1. Solution: Assume $V$ is 5-dimensional vector space with a basis $e_1$, $\cdots$, $e_5$. Define $T\in\ca L(V,V)$ by \[Te_1=e_1,Te_2=e_2,Te_3=Te_4=Te_5=0.\]Then $\mathrm{null} T=\mathrm{span}(e_3,e_4,e_5)$, hence $\dim \mathrm{null} T=3$. Similarly, $\mathrm{range} T=\mathrm{span}(e_1,e_2)$, hence $\dim \mathrm{range} T=2$.
2. Solution: Since $\m{range} S \subset \m{null} T$, it follows that $TSv=0$ for any $v\in V$. Hence for any $u\in V$, \[(ST)^2u=S[(TS)Tu]=S0=0,\]i.e. $(ST)^2=0$.
3. Solution:
(a) Note that $\m{range}T=\m{span}(v_1,\cdots,v_m)$, hence if $v_1$, $\cdots$, $v_m$ spans $V$, it follows that $T$ is surjective.
(b) Note that $z_1v_1+\cdots+z_mv_m=0$ if and only if $(z_1,\cdots,z_m)=(0,\cdots,0)$ since $v_1$, $\cdots$, $v_m$ is linearly independent. Hence $\m{null}T=\{(0,\cdots,0)\}$, then $T$ is injective.
4. Solution: Let $e_1$, $\cdots$, $e_5$ be a basis of $\R^5$ and $f_1$, $f_2$, $f_3$, $f_4$ be a basis of $\R^4 $. Define $S_1$ and $S_2$ by \[S_1e_i=0,\quad S_1e_4=f_1,\quad S_1e_5=f_2,\quad \text{for~}i=1,2,3;\] \[S_2e_i=0,\quad S_2e_3=f_3\quad S_2e_5=f_4,\quad \text{for~}i=1,2,4.\]Then it is obvious that $S_1,S_2\in \{T\in\ca L(\R^5,\R^4):\dim \m{null} T >2\}$. However, \[(S_1+S_2)e_1=0,\quad (S_1+S_2)e_2=0\]and \[(S_1+S_2)e_3=f_3,\quad (S_1+S_2)e_4=f_1,\quad (S_1+S_2)e_5=f_2+f_4.\]Then you can check that $\dim \m{null} (S_1+S_2)=2$. Hence $\{T\in\ca L(\R^5,\R^4):\dim \m{null} T >2\}$ is not closed under addition, this implies it is not a subspace of $L(\R^5,\R^4)$.
5. Solution: Let $e_1$, $e_2$, $e_3$, $e_4$ be a basis of $\R^4$. Define $T\in\ca L(\R^4,\R^4)$ by \[Te_1=e_3,Te_2=e_4,Te_3=Te_4=0.\]Then $\mathrm{null} T=\mathrm{span}(e_3,e_4)$, and $\mathrm{range} T=\mathrm{span}(e_3,e_4)$. Hence $\m{range} T = \m{null} T$.
6. Solution: By 3.22, we know that \[ \dim \m{range} T +\dim \m{null} T=\dim(\R^5)=5. \]If $\m{range} T = \m{null} T$, we will get that $\m{range} T = \m{null} T=2.5$. This is impossible since dimension is an integer.
7. Solution: Let $v_1$, $\cdots$, $v_n$ be a basis of $V$ and $w_1$, $\cdots$, $w_m$ be a basis of $W$, then we have $2\le n\le m$. Define $T_1, T_2\in\ca L(V,W)$ by \[T_1v_1=0,T_1v_i=w_i,i=2,\cdots,n\]and \[ T_2v_1=w_1,T_2v_2=0,T_2v_i=w_i,i=3,\cdots,n. \]Then $T_1,T_2\in \{T\in\ca L(V,W): T\text {~is not injective}\}$. However, we have \[ (T_1+T_2)v_1=w_1,(T_1+T_2)v_2=w_2,(T_1+T_2)v_i=2w_i,i=3,\cdots,n. \]Then by Problem 3 (b), it follows that $T_1+T_2$ is injective. Hence $\{T\in\ca L(V,W): T\text {~is not injective}\}$ is not closed under addition, which implies it is not a subspace of $\ca L(V,W)$.
8. Solution: Let $v_1$, $\cdots$, $v_n$ be a basis of $V$ and $w_1$, $\cdots$, $w_m$ be a basis of $W$, then we have $n\ge m\ge 2$. Define $T_1, T_2\in\ca L(V,W)$ by \[T_1v_1=0,T_1v_i=w_i,T_1v_j=0,i=2,\cdots,m;j=m+1,\cdots,n\]and \[ T_2v_1=w_1,T_2v_2=0,T_2v_i=w_i,T_2v_j=0,i=3,\cdots,m;j=m+1,\cdots,n. \]Then $T_1,T_2\in \{T\in\ca L(V,W): T\text {~is not surjective}\}$. However, we have \[ (T_1+T_2)v_1=w_1,(T_1+T_2)v_2=w_2,(T_1+T_2)v_i=2w_i,i=3,\cdots,m; \]Then by Problem 3 (a), it follows that $T_1+T_2$ is surjective. Hence $\{T\in\ca L(V,W): T\text {~is not surjective}\}$ is not closed under addition, which implies it is not a subspace of $\ca L(V,W)$.
9. Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 5.
10. Solution: Note that $v_1$, $\cdots$, $v_n$ spans $V$, any $v\in V$ can be written as a linear combination of $v_1$, $\cdots$, $v_n$. That is there are $a_1$, $\cdots$, $a_n\in\mb F$ such that \[v=a_1v_1+\cdots+a_nv_n.\]Since $T\in\ca L(V,W)$, it follows that \[Tv=a_1Tv_1+\cdots+a_nT_nv_n.\]Hence $\m{range}T\subset\m{span}(Tv_1,\cdots,Tv_n)$. On the other hand $Tv_1$, $\cdots$, $Tv_n$ are contained in $\m{range}T$. By the definition of span, we conclude that $Tv_1$, $\cdots$, $Tv_n$ spans $\m{range}T$.
See a similar problem in Linear Algebra Done Right Solution Manual Chapter 3 Problem 7.
11. Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 6.
12. Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 8.
13. Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 9.
14. Solution: By 3.22, we have \[\dim\m{null} T+\dim\m{range}T=\dim(\R^8)=8.\]Note that $\m{null} T = U$ and $\dim U=3$, it follows that \[\dim\m{range}T=8-\dim\m{null} T=8-3=5=\dim(\R^5).\]Therefore $T$ is surjective by Problem 1 of Exercises 2.C.
15. Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 10.
16. Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 11.
17. Solution: By 3.22, it follows that for any injective $T\in\ca L(V,W)$, we have \[\dim V=\dim \m{null}T+\dim\m{range}T=\dim\m{range}T\le\dim W.\]Hence there exists an injective linear map from $V$ to $W$, then $\dim V \le \dim W$.
If $n=\dim V \le \dim W=m$, then let $v_1$, $\cdots$, $v_n$ and $w_1$, $\cdots$, $w_m$ be the bases of $V$ and $W$, respectively. Define $T\in\ca L(V,W)$ such that \[ Tv_i=w_i,\quad i =1,\cdots,n. \]Here we use $n\le m$. Similar to Problem 3(b), we can show that $T$ is injective.
18. Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 12. It is similar to Problem 17.
19. Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 13.
20. Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 14.
21. Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 15.
22. Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 16.
23. Solution: It is obvious that $\m{range}ST\subset\m{range}S$, hence $\dim \m{range}ST \le \dim \m{range}S$. Let $u_1,\cdots,u_m$ be a basis of $\m{range}T$, then by Problem 10, we have \[ \m{range}ST=\m{span}(Su_1,\cdots,Su_m). \]Hence $\dim \m{range}ST \le m=\dim \m{range}T$. Thus \[\dim \m{range}ST \le \min\{\dim \m{range}S ,\dim \m{range}T\}.\]
24. Solution: Suppose $\m{null} T_1 \subset \m{null} T_2$, we show that there exists $S \in\ca L(W,W)$ such that $T_2=ST_1$.
Since $W$ is finite-dimensional, $\mathrm{Range}\, T_1$ is finite-dimensional. Let $T_1v_1,\dots,T_1v_n$ be a basis of $\mathrm{Range}\, T_1$, then $v_1,\dots,v_n$ is linearly independent. Let $K=\mathrm{span}(v_1,\dots,v_n)$, then $V\cong K\oplus \mathrm{Null}\, T_1$. Hence for any $v\in V$, we have $$v=\sum_{i=1}^na_i v_i+u,$$ where $a_i\in\mathbb F$ and $u\in\mathrm{Null}\, T_1$. Let $S\in\ca L(W,W)$ such that $S(T_1v_i)=T_2v_i$. Such an $S$ exists since $T_1v_1,\dots,T_1v_n$ is linearly independent.
Since $\m{null} T_1 \subset \m{null} T_2$, we have \begin{align*}T_2v&=\sum_{i=1}^na_i T_2v_i+T_2(u)\\&=\sum_{i=1}^na_i T_2v_i.\end{align*}Hence $\mathrm{Range}\, T_2$ is spanned by $T_2v_1,\dots,T_2v_n$.
On the other hand, we also have\begin{align*}S(T_1v)&=\sum_{i=1}^nS(a_i T_1v_i)+S(T_1u)\\&=\sum_{i=1}^na_i T_2v_i.\end{align*}
Hence $T_2v=ST_1v$ for all $v\in V$, i.e. $T_2=ST_1$.
Conversely, if $T_2=ST_1$, it is clear that $\m{null} T_1 \subset \m{null} T_2$.
25. Solution: If we assume $\m{range} T_1 \subset \m{range} T_2$. Let $u_1$, $\cdots$, $u_m$ be a basis of $V$, then we can find $v_1$, $\cdots$, $v_m\in V$ such that $T_1u_i=T_2v_i$ for $i=1,\cdots,m$ since $\m{range} T_1 \subset \m{range} T_2$. Define $S\in\ca L(V,V) $ by $Su_i=v_i$, then we have \[T_1u_i=T_2v_i=T_2Su_i,\quad i=1,\cdots,m,\]hence $T_1=T_2S$ by uniqueness in 3.5.
If there exists $S \in\ca L(V,V)$ 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 $\m{range} T_1 \subset \m{range} T_2$.
26. Solution: Consider Problem 10 of Exercise 2C. Note that we have \[\deg Dx^n=n-1,\]it follows that for any $j\in\mb N$, there is a polynomial $Dx^{j+1}$ with degree $j$. Note that $\m{range}D$ is a subspace, hence \[\m{span}(Dx,Dx^2,\cdots)\subset \m{range}D.\]Moreover, by Problem 10 of Exercise 2C, we have \[ \m{span}(Dx,Dx^2,Dx^3,\cdots)=\m{span}(1,x,x^2,\cdots). \]Hence $\m{span}(1,x,\cdots)\subset \m{range}D$, namely $\m{range}D=\ca P(\R)$.
27. Solution: Denote the differentiation map by $D$, then $5D^2+3D\in\ca L(\ca P(\R),\ca P(\R))$ is of the type in Problem 26. Hence $5D^2+3D$ is surjective, thus there exists $q\in \ca P(\R)$ such that $(5D^2+3D)q=p$. Note that $(5D^2+3D)q=5q”+3q’$, the proof completes.
28. Solution: For any $v\in V$, there is a uniquely linear combination \[Tv=a_1w_1+\cdots+a_mw_m,\]where $a_1,\cdots,a_m\in\mb F$ are uniquely determined by $v$, as $w_1$, $\cdots$, $w_m$ is a basis of range $T$. Denote $a_i$ by $\vp_i(v)$. Now we check that $\vp_i\in \ca L(V,\mb F)$, $i=1,\cdots,m$. For $u,v\in V$, we have \[ Tu=\vp_1(u)w_1+\cdots+\vp_m(u)w_m \] and \[ Tv=\vp_1(v)w_1+\cdots+\vp_m(v)w_m. \]Similarly, we also have \[ T(u+v)=\vp_1(u+v)w_1+\cdots+\vp_m(u+v)w_m. \]Since $T\in\ca L(V,W)$, it follows that \begin{align*} &\vp_1(u+v)w_1+\cdots+\vp_m(u+v)w_m\\=&T(u+v)=Tu+Tv\\=&\vp_1(u)w_1+\cdots+\vp_m(u)w_m+\vp_1(v)w_1+\cdots+\vp_m(v)w_m\\ =&(\vp_1(u)+\vp_1(v))w_1+\cdots+(\vp_m(u)+\vp_m(v))w_m. \end{align*} As $w_1$, $\cdots$, $w_m$ is a basis of range $T$, it follows that \[ \vp_i(u+v)=\vp_i(u)+\vp_i(v),i=1,\cdots,m. \]Similarly, by consider $T(\lambda u)=\lambda Tu$ for $\lambda\in \mb F$, it follows that \[\vp_i(\lambda u)=\lambda\vp_i(u),i=1,\cdots,m.\]Hence $\vp_i\in\ca L(V,\mb F)$.
29. Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 4.
30. Solution: If $\m{null}\vp_1=\m{null}\vp_2=V$, then $\vp_1=\vp_2=0$.
If $\m{null}\vp_1=\m{null}\vp_2\ne V$, then there is a $u\in V$ is not in null $\vp_2$, namely $\vp_2(u)\ne0$. By problem 29, we have\[V=\m{null}\vp_1\oplus\{au:a\in\mb F\}.\]Hence for any $v\in V$, $v$ can be written as $w+a_vu$ where $w\in \m{null}\vp_1$. Then $\vp_1(w)=\vp_2(w)=0$, we have \begin{align*} \frac{\vp_1(u)}{\vp_2(u)}\vp_2(v)=&\frac{\vp_1(u)}{\vp_2(u)}\vp_2(w+a_vu)=\frac{\vp_1(u)}{\vp_2(u)}a_v\vp_2(u)\\=&a_v\vp_1(u)=\vp_1(w+a_vu)\\=&\vp_1(v), \end{align*}i.e. $\vp_1=\dfrac{\vp_1(u)}{\vp_2(u)}\vp_2$. Let $c=\dfrac{\vp_1(u)}{\vp_2(u)}$, we complete the proof.
31. Solution: Let $e_1$, $e_2$, $e_3$, $e_4$, $e_5$ and $f_1$, $f_2$ be a basis of $\R^5$ and $\R^2$, respectively. Define $T_1\in\ca L(\R^5,\R^2)$ by \[T_1e_1=f_1,T_1e_2=f_2,T_1e_3=T_1e_4=T_1e_5=0.\]Define $T_2\in\ca L(\R^5,\R^2)$ by \[T_2e_1=f_1,T_2e_2=2f_2,T_2e_3=T_2e_4=T_2e_5=0.\]Then \[\m{null}T_1=\m{null}T_2=\m{span}(e_3,e_4,e_5)\]and $T_1$ is not a scalar multiple of $T_2$. Otherwise, if $T_1=cT_2$, then \[f_1=T_1e_1=cT_2e_1=cf_1.\]We get $c=1$. However, from \[f_2=T_1e_2=cT_2e_2=2cf_2,\]we get $c=1/2$.
