实对称矩阵相似对角化时涉及到的正交化和单位化怎么算？

一、题目

已知 $\boldsymbol{A}$ 是三阶实对称矩阵，若正交矩阵 $Q$ 使得 $Q^{-1} A Q=\left[\begin{array}{lll}3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 6\end{array}\right]$, 如果 $\boldsymbol{\alpha}_{1}=(1$, $0,-1)^{\mathrm{\top}}, \boldsymbol{\alpha}_{2}=(0,1,1)^{\mathrm{\top}}$ 是矩阵 $\boldsymbol{A}$ 属于特征值 $\lambda=3$ 的特征向量，则 $Q=?$

难度评级：

二、解析

首先，设特征值 $\lambda_{3} = \alpha_{3} = (x_{1}, x_{2}, x_{3})$, 则：

$$
Q = (\alpha_{1}, \alpha_{2}, \alpha_{3})
$$

而且，$\alpha_{1}, \alpha_{2}$ 和 $\alpha_{3}$ 是正交的，因此：

$$
(1, 0, -1) \begin{pmatrix}
x_{1} \\
x_{2} \\
x_{3}
\end{pmatrix} = x_{1}- x_{3} = 0 \tag{1}
$$

$$
(0, 1, 1) \begin{pmatrix}
x_{1} \\
x_{2} \\
x_{3}
\end{pmatrix} = x_{2} + x_{3} = 0 \tag{2}
$$

令 $x_{3} = 1$, 则，$x_{1} = 1$， $x_{2} = -1$

Tips:

也可以令 $x_{3}$ 等于其他数字，但是最终都可以化简成当 $x_{3}$ 等于 $1$ 时的值，所以直接令 $x_{3} = 1$ 即可。

即：

$$
\alpha_{3} = (1, -1, 1)^{\top}
$$

但是：

$$
\alpha_{1}^{\top} \alpha_{2} =
$$

$$
(1, 0, -1) \begin{pmatrix}
0 \\
1 \\
1
\end{pmatrix} = -1 \neq 0
$$

于是可知，$\alpha_{1}$ 和 $\alpha_{2}$ 不正交，所以需要正交化：

$$
\beta_{1} = \alpha_{1}
$$

$$
\beta_{2} = \alpha_{2} – \frac{(\alpha_{2}, \beta_{1})}{(\beta_{1}, \beta_{1})} \beta_{1} \Rightarrow
$$

$$
(\alpha_{2}, \beta_{1}) = \alpha_{2}^{\top} \beta_{1} = (0, 1, 1) \begin{pmatrix}
1 \\
0 \\
-1
\end{pmatrix} = -1
$$

$$
(\beta_{1}, \beta_{1}) = \beta_{1}^{\top} \beta_{1} = (1, 0, -1) \begin{pmatrix}
1 \\
0 \\
-1
\end{pmatrix} = 1 + 1 = 2
$$

于是：

$$
\beta_{2} = \alpha_{2} – \frac{-1}{2} \beta_{1} \Rightarrow
$$

$$
\beta_{2} = \begin{pmatrix}
0 \\
1 \\
1
\end{pmatrix} + \frac{1}{2} \begin{pmatrix}
1 \\
0 \\
-1
\end{pmatrix} = \begin{pmatrix}
\frac{1}{2} \\
1 \\
\frac{1}{2}
\end{pmatrix}
$$

此外：

$$
\beta_{3} = \alpha_{3} = \begin{pmatrix}
1 \\
-1 \\
1
\end{pmatrix}
$$

之后再对 $\beta_{1}, \beta_{2}, \beta_{3}$ 进行单位化：

$$
\gamma_{1} = \frac{1}{||\beta_{1}||} \beta_{1} = \frac{1}{\sqrt{2}} \begin{pmatrix}
1 \\
0 \\
-1
\end{pmatrix} = \begin{pmatrix}
\frac{1}{\sqrt{2}} \\
0 \\
\frac{-1}{\sqrt{2}}
\end{pmatrix}
$$

$$
\gamma_{2} = \frac{1}{||\beta_{2}||} \beta_{2} = \sqrt{\frac{2}{3}} \begin{pmatrix}
\frac{1}{2} \\
1 \\
\frac{1}{2}
\end{pmatrix} = \frac{2}{\sqrt{6}} \begin{pmatrix}
\frac{1}{2} \\
1 \\
\frac{1}{2}
\end{pmatrix} = \begin{pmatrix}
\frac{1}{\sqrt{6}} \\
\frac{2}{\sqrt{6}} \\
\frac{1}{\sqrt{6}}
\end{pmatrix}
$$

Tips:

拓展：《二次根式的运算法则汇总》

$$
\gamma_{3} = \frac{1}{||\beta_{3}||} \begin{pmatrix}
1 \\
-1 \\
1
\end{pmatrix} = \frac{1}{\sqrt{3}} \begin{pmatrix}
1 \\
-1 \\
1
\end{pmatrix} = \begin{pmatrix}
\frac{1}{\sqrt{3}} \\
\frac{-1}{\sqrt{3}} \\
\frac{1}{\sqrt{3}}
\end{pmatrix}
$$

于是：

$$
Q = (\gamma_{1}, \gamma_{2}, \gamma_{3}) \Rightarrow
$$

$$
Q = \begin{pmatrix}
\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{3}} \\
0 & \frac{2}{\sqrt{6}} & \frac{-1}{\sqrt{3}} \\
\frac{-1}{\sqrt{2}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{3}}
\end{pmatrix}
$$

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