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2014年武汉大学基础数学复试试题解答
时间：2014年3月22日8:30—10:30
专业：基础数学
一、（10分）已知函数$f(x)$在$(-1,1)$上连续，在除$x=0$上存在导函数
（1）若$\underset{x\to 0}{\mathop{\lim }}\,f(x)$存在，证明存在；
（2）若$\underset{x\to 0}{\mathop{\lim }}\,f(x)$不存在，则$f(0)$一定不存在吗？若不存在，说明理由；若存在，请给出反例并证明。
证明：
（1）由于$f(0)=\underset{x\to 0}{\mathop{\lim }}\,\frac{f(x)-f(0)}{x-0}=\underset{x\to 0}{\mathop{\lim }}\,\frac{f(x)-f(0)}{x}$
而$f(x)$在$(-1,0)\cup (0,1)$上可导
由微分中值定理可知：
存在$\xi $在$0$和$x$之间，使得$f(x)-f(0)=f(\xi )x$
于是\[f(0)=\underset{x\to 0}{\mathop{\lim }}\,f(\xi )\]存在
于是$f(0)$存在
（2）不一定
例如
$f(x)=\left\{\begin{array}{ll} {x^2}\sin \frac{1}{x}, \hbox{$x \ne 0$;} \\ 0, \hbox{$x = 0$.} \end{array} \right.$
显然$f(x)$在$(-1,1)$上连续，且$x\ne 0$时，$f(x)=2x\sin \frac{1}{x}-\cos \frac{1}{x}$
则$\underset{x\to 0}{\mathop{\lim }}\,f(x)$不存在
但$f(x)=\underset{x\to 0}{\mathop{\lim }}\,\frac{f(x)-f(0)}{x-0}=\underset{x\to 0}{\mathop{\lim }}\,x\sin \frac{1}{x}=0$存在
于是不一定不存在
二、（10分）证明：函数$f(x)$在$(a,b)$上一致连续的充要条件是：对任意的$\{ { {x}_{n}}\}\subset (a,b)$，只要$\{ { {x}_{n}}\}$收敛，则$f({ {x}_{n}})$收敛。
证明：
必要性：若函数$f(x)$在$(a,b)$上一致连续
则对任意的$\varepsilon 0$，存在$\delta (\varepsilon )0$，对任意的$x,x\subset (a,b)$，只要$\left| x-x \right|\delta $，则
$\left| f(x)-f(x) \right|\varepsilon $
于是当$\{ { {x}_{n}}\}$收敛时，则对任意的$\varepsilon 0$，存在$\delta (\varepsilon )0$，对任意的$x_{n}^{},x_{n}^{}\subset (a,b)$，有
$\left| x_{n}^{}-x_{n}^{} \right|\delta $
于是$\left| f(x_{n}^{})-f(x_{n}^{}) \right|\varepsilon $
由$Cauchy$收敛准则知，$f({ {x}_{n}})$收敛
充分性：对任意的$\{ { {x}_{n}}\}\subset (a,b)$，只要$\{ { {x}_{n}}\}$一致收敛，则$f({ {x}_{n}})$收敛
即则对任意的$\varepsilon 0$，存在$\delta (\varepsilon )0$，对任意的$x_{n}^{},x_{n}^{}\subset (a,b)$，只要$\left| x_{n}^{}-x_{n}^{} \right|\delta $，于是
$\left| f(x_{n}^{})-f(x_{n}^{}) \right|\varepsilon $
下证$f(x)$在$(a,b)$上一致收敛
反证：若$f(x)$在$(a,b)$上不一致收敛
则存在${ {\varepsilon }_{0}}0$，对任意的$\delta 0$，存在$\left| x-x \right|\delta $时，但有$\left| f(x)-f(x) \right|\ge { {\varepsilon }_{0}}$
取${ {\delta }_{1}}=1$，存在$x_{1}^{},x_{1}^{}\in (a,b),\left| x_{1}^{}-x_{1}^{} \right|1$，但$\left| f(x_{1}^{})-f(x_{1}^{}) \right|\ge { {\varepsilon }_{0}}$
取${ {\delta }_{2}}=\frac{1}{2}$，存在$x_{2}^{},x_{2}^{}\in (a,b),\left| x_{2}^{}-x_{2}^{} \right|\frac{1}{2}$，但$\left| f(x_{2}^{})-f(x_{2}^{}) \right|\ge { {\varepsilon }_{0}}$
$\cdots \cdots $
取${ {\delta }_{n}}=\frac{1}{n}$，存在$x_{n}^{},x_{n}^{}\in (a,b),\left| x_{n}^{}-x_{n}^{} \right|\frac{1}{n}$，但$\left| f(x_{n}^{})-f(x_{n}^{}) \right|\ge { {\varepsilon }_{0}}$
于是存在$\{ { {x}_{n}}\}$收敛，但$\{f({ {x}_{n}})\}$发散，矛盾
于是$f(x)$在$(a,b)$上一致收敛
三、（10分）设$X$是度量空间，$E$为$X$中的紧子集，$\varphi $是$E$上到自身的映射，
$d(\varphi (x),\varphi (y))d(x,y)(x,y\in E,x\ne y)$ ，求证：$\varphi $在$E$中存在唯一不动点
证明：定义$E$上的函数$f(x)=d(\varphi x,x)$
由于$\left| f(x)-f(y) \right|=\left| d(\varphi x,x)-d(\varphi y,y) \right|\le d(\varphi x,\varphi y)+d(x,y)2d(x,y)$
于是$f$是$E$上的连续映射
由于$E$是度量空间$X$的紧子集
则必有${ {x}_{0}}\in E$，使得$f({ {x}_{0}})=\underset{x\in E}{\mathop{\min }}\,f(x)$
先证$f({ {x}_{0}})=0$
反证：若$f({ {x}_{0}})\ne 0\Rightarrow \varphi { {x}_{0}}\ne { {x}_{0}}$
记${ {x}_{1}}=\varphi { {x}_{0}}$，则$\varphi { {x}_{1}}={ {\varphi }^{2}}{ {x}_{0}}$
于是
$f({ {x}_{1}})=d(\varphi { {x}_{1}},{ {x}_{1}})=d({ {\varphi }^{2}}{ {x}_{0}},\varphi { {x}_{0}})d(\varphi { {x}_{0}},{ {x}_{0}})=f({ {x}_{0}})$
这与$f({ {x}_{0}})$是$f$的最小值矛盾
于是$d(\varphi { {x}_{0}},{ {x}_{0}})=0$
即$\varphi { {x}_{0}}={ {x}_{0}}$
唯一性：设${ {x}_{2}}$是$\varphi $的另一个不动点，则$d({ {x}_{0}},{ {x}_{1}})=d(\varphi { {x}_{0}},\varphi { {x}_{1}})d({ {x}_{0}},{ {x}_{1}})$矛盾
于是$\varphi $在$E$中存在唯一不动点
四、（10分）证明积分$\int_{0}^{\text{+}\infty }{\frac{\sin xy}{x(1+y)}}dy$在$0\delta \le x+\infty $上一致收敛，但在$0x+\infty $上不一致收敛
证明：对任意的$\delta 0$，当$x\in [\delta ,+\infty )$时，有$\left| \int_{0}^{A}{\sin xydy} \right|=\left| \frac{1-\cos Ax}{x} \right|\le \frac{2}{\delta }$
而$\frac{1}{x(1+y)}$关于$y$单调递减且一致收敛于$0$
由$Dirichlet$判别法可知：$\int_{0}^{\text{+}\infty }{\frac{\sin xy}{x(1+y)}}dy$在$0\delta \le x+\infty $上一致收敛
另一方面：
取$A_{n}^{}=n(2n\pi +\frac{\pi }{4}),A_{n}^{}=n(2n\pi +\frac{\pi }{2}),{ {x}_{n}}=\frac{1}{n}$，则
$\int_{A_{n}^{}}^{A_{n}^{}}{\frac{\sin { {x}_{n}}y}{ { {x}_{n}}(1+y)}}dy=\int_{n(2n\pi +\frac{\pi }{4})}^{n(2n\pi +\frac{\pi }{2})}{\frac{n\sin \frac{y}{n}}{1+y}}dy$
$\ge \int_{n(2n\pi +\frac{\pi }{4})}^{n(2n\pi +\frac{\pi }{2})}{\frac{\sqrt{2}n}{2(1+y)}}dy$
$=\frac{\sqrt{2}}{2}n\ln [1+\frac{n\pi }{4+n(8n\pi +\pi )}]$
.而$\underset{n\to +\infty }{\mathop{\lim }}\,\frac{\sqrt{2}}{2}n\ln [1+\frac{n\pi }{4+n(8n\pi +\pi )}]=\underset{n\to +\infty }{\mathop{\lim }}\,\frac{\sqrt{2}{ {n}^{2}}\pi }{8+2n(8n\pi +\pi )}=\frac{\sqrt{2}}{16}0$
则$\int_{A_{n}^{}}^{A_{n}^{}}{\frac{\sin { {x}_{n}}y}{ { {x}_{n}}(1+y)}}dy\ge \frac{\sqrt{2}}{16}$
故$\int_{0}^{\text{+}\infty }{\frac{\sin xy}{x(1+y)}}dy$在$(0,+\infty )$上不一致收敛
五、（10分）设$f(x)=\sum\limits_{n=0}^{+\infty }{ { {a}_{n}}{ {x}^{n}}}(-1x1)$，其中
$\underset{n\to +\infty }{\mathop{\lim }}\,n{ {a}_{n}}=0$
$\underset{x\to { {1}^{-}}}{\mathop{\lim }}\,f(x)=s$.
求证：$\sum\limits_{n=0}^{+\infty }{ { {a}_{n}}}$一致收敛且和为$s$
证明：令${ {\delta }_{n}}=\frac{\sum\limits_{k=1}^{n}{\left| k{ {a}_{k}} \right|}}{n}$，下证
$\underset{n\to +\infty }{\mathop{\lim }}\,{ {\delta }_{n}}=0$
由于$\underset{n\to +\infty }{\mathop{\lim }}\,n{ {a}_{n}}=0$
则对任意的$\varepsilon 0$，总可以找到一个正整数$N(\varepsilon )$，使得当时，总有$\left| n{ {a}_{n}} \right|\frac{\varepsilon }{2}$
当$nN(\varepsilon )$就有$\left| n{ {a}_{n}} \right|\frac{\varepsilon }{2}$的最小自然数${ {N}_{1}}$，当$n{ {N}_{1}}$时，有
$\left| \frac{\sum\limits_{k=0}^{n}{k{ {a}_{k}}}}{n} \right|\le \frac{\sum\limits_{k=0}^{n}{\left| k{ {a}_{k}} \right|}}{n}=\frac{\sum\limits_{k=0}^{ { {N}_{1}}}{\left| k{ {a}_{k}} \right|}}{n}+\frac{\sum\limits_{k={ {N}_{1}}+1}^{n}{\left| k{ {a}_{k}} \right|}}{n}\frac{\sum\limits_{k=0}^{ { {N}_{1}}}{\left| k{ {a}_{k}} \right|}}{n}+\frac{(n-{ {N}_{1}})\varepsilon }{2n}\frac{\sum\limits_{k=0}^{ { {N}_{1}}}{\left| k{ {a}_{k}} \right|}}{n}+\frac{\varepsilon }{2}$
考虑到$\underset{n\to +\infty }{\mathop{\lim }}\,\frac{\sum\limits_{k=0}^{ { {N}_{1}}}{\left| k{ {a}_{k}} \right|}}{n}=0$
则对上述$\varepsilon 0$，存在${ {N}_{2}}0$，当$n{ {N}_{2}}$时，有$\frac{\sum\limits_{k=0}^{n}{\left| k{ {a}_{k}} \right|}}{n}\frac{\varepsilon }{2}$
令$N=\max \{ { {N}_{1}},{ {N}_{2}}\}$，当$nN$时，有$\frac{\sum\limits_{k=0}^{n}{\left| k{ {a}_{k}} \right|}}{n}\varepsilon $
于是$\underset{n\to +\infty }{\mathop{\lim }}\,{ {\delta }_{n}}=0$
由题可知：$\underset{n\to +\infty }{\mathop{\lim }}\,f(1-\frac{1}{n})=s$
于是存在$N0$，当$nN$时，有$\left| f(1-\frac{1}{n})-s \right|\frac{\varepsilon }{3},{ {\delta }_{n}}\frac{\varepsilon }{3},n\left| { {a}_{n}} \right|\frac{\varepsilon }{3}$
记${ {S}_{n}}=\sum\limits_{k=0}^{n}{ { {a}_{k}}}$，且${ {S}_{n}}-s=f(x)-s+\sum\limits_{k=0}^{n}{ { {a}_{k}}(1-{ {x}^{k}})-\sum\limits_{k=n+1}^{+\infty }{ { {a}_{k}}{ {x}^{k}}}}$
注意到对每个$k$以及$x\in (0,1)$，有$1-{ {x}^{k}}=(1-x)(1+x+\cdots +{ {x}^{k-1}})\le k(1-x)$
因此，当$nN$且$x\in (0,1)$时，有
$\left| { {S}_{n}}-s \right|\le \left| f(x)-s \right|+(1-x)\sum\limits_{k=0}^{n}{k\left| { {a}_{n}} \right|}+\frac{\varepsilon }{3n(1-x)}$
令$x=1-\frac{1}{n}$，则$\left| { {S}_{n}}-s \right|\frac{\varepsilon }{3}+\frac{\varepsilon }{3}+\frac{\varepsilon }{3}=\varepsilon $
即$\sum\limits_{n=0}^{+\infty }{ { {a}_{n}}}$收敛且和为$s$
六、（15分）讨论正数$\alpha 0$的初值问题$\left\{\begin{array}{ll} y = {\left| y \right|^\alpha } \\ y(0) = 0 \end{array} \right.$的解的稳定性
解：由于$\left| f(x,{ {y}_{1}})-f(x,{ {y}_{2}}) \right|=\left| { {\left| { {y}_{1}} \right|}^{\alpha }}-{ {\left| { {y}_{2}} \right|}^{\alpha }} \right|\le \left| y_{1}^{\alpha }-y_{2}^{\alpha } \right|\le { {C}_{\alpha }}{ {\left| { {y}_{1}}-{ {y}_{2}} \right|}^{\alpha }}$
令$F(r)={ {\left| r \right|}^{\alpha }}$，有
$\int_{0}^{ { {r}_{1}}}{\frac{dr}{F(r)}=}\int_{0}^{ { {r}_{1}}}{\frac{dr}{ { {\left| r \right|}^{\alpha }}}}$ $=\left\{\begin{array}{ll} \frac{1}{ {1 - \alpha }}{\left| r \right|^{1 - \alpha }}|_0^{ {r_1}}, \hbox{$\alpha \ne 1$;} \\ \ln \left| r \right||_0^{ {r_1}}, \hbox{$\alpha=1$;} \end{array} \right.$
因此当$\alpha \ge 1$时，$\int_{0}^{ { {r}_{1}}}{\frac{dr}{F(r)}=\infty }$
当$0\alpha 1$时，$\int_{0}^{ { {r}_{1}}}{\frac{dr}{F(r)}\infty }$
于是$\alpha \ge 1$时解唯一；当$0\alpha 1$时解不唯一
七、（15分）已知$A$是$n\times n$阶的非奇异复矩阵，求证：$A=UT$（其中$U$为酉矩阵，而$T$是对角线元素均大于$0$的对角矩阵）
证明：设$A={ {({ {a}_{ij}})}_{n\times n}},\left| A \right|\ne 0$，令$A=({ {\alpha }_{1}},{ {\alpha }_{2}},\cdots ,{ {\alpha }_{n}})$
其中${ {\alpha }_{i}}$为$A$的列向量，则${ {\alpha }_{1}},{ {\alpha }_{2}},\cdots ,{ {\alpha }_{n}}$线性无关
由施密特正交化，令
$\left\{\begin{array}{ll} {\beta _1} = {\alpha _1}, \\ {\beta _2} = {\alpha _2} - \frac{ {({\alpha _2},{\beta _1})}}{ {({\beta _1},{\beta _1})}}{\beta _1}, \\ \vdots \\ {\beta _n} = {\alpha _n} - \frac{ {({\alpha _n},{\beta _1})}}{ {({\beta _1},{\beta _1})}}{\beta _1} - \cdots - \frac{ {({\alpha _n},{\beta _{n - 1}})}}{ {({\beta _{n - 1}},{\beta _{n - 1}})}}{\beta _{n - 1}} \end{array} \right.$
其中$(\alpha ,\beta )=\alpha \beta $，则
$\left\{\begin{array}{ll} {\alpha _1} = {\beta _1}, \\ {\alpha _2} = \frac{ {({\alpha _2},{\beta _1})}}{ {({\beta _1},{\beta _1})}}{\beta _1} + {\beta _2}, \\ \vdots \\ {\alpha _n} = \frac{ {({\alpha _n},{\beta _1})}}{ {({\beta _1},{\beta _1})}}{\beta _1} - \cdots - \frac{ {({\alpha _n},{\beta _{n - 1}})}}{ {({\beta _{n - 1}},{\beta _{n - 1}})}}{\beta _{n - 1}} + {\beta _n} \end{array} \right.$
于是
$A=({ {\alpha }_{1}},{ {\alpha }_{2}},\cdots ,{ {\alpha }_{n}})$$ =({\beta _1},{\beta _2}, \cdots ,{\beta _n})\left(\begin{array}{cccc} 1 { {b_{12}}} \cdots { {b_{1n}}} \\ 0 1 \cdots { {b_{1n}}} \\ \vdots \vdots \ddots \vdots \\ 0 0 \cdots 1 \end{array}\right) $
其中${b_{ij}}=\left\{\begin{array}{ll} \frac{ {({\alpha _j},{\beta _i})}}{ {({\beta _i},{\beta _i})}}, \hbox{$i \ne j$;} \\ 1, \hbox{$i = j$.} \end{array} \right.$
将${ {\beta }_{i}}$单位化，即${ {\gamma }_{i}}=\frac{ { {\beta }_{i}}}{\left| { {\beta }_{i}} \right|},\left| { {\beta }_{i}} \right|=\sqrt{\beta _{i}^{}{ {\beta }_{i}}}0,i=1,2,\cdots ,n$
令$U=({ {\gamma }_{1}},{ {\gamma }_{2}},\cdots ,{ {\gamma }_{n}})$，则$U$为酉矩阵，且
$A=({\gamma _1},{\gamma _2},\cdots ,{\gamma _n})$$ \left(\begin{array}{cccc} {\left| { {\beta _1}} \right|} \\ {\left| { {\beta _2}} \right|} \\ \ddots \\ {\left| { {\beta _n}} \right|} \end{array}\right) $$ \left(\begin{array}{cccc} 1 { {b_{12}}} \cdots { {b_{1n}}} \\ 0 1 \cdots { {b_{1n}}} \\ \vdots \vdots \ddots \vdots \\ 0 0 \cdots 1 \end{array}\right) $=UT
其中$T=$$ \left(\begin{array}{cccc} {\left| { {\beta _1}} \right|} \\ {\left| { {\beta _2}} \right|} \\ \ddots \\ {\left| { {\beta _n}} \right|} \end{array}\right) $$ \left(\begin{array}{cccc} 1 { {b_{12}}} \cdots { {b_{1n}}} \\ 0 1 \cdots { {b_{1n}}} \\ \vdots \vdots \ddots \vdots \\ 0 0 \cdots 1 \end{array}\right) $$ =\left(\begin{array}{cccc} {\left| { {\beta _1}} \right|} \\ {\left| { {\beta _2}} \right|} \\ \ddots \\ {\left| { {\beta _n}} \right|} \end{array}\right) $是主对角元为正的上三角阵
八、（20分）设$V={ {M}_{2}}(C)$是二阶复方阵全体在通常运算下构成的复数域$C$上的线性空间$ A=\left(\begin{array}{cccc} a b \\ c d \end{array}\right) $$ \in V$定义$V$上的线性变换为 $f(X)=XA,\forall X\in V$
（1）求$f$在$V$的基
\[{E_{11}} = \left( {\begin{array}{*{20}{c}} 10\\ 00 \end{array}} \right),{E_{12}} = \left( {\begin{array}{*{20}{c}} 01\\ 00 \end{array}} \right),{E_{21}} = \left( {\begin{array}{*{20}{c}} 00\\ 10 \end{array}} \right),{E_{22}} = \left( {\begin{array}{*{20}{c}} 00\\ 01 \end{array}} \right)\]
下的矩阵$M;$
（2）请给出$V$关于$f$的两个非零不变子空间${ {V}_{1}},{ {V}_{2}}$，使得$V={ {V}_{1}}\oplus { {V}_{2}}$（要求给出${ {V}_{1}},{ {V}_{2}}$的基，并阐述“${ {V}_{1}},{ {V}_{2}}$关于$f$是不变的”以及“$V={ {V}_{1}}\oplus { {V}_{2}}$”的理由）；
（3）证明：存在$V$的一个基使得$f$在这个基下的矩阵为对角阵当且仅当$A$与对角阵相似.
解：
（1）由定义知：
$f({E_{11}}) $$ = \left( { {E_{11}},{E_{12}},{E_{21}},{E_{22}}} \right)\left(\begin{array}{cccc} a \\ b \\ 0 \\ 0 \end{array}\right) $,$f({E_{12}}) $$ = \left( { {E_{11}},{E_{12}},{E_{21}},{E_{22}}} \right)\left(\begin{array}{cccc} c \\ d \\ 0 \\ 0 \end{array}\right) $
$f({E_{21}}) $$ = \left( { {E_{11}},{E_{12}},{E_{21}},{E_{22}}} \right)\left(\begin{array}{cccc} 0 \\ 0 \\ a \\ b \end{array}\right) $,$f({E_{22}}) $$ = \left( { {E_{11}},{E_{12}},{E_{21}},{E_{22}}} \right)\left(\begin{array}{cccc} 0 \\ 0 \\ c \\ d \end{array}\right) $
于是$f({E_{11}},{E_{12}},{E_{21}},{E_{22}})$$ = ({E_{11}},{E_{12}},{E_{21}},{E_{22}})\left(\begin{array}{cccc} a c 0 0 \\ b d 0 0 \\ 0 0 a c \\ 0 0 b d \end{array}\right) $
即对应的矩阵为$\left(\begin{array}{cccc} a c 0 0 \\ b d 0 0 \\ 0 0 a c \\ 0 0 b d \end{array}\right) $
（2）令${ {V}_{1}}=L({ {E}_{11}},{ {E}_{12}}),{ {V}_{2}}=L({ {E}_{21}},{ {E}_{22}})$，则${ {V}_{1}},{ {V}_{2}}$是关于$f$的不变子空间，理由如下
对任意的${k_{11}}{E_{11}} + {k_{12}}{E_{12}}$$ =\left(\begin{array}{cccc} { {k_{11}}} { {k_{12}}} \\ 0 0 \end{array}\right) $$ \in V$，有
$f({k_{11}}{E_{11}} + {k_{12}}{E_{12}})$$ =\left(\begin{array}{cccc} { {k_{11}}} { {k_{12}}} \\ 0 0 \end{array}\right) $$\left(\begin{array}{cccc} a c \\ b d \end{array}\right) $$=\left(\begin{array}{cccc} {a{k_{11}} + c{k_{12}}} {b{k_{11}} + d{k_{12}}} \\ 0 0 \end{array}\right) $$ \in V$
则${ {V}_{1}}$是关于$f$的不变子空间
同理可证${ {V}_{2}}$是关于$f$的不变子空间
且$\dim{ {V}_{1}}=\dim{ {V}_{2}}=1,{ {V}_{1}}\cap { {V}_{2}}=\{0\},\dim{ {V}_{1}}+\dim{ {V}_{2}}=2$
于是${ {V}_{1}}\oplus { {V}_{2}}=V$
（3）
充分性：设$A={ {P}^{-1}}diag({ {\lambda }_{1}},{ {\lambda }_{2}})P$，其中${ {\lambda }_{1}},{ {\lambda }_{2}}$是$A$的全部特征值
则$f({ {E}_{11}}P)={ {\lambda }_{1}}{ {E}_{11}}P,f({ {E}_{12}}P)={ {\lambda }_{2}}{ {E}_{12}}P,f({ {E}_{21}}P)={ {\lambda }_{1}}{ {E}_{21}}P,f({ {E}_{22}}P)={ {\lambda }_{2}}{ {E}_{22}}P$
于是$f$在基${ {E}_{11}}P,{ {E}_{12}}P,{ {E}_{21}}P,{ {E}_{22}}P$下的矩阵为$ \left(\begin{array}{cccc} { {\lambda _1}} \\ { {\lambda _2}} \\ { {\lambda _3}} \\ { {\lambda _4}} \end{array}\right) $
必要性：若$A$不与对角阵相似
由于$A$是复矩阵，故存在可逆矩阵$Q$，使得$A=$$ {Q^{ - 1}}\left(\begin{array}{cccc} l k \\ 0 l \end{array}\right) $$Q$,$k \ne 0$
于是
$f({ {E}_{11}}Q)=l{ {E}_{11}}Q+k{ {E}_{12}}Q,f({ {E}_{12}}Q)=l{ {E}_{12}}Q,f({ {E}_{21}}Q)=l{ {E}_{21}}Q+k{ {E}_{22}}Q,f({ {E}_{22}}Q)=l{ {E}_{22}}Q$
即$f$在基${ {E}_{11}}Q,{ {E}_{12}}Q,{ {E}_{21}}Q,{ {E}_{22}}Q$下的矩阵为$ \left(\begin{array}{cccc} l \\ k l \\ l \\ k l \end{array}\right) $
而$B$的初等因子为${ {(\lambda -l)}^{2}},{ {(\lambda -l)}^{2}}$，故$B$不与对角阵相似，矛盾