题号:
6281
题型:
解答题
来源:
2023《线性代数》方阵n次方计算方法总结与典型例题求解
已知矩阵 $A=\left(\begin{array}{ccc}1 & 2 & 2 \\ 2 & 1 & -2 \\ -2 & -2 & 1\end{array}\right)$, 求 $A^n$.
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答案：
答案：
易知矩阵 $\boldsymbol{A}$ 的特征多项式为:
$$
|\lambda E-A|=\left|\begin{array}{ccc}
\lambda-1 & -2 & -2 \\
-2 & \lambda-1 & 2 \\
2 & 2 & \lambda-1
\end{array}\right|=\left|\begin{array}{ccc}
\lambda-1 & \lambda-1 & \lambda-1 \\
-2 & \lambda-1 & 2 \\
2 & 2 & \lambda-1
\end{array}\right|
$$
$$
\begin{aligned}
& =(\lambda-1)\left|\begin{array}{ccc}
1 & 1 & 1 \\
-2 & \lambda-1 & 2 \\
2 & 2 & \lambda-1 \\
1 & 0 & 1 \\
-2 & \lambda-3 & 2 \\
0 & 0 & \lambda+1
\end{array}\right|=(\lambda-1)\left|\begin{array}{ccc}
1 & 1 & 1 \\
-2 & \lambda-1 & 2 \\
0 & \lambda+1 & \lambda+1
\end{array}\right| \\
& =(\lambda-1-1)(\lambda+1)\left|\begin{array}{cc}
1 & 0 \\
-2 & \lambda-3
\end{array}\right| \\
& =(\lambda+1)(\lambda-1)(\lambda-3)
\end{aligned}
$$
令 $|\lambda E-A|=0$ ，得 $\lambda_1=-1, \lambda_2=1, \lambda_3=3$. 当特征值 $\lambda_1=-1$ 时，则解方程 $(-E-A) \boldsymbol{X}=\overline{0}$.
$$
\begin{aligned}
-E-A & =\left(\begin{array}{ccc}
-2 & -2 & -2 \\
-2 & -2 & 2 \\
2 & 2 & -2
\end{array}\right) \rightarrow\left(\begin{array}{ccc}
1 & 1 & 1 \\
0 & 0 & 1 \\
0 & 0 & -1
\end{array}\right) \\
& \rightarrow\left(\begin{array}{lll}
1 & 1 & 1 \\
0 & 0 & 1 \\
0 & 0 & 0
\end{array}\right) \rightarrow\left(\begin{array}{lll}
1 & 1 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0
\end{array}\right)
\end{aligned}
$$
解得基础解系为: $\alpha_1=(1,-1,0)^T$ ，所以矩阵 $A$ 属于特征值 $\lambda_1=-1$ 的全部特征向量为: $k_1(1,-1,0)^T, k_1 \neq 0$.
当特征值 $\lambda_2=1$ 时，则解方程 $(E-A) X=0$
$$
E-A=\left(\begin{array}{ccc}
0 & -2 & -2 \\
-2 & 0 & 2 \\
2 & 2 & 0
\end{array}\right) \rightarrow\left(\begin{array}{ccc}
0 & 1 & 1 \\
-1 & 0 & 1 \\
1 & 1 & 0
\end{array}\right) \rightarrow\left(\begin{array}{ccc}
0 & 1 & 1 \\
-1 & 0 & 1 \\
0 & 0 & 0
\end{array}\right)
$$
解得基础解系为: $\alpha_2=(1,-1,1)^T$ ，所以矩阵 $A$ 属于特征值 $\lambda_1=1$ 的 全部特征向量为: $k_2(1,-1,1)^T, k_2 \neq 0$.
当特征值 $\lambda_2=3$ 时，则解方程 $(3 E-A) X=0$
$$
3 E-A=\left(\begin{array}{ccc}
2 & -2 & -2 \\
-2 & 2 & 2 \\
2 & 2 & 2
\end{array}\right) \rightarrow\left(\begin{array}{ccc}
1 & -1 & -1 \\
-1 & 1 & 1 \\
1 & 1 & 1
\end{array}\right) \rightarrow\left(\begin{array}{lll}
0 & 1 & 1 \\
0 & 0 & 0 \\
1 & 0 & 0
\end{array}\right) .
$$
解得基础解系为: $\alpha_2=(0,1,-1)^T$ ，所以矩阵 $A$ 属于特征值 $\lambda_1=3$ 的全部特征向量为: $k_3(0,1,-1)^T, k_3 \neq 0$.
由矩阵 $A$ 可对角化条件可知: 矩阵 $A$ 可对角化且对应对角阵为
$$
\begin{aligned}
& \Lambda=\left(\begin{array}{lll}
1 & & \\
& -1 & \\
& & 3
\end{array}\right) . \\
& \text { 令 } P=\left(\alpha_1, \alpha_2, \alpha_3\right)=\left(\begin{array}{ccc}
1 & 1 & 0 \\
-1 & -1 & 1 \\
1 & 0 & -1
\end{array}\right) \text {, 由求逆矩阵的方法可知: } \\
& P^{-1}=\left(\begin{array}{ccc}
1 & 1 & 1 \\
0 & -1 & -1 \\
1 & 1 & 0
\end{array}\right) \\
&
\end{aligned}
$$
即 $P^{-1} A P=\Lambda$ ，所以
$$
\left(P^{-1} A P\right)^n=P^{-1} A^n P=\Lambda^n=\left(\begin{array}{cc}
1 \text { ath }) & \\
(-1)^n & \\
& 3^n
\end{array}\right) .
$$
由 $A^n=P \Lambda^n P^{-1}$ ，由矩阵的乘法运算法则可知:
$$
\begin{aligned}
& A^n=P \Lambda^n P^{-1} \\
& =\left(\begin{array}{ccc}
1 & 1 & 0 \\
-1 & -1 & 1 \\
1 & 0 & -1
\end{array}\right)\left(\begin{array}{ccc}
1 & & \\
& (-1)^n & \\
& & 3^n
\end{array}\right)\left(\begin{array}{ccc}
1 & 1 & 1 \\
0 & -1 & -1 \\
1 & 1 & 0
\end{array}\right) . \\
& =\left(\begin{array}{ccc}
1 & 1-(-1)^n & 1-(-1)^n \\
3^n-1 & (-1)^n+3^n-1 & (-1)^n-1 \\
1-3^n & 1-3^n & 1
\end{array}\right) \\
&
\end{aligned}
$$
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