题号:
6640
题型:
解答题
来源:
B站刘老师开讲《线性代数B》第七套期末模拟考试
求 $\boldsymbol{A}^{20}$, 其中矩阵 $\boldsymbol{A}=\left(\begin{array}{lll}2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2\end{array}\right)$.
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答案：
答案：
析 $(\lambda \boldsymbol{E}-\boldsymbol{A}) \boldsymbol{x}=\mathbf{0} \Rightarrow \mathrm{r}(\lambda \boldsymbol{E}-\boldsymbol{A}) < n$
$\Rightarrow$ 当 $\lambda \boldsymbol{E}-\boldsymbol{A}$ 中无零行, 可令其任意一行(成比例行优先)为零行且换到第 $n$ 行 (称为零行置换)
解 矩阵 $\boldsymbol{A}$ 的特征多项式为:
$$
|\lambda \boldsymbol{E}-\boldsymbol{A}|=\left|\begin{array}{ccc}
\lambda-2 & -1 & -1 \\
-1 & \lambda-2 & -1 \\
-1 & -1 & \lambda-2
\end{array}\right|
$$
=
$$
\left|\begin{array}{ccc}
\lambda-4 & -1 & -1 \\
\lambda-4 & \lambda-2 & -1 \\
\lambda-4 & -1 & \lambda-2
\end{array}\right|
$$
=
$$
(\lambda-4)\left|\begin{array}{ccc}
1 & -1 & -1 \\
1 & \lambda-2 & -1 \\
1 & -1 & \lambda-2
\end{array}\right|
$$
=
$$
=(\lambda-4)\left|\begin{array}{ccc}
1 & -1 & -1 \\
0 & \lambda-1 & 0 \\
0 & 0 & \lambda-1
\end{array}\right|
$$
=
$$
(\lambda-4)(\lambda-1)^2,
$$
所以矩阵 $\boldsymbol{A}$ 的特征值为: $\lambda_1=4, \lambda_2=\lambda_3=1$.
当 $\lambda_1=4$ 时, 解齐次线性方程组 $(4 \boldsymbol{E}-\boldsymbol{A}) \boldsymbol{x}=\mathbf{0}$, 由
$4E-A$= $ \left(\begin{array}{ccc}
x_1 & x_2 & x_3 \\
1 & 0 & -1 \\
0 & 1 & -1 \\
0 & 0 & 0
\end{array}\right),$
得同解方程组为:
$$
\left\{\begin{array}{l}
x_1=x_3, \\
x_2=x_3
\end{array} \text { ( } x_3 \text { 为自由末知量), 取 } x_3=1 \text { 得矩阵 } \boldsymbol{A} \text { 对应于特征值 } \lambda_1=4 \text { 的特征向量: } \xi_1=\left(\begin{array}{l}
1 \\
1 \\
10
\end{array}\right)\right. \text {, }
$$
当 $\lambda_2=\lambda_3=1$ 时, 解齐次线性方程组 $(E-A) x=0$, 由
$E-A$=$\left(\begin{array}{ccc}
x_1 & x_2 & x_3 \\
1 & 1 & 1 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right)$
得同解方程组为: $x_1=-x_2-x_3 \quad\left(x_2, x_3\right.$ 为自由末知量),
$$
\text { 取 }\left(\begin{array}{l}
x_2 \\
x_3
\end{array}\right)=\left(\begin{array}{l}
1 \\
0
\end{array}\right),\left(\begin{array}{l}
0 \\
1
\end{array}\right) \text { 得矩阵 } \boldsymbol{A} \text { 对应于特征值 } \lambda_2=\lambda_3=1 \text { 的特征向量: } \xi_2=\left(\begin{array}{r}
-1 \\
1 \\
0
\end{array}\right), \xi_3=\left(\begin{array}{r}
-1 \\
0 \\
1
\end{array}\right) \text {; }
$$
令 $\boldsymbol{P}=\left(\boldsymbol{\xi}_1, \boldsymbol{\xi}_2, \boldsymbol{\xi}_3\right)=\left(\begin{array}{rrr}1 & -1 & -1 \\ 1 & 1 & 0 \\ 1 & 0 & 1\end{array}\right)$, 则 $\boldsymbol{P}^{-1} \boldsymbol{A} \boldsymbol{P}=\boldsymbol{\Lambda}=\left(\begin{array}{lll}4 & & \\ & 1 & \\ & & 1\end{array}\right), \boldsymbol{A}=\boldsymbol{P} \Lambda \boldsymbol{P}^{-1}, \boldsymbol{A}^{20}=\boldsymbol{P} \boldsymbol{\Lambda}^{20} \boldsymbol{P}^{-1}$, 其中
$$
\Lambda^{20}=\left(\begin{array}{lll}
4 & & \\
& 1 & \\
& & 1
\end{array}\right)^{20}=\left(\begin{array}{ccc}
4^{20} & & \\
& 1 & \\
& & 1
\end{array}\right)
$$
$$
(\boldsymbol{P}, \boldsymbol{E})=\left(\begin{array}{rrrlll}
1 & -1 & -1 & 1 & 0 & 0 \\
1 & 1 & 0 & 0 & 1 & 0 \\
1 & 0 & 1 & 0 & 0 & 1
\end{array}\right)
$$
$$
\boldsymbol{P}^{-1}=\left(\begin{array}{ccc}
\frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\
-\frac{1}{3} & \frac{2}{3} & -\frac{1}{3} \\
-\frac{1}{3} & -\frac{1}{3} & \frac{2}{3}
\end{array}\right)=\frac{1}{3}\left(\begin{array}{rrr}
1 & 1 & 1 \\
-1 & 2 & -1 \\
-1 & -1 & 2
\end{array}\right)
$$
$$
\boldsymbol{A}^{20}=\boldsymbol{P} \boldsymbol{\Lambda}^{20} \boldsymbol{P}^{-1}= =\left(\begin{array}{rrr}
1 & -1 & -1 \\
1 & 1 & 0 \\
1 & 0 & 1
\end{array}\right)\left(\begin{array}{lll}
4^{20} & & \\
& 1 & \\
& & 1
\end{array}\right) \frac{1}{3}\left(\begin{array}{rrr}
1 & 1 & 1 \\
-1 & 2 & -1 \\
-1 & -1 & 2
\end{array}\right)
$$
$$
=\frac{1}{3}\left(\begin{array}{ccc}
4^{20}+2 & 4^{20}-1 & 4^{20}-1 \\
4^{20}-1 & 4^{20}+2 & 4^{20}-1 \\
4^{20}-1 & 4^{20}-1 & 4^{20}+2
\end{array}\right) .
$$
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