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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 $(\lambda T)’=\lambda T’$.

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

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.


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