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.

About Linearity

This website is supposed to help you study Linear Algebras. Please only read these solutions after thinking about the problems carefully. Do not just copy these solutions.
This entry was posted in Chapter 3 and tagged .