# Chapter 3 Exercise B

1. Give an example of a linear map $T$ such that $\dim \mathrm{null} T=3$ and $\dim \mathrm{range} T = 2$.

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. Suppose $V$ is a vector space and $S, T \in\ca L(V, V)$ are such that $\m{range} S \subset \m{null} T.$ Prove that $(ST)^2=0$.

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. Suppose $v_1$, $\cdots$, $v_m$ is a list of vectors in $V$. Define $T\in\ca L(\mb F^m, V )$ by $T(z_1,\cdots,z_m)=z_1v_1+\cdots+z_mv_m.$(a) What property of $T$ corresponds to $v_1$, $\cdots$, $v_m$ spanning $V$ ?(b) What property of $T$ corresponds to $v_1$, $\cdots$, $v_m$ being linearly independent?

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. Show that $\{T\in\ca L(\R^5,\R^4):\dim \m{null} T >2\}$ is not a subspace of $L(\R^5,\R^4)$.

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. Give an example of a linear map $T:\R^4\to\R^4$ such that $\m{range} T = \m{null} T.$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. Prove that there does not exist a linear map $T:\R^5\to\R^5$ such that $\m{range} T = \m{null} T.$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. Suppose $V$ and $W$ are finite-dimensional with $2\le \dim V \le \dim W$. Show that $\{T\in\ca L(V,W): T\text {~is not injective}\}$ is not a subspace of $\ca L(V,W)$.

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. Suppose $V$ and $W$ are finite-dimensional with $\dim V \ge \dim W\ge 2$. Show that $\{T\in\ca L(V,W): T\text {~is not surjective}\}$ is not a subspace of $\ca L(V,W)$.

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. Suppose $T\in\ca L(V,W)$ is injective and $v_1$, $\cdots$, $v_n$ is linearly independent in $V$. Prove that $Tv_1$, $\cdots$, $Tv_n$ is linearly independent in $W$.

Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 5.

10. Suppose $v_1$, $\cdots$, $v_n$ spans $V$ and $T\in\ca L(V,W)$. Prove that the list $Tv_1$, $\cdots$, $Tv_n$ spans $\m{range}T$.

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 $Tv=a_1Tv_1+\cdots+a_nT_n.$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. Suppose $S_1$, $\cdots$, $S_n$ are injective linear maps such that $S_1S_2\cdots S_n$ makes sense. Prove that $S_1S_2\cdots S_n$ is injective.

Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 6.

12. Suppose that $V$ is finite-dimensional and that $T\in\ca L(V,W)$. Prove that there exists a subspace $U$ of $V$ such that $U\cap \m{null}=\{0\}$ and $\m{range} T=\{Tu:u\in U\}$.

Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 8.

13. Suppose $T$ is a linear map from $\mb F^4$ to $\mb F^2$ such that $\m{null} T=\{(x_1,x_2, x_3, x_4)\in \mb F^4: x_1=5x_2~ \text{and} x_3=7x_4\}.$ Prove that $T$ is surjective.

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

14. Suppose $U$ is a 3-dimensional subspace of $\R^8$ and that $T$ is a linear map from $\R^8$ to $\R^5$ such that $\m{null} T = U$. Prove that $T$ is surjective.

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. Prove that there does not exist a linear map from $\mb F^5$ to $\mb F^2$ whose null space equals $\{(x_1, x_2, x_3, x_4, x_5)\in\mb F^5:x_1=3x_2 \text{and} x_3=x_4=x_5\}.$Solution: Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 10.

16. Suppose there exists a linear map on $V$ whose null space and range are both finite-dimensional. Prove that $V$ is finite-dimensional.

Solution: Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 11.

17. Suppose $V$ and $W$ are both finite-dimensional. Prove that there exists an injective linear map from $V$ to $W$ if and only if $\dim V \le \dim W$.

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. Suppose $V$ and $W$ are both finite-dimensional. Prove that there exists an surjective linear map from $V$ to $W$ if and only if $\dim V \ge \dim W$.

Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 12. It is similar to Problem 17.

19. Suppose $V$ and $W$ are finite-dimensional and that $U$ is a subspace of $V$. Prove that there exists $T\in\ca L(V,W)$ such that $\m{null} T = U$ if and only if $\dim U \ge \dim V – \dim W$.

Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 13.

20. Suppose $W$ is finite-dimensional and $T\in\ca L(V,W)$. Prove that $T$ is injective if and only if there exists $S\in\ca L(W,V)$ such that $ST$ is the identity map on $V$.

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

21. Suppose $W$ is finite-dimensional and $T\in\ca L(V,W)$. Prove that $T$ is surjective if and only if there exists $S\in\ca L(W,V)$ such that $TS$ is the identity map on $V$.

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

22. Suppose $U$ and $V$ are finite-dimensional vector spaces and $S\in\ca L(V,W)$ and $T\in\ca L(U,V)$. Prove that $\dim \m{null}ST \le \dim \m{null} S + \dim \m{null} T.$Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 16.

23. Suppose $U$ and $V$ are finite-dimensional vector spaces and $S\in\ca L(V,W)$ and $T\in\ca L(U,V)$. Prove that $\dim \m{range}ST \le \min\{\dim \m{range}S ,\dim \m{range}T\}.$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. Suppose $W$ is finite-dimensional and $T_1,T_2\in\ca L(V,W)$. Prove that $\m{null} T_1 \subset \m{null} T_2$ if and only if there exists $S \in\ca L(W,W)$ such that $T_2=ST_1$.

Solution: If we assume $\m{null} T_1 \subset \m{null} T_2$. Let $u_1$, $\cdots$, $u_m$ be a basis of $\m{null} T_1$, then it can be extended to a basis $u_1,\cdots,u_m,v_1,\cdots,v_n$ of $V$. Then from the process of the proof of 3.22, we know that $T_1v_1,\cdots,T_1v_n$is linear independent. Hence we can extend it to a basis $T_1v_1,\cdots,T_1v_n,\nu_1,\cdots,\nu_l$ of $W$. Now define $S\in\ca L(W,W)$ such that $ST_1v_i=T_2v_i,\quad S\nu_j=0, \quad i=1,\cdots,n;j=1,\cdots,l.$Then $ST_1v_i=T_2v_i$ and $ST_1u_j=0=T_2u_j$ since $\m{null} T_1 \subset \m{null} T_2$, hence $ST_1=T_2$ by uniqueness in 3.5.

If there exists $S \in\ca L(W,W)$ such that $T_2=ST_1$, then for any $\mu\in\m{null} T_1$, we have $T_2\mu=ST_1\mu=S0=0.$Hence $\mu\in \m{null} T_2$, it follows that $\m{null} T_1 \subset \m{null} T_2$.

25. Suppose $W$ is finite-dimensional and $T_1,T_2\in\ca L(V,W)$. Prove that $\m{range} T_1 \subset \m{range} T_2$ if and only if there exists $S \in\ca L(V,V)$ such that $T_1=T_2S$.

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. Suppose $D \in\ca L(\ca P(\R),\ca P(\R))$ is such that $\deg Dp=\deg p-1$ for every nonconstant polynomial $p\in \ca P(\R)$. Prove that $D$ is surjective. [The notation $D$ is used above to remind you of the differentiation map that sends a polynomial $p$ to $p’$. Without knowing the formula for the derivative of a polynomial (except that it reduces the degree by 1), you can use the exercise above to show that for every polynomial $q\in\ca P(\R)$, there exists a polynomial $p\in\ca P(\R)$ such that $p’=q$.]

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. Suppose $p\in \ca P(\R)$. Prove that there exists a polynomial $q\in \ca P(\R)$ such that $5q”+3q’=p$. [This exercise can be done without linear algebra, but it’s more fun to do it using linear algebra.]

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. Suppose $T \in\ca L(V,W)$ and $w_1$, $\cdots$, $w_m$ is a basis of range $T$. Prove that there exist $\vp_1$, $\cdots$, $\vp_m\in$ $\ca L(V,\mb F)$ such that $Tv=\vp_1(v)w_1+\cdots+\vp_m(v)w_m$ for every $v\in V$.

Solution: For any $v\in V$, there is a uniquely linear combination $Tv=a_1w_1+\cdots+a_mw_m,$where $a_1,\cdots,a_m\in\mb F$, 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. Suppose $\vp\in\ca L(V,\mb F)$. Suppose $u\in V$ is not in null $\vp$. Prove that $V=\m{null}\vp\oplus\{au:a\in\mb F\}.$Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 4.

30. Suppose $\vp_1$ and $\vp_2$ are linear maps from $V$ to $\mb F$ that have the same null space. Show that there exists a constant $c\in \mb F$ such that $\vp_1= c\vp_2$.

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. Give an example of two linear maps $T_1$ and $T_2$ from $\R^5$ to $\R^2$ that have the same null space but are such that $T_1$ is not a scalar multiple of $T_2$.

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 $c=1/2$.