Chapter 3 Exercise A


1. Solution: If $T$ is linear, then \[(0,0)=T(0,0,0)=(b,0)\]by 3.11, hence $b=0$. We also have \[T(1,1,1)=T(1,1,0)+T(0,0,1),\]it is equivalent to \[(1+b,6+c)=(b-2,6)+(3+b,0)=(1+2b,6).\]Thus $6+c=6$ implies $c=0$.
Conversely, if $b=c=0$, $T$ is obviously linear. See 3.4 or Problem 3.

You may consider a linear map only has linear terms, hence the terms like $xyz$ is impossible.


2. Solution: We show that if $b=c=0$, then $T$ is linear. Let $f,g\in \ca P(\R)$, then we have \[(f+g)(4)=f(4)+g(4)\]and \[(f+g)'(4)=f'(4)+g'(4).\]It is easy to check. Moreover, by linearity of integration, one has \[\int_{-1}^2x^3(f+g)(x)dx=\int_{-1}^2x^3(f(x)+g(x))dx=\int_{-1}^2x^3f(x)dx+\int_{-1}^2x^3g(x)dx.\] By the above, it follows that \begin{align*} T(f+g)=&(3(f+g)(4)+5(f+g)'(6),\int_{-1}^2x^3(f+g)(x)dx)\\ =&(3f(4)+5f'(6),\int_{-1}^2x^3f(x)dx)+(3g(4)+5g'(6),\int_{-1}^2x^3g(x)dx)\\ =&Tf+Tg. \end{align*} Similarly, we can check that $T(\lambda f)=\lambda Tf$ for any $\lambda\in \R$ and $f\in \ca P(\R)$. Conversely, denote the linear map above by $S$. Then by Problem 5, it follows that $T-S$ is a linear map. That means \[(T-S)p=(bp(1)p(2),c\sin p(0))\]is linear. Consider $f(x)=\pi/2$ and $g(x)=\pi/2$, then $f,g\in\ca P(\R)$. We have \[ (T-S)(f+g)=(b\pi^2,c\sin \pi)=(b\pi^2,0) \]and \[ (T-S)f+(T-S)g=(b\pi^2/2,c)+(b\pi^2/2,c)=(b\pi^2/2,2c). \]Thus, we should have \[(b\pi^2,0)=(b\pi^2/2,2c).\]It follows that $b=c=0$.

We do not really need Problem 5, we need them just for simplifying computation.


3. Solution: If we denote \[ T(1,0,\cdots,0)=(A_{1,1},\cdots,A_{m,1}), \]\[ T(0,1,\cdots,0)=(A_{1,2},\cdots,A_{m,2}), \]\[\cdots\cdots\]and\[ T(0,0,\cdots,0,1)=(A_{1,n},\cdots,A_{m,n}). \]Note that $(1,0,\cdots,0)$, $(0,1,\cdots,0)$, $\cdots$ and $(0,\cdots,0,1)$ is a basis of $\mb F^n$, then by the proof of 3.5 we conclude that \[ T(x_1,\cdots,x_n)=(A_{1,1}x_1+\cdots+A_{1,n}x_n,\cdots,A_{m,1}x_1+\cdots+A_{m,n}x_n). \]


4. Solution: Suppose there are $a_1$, $\cdots$, $a_m\in\mb F$ such that \[0=a_1v_1+\cdots+a_mv_m,\]then \[0=T(a_1v_1+\cdots+a_mv_m)=a_1Tv_1+\cdots+a_mTv_m.\]Note that $T v_1$, $\cdots$, $T v_m$ is linearly independent, it follows that \[ a_1=\cdots=a_m=0, \]thus $v_1$, $\cdots$, $v_m$ is linearly independent.


5. Solution: It suffices to show that $T+S$ and $\lambda T$ are linear maps provided $T$ and $S$ are linear maps, where $\lambda \in\mb F$. Then $\ca L(V,W)$ is closed under addition and scalar multiplication, hence is a vector space.
For any $u,v\in V$, we have \begin{align*} (T+S)(u+v)=&T(u+v)+S(u+v)=Tu+Tv+Su+Sv\\ =&(Tu+Su)+(Tv+Sv)=(T+S)u+(T+S)v. \end{align*} The first and last equality hold by the definitions in 3.6, the second equality holds since $T$ and $S$ are linear maps. Similarly, for $\eta\in\mb F$, \begin{align*} (T+S)(\eta u)=&T(\eta u)+S(\eta u)=\eta Tu+\eta Su\\ =&\eta(Tu+Su)=\eta(T+S)u. \end{align*} Combining these arguments, it follows that $T+S$ is a linear map. Again, for any $u,v\in V$, we have \begin{align*} (\lambda T)(u+v)=&\lambda (T(u+v))=\lambda (Tu+Tv)\\ =&\lambda (Tu)+\lambda (Tv)=(\lambda T)u+(\lambda T)v. \end{align*} The first and last equality hold by the definitions in 3.6, the second equality holds since $T$ is a linear map. Similarly, for $\eta\in\mb F$, \begin{align*} (\lambda T)(\eta u)=&\lambda(T(\eta u))=\lambda(\eta T( u))\\ =&\lambda\eta(Tu)=\eta(\lambda Tu)=\eta(\lambda T)u. \end{align*} Combining these arguments, it follows that $\lambda T$ is a linear map.


6. Solution: Associativity: by definition, for any $x\in V$, we have \[ ((T_1T_2)T_3)x=(T_1T_2)(T_3x)=T_1(T_2(T_3x)), \]while \[ (T_1(T_2T_3))x=T_1((T_2T_3)x)=T_1(T_2(T_3x)). \]Hence $((T_1T_2)T_3)x=(T_1(T_2T_3))x$ for any $x\in V$, therefore $(T_1T_2)T_3=T_1(T_2T_3)$.
Identity: for any $x\in V$, we have \[ (TI)x=T(Ix)=Tx \]while\[(IT)x=I(Tx)=Tx.\]Hence $(IT)x=(TI)x=Tx$ for any $x\in V$, therefore $IT=TI=T$.
Distributive properties: we only show $(S_1+S_2)T=S_1T+S_2T$. For any $u\in U$, we have \[ ((S_1+S_2)T)u=(S_1+S_2)(Tu)=S_1(Tu)+S_2(Tu), \]while \[ (S_1T+S_2T)u=(S_1T)u+(S_2T)u=S_1(Tu)+S_2(Tu). \] Hence $((S_1+S_2)T)u=(S_1T+S_2T)u$ for any $u\in U$, therefore $(S_1+S_2)T=S_1T+S_2T$. Similarly, we can show $S(T_1+T_2)=ST_1+ST_2$.


7. Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 1.


8. Solution: See Linear Algebra Done Right Solution Manual Chapter 3 Problem 2.


9. Solution: Consider the function defined by $\vp(a+bi)=a$, where $a,b\in\R$. If $w=\alpha_1+\beta_1 i$ and $z=\alpha_2+\beta_2 i$, where $\alpha_1$, $\alpha_2$, $\beta_1$ and $\beta_2$ are real numbers, then we have \[\vp(w+z)=\vp(\alpha_1+\beta_1 i+\alpha_2+\beta_2 i)=\alpha_1+\alpha_2=\vp(w)+\vp(z).\]However, \[i\vp(1)=i\ne \vp(i\cdot 1)=0,\]hence $\vp$ is not linear over $\C$.

For $\vp:\R\to\R$ case, please see On sort-of-linear functions.


10. Solution: Note that $U\ne V$ and $S\ne 0$, we can choose $u\in U$ such that $Su\ne0$ and $v\in V$ but $v\notin U$, then $u+v\notin U$. Otherwise \[v=(u+v)-u\in U\]will yield a contradiction. Hence $T(u+v)=0$ by definition. On the other hand, $Tu+Tv=Su\ne 0$. It follows that $T(u+v)\ne Tu+Tv$, hence $T$ is not a linear map on $V$.


11. Solution: Let $u_1$, $\cdots$, $u_m$ be a basis of $U$, then we can extend it to a basis of $V$ by 2.33. That is $u_1$, $\cdots$, $u_m$, $v_{m+1}$, $\cdots$, $v_n$ is a basis of $V$. Define $T\in \ca L(V,W)$ as below \[Tu_i=Su_i,\quad Tv_{j}=0, \quad 1\le i\le m, m+1\le j \le n.\]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*}


12. Solution: Here, we use Problem 14 of Exercises 2.A. Hence there is a sequence $W_1$, $W_2$, $\cdots$ of vectors in $W$ such that $W_1$, $W_2$, $\cdots$, $W_m$ is linearly independent for every positive integer $m$. Consider $T_i\in \ca L(V,W)$ such that $T_i(v_1)=w_i$, where $v_1$, $v_2$, $\cdots$, $v_n$ is a basis of $V$. The existence of $T_i$ are guaranteed by 3.5(not unique). Then we show that $T_1$, $\cdots$, $T_m$ is linearly independent for every positive integer $m$. Suppose there are $a_1$, $\cdots$, $a_m\in\mb F$ such that \[a_1T_1+\cdots+a_mT_m=0.\]Then we have $(a_1T_1+\cdots+a_mT_m)(v_1)=0$, i.e. \[a_1w_1+\cdots+a_mw_m=0.\]Since $W_1$, $W_2$, $\cdots$, $W_m$ is linearly independent, it follows that $a_1=\cdots=a_m=0$. Thus $T_1$, $\cdots$, $T_m$ is linearly independent. Again by Problem 14 of Exercises 2.A, it follows that $\ca L(V,W)$ is infinite-dimensional.


13. Solution: Because $v_1$, $v_2$, $\cdots$, $v_m$ is a linearly dependent, there exist $a_1$, $\cdots$, $a_m\in\mb F$ such that \[a_1v_1+\cdots+a_mv_m=0\]and some $a_i\ne 0$(this $i$ is fixed). Then let $w_i\ne 0$ while $w_j=0$ if $j\ne i$, where $w_1$, $w_2$, $\cdots$, $w_m\in W$. We will show there is no $T\in\ca L(V,W) $ satisfies $Tv_k= w_k$ for each $k=1,\cdots,m$. Otherwise, we have \[0=T(a_1v_1+\cdots+a_mv_m)=a_1w_1+\cdots+a_mw_m=a_iw_i.\]Notice that $a_i\ne0$ and $w_i\ne 0$ by our choice, we get a contradiction.


14. Solution: Let $e_1$, $\cdots$, $e_n$ be a basis of $V$, define $T\in \ca L(V,V)$ such that \[Te_1=e_2,\quad Te_2=e_1,\quad Te_{i}=e_{i},\quad\text{for}\quad i\ge 2,\]and $S\in \ca L(V,V)$ such that \[Se_1=e_1,\quad Se_2=2e_2,\quad Se_{i}=e_{i},\quad\text{for}\quad i\ge 2.\]The existences of $S$ and $T$ are guaranteed by 3.5(unique). Then \[STe_1=Se_2=2e_2,\quad TSe_1=Te_1=e_2.\]Hence $STe_1\ne TSe_1$, it implies $ST\ne TS$.

Linearity

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