武汉大学2014年高代试题

陈洪葛 posted @ Mar 02, 2014 06:02:02 PM in 高等代数 , 885 阅读

1.设$A = \left( {\begin{array}{*{20}{c}}
1&2&0&0\\
1&3&0&0\\
0&0&0&2\\
0&0&{ - 1}&0
\end{array}} \right)$且$\left[\left(\frac{1}{2}A\right)^{*}\right]^{-1}BA=6AB+12E$,求$B$.


解:我们有$A^{*}A=AA^{*}=|A|E$,而
\[A = \left( {\begin{array}{*{20}{c}}
X&{}\\
{}&Y
\end{array}} \right)\]
其中
\[ X = \left( {\begin{array}{*{20}{c}}
1&2\\
1&3
\end{array}} \right)\qquad Y = \left( {\begin{array}{*{20}{c}}
0&2\\
{-1}&0
\end{array}} \right)\]

\[ |A|=|X||Y|=1\cdot 2=2\]

\[ BA=3\cdot |A|B+6A^{*}=6B+6A^{*}\]
\[\Rightarrow B(A-6E)=6A^{*}\]
\[\Rightarrow B(A-6E)A=12E \]
\[ B=12[(A-6E)A]^{-1} \]

\[\left( {A - 6E} \right)A = \left( {\begin{array}{*{20}{c}}
{ - 5}&2&{}&{}\\
1&{ - 3}&{}&{}\\
{}&{}&{ - 6}&2\\
{}&{}&{ - 1}&{ - 6}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
1&2&{}&{}\\
1&3&{}&{}\\
{}&{}&0&2\\
{}&{}&{ - 1}&0
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{ - 3}&{ - 4}&{}&{}\\
{ - 2}&{ - 7}&{}&{}\\
{}&{}&{ - 2}&{ - 12}\\
{}&{}&6&{ - 2}
\end{array}} \right)\]
\[{\left( {\begin{array}{*{20}{c}}
{ - 3}&{ - 4}&{}&{}\\
{ - 2}&{ - 7}&{}&{}\\
{}&{}&{ - 2}&{ - 12}\\
{}&{}&6&{ - 2}
\end{array}} \right)^{ - 1}} = \left( {\begin{array}{*{20}{c}}
{ - \frac{7}{{13}}}&{\frac{4}{{13}}}&{}&{}\\
{\frac{2}{{13}}}&{ - \frac{3}{{13}}}&{}&{}\\
{}&{}&{ - \frac{1}{{38}}}&{\frac{6}{{38}}}\\
{}&{}&{ - \frac{3}{{38}}}&{ - \frac{1}{{38}}}
\end{array}} \right)\]
所以
\[B = 12\left( {\begin{array}{*{20}{c}}
{ - \frac{7}{{13}}}&{\frac{4}{{13}}}&{}&{}\\
{\frac{2}{{13}}}&{ - \frac{3}{{13}}}&{}&{}\\
{}&{}&{ - \frac{1}{{38}}}&{\frac{6}{{38}}}\\
{}&{}&{ - \frac{3}{{38}}}&{ - \frac{1}{{38}}}
\end{array}} \right)\]
二、计算$D = \left| {\begin{array}{*{20}{c}}
{{s_0}}&{{s_1}}& \cdots &{{s_{n - 1}}}\\
{{s_1}}&{{s_2}}& \cdots &{{s_n}}\\
 \vdots & \vdots & \cdots & \vdots \\
{{s_n}}&{{s_{n + 1}}}& \cdots &{{s_{2n - 1}}}
\end{array}\begin{array}{*{20}{c}}
1\\
x\\
 \vdots \\
{{x^n}}
\end{array}} \right|$,其中${s_k} = x_1^k + x_2^k +  \cdots x_n^k$


解:注意到
\begin{align*}
D &= \left| {\begin{array}{*{20}{c}}
{1 + 1 +  \cdots  + 1}&{{x_1} + {x_2} +  \cdots  + {x_n}}& \cdots &{x_1^{n - 1} + x_2^{n - 1} +  \cdots  + x_n^{n - 1}}&1\\
{{x_1} + {x_2} +  \cdots  + {x_n}}&{x_1^2 + x_2^2 +  \cdots  + x_n^2}& \cdots &{x_1^n + x_2^n +  \cdots  + x_n^n}&x\\
 \vdots & \vdots & \cdots & \vdots & \vdots \\
 \vdots & \vdots & \cdots & \vdots & \vdots \\
{x_1^n + x_2^n +  \cdots  + x_n^n}&{x_1^{n + 1} + x_2^{n + 1} +  \cdots  + x_n^{n + 1}}& \cdots &{x_1^{2n - 1} + x_2^{2n - 1} +  \cdots  + x_n^{2n - 1}}&{{x^n}}
\end{array}} \right|\\
 &= \left| {\begin{array}{*{20}{c}}
1&1& \cdots &1&1\\
{{x_1}}&{{x_2}}& \cdots &{{x_n}}&x\\
{x_1^2}&{x_2^2}& \cdots &{x_n^2}&{{x^2}}\\
 \vdots & \vdots & \cdots & \vdots & \vdots \\
{x_1^n}&{x_2^n}& \cdots &{x_n^n}&{{x^n}}
\end{array}} \right| \cdot \left| {\begin{array}{*{20}{c}}
1&{{x_1}}&{x_1^2}& \cdots &{x_1^{n - 1}}&0\\
1&{{x_2}}&{x_2^2}& \cdots &{x_2^{n - 1}}&0\\
1&{{x_3}}&{x_3^2}& \cdots &{x_3^{n - 1}}&0\\
 \vdots & \vdots & \vdots & \cdots & \vdots &0\\
1&{{x_n}}&{x_n^2}& \cdots &{x_n^{n - 1}}&0\\
0&0&0& \cdots &0&1
\end{array}} \right|\\
&=V(x_{1},x_{2},\cdots,x_{n},x)\cdot V(x_{1},x_{2},\cdots,x_{n})
\end{align*}
三、有$\alpha_{1},\alpha_{2},\cdots,\alpha_{s},\alpha_{s+1}$,且$\beta_{i}=\alpha_{i}+t_{i}\alpha_{s+1},i=1,2,\cdots,s$, 证明如果$\beta_{1},\beta_{2},\cdots,\beta_{s}$,线性无关,则$\alpha_{1},\alpha_{2},\cdots,\alpha_{s+1}$必定线性无关。


五、设$sl_{n}(F)$是$M(F)$上$A,B$矩阵满足$AB-BA$生成的子空间,证明$\dim(sl_{n}(F))=n^2-1$.


证明:
\[ sl_{n}(F)=\{AB-BA|A,B\in M(F)\} \]
对任意的$X\in sl_{n}(F)$,存在$A,B\in M(F)$,使得
\[ X=AB-BA \]
我们注意到
\[ \text{tr}(X)=\text{tr}(AB-BA)=0 \]
假设
\[X = \left( {\begin{array}{*{20}{c}}
{{a_{11}}}&{{a_{12}}}& \cdots & \cdots &{{a_{1n}}}\\
{{a_{21}}}&{{a_{22}}}& \cdots & \cdots &{{a_{2n}}}\\
 \vdots & \vdots & \vdots & \vdots & \vdots \\
 \vdots & \vdots & \vdots & \vdots & \vdots \\
{{a_{n1}}}&{{a_{n2}}}& \cdots & \cdots &{{a_{nn}}}
\end{array}} \right)\]
则有$a_{11}+a_{22}+\cdots+a_{nn}=0$,于是,当$a_{22},a_{33},\cdots,a_{nn}$确定了,$a_{11}$自然就确定了。下证明
\[{E_{ij}} = \left( {\begin{array}{*{20}{c}}
{}&{}&{}&{}&{}\\
{}&{}&{}&{}&{}\\
{}&{}&{}&1&{}\\
{}&{}&{}&{}&{}\\
{}&{}&{}&{}&{}
\end{array}} \right)(i\neq j)\]
\[{E_{ii}} = \left( {\begin{array}{*{20}{c}}
{ - 1}&{}&{}&{}&{}&{}\\
{}& \ddots &{}&{}&{}&{}\\
{}&{}&1&{}&{}&{}\\
{}&{}&{}& \ddots &{}&{}\\
{}&{}&{}&{}& \ddots &{}\\
{}&{}&{}&{}&{}&0
\end{array}} \right)\qquad (i\geq 2)\]
这里$E_{ij}$表示第$i$行第$j$列为$1$,其他为$0$的矩阵,$E_{ii}$表示第一行第一列为$-1$,第$i$行第$i$列为$1$,其余元素为$0$,的矩阵。不难计算发现所有的这2类矩阵数目为$n^2-1$,下证明它们可以构成$sl_{n}(F)$的一组基。
\[ \sum_{ij\neq 1}k_{ij}E_{ij}=0\]
\[\left( {\begin{array}{*{20}{c}}
{ - \sum\limits_{i = 2}^n {{k_{ii}}} }&{{k_{12}}}&{{k_{13}}}& \cdots & \cdots &{{k_{1n}}}\\
{{k_{21}}}&{{k_{22}}}&{{k_{23}}}& \cdots & \cdots &{{k_{2n}}}\\
 \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\
 \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
 \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
{{k_{n1}}}&{{k_{n2}}}& \cdots & \cdots & \cdots &{{k_{nn}}}
\end{array}} \right) = 0\]
显然有$k_{ij}=0$,因此,这$n^2-1$个矩阵线性无关,而对任意的$X\in sl_{n}(F)$,有
\[ X=\sum_{ij\neq 1}a_{ij}E_{ij}\]
故$E_{12},\cdots,E_{1n},\cdots E_{nn}$为$sl_{n}(F)$的一组基。马上得到$\dim(sl_{n}(F))=n^2-1$。

 


六、设数域$K$上的$n$维线性空间$V$到$m$维线性空间上的所有线性映射组成线性空间$Hom_{k}(V,V')$.证明
(1)$Hom_{k}(V,V')$是线性空间;
(2)$Hom_{k}(V,V')$的维数为$mn$.


证明:(1)先验证线性空间的8条运算法则。对任意的$\alpha\in V$,及$\forall f,g,h\in Hom_{k}(V,V')$
1
$(f+g)(\alpha)=f(\alpha)+g(\alpha)=g(\alpha)+f(\alpha)=(g+f)(\alpha)\qquad \forall f,g\in Hom_{k}(V,V')$
2
$((f+g)+h)(\alpha)=(f+g)(\alpha)+h(\alpha)=f(\alpha)+g(\alpha)+h(\alpha)=f(\alpha)+(g+h)(\alpha)=(f+(g+h))(\alpha) $
3在$Hom_{k}(V,V')$中存在一个变换$0$,使得$f+0=f,\forall f\in Hom_{k}(V,V')$.
4.对于$Hom_{k}(V,V')$中的每个元素$f$,都存在$-f\in Hom_{k}(V,V')$,使得$f+(-f)=0$.
5.对于$Hom_{k}(V,V')$中的每个元素$f$,有$1f=f$
6.$k(lf)(\alpha)=klf(\alpha)=(kl)f(\alpha),\forall k,l\in K,\forall \alpha\in V$.
7.$(k+l)f(\alpha)=kf(\alpha)+lf(\alpha),\forall k,l \in K,\forall \alpha \in V$.
8.$(k(f+g))(\alpha)=kf(\alpha)+kg(\alpha),\forall k\in K,\forall \alpha \in V$.
所以$Hom_{k}(V,V')$是线性空间;
(2)我们在$V$中找一组基,记为$\eta_{1},\eta_{2},\cdots,\eta_{n}$,$V'$中找一组基,记为$\nu_{1},\nu_{2},\cdots,\nu_{m}$,则对任意的$\alpha\in V$,
\[ \alpha=\sum_{k=1}x_{k}\eta_{k} \]
\[ f(\alpha)=\sum_{k=1}^{n}x_{k}f(\eta_{k})\]
而对于$f(\eta_{k})\in V'$,
\[ f(\eta_{k})=\sum_{j=1}^{m}y_{kj}\nu_{j}\]
所以
\[ f(\alpha)=\sum_{k=1}^{n}\sum_{j=1}^{m}x_{k}y_{kj}\nu_{j}\]
\[ f(\eta_{1},\eta_{2},\cdots,\eta_{n})=(\nu_{1},\nu_{2},\cdots,\nu_{m})\left( {\begin{array}{*{20}{c}}
{{y_{11}}}&{{y_{12}}}& \cdots & \cdots & \cdots &{{y_{1n}}}\\
{{y_{21}}}&{{y_{22}}}& \cdots & \cdots & \cdots &{{y_{2n}}}\\
 \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
 \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
 \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
{{y_{m1}}}&{{y_{m2}}}& \cdots & \cdots & \cdots &{{y_{mn}}}
\end{array}} \right)\]

\[ A=\left( {\begin{array}{*{20}{c}}
{{y_{11}}}&{{y_{12}}}& \cdots & \cdots & \cdots &{{y_{1n}}}\\
{{y_{21}}}&{{y_{22}}}& \cdots & \cdots & \cdots &{{y_{2n}}}\\
 \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
 \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
 \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
{{y_{m1}}}&{{y_{m2}}}& \cdots & \cdots & \cdots &{{y_{mn}}}
\end{array}} \right)\]
构造映射$\phi:f\to A$,则$\varphi$是从$Hom_{n}(V,V')$到$M_{m\times n}$上的映射。则有
\[ \phi(f+g)=\phi(f)+\phi(g) \]
\[ \phi(kf)=k\phi(f) \]
对于任意的$f,g$满足
\[ \phi(f)=\phi(g)\]
则有
\[ f=g \]
对任意的$A\in M_{m\times n}$,存在$f\in Hom_{n}(V,V')$,使得$\phi(f)=A$,所以
\[ Hom_{n}(V,V')\cong M_{m\times n}\]
而我们知道$\dim(M_{m\times n})=mn$,故$Hom_{k}(V,V')$的维数为$mn$.

七、已知
\[F = \left( {\begin{array}{*{20}{c}}
0&{}&{}&{}&{}&{ - {c_0}}\\
1&0&{}&{}&{}&{ - {c_1}}\\
{}&1&0&{}&{}& \vdots \\
{}&{}&1& \ddots &{}&{ - {c_{n - 3}}}\\
{}&{}&{}& \ddots &0&{ - {c_{n - 2}}}\\
{}&{}&{}&{}&1&{ - {c_{n - 1}}}
\end{array}} \right)\]
(1)求$F$的特征多项式$f(x)$与最小多项式$m(x)$;
(2)求所有与$F$可交换的矩阵.


八、设$\varphi$是复数域上的线性变换,$\epsilon$为恒等变换,$\lambda_{0}$为$\varphi$的一个特征值,$\lambda_{0}$在$\varphi$的最小多项式中的重数$$m_{0}=\min_{k}\{k\in\mathbf{N}^{+}|\text{ker}(\lambda_{0}\epsilon-\varphi)^{k}=\text{ker}(\lambda_{0}\epsilon-\varphi)^{k+1}\}$$

 


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