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.

## lyk2527

26 Dec 2021For problem 16, S is any linear map belong L(v), but it proved that when S is the particular linear map which is S(av+bTv+c1u1+...+cnun)=bv. Is that right?

If we suppose T=0, then for any S belong L(v), we have ST=0, TS=0, hence ST=TS. Is that correct?

I can not understand the answer, It's regardful if someone reply me.

## lyk2527

26 Dec 2021T=0=0I, that's a scalar multiple of identity.

But i still can not understand the implacation.

## Ng Shao Qian

27 Aug 2021I think there is a little problem in your solution #5

In line 3, we can't ensure that T1w_1 , ... , T1w_n is linearly independent as it requires T1 to be injective in the first place (Exercise 3B Q9). Indeed, we can only show that if Tv_1 , ... , Tv_n is linearly independent, then so is v_1 , ... , v_n (Exercise 3A Q4), its converse is not necessarily true.

I suggest my solution here, kindly correct me if I am wrong.

If range T1 = range T2, let v_1 , ... , v_n be basis of V. Then, there exists a linear map S s.t. Sv_1 = u_1 != 0, and Sv_i = u_i not in span(u_1 , ... , u_i-1). Again, since v_1 , ... , v_n is basis of V and u_1 , ... , u_n is some list in V, there exists T1v_i = T2u_i = T2Sv_i. This follows that T1v = T2Sv for any vector v in V.

Next, we want to show that S is invertible. From the corollary of Linear Dependence Lemma, v_1 , ... , v_n , w is linearly independent iff w is not in span(v_1 , ... , v_n). Hence in our case, u_1 , ... , u_n is linearly independent in range S (and also basis of S). Because range S is a subspace of V, then u_1 , ... , u_n can be extended to a basis with length = n of V. However, our case is trivial, no new vector is needed to be added. So, range S = V since they share same basis, which means that S is surjective. From theorem 3.69, S is invertible.

## efelmzmg

1 Nov 2021Please look at 3.22.

## Curtis CHEE

8 Apr 2021For #8 why is W finite dimensional?

## Halarious

14 Jul 2021Surjectivity of T from a finite dimensional space implies this I believe.

## Kevin

11 Feb 2021For #2, can we just conclude that the set of non invertible operators on V is not a subspace of L(V) since there are no inverse elements? If T in this set set and there exists S such that TS=I, then T is invertible, which yields a contradiction. Am I doing something wrong?

## Ng Shao Qian

27 Aug 2021For your example, there doesn't exist such S because T is in the set of non-invertible operators, you can't make an assumption with both arguments never happen together.

In fact, to prove that the set of non-invertible operators is not a subspace of L(V), you can show that the set doesn't fulfill one of the subspace properties: additive identity, closed under addition property, or closed under scalar multiplication property.

## 4da

5 Dec 2020I don't see how when can directly conclude form ex.9 that TU is invertible.

I see solution as follows: $STU$ = I$ => $T(US)$ (by ex. 10).

Then, for any $v \in V$ we see that $v \in $ range($T$). Thus, T is surjective and that implies that T is invertible because $V$ is finite-dimensional.

After that it is easy to see that $(US)^-1$ is the only inverse of $T$.

## George

30 Sep 2020This is my proof for number 17.

Suppose E is not the zero transform.

Suppose then that T is part of E, and S is not part of E.

The only way ST would be part of E if you can express S as a linear sum of a basis of transforms in E (since ST eventually transforms a vector v by S). But that would mean that S is part of E, which is a contradiction.

The proof is obvious if E =0. if T is part of E, then T=0. ST=0 and TS=0, thus ST is part of E.

Would appreciate any feedback.

## Marie

27 Jul 2020For Q3, to prove the backward direction, can we not just define T as it is defined here in Q.11 of 3A https://linearalgebras.com/3a.html ? So define T as T(uj)=S(uj) and T(vi)=0. Then we can say that since S is injective, T is injective and given dimV is finite, T is invertible. Just wondering if this method checks out.

## Marie

27 Jul 2020For Q.9, when proving in the forward direction, may I use this logic to show T is injective:

Assume v is in nullT. Then Tv=0.

ST(v) = S(Tv) = S(0) = ST(0) where the last equality holds because any linear map maps 0 to 0.

Since ST is injective, ST(v) = ST(0) implies v = 0.

?

## Yuheng

20 Jul 2020I have another solution for 3.D.6.

Suppose T1=ST2R. Take basis of null T1 as v1,v2...vn. Then Rv1, Rv2...Rvn are in null T2 (by injectivity of S) and they are linearly independent (by injectivity of R). So dim null T1 = dim Null T2. So they equal.

Suppose dim null T1 = dim null T2. We just need to prove there exist invertible operators S, R such that ST1=T2R. Take basis of nullT1 v1...vn. Extend it to v1...vn, u1...um. Take basis of null T2 w1...wn. Let Rvi=wi for i=1...n (so we have dealt with null T1.) Extend w1...wn to w1...wn, p1...pm. Let Rui=pi for I=1...m. By now R is a well-defined isomorphism on V because it maps a basis to another basis. Define S as a FUNCTION from range T1 to W by ST1v=T2Rv for each v in V. Then it's easy to check that S is an injective linear map. Then by 3.D.3, there exists an invertible operator S' on W such that S'u=Su for all u in range T1, and we are done.

I think, after dealing with nullT1, this way of directly defining S as a function and then coming back to prove its property might by more natural.

## Yuheng

19 Jul 2020For 4, the proof can be done in a more natural way, using 3.D.3.

Suppose null T1 = null T2. Define S as a FUNCTION from range T2 to W by ST2v = T1v for each v in V. Then we prove S is a linear map, which is easy by checking additivity and homogeneity.

Then we prove S is injective, suppose ST2v=0, then T1v=0, then T2v=0 by assumption. So S is injective.

Then because S is an injective linear map from range T2 to W, rangeT2 is a subspace of W, by 3.D.3, there exists an invertible operator S' on W such that Su=Wu for all u in range T2, and we're done.

## Yuheng

19 Jul 2020I meant S'u=Su in the end.

## Subhasish Mukherjee

27 May 2020I think you made a mistake in problem 17. I think you meant delta_ii = 1

## Austin Watkins

26 May 2020For problem 17, I didn't understand your indexing. But I did come up with this proof by contradiction:

Suppose E != {0} and S in L(V)

Case 1 - Identity map in E:

(see end of your proof)

Case 2 - Identity map not in E:

As E is non-empty it has some non-zero T function.

But if the identity map I is not in E, then TI is not in E or IT is not in E.

Which implies T is not in E. A contradiction.

## Leonardo Sebastian

16 Jul 2020I think this doesn't work, I tried it also, but it doesn't work since TI and IT are both in E. What is not in E is the identity map alone i think. Another way to think about it is realizing that you can "evaluate" the statement: For some T in E and I in L(V) then (TI) is in E, but TI is T which states T is in E which is redundant. We do not need I to be in E for TI or IT to make sense.

## samir

10 May 2020For problem 1, the document refers to the solutions manual, chapter 3 problem 22. But note that the proof given there is for S and T being in L(V). The statement of the linked problem does not work when T:U -> V, and S:V -> W, because it is possible for ST to be invertible, but S,T to not be individually invertible.

## Linearity

10 May 2020I meant only one direction (not if and only if statement), $S$ and $T$ invertible then $(ST)^{-1}=T^{-1}S^{-1}$. The proof is the same, no matter it is in $\mathcal L(V)$ or $\mathcal L(U,V)$.

## sam

21 Feb 2020I think you forgot to show that T defined as such is indeed invertible on V in problem 3.

## Linearity

27 Feb 2020I showed, see the words "by 3.69".

## 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.

## lyk2527

26 Dec 2021"every vector in null S is a scalar multiple of v", how to prove it? What about if dim null S>=2?

## 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.

## Curtis CHEE

9 Apr 2021Is I in E since E is a subspace of L(V)?

## 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?

## Xinyu Jian

12 Oct 2020I think this solution is also correct.