线性代数——实对成矩阵的对角化例子

文章目录

题目
解答过程
- 1、求 A 的特征值
- 2、得到对角化矩阵
- 3、求 A 的特征向量
- 4、将特征向量正交单位化
- 5、计算 U − 1 U^{-1} U−1

题目

将下列矩阵对角化
A = [ 0 0 0 0 0 0 − 2 i θ 0 0 − 2 i θ 0 0 0 0 0 0 ] \begin{aligned} A&=\left[\begin{array}{ll} 0 & 0& 0& 0 \\ 0 & 0& -2i\theta& 0 \\ 0 & -2i\theta& 0& 0 \\ 0 & 0& 0& 0 \end{array}\right] \end{aligned} A=⎣⎢⎢⎡000000−2iθ00−2iθ000000⎦⎥⎥⎤

解答过程

1、求 A 的特征值

解方程 ∣ A − λ ∣ = 0 |A-\lambda|=0 ∣A−λ∣=0 :
A = ∣ − λ 0 0 0 0 − λ − 2 i θ 0 0 − 2 i θ − λ 0 0 0 0 − λ ∣ = 0 \begin{aligned} A&=\left|\begin{array}{ll} -\lambda& 0& 0& 0 \\ 0 & -\lambda& -2i\theta& 0 \\ 0 & -2i\theta& -\lambda& 0 \\ 0 & 0& 0& -\lambda \end{array}\right|=0 \end{aligned} A=∣∣∣∣∣∣∣∣−λ0000−λ−2iθ00−2iθ−λ0000−λ∣∣∣∣∣∣∣∣=0 得到：
λ 1 = λ 2 = 0 , λ 3 = 2 i θ λ 4 = − 2 i θ \begin{aligned} \lambda_1&=\lambda_2=0, \\ \lambda_3&=2i\theta \\ \lambda_4&=-2i\theta \end{aligned} λ1λ3λ4=λ2=0,=2iθ=−2iθ

2、得到对角化矩阵

D = [ λ 1 0 0 0 0 λ 2 0 0 0 0 λ 3 0 0 0 0 λ 4 ] = [ 0 0 0 0 0 0 0 0 0 0 2 i θ 0 0 0 0 − 2 i θ ] \begin{aligned} D&=\left[\begin{array}{ll} \lambda_1 & 0& 0& 0 \\ 0 & \lambda_2& 0& 0 \\ 0 & 0& \lambda_3& 0 \\ 0 & 0& 0& \lambda_4 \end{array}\right] \end{aligned} =\left[\begin{array}{ll} 0 & 0& 0& 0 \\ 0 & 0& 0& 0 \\ 0 & 0& 2i\theta& 0 \\ 0 & 0& 0& -2i\theta \end{array}\right] D=⎣⎢⎢⎡λ10000λ20000λ30000λ4⎦⎥⎥⎤=⎣⎢⎢⎡00000000002iθ0000−2iθ⎦⎥⎥⎤

3、求 A 的特征向量

λ 1 = λ 2 = 0 \lambda_1=\lambda_2=0 λ1=λ2=0：

解方程：
[ 0 0 0 0 0 0 − 2 i θ 0 0 − 2 i θ 0 0 0 0 0 0 ] [ x 1 x 2 x 3 x 4 ] = [ 0 0 0 0 ] \begin{aligned} \left[\begin{array}{ll} 0 & 0& 0& 0 \\ 0 & 0& -2i\theta& 0 \\ 0 & -2i\theta& 0& 0 \\ 0 & 0& 0& 0 \end{array}\right] \left[\begin{array}{ll} x_1\\ x_2 \\ x_3 \\ x_4 \end{array}\right]= \left[\begin{array}{ll} 0\\ 0 \\ 0 \\ 0 \end{array}\right] \end{aligned} ⎣⎢⎢⎡000000−2iθ00−2iθ000000⎦⎥⎥⎤⎣⎢⎢⎡x1x2x3x4⎦⎥⎥⎤=⎣⎢⎢⎡0000⎦⎥⎥⎤得到对应的基础解系：

η 1 = [ 1 0 0 0 ] , η 2 = [ 0 0 0 1 ] \eta_1=\left[\begin{array}{ll} 1\\ 0 \\ 0 \\ 0 \end{array}\right], \eta_2=\left[\begin{array}{ll} 0\\ 0 \\ 0 \\ 1 \end{array}\right] η1=⎣⎢⎢⎡1000⎦⎥⎥⎤,η2=⎣⎢⎢⎡0001⎦⎥⎥⎤ λ 3 = 2 i θ \lambda_3=2i\theta λ3=2iθ：

解方程：
[ − 2 i θ 0 0 0 0 − 2 i θ − 2 i θ 0 0 − 2 i θ − 2 i θ 0 0 0 0 − 2 i θ ] [ x 1 x 2 x 3 x 4 ] = [ 0 0 0 0 ] \begin{aligned} \left[\begin{array}{ll} -2i\theta & 0& 0& 0 \\ 0 & -2i\theta& -2i\theta& 0 \\ 0 & -2i\theta& -2i\theta& 0 \\ 0 & 0& 0& -2i\theta \end{array}\right] \left[\begin{array}{ll} x_1\\ x_2 \\ x_3 \\ x_4 \end{array}\right]= \left[\begin{array}{ll} 0\\ 0 \\ 0 \\ 0 \end{array}\right] \end{aligned} ⎣⎢⎢⎡−2iθ0000−2iθ−2iθ00−2iθ−2iθ0000−2iθ⎦⎥⎥⎤⎣⎢⎢⎡x1x2x3x4⎦⎥⎥⎤=⎣⎢⎢⎡0000⎦⎥⎥⎤得到对应的基础解系：

η 3 = [ 0 1 − 1 0 ] \eta_3=\left[\begin{array}{ll} 0\\ 1 \\ -1 \\ 0 \end{array}\right] η3=⎣⎢⎢⎡01−10⎦⎥⎥⎤ λ 4 = − 2 i θ \lambda_4=-2i\theta λ4=−2iθ：

解方程：
[ 2 i θ 0 0 0 0 2 i θ − 2 i θ 0 0 − 2 i θ 2 i θ 0 0 0 0 2 i θ ] [ x 1 x 2 x 3 x 4 ] = [ 0 0 0 0 ] \begin{aligned} \left[\begin{array}{ll} 2i\theta & 0& 0& 0 \\ 0 & 2i\theta& -2i\theta& 0 \\ 0 & -2i\theta& 2i\theta& 0 \\ 0 & 0& 0& 2i\theta \end{array}\right] \left[\begin{array}{ll} x_1\\ x_2 \\ x_3 \\ x_4 \end{array}\right]= \left[\begin{array}{ll} 0\\ 0 \\ 0 \\ 0 \end{array}\right] \end{aligned} ⎣⎢⎢⎡2iθ00002iθ−2iθ00−2iθ2iθ00002iθ⎦⎥⎥⎤⎣⎢⎢⎡x1x2x3x4⎦⎥⎥⎤=⎣⎢⎢⎡0000⎦⎥⎥⎤得到对应的基础解系：

η 4 = [ 0 1 1 0 ] \eta_4=\left[\begin{array}{ll} 0 \\ 1 \\ 1 \\ 0 \end{array}\right] η4=⎣⎢⎢⎡0110⎦⎥⎥⎤

4、将特征向量正交单位化

U = [ 1 0 0 0 0 0 2 2 2 2 0 0 − 2 2 2 2 0 1 0 0 ] U=\left[\begin{array}{ll} 1 & 0& 0& 0 \\ 0 & 0& \frac{\sqrt{2}}{2}& \frac{\sqrt{2}}{2} \\ 0 & 0& -\frac{\sqrt{2}}{2}& \frac{\sqrt{2}}{2} \\ 0 & 1& 0& 0 \end{array}\right] U=⎣⎢⎢⎡10000001022 −22 0022 22 0⎦⎥⎥⎤

5、计算 U − 1 U^{-1} U−1

U − 1 = U † = [ 1 0 0 0 0 0 0 1 0 2 2 − 2 2 0 0 2 2 2 2 0 ] U^{-1}=U^{\dagger}=\left[\begin{array}{ll} 1 & 0& 0& 0 \\ 0 & 0& 0& 1 \\ 0 & \frac{\sqrt{2}}{2}& -\frac{\sqrt{2}}{2}& 0 \\ 0 & \frac{\sqrt{2}}{2}& \frac{\sqrt{2}}{2}& 0 \end{array}\right] U−1=U†=⎣⎢⎢⎡10000022 22 00−22 22 0100⎦⎥⎥⎤