } \varphi \in U^0\}$.

Suppose that $u \in U$. By definition, $\varphi(u) = 0$ for all $\varphi \in U^0$. Thus $u \in B$ and, therefore, $U \subset B$.

For the inclusion in the other direction, we will prove the contrapositive. Suppose that $v \notin U$. Since $0 \in U$, it follows that $v \neq 0$. Let $u_1, \dots, u_n$ be a basis of $U$. We have that $v, u_1, \dots, u_n$ is a linearly indepedent list in $V$. Extend it to a basis $v, u_1, \dots, u_n, v_1, \dots, v_m$ of $V$ and let $\varphi, \psi_1, \dots, \psi_n, \varphi_1, \dots, \varphi_m$ be its dual basis. It is easy to see that $\varphi, \varphi_1, \dots, \varphi_m$ is a basis of $U^0$. But $\varphi(v) = 1$, thus $v \notin B$.

By modus tollens, $v \in B$ implies $v \in U$. Therefore $B \subset U$.

26. Solution: Let $U$ be the subspace of $V$ such that

$$U = \bigcap\limits_{\varphi \in \Gamma} \operatorname{null} \varphi$$

We have that

\begin{aligned} U &= \{v \in V: v \in U\}\\ &= \{v \in V: v \in \bigcap\limits_{\varphi \in \Gamma} \operatorname{null} \varphi\}\\ &= \{v \in V: v \in \operatorname{null} \varphi \text{ for every } \varphi \in \Gamma\}\\ &= \{v \in V: \varphi(v) = 0 \text{ for every } \varphi \in \Gamma\}\\ \end{aligned}

By Exercise 25, $\Gamma = U^0$. Therefore

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

27. Solution: We have

\begin{aligned} \operatorname{range} T &= \{p \in \mathcal{P_5}(\mathbb{R}): \psi(p) = 0 \text{ for every } \psi \in (\operatorname{range} T)^0\}\\ &= \{p \in \mathcal{P_5}(\mathbb{R}): \psi(p) = 0 \text{ for every } \psi \in \operatorname{null} T’\}\\ &= \{p \in \mathcal{P_5}(\mathbb{R}): \psi(p) = 0 \text{ for every } \psi \in \operatorname{span}(\varphi)\}\\ &= \{p \in \mathcal{P_5}(\mathbb{R}): \varphi(p) = 0\}\\ &= \{p \in \mathcal{P_5}(\mathbb{R}): p(8) = 0\}\\ \end{aligned}

Where the first equation follows from Exercise 25 and the second from 3.107.

28. Solution: Similarly to Exercise 27, we have

\begin{aligned} \operatorname{range} T &= \{v \in V: \psi(v) = 0 \text{ for every } \psi \in (\operatorname{range} T)^0\}\\ &= \{v \in V: \psi(v) = 0 \text{ for every } \psi \in \operatorname{null} T’\}\\ &= \{v \in V: \psi(v) = 0 \text{ for every } \psi \in \operatorname{span}(\varphi)\}\\ &= \{v \in V: \varphi(v) = 0\}\\ &= \{v \in V: v \in \operatorname{null} \varphi\}\\ &= \operatorname{null} \varphi. \end{aligned}

29. Solution: This is almost the same as Exercise 28. Just use 3.109 instead.

30. Solution: We have

\begin{aligned} \operatorname{dim}(\operatorname{null} \varphi_1 \cap \dots \cap \operatorname{null} \varphi_m) &= \operatorname{dim} V – \operatorname{dim}((\operatorname{null} \varphi_1 \cap \dots \cap \operatorname{null} \varphi_m)^0)\\ &= \operatorname{dim} V – \operatorname{dim}((\operatorname{null} \varphi_1)^0 + \dots + (\operatorname{null} \varphi_m)^0)\\ &= \operatorname{dim} V – \operatorname{dim}(\operatorname{span}(\varphi_1) + \dots + \operatorname{span}(\varphi_m))\\ &= \operatorname{dim} V – \operatorname{dim}(\operatorname{span}(\varphi_1, \dots, \varphi_m))\\ &= \operatorname{dim} V – m\\ \end{aligned}

Where the first equation follows from 3.106, the second from Exercise 23, the third from Theorem 1 in Chapter 3 notes, the fourth from definition of sum of subspaces and the fifth because $\varphi_1, \dots, \varphi_m$ is linearly independent.

31. Solution:

By Exercise 1, all the $\varphi$’s are surjective. Consider the following process

• Step 1.

Choose $v_1 \in V$ such that $\varphi_1(v_1) = 1$.

• Step j.

If $j = n + 1$, stop the process. By the contrapositive of the statement in Theorem 2 of Chapter 3 notes, it follows that there is a vector $v_j \in V$ such that $v_j \in \bigcap_{1 \le k \le n, k \neq j}$ and $v_j \notin \operatorname{null} \varphi_j$. Because both these subspaces are closed under scalar multiplication and because $\varphi_j$ is surjective, we can assume without loss of generality that $\varphi_j(v_j) = 1$.

After step $n$ the process stops and we will have a list $v_1, \dots, v_n$ such that $\varphi_j(v_k) = 1$ if $j = k$ and $\varphi_j(v_k) = 0$ if $j \neq k$. We but need to prove that $v_1, \dots, v_n$ is linearly independent, since it already has length $\operatorname{dim} V$.

Suppose there are $a_1, \dots, a_n \in \mathbb{F}$ such that

$$a_1 v_1 + \dots + a_n v_n = 0$$

Applying $\varphi_j$ to both sides of the equation above gives $a_j = 0$, for each $j = 1, \dots, n$. Hence $v_1, \dots, v_n$ is linearly independent and, therefore, a basis of $V$.

33. Solution:

Let $t \in \mathcal{L}(\mathbb{F}^{m,n}, \mathbb{F}^{n,m})$ denote the linear map that takes a matrix to its transpose. For this exercise, assume $1 \le k \le m$ and $1 \le j \le n$.

Suppose $A, C \in \mathbb{F}^{m, n}$. Then

\begin{aligned} (t(A + C))_{k, j} &= ((A + C)^T)_{k, j}\\ &= (A + C)_{j, k}\\ &= A_{j, k} + C_{j, k}\\ &= (A^T)_{k, j} + (C^T)_{k, j}\\ &= (t(A))_{k, j} + (t(C))_{k, j}\\ \end{aligned}

Let $\lambda \in \mathbb{F}$. We have

\begin{aligned} (t(\lambda A))_{k, j} &= ((\lambda A)^T)_{k, j}\\ &= (\lambda A)_{j, k}\\ &= \lambda (A_{j, k})\\ &= \lambda (A^T)_{k, j}\\ &= \lambda (t(A))_{k, j}\\ \end{aligned}

Therefore $t$ is indeed a linear map.

Since $\operatorname{dim}(\mathbb{F}^{m, n}) = \operatorname{dim}(\mathbb{F}^{n, m})$, to prove that $t$ is invertible we only need to show that it is injective.

Suppose $t(A) = 0$ for some $A \in F^{m,n}$ (here 0 denotes a matrix in $\mathbb{F}^{n, m})$ with 0 in all entries). We have that

$$0 = (t(A))_{k, j} = (A^T)_{k, j} = A_{j,k}$$

Because $A$ has zero in all its entries, it follows that $A = 0$ and, therefore, $\operatorname{null} t = \{ 0 \}$, which implies that $t$ is injective.

34. Solution: (a) Given $k_1,k_2\in\mb F$ and $v_1,v_2\in V$. For any $\vp\in V’$, we have\begin{align*}(\Lambda(k_1v_1+k_2v_2))(\vp)=&\, \vp(k_1v_1+k_2v_2)\\=&\, k_1\vp (v_1)+k_2\vp(v_2)\\=&\, k_1(\Lambda v_1)(\vp)+k_2(\Lambda v_2)(\vp)\\ =&\, (k_1\Lambda v_1+k_2\Lambda v_2)(\vp).\end{align*}Since this is true for any $\vp$, it follows that $$\Lambda(k_1v_1+k_2v_2)=k_1\Lambda v_1+k_2\Lambda v_2.$$Hence $\Lambda$ is a linear map from $V$ to $V^{\prime\prime}$.

(b) For any given $v\in V$, $(T^{\prime\prime}\circ \lambda) v=T^{\prime\prime}(\Lambda v)$ and $(\Lambda \circ T)v=\Lambda(Tv)$ are elements of $V^{\prime\prime}$. To show they are equal, it suffices to show that for any $f\in V’$ we have$$(T^{\prime\prime}(\Lambda v))f=(\Lambda(Tv))f.$$To see this, we have\begin{align*}&\,(T^{\prime\prime}(\Lambda v))f\\ \text{by the definition of dual map, 3.99}\quad =&\,(\Lambda v)(T’f)\\ \text{by the definition of }\Lambda \quad=&\, (T’f)v\\ \text{by the definition of dual map, 3.99}\quad =&\, f(Tv).\end{align*}On the other hand, by the definition of $\Lambda$, we also have$$(\Lambda(Tv))f=f(Tv).$$Hence we have $T^{\prime\prime}(\Lambda v)=\Lambda(Tv)$, therefore$$(T^{\prime\prime}\circ \lambda) v=(\Lambda \circ T)v.$$As the vector $v$ is chosen arbitrarily, we prove that $T^{\prime\prime}\circ \lambda=\Lambda \circ T$.

(c) Since $V$ is finite-dimensional, by 3.95, we have $\dim V=\dim V’ =\dim V^{\prime\prime}$. Hence it suffices to show that $\Lambda$ is injective. Suppose $\Lambda v=0$, then for any $f\in V’$ we have $$(\Lambda v)f=f(v)=0.$$Let $U=\{v\}$ as in Problem 18, by our assumption we have $U^0=V’$, hence it follows from Problem 18 that $U=\{0\}$. Therefore $v=0$, which implies that $\Lambda$ is injective.

35. Solution: Define $\tau \in \mathcal{L}((\mathcal{P}(\mathbb{R}))’, \mathbb{R}^\infty)$ by

$$\tau(\sigma) = (\sigma(1), \sigma(x), \sigma(x^2), \dots)$$

You can check that $\tau$ is indeed linear. To prove $\tau$ is injective, suppose there are $\sigma, \eta \in (\mathcal{P}(\mathbb{R}))’$ such that $\tau(\sigma) = \tau(\eta)$. By definition of $\tau$ we have that $\sigma(x^j) = \eta(x^j)$. Let $p \in \mathcal{P}(\mathbb{R})$. Clearly $p$ takes the form

$$p(x) = a_0 + a_1 x + \dots + a_m x^m$$

For some $a_0, \dots, a_m \in \mathbb{F}$ where $\deg p = m$. Then

\begin{aligned} \sigma(p) &= \sigma(a_0 + a_1 x + \dots + a_m x^m)\\ &= \sigma(a_0) + \sigma(a_1 x) + \dots + \sigma(a_m x^m)\\ &= a_0 \sigma(1) + a_1 \sigma(x) + \dots + a_m \sigma(x^m)\\ &= a_0 \eta(1) + a_1 \eta(x) + \dots + a_m \eta(x^m)\\ &= \eta(a_0) + \eta(a_1 x) + \dots + \eta(a_m x^m)\\ &= \eta(a_0 + a_1 x + \dots + a_m x^m)\\ &= \eta(p)\\ \end{aligned}

Because $p$ was arbitrary, it follows that $\sigma = \eta$ and thus $\tau$ is injective.

To prove surjectivity, let $a = (a_0, a_1, a_2, \dots)\in \mathbb{R}^\infty$. Define $\sigma \in (\mathcal{P}(\mathbb{R}))’$ by

$$\sigma(x^n) = a_n$$

You can check that $\sigma$ is also linear. By definition of $\tau$ we have that $\tau(\sigma) = a$ and, because $a$ was arbitrary, it follows that $\tau$ is surjective.

Hence $\tau$ is an isomorphism between $(\mathcal{P}(\mathbb{R}))’$ and $\mathbb{R}^\infty$.

36. Solution: (a) We have

\begin{aligned} \operatorname{null} i’ &= \{\varphi \in V’: i'(\varphi) = 0\}\\ &= \{\varphi \in V’: i'(\varphi)(u) = 0 \text{ for every } u \in U\}\\ &= \{\varphi \in V’: \varphi \circ i(u) = 0 \text{ for every } u \in U\}\\ &= \{\varphi \in V’: \varphi(u) = 0 \text{ for every } u \in U\}\\ &= \{\varphi \in V’: \varphi \in U^0\}\\ &= U^0\\ \end{aligned}

(b)

##### SHARE ON 