Chapter 3 Exercise B


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 \[Tv=a_1Tv_1+\cdots+a_nT_n.\]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: Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 10.


16. Solution: 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: 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. 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 $a_1,\cdots,a_m\in\mb F$, 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 $c=1/2$.

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.

You may also like...

Leave a Reply

Your email address will not be published. Required fields are marked *