矩阵 $\mathbf{M}$ 定义为

已知 8 是 $\mathbf{M}$ 的一个特征值，

(a) 求一个与特征值 8 对应的特征向量。

(b) 求 $\mathbf{M}$ 的另外两个特征值。

(c) 因此，求一个正交矩阵 $\mathbf{P}$ 和一个对角矩阵 $\mathbf{D}$，使得 $\mathbf{P}^{\mathsf T}\mathbf{M}\mathbf{P}=\mathbf{D}$。

与特征值 8 对应的特征向量满足 $(M-8I)\mathbf{v}=\mathbf0$。联立所得方程，取一个最简单的非零解即可。

Let

For the eigenvalue 8,

so

The independent equations are

and

Writing $u=x+y$, these become $2u+z=0$ and $u+3z=0$. Hence $z=0$ and $u=0$, so $y=-x$.

Taking $x=1$, an eigenvector is

由 $\det(M-\lambda I)=0$ 建立特征方程。题目已给出 8 是特征值，因此三次多项式应含因子 $\lambda-8$；除去此因子即可求另外两个特征值。

The characteristic equation is

Expanding the determinant,

Collecting the terms gives

Multiplying by $-1$,

Since 8 is already known to be an eigenvalue,

Therefore the other two eigenvalues are

实对称矩阵可由正交矩阵对角化。分别求特征值 3、6 的特征向量，并与 (a) 中特征值 8 的向量一起单位化；把这些单位特征向量按列组成 $P$，同时按相同顺序把特征值放入 $D$。

For $\lambda=3$, $(M-3I)\mathbf{v}=\mathbf0$ gives

Subtracting the second equation from the first gives $5x-5y=0$, so $x=y$. The first equation then gives $z=x$. Taking $x=1$, an eigenvector is

For $\lambda=6$, $(M-6I)\mathbf{v}=\mathbf0$ gives

The first two equations give $z=-2y=-2x$, so $x=y$ and $z=-2x$. These also satisfy the third equation. Taking $x=1$, an eigenvector is

Together with the eigenvector from part (a), take

Their scalar products are all zero:

Their lengths are $|\mathbf{v}_8|=\sqrt2$, $|\mathbf{v}_3|=\sqrt3$ and $|\mathbf{v}_6|=\sqrt6$. Dividing by these lengths gives the corresponding unit eigenvectors

and

Using the order $8,3,6$, we may therefore take

and

The columns of $P$ are orthonormal, so $P$ is orthogonal, and the matching column order gives $P^{\mathsf T}MP=D$.