题目

Problem

M=\begin{pmatrix} 1&2&k\\ -1&-3&4\\ 2&6&-8 \end{pmatrix}

where $k$ is a constant

Given that $M$ has a repeated eigenvalue, determine

(i) the possible values of $k$ ,

(ii) all corresponding eigenvalues of $M$ for each value of $k$ .

(7)

题目中文翻译

M=\begin{pmatrix} 1&2&k\\ -1&-3&4\\ 2&6&-8 \end{pmatrix}

其中 $k$ 为常数。

已知 $M$ 有一个重根特征值，求

(i) $k$ 的可能值，

(ii) 对每个 $k$ 值， $M$ 的全部对应特征值。

(7)

解答

展开

先展开特征行列式。所得三次式含有因子 $\lambda$ ，所以 $0$ 恒为一个特征值。要出现重根，只有两种可能： $0$ 也是剩余二次因式的根，或者二次因式本身有重根。分别处理这两种情况，并把每个 $k$ 对应的全部特征值列出。

展开

The characteristic equation is

\det(M-\lambda I)=0.

Expanding along the first row,

\begin{align*} \det(M-\lambda I) =&\,(1-\lambda) \bigl[(-3-\lambda)(-8-\lambda)-24\bigr] \\ &\,-2\bigl[(-1)(-8-\lambda)-8\bigr] \\ &\,+k\bigl[(-1)(6)-2(-3-\lambda)\bigr] \\ =&\,(1-\lambda)(\lambda^2+11\lambda) -2\lambda+2k\lambda \\ =&\,-\lambda^3-10\lambda^2 +(9+2k)\lambda \\ =&\,-\lambda \bigl(\lambda^2+10\lambda-9-2k\bigr). \end{align*}

Hence one eigenvalue is always $\lambda=0$ .

For $0$ to be a repeated eigenvalue, it must also satisfy the quadratic factor. Thus

-9-2k=0 \quad\Longrightarrow\quad k=-\frac92.

In this case,

\lambda^2+10\lambda =\lambda(\lambda+10),

so the eigenvalues are

\boxed{0,\ 0,\ -10}.

Alternatively, the quadratic factor may itself have a repeated root. Its discriminant must then be zero:

\begin{align*} 10^2-4(1)(-9-2k) =&\,0, \\ 136+8k =&\,0, \\ k=&\,-17. \end{align*}

The quadratic becomes

\lambda^2+10\lambda+25 =(\lambda+5)^2,

so the eigenvalues are

\boxed{0,\ -5,\ -5}.

Therefore,

\boxed{ \begin{array}{c|c} k & \text{eigenvalues of }M\\ \hline -\dfrac92 & 0,\ 0,\ -10\\[2mm] -17 & 0,\ -5,\ -5 \end{array} }.