3.2 The Algebra of Linear Transformations

这一节内容非常丰富。Theorem 4提出了linear transformation的线性组合是什么，以及所有的linear transformation是一个vector space。Theorem 5是说明 $L(V,W)$和 $V,W$的dimension之间的关系。Theorem 6定义了linear transformation的composition。在同一个space上的linear transformation称为operator，有翻译为算子的。EXAMPLE8-10都是很好的例子，其揭示了linear operator可以是多样的。之后定义invertible的概念，注意invertible是在两个不同的space中可以成为单位算子，Theorem 7说明inverse也是linear transformation。然后是non-singular的概念，很类似于函数中的injective，Theorem 8说明non-singular的线性变换可以保持线性无关性。EXAMPLE11说明non-singular和onto之间没有必然的关系，EXAMPLE12则是一个同时non-singular和onto（从而invertible）的例子。Theorem 9说明，如果 $\dim V=\dim W$，那么从 $V$到 $W$的线性变换 $T$满足non-singular、onto和invertible三者成立其一则另外两者也成立。最后给出了group和commutative group的定义。

Exercises

1. Let $T$ and $U$ be the linear operators on $R^2$ defined by
   $T(x_1,x_2)=(x_2,x_1)\quad\text{and}\quad U(x_1,x_2)=(x_1,0)$
   ( a ) How would you describe $T$ and $U$ geometrically?
   ( b ) Give rules like the ones definning $T$ and $U$ for each of the transformations $(U+T),UT,TU,T^2,U^2$.
   Solution:
   ( a ) $T$ is a symmetric transformation with respect to the line $y=x$, $U$ is the projection on $x$-axis.
   ( b ) We have
   $\begin{aligned}(U+T)(x_1,x_2 )&=(x_1+x_2,x_1 ) \\ UT(x_1,x_2 )&=(x_2,0) \\ TU(x_1,x_2 )&=(0,x_1 ) \\ T^2 (x_1,x_2 )&=(x_1,x_2 ) \\ U^2 (x_1,x_2 )&=(x_1,0)\end{aligned}$

2. Let $T$ be the (unique) linear operator on $C^3$ for which
   $T\epsilon_1 =(1,0,i),\quad T\epsilon_2 =(0,1,1),\quad T\epsilon_3 =(i,1,0).$
   Is $T$ invertible?
   Solution: No, since $T(ϵ_3-iϵ_1 )=(i,1,0)-(i,0,-1)=(0,1,1)=Tϵ_2$, thus
   $T(ϵ_3-iϵ_1-ϵ_2 )=0$
   but $ϵ_3-iϵ_1-ϵ_2=(-i,-1,1)≠0$.

3. Let $T$ be the linear operator on $R^3$ defined by
   $T(x_1,x_2,x_3)=(3x_1,x_1-x_2,2x_1+x_2+x_3)$
   Is $T$ invertible? If so, find a rule for $T^{-1}$ like the one which defines $T$.
   Solution: $T$ is invertible, the rule for $T^{-1}$ can be described as
   $T^{-1} (x_1,x_2,x_3 )=\left(\frac{1}{3} x_1,\frac{1}{3} x_1-x_2,-x_1+x_2+x_3 \right)$

4. For the liear operator $T$ of Exercise 3, prove that
   $(T^2-I)(T-3I)=0.$
   Solution: Since
   $\begin{aligned}(T-3I)(x_1,x_2,x_3 )&=T(x_1,x_2,x_3 )-3(x_1,x_2,x_3 )\\&=(0,x_1-4x_2,2x_1+x_2-2x_3 )\end{aligned}$
   and
   $\begin{aligned}&{\quad}(T^2-I)(x_1,x_2,x_3 )\\&=T^2 (x_1,x_2,x_3 )-(x_1,x_2,x_3 )\\&=T(3x_1,x_1-x_2,2x_1+x_2+x_3 )-(x_1,x_2,x_3 )\\&=(9x_1,3x_1-(x_1-x_2 ),6x_1+(x_1-x_2 )+(2x_1+x_2+x_3 ))-(x_1,x_2,x_3 )\\&=(9x_1,2x_1+x_2 ,9x_1+x_3)-(x_1,x_2,x_3 )\\&=(8x_1,2x_1,9x_1)\end{aligned}$
   So
   $(T^2-I)(T-3I)(x_1,x_2,x_3 )=(T^2-I)(0,x_1-4x_2,2x_1+x_2-2x_3 )=(0,0,0)$

5. Let $C^{2\times 2}$ be the complex vector space of $2\times 2$ matrices with complex entries. Let
   $B=\begin{bmatrix}1&-1\\-4&4\end{bmatrix}$
   and let $T$ be the linear operator on $C^{2\times 2}$ defined by $T(A)=BA$. What is the rank of $T$? Can you describe $T^2$?
   Solution: If $A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$, then $T(A)=BA=\begin{bmatrix}1&-1\\-4&4\end{bmatrix}\begin{bmatrix}a&b\\c&d\end{bmatrix}=\begin{bmatrix}a-c&b-d\\4c-4a&4d-4b\end{bmatrix}$
   the rank of $T$ is 2. Also we have $\begin{aligned}T^2(A)=T(T(A))&=\begin{bmatrix}1&-1\\-4&4\end{bmatrix}\begin{bmatrix}a-c&b-d\\4c-4a&4d-4b\end{bmatrix}\\&=\begin{bmatrix}5a-5c&5b-5d\\20c-20a&20d-20b\end{bmatrix}\end{aligned}$.

6. Let $T$ be a linear transformation from $R^3$ into $R^2$, and let $U$ be a linear transformation from $R^2$ into $R^3$. Prove that the transformation $UT$ is not invertible. Generalize the theorem.
   Solution: Let $a=Tϵ_1,b=Tϵ_2,c=Tϵ_3$, then $a,b,c∈R^2$, thus are linearly dependent, without loss of generality, we suppose $c=k_1 a+k_2 b,k_1,k_2∈R$, then $k_1 a+k_2 b-c=0$, or
   $k_1 Tϵ_1+k_2 Tϵ_2-Tϵ_3=T(k_1 ϵ_1+k_2 ϵ_2-ϵ_3 )=0$
   Notice $k_1 ϵ_1+k_2 ϵ_2-ϵ_3≠0$, we can have
   $UT(k_1 ϵ_1+k_2 ϵ_2-ϵ_3 )=U(0)=0$
   thus $UT$ is not invertible.

7. Find two linear operators $T$ and $U$ on $R^2$ such that $TU=0$ but $UT\neq 0$.
   Solution:
   Let $U(x_1,x_2 )=(x_1,0),T(x_1,x_2 )=(x_2,0)$, then
   $TU(x_1,x_2 )=T(x_1,0)=(0,0),∀x_1,x_2∈R\\UT(0,1)=U(1,0)=(1,0)$

8. Let $V$ be a vector space over the field $F$ and $T$ a linear operator on $V$. If $T^2=0$, what can you say about the relation of the range of $T$ to the null space of $T$? Give an example of a linear operator $T$ on $R^2$ such that $T^2=0$ but $T\neq 0$.
   Solution: If $T^2=0$, then for any $α∈\text{range }T$, we have $α=Tβ,β∈V$, thus $Tα=T^2 β=0$, so $α$ also belongs to the null space of $T$, it follows that $\text{range }T$ is a subspace of the null space of $T$.
   $T(x_1,x_2 )=(x_2,0)$ is an example of $T^2=0$, while $T≠0$.

9. Let $T$ be a linear operator on the finite-dimensional space $V$. Suppose there is a linear operator $U$ on $V$ such that $TU=I$. Prove that $T$ is invertible and $U=T^{-1}$. Give an example which shows that this is false when $V$ is not finite dimensional.
   Solution: For any $α∈V$, we have $α=(TU)(α)=T(Uα)$, this means $α∈\text{range }T$, so the range of $T$ is $V$, by Theorem 9, $T$ is invertible. To see $U=T^{-1}$, notice that $T^{-1} T=I$ and $TU=I$, thus
   $U=IU=(T^{-1} T)U=T^{-1} (TU)=T^{-1} I=T^{-1}$
   If $T=D$ on the space of polynomial functions, then let $U$ be the integrable operator on the space of polynomial functions, we have $DU=I$, but $D$ is not invertible.

10. Let $A$ be an $m\times n$ with entries in $F$ and let $T$ be the linear transformation from $F^{n\times 1}$ into $F^{m\times 1}$ defined by $T(X)=AX$. Show that if $m<n$ it may happen that $T$ is onto without being non-singular. Similarly, show that if $m>n$ we may have $T$ non-singular but not onto.
    Solution: If $m<n$, then the system $AX=b$ can have a solution if rank $A=m$, so $T$ can be onto, but $AX=0$ must have an nonzero solution, since the solution space has dimension $n-m≥1$.
    If $m>n$, then let $AX=x_1 A_1+\dots+x_n A_n$, each $A_i$ a $m×1$ vector, as long as those $A_i$ are linearly independent, $AX=0$ will mean $x_i=0$ and $X=0$, so $T$ is non-singular, but $\dim F^{m×1}=m$ and $A_1,\dots,A_n$ cannot span $F^{m×1}$, thus $T$ is not onto.

11. Let $V$ be a finite-dimensional vector space and let $T$ be a linear operator on $V$. Suppose that $\text{rank }(T^2)=\text{rank }(T)$. Prove that the range and null space of $T$ are disjoint, i.e., have only the zero vector in common.
    Solution: Assume $α≠0,α∈\text{range }T∩\text{null }T$, then $Tα=0$, and $α$ is a linear independent set in range $T$, thus can be expanded to a basis of range $T$, namely $(α,β_1,\dots,β_n )$, so $\text{rank }(T)=n+1$.
    Notice that $\text{range }T^2$ is the set of vectors $T^2 β,β∈V$, and $T^2 β=T(Tβ)$, as $Tβ∈\text{range }T$, we have $Tβ=yα+∑_{i=1}^ny_i β_i$, in which $y,y_1,\cdots,y_n$ are scalars. So $T^2 β=T(yα+∑_{i=1}^ny_i β_i)=∑_{i=1}^ny_i Tβ_i$, which means $Tβ_1,…,Tβ_n$ spans range $T^2$, so $\text{rank }(T^2 )≤n$, a contradiction.

12. Let $p,m$ and $n$ be positive integers and $F$ a field. Let $V$ be the space of $m\times n$ matrices over $F$ and $W$ the space of $p\times n$ matrices over $F$. Let $B$ be a fixed $p\times m$ matrix and let $T$ be the linear transformation from $V$ into $W$ defined by $T(A)=BA$. Prove that $T$ is invertible if and only if $p=m$ and $B$ is an invertible $m\times m$ matrix.
    Solution: If $p=m$ and $B$ is an invertible $m×m$ matrix, then define $U(A)=B^{-1} A$, we have $TU(A)=T(B^{-1} A)=B(B^{-1}A)=(BB^{-1})A=A$ and $UT(A)=U(BA)=B^{-1} (BA)=(BB^{-1})A=A$. Thus $T$ is invertible and $T^{-1} (A)=B^{-1} A$.
    Conversely, if $T$ is invertible, then $T$ is non-singular and $\text{range }T=W$, use Exercise 3.2.10 we know if $p>m$, then $T$ cannot be onto, if $p<m$, then $T$ cannot be non-singular. Thus $p=m$ and $B$ is a $m×m$ matrix, if B is not invertible then the column vectors of $B$ are linearly dependent, thus there is $x_1,\dots,x_m$ not all 0 such that $x_1 b_1+\dots+x_m b_m=0$, then let $A$ be a matrix with $x_1,\dots,x_m$ be in the first column and 0 otherwise, we have $T(A)=0$, this contradicts $T$ being non-singular. Thus $B$ must be invertible.