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## Chapter 3 Exercise D

1. Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 22. It is almost the same.

2. Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 25.

3. Solution: If there exists an invertible operator $T\in\ca L(V)$ such that $Tu=Su$ for every $u\in U$, then $S$ is injective since $T$ is injective.

If $S$ is injective. Assume $u_1$, $\cdots$, $u_m$ is a basis of $U$, we can extend it to a basis of $V$ as $u_1$, $\cdots$, $u_m$, $v_{m+1}$, $\cdots$, $v_{n}$. Since $S$ is injective, by Problem 9 of Exercises 3B, we have $Su_1$, $\cdots$, $Su_m$ is linearly independent in $V$. Hence we can extend it to a basis of $V$ as $Su_1$, $\cdots$, $Su_m$, $w_{m+1}$, $\cdots$, $w_{n}$. Define $T\in \ca L(V)$ as below $Tu_i=Su_i,\quad Tv_{j}=w_j, \quad 1\le i\le m, m+1\le j \le n.$The existence of $T$ is guaranteed by 3.5(unique). Then for any $u=a_1u_1+\cdots+a_mu_m$, $a_i\in\mb F$, we have \begin{align*} Tu=&T(a_1u_1+\cdots+a_mu_m)\\ =&a_1Tu_1+\cdots+a_mTu_m\\ =&a_1Su_1+\cdots+a_mSu_m\\ =&S(a_1u_1+\cdots+a_mu_m)=Su. \end{align*}Moreover, $T$ is surjective by Problem 10 of Exercises 3B hence invertible by 3.69.

Compare this problem with Problem 11 of Exercises 3A.

4. Solution: If we assume $\m{null} T_1 =\m{null} T_2$. Since $W$ is finite-dimensional, so is range $T_2$. Let $w_1$, $\cdots$, $w_n$ be a basis of $\m{range} T_2$, then there exist $v_1$, $\cdots$, $v_n\in V$ such that $T_2v_i=w_i,\quad i=1,\cdots,n.$Now we will show that $V=\m{null}T_2\oplus \m{span}(v_1,\cdots,v_n)$. For any $v\in V$, we $T_2v=a_1w_1+\cdots+a_nw_n$for some $a_1$, $\cdots$, $a_n\in\mb F$. Hence $T_2(v-a_1v_1-\cdots-a_nv_n)=0,$namely $v=(v-a_1v_1-\cdots-a_nv_n)+(a_1v_1+\cdots+a_nv_n),$this implies $V=\m{null}T_2 +\m{span}(v_1,\cdots,v_n)$. Moreover, if $a_1v_1+\cdots+a_nv_n\in \m{null}T_2$, then we have $T_2(a_1v_1+\cdots+a_nv_n)=a_1w_1+\cdots+a_nw_n=0.$Note that $w_1$, $\cdots$, $w_n$ is linearly independent, it follows $a_1=\cdots=a_n=0$. Thus we have$V=\m{null}T_2 \oplus\m{span}(v_1,\cdots,v_n).$ Similarly, $T_1v_1$, $\cdots$, $T_1v_n$ is linearly independent. For if $a_1T_1v_1+\cdots+a_nT_1v_n=0$, we have $T_1(a_1v_1+\cdots+a_nv_n)=0.$Note that $\m{null} T_1 =\m{null} T_2$, it follows that $0=T_2(a_1v_1+\cdots+a_nv_n)=a_1w_1+\cdots+a_nw_n.$Thus $a_1=\cdots=a_n=0$. Now extend $w_1$, $\cdots$, $w_n$ to a basis of $W$ as $w_1$, $\cdots$, $w_n$, $e_1$, $\cdots$, $e_m$ and $T_1v_1$, $\cdots$, $T_1v_n$ to a basis of $W$ as $T_1v_1$, $\cdots$, $T_1v_n$, $f_1$, $\cdots$, $f_m$. Define $S\in\ca L(W)$ by $Sw_i=T_1v_i,Se_j=f_j,i=1,\cdots,n;j=1,\cdots,m.$Note that $V=\m{null}T_2\oplus \m{span}(v_1,\cdots,v_n),$any $v\in V$ can be expressed as $v=v_{\m{null}}+a_1v_1+\cdots+a_nv_n,$where $v_{\m{null}}\in\m{null}T_2=\m{null}T_1$ and $a_1$, $\cdots$, $a_n\in\mb F$. Hence we have \begin{align*} ST_2(v)=&ST_2(v_{\m{null}}+a_1v_1+\cdots+a_nv_n)\\ =&ST_2(a_1v_1+\cdots+a_nv_n)\\ =&S(a_1w_1+\cdots+a_nw_n)\\ =&a_1T_1v_1+\cdots+a_nT_1v_n\\ =&T_1(a_1v_1+\cdots+a_nv_n)\\ =&T_1(v_{\m{null}}+a_1v_1+\cdots+a_nv_n)=T_1(v) \end{align*}namely $ST_2=T_1$. Moreover, $S$ is surjective by Problem 10 of Exercises 3B hence invertible by 3.69.

If there exists an invertible operator $S\in\ca L(W)$ such that $ST_2=T_1$, then for any $\mu\in\m{null} T_1$, we have $ST_2\mu=T_1\mu=0.$As $S$ is invertible, we have $T_2\mu=0$. Hence $\mu\in \m{null} T_2$, it follows that $\m{null} T_1 \subset \m{null} T_2$. Similarly, consider $T_2=S^{-1}T_1$, $\m{null} T_2 \subset \m{null} T_1$. Thus we conclude null $T_1$= null $T_2$.

Compare this problem with Problem 24 of Exercises 3B.

5. Solution: If we assume $\m{range} T_1 = \m{range} T_2$. Let $u_1$, $\cdots$, $u_m$ be a basis of null $T_1$, then we can extend it to a basis of $V$ as $u_1$, $\cdots$, $u_m$, $w_1$, $\cdots$, $w_n$. Then range $T_1$ is $\m{span}(T_1w_1,\cdots,T_1w_n)$ and $T_1w_1,\cdots,T_1w_n$ is linearly independent. There exist $v_1$, $\cdots$, $v_n\in V$ such that $T_1w_i=T_2v_i$ for $i=1,\cdots,n$ since $\m{range} T_1=\m{range} T_2$. As $T_1w_1,\cdots,T_1w_n$ is linearly independent, it follows that $v_1$, $\cdots$, $v_n$ is linearly independent by Problem 4 of Exercises 3A. Note that range $T_1=$ range $T_2$ implies null $T_1$ and null $T_2$ have the same dimension. Let $\zeta_1$, $\cdots$, $\zeta_m$ be a basis of null $T_2$, then $\zeta_1$, $\cdots$, $\zeta_m$, $v_1$, $\cdots$, $v_n$ is a basis of $V$ by the proof of 3.22. Define $S\in\ca L(V)$ by $Su_i=\zeta_i$ and $Sw_j=v_j$, then we have $T_1w_j=T_2v_j=T_2Sw_j,\quad j=1,\cdots,n,$and $T_1u_i=0=T_2\zeta_i=T_2Su_i,\quad i =1,\cdots, m,$hence $T_1=T_2S$ by uniqueness in 3.5. Moreover, $S$ is surjective by Problem 10 of Exercises 3B hence invertible by 3.69.

If there exists an invertible operator $S\in\ca L(V)$ such that $T_1=T_2S$, then for any $\mu\in V$, we have $T_1\mu=T_2S\mu\in \m{range} T_2.$Hence $\m{range} T_1 \subset \m{range} T_2$. As $S$ is invertible, we have $T_2=T_1S^{-1}$. Similarly, we conclude $\m{range} T_1 \subset \m{range} T_2$. Thus range $T_1=$ range $T_2$.

Compare this problem with Problem 25 of Exercises 3B.

6. Solution: If there exist invertible operators $R\in\ca L(V)$ and $S\in\ca L(W)$ such that $T_1=ST_2R$, then $S^{-1}T_1=T_2R$. Hence null $T_1$= null $T_2R$ by Problem 4. Note that we have range $T_2R$= range $T_2$ by Problem 5, it follows that $\dim\m{null} T_2R=\dim V-\dim\m{range} T_2R=\dim V-\dim\m{range} T_2=\dim\m{null} T_2.$Hence $\dim\m{null} T_1=\dim\m{null} T_2$.

Conversely, if $\dim$ null $T_1= \dim$ null $T_2$. Let $u_1$, $\cdots$, $u_m$ be a basis of null $T_1$, then we can extend it to a basis of $V$ as $u_1$, $\cdots$, $u_m$, $w_1$, $\cdots$, $w_n$. Let $v_1$, $\cdots$, $v_m$ be a basis of null $T_2$, then we can extend it to a basis of $V$ as $v_1$, $\cdots$, $v_m$, $\zeta_1$, $\cdots$, $\zeta_n$. Then $T_1w_1$, $\cdots$, $T_1w_n$ is linearly independent in $W$, hence we can extend it to a basis of $V$ as $T_1w_1$, $\cdots$, $T_1w_n$, $\alpha_1$, $\cdots$, $\alpha_l$. Similarly, $T_2\zeta_1$, $\cdots$, $T_2\zeta_n$ is linearly independent in $W$, hence we can extend it to a basis of $V$ as $T_2\zeta_1$, $\cdots$, $T_2\zeta_n$, $\beta_1$, $\cdots$, $\beta_l$. Define $R\in\ca L(V)$ by $Ru_i=v_i,Rw_j=\zeta_j,i=1,\cdots,m;j=1,\cdots,n.$Define $S\in\ca L(W)$ by $ST_2\zeta_j=T_1w_j,S\beta_k=\alpha_k,j=1,\cdots,n;k=1,\cdots,l.$Since $S$ and $T$ map basis to basis, hence $S$ and $T$ are invertible(surjective). Moreover, it is easy to check $T_1u_i=0=ST_2Ru_i$ and $T_1w_j=ST_2\zeta_j=ST_2Rw_j,$hence $T_1=ST_2R$.

Compare this problem with Problem 4 and Problem 5.

7. Solution: a) Let $T,S\in E$, then $(T+S)v=Tv+Sv=0+0=0,$hence $T+S\in E$, namely $E$ is closed under addition. For every $\lambda\in \mb F$, $(\lambda T)v=\lambda(Tv)=\lambda 0=0,$hence $\lambda T\in E$, namely $E$ is closed under scalar multiplication. Therefore $E$ is a subspace of $\ca L(V,W)$.

(b) Since $v\ne 0$, we can extend it to a basis of $V$, namely $v$, $v_2$, $\cdots$, $v_n$. Let $w_1$, $\cdots$, $w_m$ be a basis of $W$. Under these bases, we have a isometric between $\ca L(V,W)$ and $\mb F^{m,n}$ by 3.60. Moreover, $Tv=0$ if and only if the first column vector of $\ca M(T)$ is zero. Hence $\dim E$ is exactly all matrices in $\mb F^{m,n}$ such the first column vector is zero. Now , it is easily seen that$\dim E=m(n-1)=\dim W(\dim V-1).$Or it is equivalent to the dimension all linear map from $\m{span}(v_2,\cdots,v_n)$ to $W$. Then we can use 3.61.

8. Solution: Let $w_1$, $\cdots$, $w_n$ be a basis of $W$. Since $T$ is surjective, there exist $v_1$, $\cdots$, $v_n$ such that $Tv_i=w_i$. Moreover, by Problem 4 of Exercises 3A, it follows that $v_1$, $\cdots$, $v_n$ is linearly independent. Consider $U=\m{span}(v_1,\cdots,v_n)$, then $T|_U$ maps a basis of $U$ to a basis of $W$. Hence $T|_U$ is an isomorphism of $U$ onto $W$.

9. Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 22.

10. Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 23.

11. Solution: By Problem 9, we have $TU$ is invertible. Again by Problem 9, we have $T$ is invertible. Multiply both side of $STU=I$ by $S^{-1}$ on left, we get $TU=S^{-1}$. Multiply both side of $TU=S^{-1}$ by $S$ on right, we have $TUS=I$. Multiply both side of $TUS=I$ by $T^{-1}$ on left, we get $US=T^{-1}$.

12. Solution: Consider $V=\C^\infty$. Define $T$, $S$, $U$ by $T(z_1,z_2,z_3,\cdots)=(0,z_1,z_2,z_3,\cdots)$$S(z_1,z_2,z_3,\cdots)=(z_2,z_3,z_4,\cdots)$and $U=I$. Then $STU=I$. However $T$ is not surjective.

13. Solution: Since $V$ is finite-dimensional, the surjectivity of $RST$ implies $RST$ is invertible. Hence $S$ is invertible by Problem 11.

14. Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 20.

15. Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 21.

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

Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 24.

17. Solution: Let $e_1$, $\cdots$, $e_n$ be a basis of $V$. Suppose $E\ne \{0\}$, then there is nonzero $T\in E$. This implies there is some $s\in\{1,\cdots,n\}$ such that $Te_s\ne 0$. Let $Te_s=a_1e_1+\cdots+a_ne_n,\quad a_j\in\mb F,j=1,\cdots,n.$Since $Te_s\ne 0$, there is some $t\in\{1,\cdots,n\}$ such that $a_t\ne 0$. Define $E_{ij}\in \ca L(V)$ by $E_{ij}e_k=\delta_{ik}e_j$, where $\delta_{ik}=0$ if $i\ne k$ and $\delta_{ii}=1$ for every $i$.

By definition, for every $i\in\{1,\cdots,n\}$, we have $E_{ti}TE_{is}e_j=a_t\delta_{ij}e_i,\quad j=1,\cdots,n.$By assumption, $E_{ti}TE_{is}\in E$. Note that $E$ is a subspace of $\ca L(V)$. It follows that $(E_{t1}TE_{1s}+\cdots+E_{tn}TE_{ns})e_j=a_te_j,\quad j=1,\cdots,n,$and $(E_{t1}TE_{1s}+\cdots+E_{tn}TE_{ns})\in E$. This implies $a_jI\in E$, hence $I\in E$. Therefore for any $S\in \ca L(V)$, $S=SI\in E$, namely $E=\ca L(V)$.

18. Solution: For given $v\in V$, define $\vp_v:\mb F\to V$ by $\vp_v(\lambda)=\lambda v$. Then $\vp_v\in \ca L(\mb F, V)$ (check it). Hence we can define $\vp: V\to \ca L(\mb F, V)$ by $v\mapsto \vp_v$. It suffices to show $\vp$ is an isomorphic from $V$ to $\ca L(\mb F, V)$. First, we should check $\vp$ is linear. For every $v_1,v_2\in V$ and $\lambda,a\in\mb F$, we have \begin{align*} \vp_{v_1+\lambda v_2}(a)=&a(v_1+\lambda v_2)=av_1+\lambda av_2\\=&\vp_{v_1}(a)+(\lambda\vp_{v_2})(a)=(\vp_{v_1}+\lambda\vp_{v_2})(a). \end{align*}Hence $\vp$ is linear. Then $\vp_v\equiv 0$ imples $av=0$ for every $a\in\mb F$, thus $v=0$. We conclude injectivity. For any $f\in \ca L(\mb F, V)$, if $f(1)=v$ then $f(\lambda)=\lambda v=\vp_v(\lambda)$. Hence every $f\in \ca L(\mb F, V)$ can be express as $\vp_{f(1)}$, namely $\vp$ is surjective.

19. Solution:

(a) Since $\deg Tp\le \deg p$ for every nonzero polynomial $p\in\ca P(\R)$, we have $T|_{\ca P_n(\R)}:\ca P_n(\R)\to \ca P_n(\R)$ for any $n\in \mb N^+$. Note that $T$ is injective, hence $T|_{\ca P_n(\R)}:\ca P_n(\R)\to \ca P_n(\R)$ is injective. Note that $\ca P_n(\R)$ is finite-dimensional, we conclude $T|_{\ca P_n(\R)}:\ca P_n(\R)\to \ca P_n(\R)$ is surjective by 3.56. Hence $T$ is surjective as any polynomial must be contained in some $\ca P_n(\R)$.

(b) We argue it by induction on $\deg p$. It is true for $\deg =0$. Suppose it is true for $n$. As $T|_{\ca P_n(\R)}:\ca P_n(\R)\to \ca P_n(\R)$ is surjective, every polynomial $p$ with degree $\le n$ can be attained by $Tq$ for some $q\in \ca P_n(\R)$. Moreover, $T|_{\ca P_{n+1}(\R)}:\ca P_{n+1}(\R)\to \ca P_{n+1}(\R)$ is surjective, every polynomial $p$ with degree $\le n+1$ can be attained by $Tq$ for some $q\in \ca P_{n+1}(\R)$. If there exist $p\in \ca P_{n+1}(\R)$ with degree $n+1$ such that $\deg Tp<n+1$. Then there exist some $q\in \ca P_n(\R)$ such that $Tq=Tp$ since $T|_{\ca P_n(\R)}:\ca P_n(\R)\to \ca P_n(\R)$ is surjective. Hence $Tp=Tq$. Note that $T$ is injective and $p\ne q$ (different degrees), we get a contradiction. Hence $\deg Tp= \deg p$ for every nonzero $p\in\ca P_{n+1}(\R)$ with degree $n+1$. The proof is complete.

20. Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 26.