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 $c=1/2$.

## Marie

24 Jul 2020Is there a way to approach Q.4 using the fundamental theorem of linear maps (3.22)?

## Ijen

24 Jul 2020For #29, I have the following solution:

dim V=dim ker phi+dim range phi

As range phi=F, dim range phi=1, so we have dim ker phi=dim V-1

Therefore if we are to extend a basis of ker phi to a basis of V, arbitrary single vector which is not in ker phi would be OK.

Is this right? Thanks for reading.

## Shawn

17 Jul 2020In #24, S is defined as $S(T_1v_i) = T_2v_i$, thus it should be $S \in L(range T_1, W)$, not $S \in L(W, W)$. However, $S$ could be extended to S' such that $S' \in L(W, W)$ using similar approach in #20.

## Yuheng

18 Jul 2020Yes, I think he is just using 3.5 in a simplified way. To define a linear map, we only need to define it on a linearly independent list of vectors (in this case, the basis of rangeT1), and the rest may be ignored if they don't matter.

## George

18 Jun 2020For number 5, can't the definition be T(e1) = e1 and T(e2)=e2, while T(e3)=0 and T(e4)= 0; or am I missing something?

## George

18 Jun 2020Oh nevermind, got it. Though we only had to make the dimensions equal.

## Tom C

4 May 2020In #30, shouldn't the book say that for fixed u, there is a constant c such that phi1 = c * phi2? As the book currently states, c seems to be a function of u, not a constant. For instance, if phi1 and phi2 are linear maps from the polynomials to F, and null phi1 = null phi2 = span(x^2, x^3, x^4, ...), and phi1(1) = 1, phi1(x) = 2, phi2(1) = 2, phi2(x) = 1, then they have the same null space but phi1(1) = 1/2 * phi2(1) but phi1(x) = 2 * phi2(x). In the u = 1 case we have c = 1/2, but in the u = x case c = 2 because c is a function of u, not a constant as claimed. That is correct, no?

## Linearity

5 May 2020$u$ is fixed and so is $c=\dfrac{\varphi_1(u)}{\varphi_2(u)}$. Your example does not define a linear function whose null space is the as the one described by you. To be more precise, a null space is of codim 1 but your example has codim 2.

## Tom C

4 May 2020In #29 it seems to me that the book omits the fact that "u is not in the null space of phi" is not a sufficient condition for {au such that a is in F} to be a subspace. I think there also needs to be a condition that u is not decomposible into a vector in the null space and another vector not in the null space. For instance, if phi was a linear map that took the coefficient of x from a polynomial p, and all the other terms were in the null space, one could express (1 + ax + bx^2) as either (ax) + (1 + bx^2) or as (ax + bx^2) + (1). Neither (ax) nor (ax + bx^2) is in the null space, but (ax + bx^2) - (ax) = (bx^2) is in the null space, so the set defined in the book is not closed under addition, and the sum is not a direct sum. If the condition is added, I think it all works out.

## Linearity

5 May 2020{au such that a is in F} is always a subspace(note $u$ is fixed). Period.

## Tom C

5 May 2020Ah, I see. That makes sense for u to be fixed. I had not read it that way. Thank you.

## Sam Tay

18 Oct 2018The definition of S is too loose in Problem 25. In particular, your version is not guaranteed to be well defined (consider when some of those u_i, v_i are in null spaces). You'll find that you need to have a basis of null T2, extend it to a basis of V, and map S(u_i) to v where v is in the span of the extension list, and has no part in the null space of T2, because that choice of v is unique. In particular, there may be many more such v where T1(u_i) = T2(v), but only one of them is in the span of the extension list.

Edit: Come to think of it, I guess it doesn't matter that you pick the unique v in that span. So long as you choose one of those v_i, the proof holds. So never mind.

## MetaMusic

30 Jun 2017You assumed V is a finite-dimensional vector space for #24 and you wrote W instead of V for #25

## Tobias Pielok

22 Apr 201724) It is not stated that V is finite-dimensional => null T1 doesn't have to be finite-dimensional => it is possible that there is no basis for null T1. But I think it is possible to construct a general proof with the range, which is finite dimensional.

## Célio Passos

26 Jul 2017You can use the same idea from the solution to Exercise 4 in 3.D.

## Sam Tay

17 Oct 2018Do you think this is a mistake in the book? Or is there a proof the author had in mind that only uses the results up to 3.B.24? I've been stuck on this one for a while.

## Sam Tay

17 Oct 2018I figured it out! Of course it comes to me after caving to look at the solutions here, and finding none. Use a direct sum. Construct U such that V = U ⊕ null T1, note that U is finite and has a basis, and then work with the linear map T1 restricted to domain U. This restricted linear map is injective and allows one to construct the well-defined linear map S.

## Elliott Mestas

27 Jan 2017I believe you can shorten the proofs to both 17 and 18 (proving the "=>" direction) by using the contrapositive statements of 3.23 and 3.24