1. Suppose $T\in\ca L(U, V)$ and $S\in\ca L(V, W)$ are both invertible linear maps. Prove that $ST\in\ca L(U, W)$ / is invertible and that $(ST)^{-1}=T^{-1}S^{-1}$.

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

2. Suppose $V$ is finite-dimensional and $\dim V > 1$. Prove that the set of noninvertible operators on $V$ is not a subspace of $\ca L(V) $.

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

3. Suppose $V$ is finite-dimensional, $U$ is a subspace of $V$, and $S\in\ca L(U,V)$. Prove there exists an invertible operator $T\in\ca L(V)$ such that $Tu=Su$ for every $u\in U$ if and only if $S$ is injective.

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. Suppose $W$ is finite-dimensional and $T_1, T_2\in\ca L(V,W)$. Prove that null $T_1$= null $T_2$ if and only if there exists an invertible operator $S\in\ca L(W)$ such that $T_1=ST_2$.

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. Suppose $V$ is finite-dimensional and $T_1,T_2\in \ca L(V,W)$. Prove that range $T_1=$ range $T_2$ if and only if there exists an invertible operator $S\in\ca L(V) $ such that $T_1=T_2S$.

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. Suppose $V$ and $W$ are finite-dimensional and $T_1,T_2\in\ca L(V,W)$. Prove that there exist invertible operators $R\in\ca L(V)$ and $S\in\ca L(W)$ such that $T_1=ST_2R$ if and only if $\dim$ null $T_1= \dim$ null $T_2$.

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. Suppose $V$ and $W$ are finite-dimensional. Let $v\in V$. Let \[E=\{T \in\ca L(V,W): T v= 0\}.\] (a) Show that $E$ is a subspace of $\ca L(V,W)$. (b) Suppose $v\ne 0$. What is $\dim E$?

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. Suppose $V$ is finite-dimensional and $T: V \to W$ is a surjective linear map of $V$ onto $W$. Prove that there is a subspace $U$ of $V$ such that $T|_U$ is an isomorphism of $U$ onto $W$. (Here $T|_U$ means the function $T$ restricted to $U$. In other words, $T|_U$ is the function whose domain is $U$, with $T|_U$ defined by $T|_U(u)=Tu$ for every $u\in U$.)

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. Suppose $V$ is finite-dimensional and $S, T \in\ca L(V)$. Prove that $ST$ is invertible if and only if both $S$ and $T$ are invertible.

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

10. Suppose $V$ is finite-dimensional and $S, T \in\ca L(V)$. Prove that $ST=I$ if and only if $TS=I$.

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

11. Suppose $V$ is finite-dimensional and $S, T ,U\in\ca L(V)$ and $ST U = I$. Show that $T$ is invertible and that $T^{-1}=US$.

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. Show that the result in the previous exercise can fail without the hypothesis that $V$ is finite-dimensional.

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. Suppose $V$ is a finite-dimensional vector space and $R,S, T\in\ca L(V)$ are such that $RST$ is surjective. Prove that $S$ is injective.

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

14. Suppose $v_1$, $\cdots$, $v_n$ is a basis of $V$. Prove that the map $T: V\to \mb F^{n,1}$ defined by \[Tv=\ca M(v)\] is an isomorphism of $V$ onto $\mb F^{n,1}$, here $\ca M(v)$ is the matrix of $v \in V$ with respect to the basis $v_1$, $\cdots$, $v_n$.

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

15. Prove that every linear map from $\mb F^{n,1}$ to $\mb F^{m,1}$ is given by a matrix multiplication. In other words, prove that if $T\in\ca L(\mb F^{n,1},\mb F^{m,1})$, then there exists an $m$-by-$n$ matrix $A$ such that $Tx=Ax$ for every $x\in \mb F^{n,1}$.

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. Suppose $V$ is finite-dimensional and $E$ is a subspace of $\ca L(V)$ such that $ST \in E$ and $TS \in E$ for all $S\in \ca L(V)$ and all $T \in E$. Prove that $E=\{0\}$ or $E=\ca L(V)$.

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 $\delta_{ii}=0$ 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. Show that $V$ and $\ca L(\mb F, V)$ are isomorphic vector spaces.

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. Suppose $T\in\ca L(\ca P(\R))$ is such that $T$ is injective and $\deg Tp\le \deg p$ for every nonzero polynomial $p\in\ca P(\R)$.

(a) Prove that $T$ is surjective.

(b) Prove that $\deg Tp= \deg p$ for every nonzero $p\in\ca P(\R)$.

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. Suppose $n$ is a positive integer and $A_{i,j}\in\mb F$ for $i, j = 1,\cdots,n$. Prove that the following are equivalent (note that in both parts below, the number of equations equals the number of variables):

(a) The trivial solution $x_1=\cdots=x_n=0$ is the only solution to the homogeneous system of equations \[\sum_{k=1}^nA_{1,k}x_k=0\] \[\cdots\] \[\sum_{k=1}^nA_{n,k}x_k=0.\](b) For every $c_1,\cdots,c_n\in\mb F$, there exists a solution to the system of equations\[\sum_{k=1}^nA_{1,k}x_k=c_1\] \[\cdots\] \[\sum_{k=1}^nA_{n,k}x_k=c_n.\] Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 26.

## David

3 Sep 2019For number 8, I think it’s worth mentioning the intuition behind the exercise. That is, you’re “removing” the nullspace of T from V, leaving behind a subspace U on which T|u is still surjective but is now injective as well.

## Zerong Xi

24 Jul 2019In exercise 6, it’s unnecessary to convert $T_1=ST_2R$ into $S^{-1}T_1=T_2R$, so to infer null$T_1$=null$T_2R$.

## Eli Bashwinger

29 Sep 2018For problem 18, why not just use the fact that L(F,V) and V are both vectors spaces over the same field and of the same dimension? Using these facts, it follows immediately from theorems 3.61 and 3.59 that L(F,V) and V are isomorphic.

## Eli Bashwinger

29 Sep 2018Ah! Perhaps you didn’t use those two theorems because the problem statement does not specify whether V is finite dimensional. Is that why?

## zagortenay333

18 Sep 2018Seriously, by what line of reasoning did you produce the solution for exercise 4, and how long did it take you to do it?

It just seems impossible to come up with that.. Baahhh, so frustrating…

Is this pattern of not being sure whether a space V is finite-dim and, therefore, decomposing it as a direct sum of a span and a kernel a recurring thing? Did you see it somewhere else or did you just produce it out of thin air for the purposes of the exercise?

## Sam Tay

30 Oct 2018I actually ran into this problem in 3.B.24 as well, and have since had to use this fact regularly in many problems. I think it may have been an oversight in the author’s exposition, as it would have been a good stepping stone problem to cite for these.

## dork forces

24 Jan 2019..

## David

3 Sep 2019I was thinking the EXACT same thing.

## Célio Passos

13 Jun 2017I have another proof for the converse in exercise 16.

The problem says that if v is in V, then S(Tv) = T(Sv) for every S in L(V). If v is in null S, then T(Sv) = T(0) = 0 = S(Tv) which implies Tv is in null S. Let S in L(V) such that v is a basis for null S (one can easily show a map like this exists for every v). We have that Tv is also in null S, but every vector in null S is a scalar multiple of v, thus Tv = cv = cIv for some scalar c. We but need to show that this c is the same for any other vector in V.

Let u = dv, where d is a scalar. Because T is linear, we have that Tu = dTv = dcv = cu, hence c is the same for all multiples of v. Let w in V such that w is not a scalar multiple of v, that is v and w are linearly independent, and Tw = aw, where a is a scalar. We have that T(v + w) = b(v + w) for some scalar b, but also T(v + w) = Tv + Tw = cv + aw. Thus bv + bw = cv + aw, which implies (b – c)v + (b – a)w = 0. Because v and w are linearly independent we have that c = a = b. The same can be done for every vector that is not a multiple of v, hence the scalar is the same for all vectors.

Therefore T is a scalar multiple of the identity.

## Elliott Mestas

3 Feb 2017For exercise 17, I think I have a shorter proof:

Consider an invertible map in E, call it T (it must exist by exercise 2). Then T T^-1 = I is in E by definition.

But if I is in E, then S*I = S is in E for all S in L(V). Therefore, E = L(V).

## Mohammad Rashidi

3 Feb 2017You cannot conclude the existence of T from exercise 2.

## statssigsushi

18 Oct 2015For problem 3,can you just define T by T(ui) = S(ui) for bases of U and T(vi) = vi for the extended part of basis of V and prove that only {0} is in null T?