2021年考研数二第22题解析：相似对角化

一、题目

难度评级：

二、解析

由 $\begin{vmatrix} \lambda\boldsymbol{E}-\boldsymbol{A} \end{vmatrix} = \begin{vmatrix} \lambda-2 & -1 & 0 \\ -1 & \lambda-2 & 0 \\ -1 & -a & \lambda-b \end{vmatrix} = \left(\lambda-b\right)\left(\lambda-3\right)\left(\lambda-1\right) = 0$ 可知，只有当 $b = 3$ 或者 $b = 1$时，矩阵 $\boldsymbol{A}$ 才会仅有两个不同的特征值.

接下来我们需要分情况讨论——

$\textcolor{lightgreen}{\blacktriangleright}$ 当 $b = 3$ 时，矩阵 $\boldsymbol{A}$ 对应的特征值为：

$$
\lambda_{1,2} = 3 , \ \lambda_{3} = 1
$$

接着，又由矩阵 $\boldsymbol{A}$ 可以相似对角化知，二重根 $\lambda_{1,2} = 3$ 所对应特征值至少存在两个线性无关的特征向量知：

$$
\mathbf{r} \left( \boldsymbol{A} \right) < 3
$$

又因为：

$$
\begin{aligned}
\left(3\boldsymbol{E}-\boldsymbol{A}\right) & = \begin{pmatrix}
3 & 0 & 0 \\
0 & 3 & 0 \\
0 & 0 & 3
\end{pmatrix} – \begin{pmatrix} 2 & 1 & 0 \\ 1 & 2 & 0 \\ 1 & a & b \end{pmatrix} \\ \\
& = \begin{pmatrix}
3 & 0 & 0 \\
0 & 3 & 0 \\
0 & 0 & 3
\end{pmatrix} – \begin{pmatrix} 2 & 1 & 0 \\ 1 & 2 & 0 \\ 1 & a & 3 \end{pmatrix} \\ \\
& = \left(\begin{array}{ccc}1 & -1 & 0 \\ -1 & 1 & 0 \\ -1 & -a & 0\end{array}\right)
\end{aligned}
$$

所以：

$$
a = -1
$$

此时：

$$
\begin{aligned}
\textcolor{lightgreen}{ \left(3\boldsymbol{E}-\boldsymbol{A}\right) } & = \left(\begin{array}{ccc}1 & -1 & 0 \\ -1 & 1 & 0 \\ -1 & 1 & 0\end{array}\right) \\ \\
& = \left(\begin{array}{ccc}1 & -1 & 0 \\ -1 & 1 & 0 \\ 0 & 0 & 0\end{array}\right) \\ \\
& = \textcolor{lightgreen}{ \left(\begin{array}{ccc}1 & -1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right) } \\ \\ \\
\textcolor{pink}{ \left(\boldsymbol{E}-\boldsymbol{A}\right) } & = \begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{pmatrix} – \begin{pmatrix} 2 & 1 & 0 \\ 1 & 2 & 0 \\ 1 & a & b \end{pmatrix} \\ \\
& = \begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{pmatrix} – \begin{pmatrix} 2 & 1 & 0 \\ 1 & 2 & 0 \\ 1 & -1 & 3 \end{pmatrix} \\ \\
& = \begin{pmatrix}
-1 & -1 & 0 \\
-1 & -1 & 0 \\
-1 & 1 & -2
\end{pmatrix} \\ \\
& = \textcolor{pink}{ \begin{pmatrix}
-1 & -1 & 0 \\
0 & 0 & 0 \\
0 & 2 & -2
\end{pmatrix} }
\end{aligned}
$$

由于“$\alpha$ 是属于特征值 $\lambda$ 的特征向量 $\Leftrightarrow$ $\alpha$ 是齐次方程组 $\left( \lambda \boldsymbol{E} – \boldsymbol{A} \right) \boldsymbol{x} = 0$ 的非零解”，所以——

此时，$\lambda_{1}=\lambda_{2}=3$ 所对应特征向量为：

$$
\begin{aligned}
& \ \textcolor{lightgreen}{ \left(\begin{array}{ccc}1 & -1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right) } \begin{pmatrix}
x_{1} \\
x_{2} \\
x_{3}
\end{pmatrix} = 0 \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \boldsymbol{\alpha}_{1} = \left(\begin{array}{c}1 \\ 1 \\ 0\end{array}\right) \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \boldsymbol{\alpha}_{2} = \left(\begin{array}{c}0 \\ 0 \\ 1\end{array}\right)
\end{aligned}
$$

$\lambda_{3}=1$ 所对应的特征向量为：

$$
\begin{aligned}
& \ \textcolor{pink}{ \begin{pmatrix}
-1 & -1 & 0 \\
0 & 0 & 0 \\
0 & 2 & -2
\end{pmatrix} } \begin{pmatrix}
x_{1} \\
x_{2} \\
x_{3}
\end{pmatrix} = 0 \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \boldsymbol{\alpha}_{3}=\left(\begin{array}{c}-1 \\ 1 \\ 1\end{array}\right)
\end{aligned}
$$

综上可知：

$$
\boldsymbol{P}^{-1}\boldsymbol{A}\boldsymbol{P}=\left(\begin{array}{ccc}3 & & \\ & 3 & \\ & & 1\end{array}\right)
$$

$\textcolor{lightgreen}{\blacktriangleright}$ 当 $b = 1$ 时，矩阵 $\boldsymbol{A}$ 对应的特征值为：

$$
\lambda_{1,2} = 1 , \ \lambda_{3} = 3
$$

接着，又由矩阵 $\boldsymbol{A}$ 可以相似对角化知，二重根 $\lambda_{1,2} = 1$ 所对应特征值至少存在两个线性无关的特征向量知：

$$
\mathbf{r} \left( \boldsymbol{A} \right) < 3
$$

又因为：

$$
\begin{aligned}
\left(\boldsymbol{E}-\boldsymbol{A}\right) & = \begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{pmatrix} – \begin{pmatrix} 2 & 1 & 0 \\ 1 & 2 & 0 \\ 1 & a & b \end{pmatrix} \\ \\
& = \begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{pmatrix} – \begin{pmatrix} 2 & 1 & 0 \\ 1 & 2 & 0 \\ 1 & a & 1 \end{pmatrix} \\ \\
& = \left(\begin{array}{ccc}-1 & -1 & 0 \\ -1 & -1 & 0 \\ -1 & -a & 0\end{array}\right)
\end{aligned}
$$

所以：

$$
a = 1
$$

此时：

$$
\begin{aligned}
\textcolor{lightblue}{ \left(\boldsymbol{E}-\boldsymbol{A}\right) } & = \left(\begin{array}{ccc}-1 & -1 & 0 \\ -1 & -1 & 0 \\ -1 & -1 & 0\end{array}\right) \\ \\
& = \left(\begin{array}{ccc}-1 & -1 & 0 \\ -1 & -1 & 0 \\ 0 & 0 & 0\end{array}\right) \\ \\
& = \textcolor{lightblue}{ \left(\begin{array}{ccc}-1 & -1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right) } \\ \\ \\
\textcolor{orange}{\left( 3 \boldsymbol{E} – \boldsymbol{A} \right)} & = \begin{pmatrix}
3 & 0 & 0 \\
0 & 3 & 0 \\
0 & 0 & 3
\end{pmatrix} – \begin{pmatrix} 2 & 1 & 0 \\ 1 & 2 & 0 \\ 1 & a & b \end{pmatrix} \\ \\
& = \begin{pmatrix}
3 & 0 & 0 \\
0 & 3 & 0 \\
0 & 0 & 3
\end{pmatrix} – \begin{pmatrix} 2 & 1 & 0 \\ 1 & 2 & 0 \\ 1 & 1 & 1 \end{pmatrix} \\ \\
& = \begin{pmatrix}
1 & -1 & 0 \\
-1 & 1 & 0 \\
-1 & -1 & 2
\end{pmatrix} \\ \\
& = \textcolor{orange}{ \begin{pmatrix}
1 & -1 & 0 \\
0 & 0 & 0 \\
-2 & 0 & 2
\end{pmatrix} }
\end{aligned}
$$

由于“$\alpha$ 是属于特征值 $\lambda$ 的特征向量 $\Leftrightarrow$ $\alpha$ 是齐次方程组 $\left( \lambda \boldsymbol{E} – \boldsymbol{A} \right) \boldsymbol{x} = 0$ 的非零解”，所以——

此时，$\lambda_{1}=\lambda_{2}=1$ 所对应特征向量为：

$$
\begin{aligned}
& \ \textcolor{lightblue}{ \left(\begin{array}{ccc}-1 & -1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right) } \begin{pmatrix}
x_{1} \\
x_{2} \\
x_{3}
\end{pmatrix} = 0 \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \boldsymbol{\beta}_{1} = \left(\begin{array}{c}-1 \\ 1 \\ 0\end{array}\right) \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \boldsymbol{\beta}_{2} = \left(\begin{array}{c}0 \\ 0 \\ 1\end{array}\right)
\end{aligned}
$$

$\lambda_{3}=3$ 所对应的特征向量为：

$$
\begin{aligned}
& \ \textcolor{orange}{ \begin{pmatrix}
1 & -1 & 0 \\
0 & 0 & 0 \\
-2 & 0 & 2
\end{pmatrix} } \begin{pmatrix}
x_{1} \\
x_{2} \\
x_{3}
\end{pmatrix} = 0 \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \boldsymbol{\alpha}_{3} = \left(\begin{array}{c}1 \\ 1 \\ 1\end{array}\right)
\end{aligned}
$$

此时：

$$
\boldsymbol{P}^{-1}\boldsymbol{A}\boldsymbol{P}=\left(\begin{array}{ccc}1 & & \\ & 1 & \\ & & 3\end{array}\right)
$$

三、补充

[1]. 求解矩阵相似对角化中可逆矩阵 P 的步骤
[2]. 矩阵 A 相似对角化中的可逆矩阵 P 为什么是由矩阵 A 的特征向量组成的？

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