$\lambda_1=-1, \lambda_2=1, \lambda_3=b$
$$A=\left[\begin{array}{ccc} 1 & 0 & -a \\ 0 & 1 & 0 \\ -a & 0 & 1 \end{array}\right]$$

$$\begin{gathered} \because \operatorname{tr}(A)=3=b \quad \therefore b=3 \\ \because|A|=\left|\begin{array}{cc} 1 & -a \\ -a & 1 \end{array}\right|=1-a^2=-3 \\ \therefore a^2=4 \quad a > 0 \quad \therefore a=2 \end{gathered}$$

(1) 当 $\lambda_1=-1$ 时，求 $(A+E)x=0$
$$(A+E)=\left[\begin{array}{ccc} 2 & 0 & -2 \\ 0 & 2 & 0 \\ -2 & 0 & 2 \end{array}\right] \sim\left[\begin{array}{ccc} 1 & 0 & -1 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{array}\right]$$

$$p_1=\left[\begin{array}{l} 1 \\ 0 \\ 1 \end{array}\right]$$

(2) 当 $\lambda_1=1$ 时，求 $(A-E)x=0$

$$A-E=\left[\begin{array}{ccc} 0 & 0 & -2 \\ 0 & 0 & 0 \\ -2 & 0 & 0 \end{array}\right] \sim\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{array}\right]$$

$P_2=\left[\begin{array}{l} 0 \\ 1 \\ 0 \end{array}\right]$

(3) 当 $\lambda_3=3$ 时，求 $(A-3E)x=0$

$$A-3 E=\left[\begin{array}{ccc} -2 & 0 & -2 \\ 0 & -2 & 0 \\ -2 & 0 & -2 \end{array}\right] \sim\left[\begin{array}{lll} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{array}\right]$$

$P_3=\left[\begin{array}{c} -1 \\ 0 \\ 1 \end{array}\right]$

\begin{aligned} q_1=\frac{\sqrt{2}}{2}\left[\begin{array}{l} 1 \\ 0 \\ 1 \end{array}\right], q_2=\left[\begin{array}{l} 0 \\ 1 \\ 0 \end{array}\right], q_3=\frac{\sqrt{2}}{2}\left[\begin{array}{c} -1 \\ 0 \\ 1 \end{array}\right] \\ & \end{aligned}

①点击 首页查看更多试卷和试题 , 点击查看 本题所在试卷