3.4 Reprsentation of Transformations by Matrices

这一节通过Theorem 11和Theorem 12 将linear transformation和矩阵建立了一对一的关联，此外Theorem 13说明linear transformation的composition与矩阵乘法也有天然的联系。Theorem 14则说明在linear operator的情况下，同一个linear transformation不同basis的矩阵存在相似（similar）的关系，且这个导致相似的矩阵实质上就是一组基在另一组基下的矩阵。

Exercises

1. Let $T$ be the linear operator on $C^2$ defined by $T(x_1,x_2)=(x_1,0)$. Let $\mathfrak B$ be the stantard ordered basis for $C^2$ and let $\mathfrak B'=\{\alpha_1,\alpha_2\}$ be the ordered basis defined by $\alpha_1=(1,i),\alpha_2=(-i,2)$.
   ( a ) What is the matrix of $T$ relative to the pair $\mathfrak B,\mathfrak B'$?
   ( b ) What is the matrix of $T$ relative to the pair $\mathfrak B',\mathfrak B$?
   ( c ) What is the matrix of $T$ in the ordered basis $\mathfrak B'$?
   ( d ) What is the matrix of $T$ in the ordered basis $\{\alpha_2,\alpha_1\}$?
   Solution:
   ( a ) $Tϵ_1=2α_1-iα_2,Tϵ_2=0α_1+0α_2$, thus the matrix of $T$ relative to the pair $\mathfrak B,\mathfrak B'$ is $\begin{bmatrix}2&0\\-i&0\end{bmatrix}$.
   ( b ) $Tα_1=ϵ_1,Tα_2=-iϵ_1$, thus the matrix of $T$ relative to the pair $\mathfrak B',\mathfrak B$ is $\begin{bmatrix}1&-i\\0&0\end{bmatrix}$.
   ( c ) $Tα_1=2α_1-iα_2,Tα_2=-2iα_1-α_2$, thus $[T]_{\mathfrak B'}=\begin{bmatrix}2&-2i\\-i&-1\end{bmatrix}$.
   ( d ) $Tα_2=-α_2-2iα_1,Tα_1=-iα_2+2α_1$, thus $[T]_{\{α_2,α_1\}}=\begin{bmatrix}-1&-i\\-2i&2\end{bmatrix}$.

2. Let $T$ be the linear transformation from $R^3$ into $R^2$ defined by
   $T(x_1,x_2,x_3)=(x_1+x_2,2x_3-x_1)$
   ( a ) If $\mathfrak B$ is the standard ordered basis for $R^3$ and $\mathfrak B'$ is the standard ordered basis for $R^2$, what is the matrix of $T$ relative to the pair $\mathfrak B,\mathfrak B'$?
   ( b ) If $\mathfrak B=\{\alpha_1,\alpha_2,\alpha_3\}$ and $\mathfrak B'=\{\beta_1,\beta_2\}$, where
   $\alpha_1=(1,0,-1),\quad\alpha_2=(1,1,1),\quad\alpha_3=(1,0,0),\quad\beta_1=(0,1),\quad\beta_2=(1,0)$
   what is the matrix of $T$ relative to the pair $\mathfrak B,\mathfrak B'$?
   Solution:
   ( a ) We let $ϵ_1,ϵ_2,ϵ_3$ be the standard ordered basis for $R^3$ and $ϵ_1',ϵ_2'$ be the standard ordered basis for $R^2$, then
   $Tϵ_1=(1,-1),Tϵ_2=(1,0),Tϵ_3=(0,2)$
   thus the matrix of $T$ relative to the pair $\mathfrak B',\mathfrak B$ is
   $A=\begin{bmatrix}1&1&0\\-1&0&2\end{bmatrix}$
   ( b ) A direct calculation shows
   $Tα_1=(1,-3)=-3β_1+β_2\quad Tα_2=(2,1)=β_1+2β_2,\quad Tα_3=(1,-1)=-β_1+β_2$
   thus the matrix of $T$ relative to the pair $\mathfrak B',\mathfrak B$ is
   $A=\begin{bmatrix}-3&1&-1\\1&2&1\end{bmatrix}$

3. Let $T$ be a linear operator on $F^n$, let $A$ be the matrix of $T$ in the standard ordered basis for $F^n$, and let $W$ be the subspace of $F^n$ spanned by the column vectors of $A$. What does $W$ have to do with $T$?
   Solution: By the condition given, if we write $A=\begin{bmatrix}A_{11}&\cdots&A_{1n}\\\vdots&\ddots&\vdots\\A_{n1}&\cdots&A_{nn}\end{bmatrix}$, then $Tϵ_j=∑_{i=1}^nA_{ij} ϵ_i=A_j$, the $j$-th column vector of $A$, thus if $W$ is spanned by $A_1,\dots,A_n$, it is easy to see $W=\text{range }T$.

4. Let $V$ be a two-dimensional vector space over the field $F$, and let $\mathfrak B$ be an ordered basis for $V$. If $T$ is a linear operator on $V$ and
   $[T]_{\mathfrak B}=\begin{bmatrix}a&b\\c&d\end{bmatrix}$
   prove that $T^2-(a+d)T+(ad-bc)I=0$.
   Solution: We write $\mathfrak B=\{α_1,α_2\}$, then for any $α∈V$, we have $α=x_1 α_1+x_2 α_2,x_1,x_2∈F$, notice that
   $[Tα_1 ]_{\mathfrak B}=[T]_{\mathfrak B} [α_1 ]_{\mathfrak B}=\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix}=\begin{bmatrix}a\\c\end{bmatrix},$
   $[Tα_2 ]_{\mathfrak B}=[T]_{\mathfrak B} [α_2 ]_{\mathfrak B}=\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}0\\1\end{bmatrix}=\begin{bmatrix}b\\d\end{bmatrix},$
   $[T^2 α_1 ]_{\mathfrak B}=[T]_{\mathfrak B} [Tα_1 ]_{\mathfrak B}=\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}a\\c\end{bmatrix}=\begin{bmatrix}a^2+bc\\ac+cd\end{bmatrix},$
   $[T^2 α_2 ]_{\mathfrak B}=[T]_{\mathfrak B} [Tα_2 ]_{\mathfrak B}=\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}b\\d\end{bmatrix}=\begin{bmatrix}ab+bd\\bc+d^2\end{bmatrix}$
   thus
   $\begin{aligned}&\quad(T^2-(a+d)T+(ad-bc)I) α_1\\&=T^2 α_1-(a+d)Tα_1+(ad-bc) α_1\\&=(a^2+bc) α_1+(ac+cd) α_2-(a+d)(aα_1+cα_2 )+(ad-bc) α_1\\&=(a^2+bc-a^2-ad+ad-bc) α_1+(ac+cd-ac-cd) α_2\\&=0\end{aligned}$
   $\begin{aligned}&\quad (T^2-(a+d)T+(ad-bc)I) α_2\\&=T^2 α_2-(a+d)Tα_2+(ad-bc) α_2\\&=(ab+bd) α_1+(bc+d^2 ) α_2-(a+d)(bα_1+dα_2 )+(ad-bc) α_2\\&=(ab+bd-ab-bd) α_1+(bc+d^2-ad-d^2+ad-bc) α_2\\&=0\end{aligned}$
   Now we have $(T^2-(a+d)T+(ad-bc)I) α_1=(T^2-(a+d)T+(ad-bc)I) α_2=0$, and so
   $\begin{aligned}&\quad (T^2-(a+d)T+(ad-bc)I)α\\&=(T^2-(a+d)T+(ad-bc)I)(x_1 α_1+x_2 α_2 )\\&=x_1 (T^2-(a+d)T+(ad-bc)I) α_1+x_2 (T^2-(a+d)T+(ad-bc)I) α_2\\&=0\end{aligned}$

5. Let $T$ be the linear operator on $R^3$, the matrix of which in the standard ordered basis is
   $A=\begin{bmatrix}1&2&1\\0&1&1\\-1&3&4\end{bmatrix}$
   Find a basis for the range of $T$ and a basis for the null space of $T$.
   Solution: Using Exercise 3 we know the range of $T$ is spanned by the column vectors of $A$, using elementary column operations we have
   $A=\begin{bmatrix}1&2&1\\0&1&1\\-1&3&4\end{bmatrix}→\begin{bmatrix}1&0&0\\0&1&1\\-1&5&5\end{bmatrix}→\begin{bmatrix}1&0&0\\0&1&0\\-1&5&0\end{bmatrix}$
   so a basis for the range of $T$ is $(1,0,-1),(0,1,5)$, notice that $Tϵ_3-Tϵ_1=Tϵ_2-2Tϵ_1$, thus
   $T(ϵ_3+ϵ_1-ϵ_2 )=T(1,-1,1)=0$
   and the dimension of the null space of $T$ is $1$, so a basis for the null space of $T$ is $(1,-1,1)$.

6. Let $T$ be the linear operator on $R^2$ defined by
   $T(x_1,x_2)=(-x_2,x_1)$
   ( a ) What is the matrix of $T$ in the standard ordered basis for $R^2$?
   ( b ) What is the matrix of $T$ in the ordered basis $\mathfrak B=\{\alpha_1,\alpha_2\}$, where $\alpha_1=(1,2)$ and $\alpha_2=(1,-1)$?
   ( c ) Prove that for every real number $c$ the operator $(T-cI)$ is invertible.
   ( d ) Prove that if $\mathfrak B$ is any ordered basis for $R^2$ and $[T]_{\mathfrak B}=A$, then $A_{12}A_{21}\neq 0$.
   Solution:
   ( a ) $Tϵ_1=(0,1)=ϵ_2,Tϵ_2=(-1,0)=-ϵ_1$, so the matrix of $T$ in the standard ordered basis for $R^2$ is $\begin{bmatrix}0&-1\\1&0\end{bmatrix}$.
   ( b ) We have
   $Tα_1=(-2,1)=-\frac{1}{3} α_1-\frac{5}{3} α_2,\quad Tα_2=(1,1)=\frac{2}{3}α_1+\frac{1}{3}α_2$
   thus
   $[T]_{\mathfrak B}=\begin{bmatrix}-1/3&2/3\\-5/3&1/3\end{bmatrix}$
   ( c ) The matrix of $T-cI$ in the standard ordered basis for $R^2$ is $\begin{bmatrix}-c&-1\\1&-c\end{bmatrix}$, thus
   $(T-cI) ϵ_1=(-c,1),\quad (T-cI) ϵ_2=(-1,-c)$
   since $(-c,1),(-1,-c)$ are linearly independent for any $c∈R$, thus a basis of $R^2$, this means $T-cI$ is invertible.
   ( d ) Let $\mathfrak B'=\{ϵ_1,ϵ_2\}$, then $[T]_{\mathfrak B' }=\begin{bmatrix}0&-1\\1&0\end{bmatrix}$, given any $\mathfrak B$, we can find an invertible $P$ s.t.
   $[T]_{\mathfrak B}=A=P[T]_{\mathfrak B'} P^{-1}$
   we can assume $P=\begin{bmatrix}a&b\\c&d\end{bmatrix}$, then $P^{-1}=\frac{1}{ad-bc} \begin{bmatrix}d&-b\\-c&a\end{bmatrix}$, obviously $a,b,c,d$ cannot be all zero.
   $\begin{aligned}A&=\frac{1}{ad-bc}\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}0&-1\\1&0\end{bmatrix}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}\\&=\frac{1}{ad-bc}\begin{bmatrix}b&-a\\d&-c\end{bmatrix}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}\\&=\frac{1}{ad-bc}\begin{bmatrix}bd+ac&-b^2-a^2\\d^2+c^2&-bd-ac\end{bmatrix}\end{aligned}$
   thus $A_{12} A_{21}=-(a^2+b^2 )(c^2+d^2 )\neq 0$.

7. Let $T$ be the linear operator on $R^3$ defined by
   $T(x_1,x_2,x_3)=(3x_1+x_3,-2x_1+x_2,-x_1+2x_2+4x_3)$.
   ( a ) What is the matrix of $T$ in the standard ordered basis for $R^3$?
   ( b ) What is the matrix of $T$ in the ordered basis $\{\alpha_1,\alpha_2,\alpha_3\}$, where $\alpha_1=(1,0,1),\alpha_2=(-1,2,1),\alpha_3=(2,1,1)$?
   ( c ) Prove that $T$ is invertible and give a rule for $T^{-1}$ like the one which defines $T$.
   Solution:
   ( a ) The matrix of $T$ in the standard ordered basis for $R^3$ is $\begin{bmatrix}3&0&1\\-2&1&0\\-1&2&4\end{bmatrix}$.
   (b) We form $P$ where $P_j=[α_j ]_{\{ϵ_1,ϵ_2,ϵ_3\}}$, it is easy to see $P=\begin{bmatrix}1&-1&2\\0&2&1\\1&1&1\end{bmatrix}$, and we perform
   $\begin{aligned}\begin{bmatrix}1&-1&2&1&0&0\\0&2&1&0&1&0\\1&1&1&0&0&1\end{bmatrix}&→\begin{bmatrix}1&-1&2&1&0&0\\0&2&1&0&1&0\\0&2&-1&-1&0&1\end{bmatrix}\\&→\begin{bmatrix}1&-1&2&1&0&0\\0&2&1&0&1&0\\0&0&-2&-1&-1&1\end{bmatrix}\\&→\begin{bmatrix}1&-1&0&0&-1&1\\0&2&0&-1/2&1/2&1/2\\0&0&1&1/2&1/2&-1/2\end{bmatrix}\\&→\begin{bmatrix}1&0&0&-1/4&-3/4&5/4\\0&1&0&-1/4&1/4&1/4\\0&0&1&1/2&1/2&-1/2\end{bmatrix}\end{aligned}$
   thus
   $P^{-1}=1/4 \begin{bmatrix}-1&-3&5\\-1&1&1\\2&2&-2\end{bmatrix}$
   and then
   $\begin{aligned}[T]_{\{α_1,α_2,α_3\}} =P^{-1} [T]_{\{ϵ_1,ϵ_2,ϵ_3\}} P&=\frac{1}{4} \begin{bmatrix}-1&-3&5\\-1&1&1\\2&2&-2\end{bmatrix}\begin{bmatrix}3&0&1\\-2&1&0\\-1&2&4\end{bmatrix}\begin{bmatrix}1&-1&2\\0&2&1\\1&1&1\end{bmatrix}\\&=\frac{1}{4} \begin{bmatrix}-2&7&19\\-6&3&3\\4&-2&-6\end{bmatrix}\begin{bmatrix}1&-1&2\\0&2&1\\1&1&1\end{bmatrix}\\&=\frac{1}{4}\begin{bmatrix}17&35&22\\-3&15&-6\\-2&-14&0\end{bmatrix}\end{aligned}$
   ( c ) It is enough to prove $[T]_{\{ϵ_1,ϵ_2,ϵ_3\}}$ is invertible, this is true since
   $\begin{bmatrix}3&0&1\\-2&1&0\\-1&2&4\end{bmatrix}^{-1}=\begin{bmatrix}4/9&2/9&-1/9\\8/9&13/9&-2/9\\-1/3&-2/3&1/3\end{bmatrix}$
   since we know $[T^{-1}]_{\{ϵ_1,ϵ_2,ϵ_3\}}=([T]_{\{ϵ_1,ϵ_2,ϵ_3\}})^{-1}$, it’s able to describe
   $T^{-1} (x_1,x_2,x_3 )=\left(\frac{4}{9} x_1+\frac{2}{9} x_2-\frac{1}{9} x_3,\frac{8}{9} x_1+\frac{13}{9} x_2-\frac{2}{9} x_3,-\frac{1}{3} x_1-\frac{2}{3} x_2+\frac{1}{3} x_3\right)$

8. Let $\theta$ be a real number. Prove that the following two matrices are similar over the field of complex numbers:
   $\begin{bmatrix}\cos{\theta}&\sin{\theta}\\\sin{\theta}&\cos{\theta}\end{bmatrix},\quad \begin{bmatrix}e^{i{\theta}}&0\\0&e^{-i\theta}\end{bmatrix}$
   Solution: Let $T$ be the linear operator on $C^2$ which is represented by the first matrix in the standard ordered basis, then $T(1,0)=(\cos⁡θ,\sin⁡θ), T(0,1)=(-\sin⁡θ,\cos⁡θ)$, let $α_1=(i,1),α_2=(1,i)$, then $Tα_1=i(\cos⁡θ,\sin⁡θ )+(-\sin⁡θ,\cos⁡θ )=(i \cos⁡θ-\sin⁡θ,i \sin⁡θ+\cos⁡θ )=e^{iθ} (i,1)=e^{iθ} α_1$
   and similarly we can see $Tα_2=e^{-iθ} α_2$, it is easy to see $\{α_1,α_2 \}$ are linearly independent, thus a basis of $C^2$, since $[T]_{\{α_1,α_2\}} =\begin{bmatrix}e^{iθ}&0\\0&e^{-iθ} \end{bmatrix}$, the two matrices are similar, with $P=\begin{bmatrix}i&1\\1&i\end{bmatrix}$ and
   $[T]_{\{α_1,α_2\}} =P^{-1} \begin{bmatrix}\cos⁡θ&-\sin⁡θ\\\sin⁡θ&\cos⁡θ \end{bmatrix}P$

9. Let $V$ be a finite-dimensional vector space over the field $F$ and let $S$ and $T$ be linear operators on $V$. We ask: When do there exist ordered bases $\mathfrak B$ and $\mathfrak B'$ for $V$ such that $[S]_{\mathfrak B}=[T]_{\mathfrak B'}$? Prove that such bases exist if and only if there is an invertible linear operator $U$ on $V$ such that $T=USU^{-1}$.
   Solution: If $[S]_{\mathfrak B}=[T]_{\mathfrak B'}$, then let $U$ be the linear operator that carries ${\mathfrak B}$ onto ${\mathfrak B'}$, i.e., if we let
   ${\mathfrak B}=\{a_1,\dots,a_n \},\quad {\mathfrak B'}=\{b_1,\dots,b_n \}$
   and define $U$ by $Ua_i=b_i,i=1,\dots,n$. Then $U$ is invertible since it carries a basis onto another basis, and $U^{-1} b_i=a_i,i=1,\dots,n$. If we denote $[S]_{\mathfrak B}=[T]_{\mathfrak B'}=A=(A_{ij})$, then by definition we have $Sa_j=∑_{i=1}^nA_{ij} a_i ,Tb_j=∑_{i=1}^nA_{ij} b_i$, so
   $USU^{-1}(b_j )=USa_j=U\left(∑_{i=1}^nA_{ij} a_i \right)=∑_{i=1}^nA_{ij} Ua_i=∑_{i=1}^nA_{ij}b_i=Tb_j,\quad j=1,\dots,n$
   Since $USU^{-1}$ and $T$ are equal on a basis of $V$, we have $T=USU^{-1}$.
   Conversely, if $T=USU^{-1}$ for some invertible $U$, we let $B=\{a_1,\dots,a_n \}$ be an ordered basis for $V$, and $\mathfrak B'=\{Ua_1,\dots,Ua_n \}$, since $U$ is invertible, $\mathfrak B'$ is a basis for $V$. Notice that if $α∈V$, then
   $α=k_1 a_1+\dots+k_n a_n,\quad k_1,\dots,k_n∈F$
   thus $[α]_{\mathfrak B}=\begin{bmatrix}k_1\\\vdots\\k_n\end{bmatrix}$, and $Uα=k_1 Ua_1+\dots+k_n Ua_n$, so $[Uα]_{\mathfrak B'}=\begin{bmatrix}k_1\\\vdots\\k_n\end{bmatrix}$, so we have $[α]_{\mathfrak B}=[Uα]_{\mathfrak B'}$, from this and the fact that $TU=US$ we can have
   $[S]_{\mathfrak B} [α]_{\mathfrak B}=[Sα]_{\mathfrak B}=[USα]_{\mathfrak B'}=[TUα]_{\mathfrak B'}=[T]_{\mathfrak B'} [Uα]_{\mathfrak B'}=[T]_{\mathfrak B'} [α]_{\mathfrak B}$
   and it must follow that $[S]_{\mathfrak B}=[T]_{\mathfrak B'}$.
   [Alternatively, one easier proof is using Theorem 14 and the consequence of Theorem 13, since in this case we have
   $[T]_{\mathfrak B'}=[U^{-1}]_{\mathfrak B} [T]_{\mathfrak B} [U]_{\mathfrak B}=[U^{-1} TU]_{\mathfrak B}=[U^{-1}(USU^{-1})U]_{\mathfrak B}=[S]_{\mathfrak B}$
   ].

10. We have seen that the linear operator $T$ on $R^2$ defined by $T(x_1,x_2)=(x_1,0)$ is represented in the standard ordered basis by the matrix
    $A=\begin{bmatrix}1&0\\0&0\end{bmatrix}.$
    This operator satisfies $T^2=T$. Prove that if $S$ is a linear operator on $R^2$ such that $S^2=S$, then $S=0$,or $S=I$, or there is an ordered basis $\mathfrak B$ for $R^2$ such that $[S]_{\mathfrak B}=A$ (above).
    Solution: If $S=0$ or $S=I$, we obviously have $S^2=S$, now suppose $S\neq 0,S\neq I$, but $S^2=S$, then it is able to find $α,β∈R^2$, s.t. $Sα\neq 0,Sβ\neq β$, since $Sα∈\text{range }S$, we have $\dim⁡ \text{range }S≥1$, also since $S(Sβ-β)=S^2 β-Sβ=Sβ-Sβ=0$, we know $Sβ-β∈\text{null }S$, and $\dim⁡ \text{null }S≥1$, as we discuss in $R^2$, $\dim \text{range }S+\dim \text{null }S=2$, thus $\dim \text{range }S=\dim \text{null }S=1$. Now if we let $a=Sα,b=Sβ-β$, and ${\mathfrak B}=\{a,b\}$, then ${\mathfrak B}$ is an ordered basis for $R^2$ and $[S]_{\mathfrak B}=A$.

11. Let $W$ be the space of all $n\times 1$ column matrices over a field $F$. If $A4 is an $n\times n$ matrix over $F$, then $A$ defines a linear operator $L_A$ on $W$ through left multiplicaition: $L_A(X)=AX$. Prove that every linear operator on $W$ is left multiplication by some $n\times n$ matrix, i.e., is $L_A$ for some $A$.
    Solution: If $T$ is a linear operator on $W$, let $\mathfrak B'=\{ϵ_1,\dots,ϵ_n \}$ be the standard basis on $W$, for each $X=\begin{bmatrix}x_1\\\vdots\\x_n \end{bmatrix}$, we have $X=∑_{j=1}^nx_j ϵ_j$, and if we define $A:=[T]_{\mathfrak B'}$, then $T(X)=T(∑_{j=1}^nx_j ϵ_j )=∑_{j=1}^nx_j Tϵ_j =∑_{j=1}^nx_j ∑_{i=1}^nA_{ij} ϵ_i =∑_{i=1}^n(∑_{j=1}^nA_{ij} x_j )ϵ_i =AX$, thus $T=L_A$.
    For the second question, let $\mathfrak B=\{a_1,\dots,a_n \}$, if $[α]_{\mathfrak B}=[β]_{\mathfrak B}=\begin{bmatrix}x_1\\\vdots \\x_n\end{bmatrix}$, then $α=β=∑_{j=1}^nx_j a_j$ , this shows $(Uα=Uβ)⇒(α=β)$, so $U$ is injective. Also for any $X=\begin{bmatrix}x_1\\\vdots\\x_n \end{bmatrix}∈W$, define $α=∑_{j=1}^nx_j a_j$, then $Uα=X$, thus $U$ is surjective, combined we show $U$ is an isomorphism.
    If $T$ is a linear operator on $V$, then let $X=\begin{bmatrix}x_1\\\vdots\\x_n\end{bmatrix},Y=\begin{bmatrix}y_1\\\vdots\\y_n \end{bmatrix}∈W$, by definition we have
    $\begin{aligned}UTU^{-1} (cX+Y)&=UT(U^{-1} (cX+Y))\\&=UT\left(∑_{j=1}^n(cx_j+y_j)a_j \right)=U\left(T\Big(∑_{j=1}^n(cx_j+y_j)a_j \Big)\right)\\&=U\left(∑_{j=1}^n(cx_j+y_j)Ta_j \right)=∑_{j=1}^n(cx_j+y_j)UTa_j \\&=c∑_{j=1}^nx_j UTa_j +∑_{j=1}^ny_j UTa_j =c\left(UT∑_{j=1}^nx_j a_j \right)+UT\left(∑_{j=1}^ny_j a_j \right)\\&=c(UT(U^{-1}(X)))+(UT(U^{-1}(Y)))=c(UTU^{-1})(X)+(UTU^{-1})(Y)\end{aligned}$
    thus $UTU^{-1}$ is a linear operator on $W$.
    If $T$ is a linear operator on $V$, then let $C=[T]_{\mathfrak B}$, i.e. $Ta_j=∑_{j=1}^nC_{ij} a_i$, in the first part we proved $A=[UTU^{-1}]_{\mathfrak B'}$, to compute $A$, we see that $UTU^{-1}(ϵ_j )=UTa_j=U(∑_{j=1}^nC_{ij}a_i)=∑_{j=1}^nC_{ij}Ua_i =∑_{j=1}^nC_{ij}ϵ_j$ , thus $[UTU^{-1}]_{\mathfrak B'}=C=[T]_{\mathfrak B}$, or $A=[T]_{\mathfrak B}$.

12. Let $V$ be an $n$-dimensional vector space over the field $F$, and let $\mathfrak B=\{\alpha_1,\dots,\alpha_n\}$ be an ordered basis for $V$.
    ( a ) According th Theorem 1, there is a unique linear operator $T$ on $V$ such that
    $T{\alpha}_j={\alpha}_{j+1},\qquad j=1,\dots,n-1,\qquad T{\alpha}_n=0.$
    What is the matrix $A$ of $T$ in the ordered basis $\mathfrak B$?
    ( b ) Prove that $T^n=0$ but $T^{n-1}\neq 0$.
    ( c ) Let $S$ be any linear operator on $V$ such that $S^n=0$ but $S^{n-1}\neq 0$. Prove that there is an ordered basis $\mathfrak B'$ for $V$ such that the matrix of $S$ in the ordered basis $\mathfrak B'$ is the matrix $A$ of part (a).
    ( d ) Prove that if $M$ and $N$ are $n\times n$ matrices over $F$ such that $M^n=N^n=0$ but $M^{n-1}\neq 0\neq N^{n-1}$, then $M$ and $N$ are similar.
    Solution:
    ( a ) A direct computation shows $A=\begin{bmatrix}0&0&\dots&0\\1&0& & \\ &\ddots&\ddots& \\0& &1&0\end{bmatrix}$.
    ( b ) We have $T^k α_{n+1-k}=0,k=1,\dots,n$, thus $T^n=0$, but $T^{n-1}α_1=α_n\neq 0$.
    ( c ) It is able to choose $α$ s.t. $S^{n-1}α\neq 0$ but $S^n α=0$, notice $α\neq 0$, and $\{α,Sα,\dots,S^{n-1}α\}$ is linearly independent, for if we have
    $k_1 α+k_2 Sα+\dots+k_n S^{n-1} α=0$
    then $S^{n-1}(k_1 α+k_2 Sα+\dots+k_n S^{n-1}α)=k_1 S^{n-1}α=0$, thus $k_1=0$, the above becomes
    $k_2 Sα+\dots+k_n S^{n-1}α=0$
    then $S^{n-2}(k_2 Sα+\dots+k_n S^{n-1}α)=k_2 S^{n-1}α=0$, thus $k_2=0$, continue this step we eventually have $k_1=\dots=k_n=0$. Thus we can define $\mathfrak B'=\{α,Sα,\dots,S^{n-1} α\}$, and $[S]_{\mathfrak B'}=A$.
    ( d ) Let $T$ and $S$ be linear operators which satisfies $[T]_{\mathfrak B}=M,[S]_{\mathfrak B}=N$, in which ${\mathfrak B}=\{ϵ_1,\dots,ϵ_n \}$. From ( c ) we can find two ordered basis ${\mathfrak B}_1$ and ${\mathfrak B}_2$ s.t. $[T]_{\mathfrak B_1}=[S]_{\mathfrak B_2}=A$, by Theorem 14, let $P$ be the $n\times n$ matrix with columns $P_j=[ϵ_j ]_{\mathfrak B_1}$, and $Q$ be the $n\times n$ matrix with columns $Q_j=[ϵ_j ]_{\mathfrak B_2}$, then
    $M=[T]_{\mathfrak B}=P^{-1} AP,\quad N=[S]_{\mathfrak B}=Q^{-1}AQ$
    thus $M=P^{-1} QAQ^{-1}P=(Q^{-1}P)^{-1}A(Q^{-1}P)$.

13. Let $V$ and $W$ be finite-dimensional vector spaces over the field $F$ and let $T$ be a linear transformation from $V$ into $W$. If
    $\mathfrak B=\{\alpha_1,\dots,\alpha_n\}\text{ and }\mathfrak B'=\{\beta_1,\dots,\beta_m\}$
    are ordered base for $V$ and $W$, respectively, define the linear transformations $E{p,q}$ as in the proof of Theorem 5: $E^{p,q}(\alpha_i)=\delta_{iq}\beta_p$. Then the $E{p,q},1\leq p\leq m,1\leq q\leq n$, form a basis for $L(V,W)$, and so
    $T=\sum_{p=1}^m\sum_{q=1}^nA_{pq}E^{p,q}$
    for certain scalars $A_{pq}$ (the coordinates of $T$ in this basis for $L(V,W)$). Show that the matrix $A$ with entries $A(p,q)=A_{pq}$ is precisely the matrix of $T$ relative to the pair $\mathfrak B,\mathfrak B'$.
    Solution: $Tα_j=(∑_{p=1}^m∑_{q=1}^nA_{pq} E^{p,q})(α_j )=∑_{p=1}^m∑_{q=1}^nA_{pq} E^{p,q}(α_j)=∑_{p=1}^m∑_{q=1}^nA_{pq} δ_{jq} β_p=∑_{p=1}^mA_{pj} β_p$, and the conclusion follows.