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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 \[Tv=a_1Tv_1+\cdots+a_nT_nv_n.\]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 $c=1/2$.


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This Post Has 18 Comments

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

  2. For #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.

  3. In #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.

    1. Yes, 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.

  4. For 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?

    1. Oh nevermind, got it. Though we only had to make the dimensions equal.

  5. In #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?

    1. $u$ 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.

  6. In #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.

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

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

  7. The 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.

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

  9. 24) 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.

    1. You can use the same idea from the solution to Exercise 4 in 3.D.

      1. Do 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.

        1. I 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.

  10. I 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

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