3.3 Isomorphism

这一节把isomorphism单独阐述，即既是一对一又是onto的，也就是一个bijection。其本质可以用文中的一句话概括：isomorphism is an equivalence relation on the class of vector spaces。Theorem 10说明了所有n维vector space都和 $F^n$是isomorphic的，实际上，isomorphism是“dimension preserving”。

Exercises

1. Let $V$ be the set of complex numbers and let $F$ be the field of real numbers. With the usual operations, $V$ is a vector space over $F$. Describe explicitly an isomorphism of this space onto $R^2$.
   Solution: For any $c=a+bi∈V$, define $U(c)=(a,b)$, then $U$ is an isomorphism of $V$ onto $R^2$.

2. Let $V$ be a vector space over the field of complex numbers, and suppose there is an isomorphism $T$ of $V$ onto $C^3$. Let $\alpha_1,\alpha_2,\alpha_3,\alpha_4$ be vectors in $V$ such that
   $T\alpha_1=(1,0,i),\qquad T\alpha_2=(-2,1+i,0),\\T\alpha_3=(-1,1,1),\qquad T\alpha_4=(\sqrt2,i,3).$
   ( a ) Is $\alpha_1$ in the subspace spanned by $\alpha_2$ and $\alpha_3$?
   ( b ) Let $W_1$ be the subspace spanned by $\alpha_1$ and $\alpha_2$, and let $W_2$ be the subspace spanned by $\alpha_3$ and $\alpha_4$. What is the intersection of $W_1$ and $W_2$?
   ( c ) Find a basis for the subspace of $V$ spanned by the four vectors $\alpha_j$.
   Solution:
   ( a ) It is easy to see $Tα_2,Tα_3$ are linearly independent, thus $α_2,α_3$ are linearly independent, since
   $\begin{bmatrix}1&0&i\\-2&1+i&0\\-1&1&1\end{bmatrix}→\begin{bmatrix}1&0&i\\0&1+i&2i\\0&1&1+i\end{bmatrix}→\begin{bmatrix}1&0&i\\0&1&1+i\\0&0&0\end{bmatrix}$
   $Tα_1,Tα_2,Tα_3$ are linearly dependent, thus $α_1$ is in the subspace spanned by $α_2,α_3$.
   ( b ) From ( a ) we have $(1+i)(Tα_2+2Tα_1 )=Tα_3+Tα_1$, thus we have
   $(1+i)(α_2+2α_1 )=α_3+α_1$
   Since $Tα_4$ is not in the span of $Tα_1,Tα_2$, we have $W_1\cap W_2=\{kα_3:k∈C\}$.
   ( c ) One basis can be $(α_1,α_2,α_4)$.

3. Let $W$ be the set of all $2\times 2$ complex Hermitian matrices, that is, the set of $2\times 2$ complex matrices $A$ such that $A_{ij}=\overline{A_{ji}}$ (the bar denoting complex conjugation). As we pointed out in Example 6 of Chapter 2, $W$ is a vector space over the field of real numbers, under the usual operations. Verify that
   $(x,y,z,t)\to\begin{bmatrix}t+x&y+iz\\y-iz&t-x \end{bmatrix}$
   is an isomorphism of $R^4$ onto $W$.
   Solution: Denote $T(x,y,z,t)=\begin{bmatrix}t+x&y+iz\\y-iz&t-x\end{bmatrix}$, if $U(x,y,z,t)=0=\begin{bmatrix}0&0\\0&0\end{bmatrix}$, then it is easy to see $t+x=t-x=0$, thus $t=x=0$, and $y+iz=0$ means $z=0$ and $y=0$. So $T$ is one-one. Next let any $A∈W$, then $A=\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix}$, since $\overline{a_{11}}=a_{11}$ we know $a_{11}∈R$, also $a_{22}∈R$, then we let
   $t=\frac{a_{11}+a_{22}}{2},\quad x=\frac{a_{11}-a_{22}}{2},\quad y=\Re(a_{12}),\quad z=\Im(a_{12})$
   it is easy to see $T(x,y,z,t)=A$.

4. Show that $F^{m\times n}$ is isomorphic to $F^{mn}$.
   Solution: For any $A=\begin{bmatrix}a_{11}&\cdots&a_{1n}\\\vdots&\ddots&\vdots\\a_{m1}&\cdots&a_{mn}\end{bmatrix}∈F^{m×n}$, we define $T(A)=(a_{11},\dots,a_{1n},\dots,a_{m1},\dots,a_{mn})$, i.e., $T(A)$ is a sequence of ordered list of the row vectors of $A$. $T$ is an isomorphism from $F^{m×n}$ to $F^{mn}$.

5. Let $V$ be the set of complex numbers regarded as a vector space over the field of real numbers (Exercise 1). We define a function $T$ from $V$ into the space of $2\times 2$ real matrices, as follows. If $z=x+iy$ with $x$ and $y$ real numbers, then
   $T(z)=\begin{bmatrix}x+7y&5y\\-10y&x-7y \end{bmatrix}.$
   ( a ) Verify that $T$ is a one-one (real) linear transformation of $V$ into the space of $2\times 2$ real matrices.
   ( b ) Verify that $T(z_1z_2)=T(z_1)T(z_2)$.
   ( c ) How would you describe the range of $T$?
   Solution:
   ( a ) It is enough to show $T(z)=0$ means $z=0$, it is easy to see
   $\begin{bmatrix}x+7y&5y\\-10y&x-7y\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix} ⇒x=0,y=0$
   ( b ) Let $z_1=a+bi,z_2=c+di$, then $z_1 z_2=(ac-bd)+(ad+bc)i$, so
   $T(z_1 z_2 )=\begin{bmatrix}ac-bd+7(ad+bc)&5(ad+bc)\\-10(ad+bc)&ac-bd-7(ad+bc)\end{bmatrix}$
   $\begin{aligned}T(z_1)T(z_2)&=\begin{bmatrix}a+7b&5b\\-10b&a-7b\end{bmatrix}\begin{bmatrix}c+7d&5d\\-10d&c-7d\end{bmatrix} \\&=\begin{bmatrix}(a+7b)(c+7d)-50bd&5d(a+7b)+5b(c-7d)\\-10b(c+7d)-10d(a-7b)&(a-7b)(c-7d)-50bd\end{bmatrix} \\&=T(z_1 z_2)\end{aligned}$
   ( c ) The range of $T$ is the subspace of $2×2$ matrices $A$ such that $A_{21}=-2A_{12}$.

6. Let $V$ and $W$ be finite-dimensional vector spaces over the field $F$. Prove that $V$ and $W$ are isomorphic if and only if $\dim V=\dim W$.
   Solution: If $V$ and $W$ are isomorphic, then let $T$ be an isomorphism from $V$ to $W$, and $\{a_1,\dots,a_n \}$ be a basis of $V$, so $\dim ⁡V=n$, consider $\{Ta_1,\dots,Ta_n \}$, for any $w∈W,w=Tv,v∈V$, then we can find scalars $x_1,\dots,x_n$ such that $v=\sum_{i=1}^nx_i a_i$, so $w=T(\sum_{i=1}^nx_i a_i)=\sum_{i=1}^nx_i Ta_i$, so $\{Ta_1,\dots,Ta_n \}$ spans $W$. If $\sum_{i=1}^nx_i Ta_i=0$, then $T(\sum_{i=1}^nx_i a_i)=0$ since $T$ is non-singular, then
   $x_i=0,\quad i=1,\dots,n$
   which means $\{Ta_1,\dots,Ta_n \}$ is linearly independent, we can conclude $\{Ta_1,\dots,Ta_n \}$ is a basis of $W$, and then $\dim ⁡W=n=\dim ⁡V$.
   Conversely, if $\dim ⁡V=\dim ⁡W:=n$, then $V$ and $W$ are isomorphic to $F^n$, by Theorem 10, let $T:V→F^n$ and $U:W→F^n$ be two isomorphism, then $U^{-1} T:V→W$ is an isomorphism.

7. Let $V$ and $W$ be vector spaces over the field $F$ and let $U$ be an isomorphism of $V$ onto $W$. Prove that $T\to UTU^{-1}$ is an isomorphism of $L(V,V)$ onto $L(W,W)$.
   Solution: Denote $M(T)=UTU^{-1}$, if $M(T)=0$, then $UTU^{-1}(w)=0$ for all $w∈W$, since $U$ is non-singular, we have $TU^{-1}(w)=0$ for all $w∈W$, given any $v∈V$, we can have some $w∈W$ s.t. $U^{-1}(w)=v$, so $Tv=0$ for all $v∈V$, thus $T$ is the zero transformation.
   For any $M'∈L(W,W)$, if we choose a basis $w_1,…,w_n$ of $W$, then $M'$ can be written as
   $M'(w_i)=a_{i1} w_1+⋯+a_{in} w_n,\quad i=1,…,n$
   Notice that $U^{-1} (w_1),…,U^{-1}(w_n)$ is a basis for $V$, so if we define
   $T(U^{-1}(w_i))=a_{i1} U^{-1}(w_1)+\cdots+a_{in}U^{-1}(w_n),\quad i=1,…,n$
   then $T∈L(V,V)$, and we have
   $\begin{aligned}UTU^{-1}(w_i )&=U(a_{i1}U^{-1}(w_1)+\cdots+a_{in}U^{-1}(w_n))\\&=a_{i1} w_1+\cdots+a_{in}w_n,\quad i=1,…,n\end{aligned}$
   this means $M(T)=UTU^{-1}=M'$, thus $M(T)$ is onto and the proof is complete.