EE263 Homework 7

课程主页：https://see.stanford.edu/Course/EE263

这次回顾EE263作业7。

10.9

(a)由行列式展开的特点，构成$s^n$的项为

$$\prod_{i=1}^n(s-a_{ii})$$

所以$s^n$的系数为$1$。

(b)由行列式展开的特点，构成$s^{n-1}$的项为

$$\prod_{i=1}^n(s-a_{ii})$$

所以$s^{n-1}$的系数为

$$-\sum_{i=1}^n a_{ii}=-\operatorname{Tr} A$$

(c)常数项为$s=0$时的多项式的值，即

$$\mathcal{X}(0)=\operatorname{det}(-A)$$

(d)第一个等式由定义即可，所以

$$\begin{aligned} a_{n-1}&=-\sum_{i=1}^{n} \lambda_{i} \\ a_{0}&=\prod_{i=1}^{n}\left(-\lambda_{i}\right) \end{aligned}$$

10.11

由条件可得

$$A=P^{-1}\Lambda P=\sum_{k=1}^n \lambda_k p_k q_k^T$$

(a)

range：

$$\begin{aligned} \mathcal{R}\left(R_{k}\right) &=\left\{R_{k} x | x \in \mathrm{C}^{n}\right\}\\ &=\left\{p_{k} q_{k}^{T} x | x \in \mathrm{C}^{n}\right\} \\ &=\left\{\alpha p_{k} | \alpha \in \mathrm{C}\right\}\\ &=\operatorname{span}\left\{p_{k}\right\} \end{aligned}$$

因此

$$\text{rank}(R_k) = 1$$

利用正交矩阵的特点可得正交补空间为

$$\mathcal{N}\left(R_{k}\right)=\operatorname{span}\left\{p_{i}, i \neq k\right\}$$

(b)因为

$$PQ=QP=I$$

所以

$$q_i^T p_j=\delta_{ij}$$

那么对于$i\neq j$

$$\begin{aligned} R_{i} R_{j} &=p_iq_i^T p_j q_j^T\\ &=p_i(q_i^T p_j)q_j^T\\ &=0 \end{aligned}$$

此外

$$\begin{aligned} R_{i} R_{i} &=p_iq_i^T p_i q_i^T\\ &=p_i(q_i^T p_i)q_i^T\\ &=p_iq_i^T\\ &=R_i \end{aligned}$$

(c)

$$\begin{aligned} (s I-A)^{-1} &=( P^{-1}sIP-P^{-1}\Lambda P)^{-1}\\ &=P^{-1}(sI-\Lambda)^{-1} P\\ &=\sum_{k=1}^{n} \frac{R_{k}}{s-\lambda_{k}} \end{aligned}$$

(d)因为

$$PQ=QP=I$$

所以

$$\sum_{k=1}^n R_k =\sum_{k=1}^n p_{k} q_{k}^{T}=I$$

(e)特征值为

$$\begin{aligned} \lambda_1& =1\\ \lambda_2 &=-2 \end{aligned}$$

对应的特征向量为

$$\begin{aligned} p_1 &=\frac 1 {\sqrt{10}}\left[ \begin{matrix} 3\\ 1 \end{matrix} \right]\\ p_2 &= \left[ \begin{matrix} 0\\ 1 \end{matrix} \right] \end{aligned}$$

即

$$P=\left[ \begin{matrix} \frac{3}{\sqrt{10}} & 0\\ \frac{1}{\sqrt{10}} &1 \end{matrix} \right]$$

所以

$$\begin{aligned} Q^{-1}&=\left[ \begin{matrix} \frac{\sqrt{10}}{3} & 0 \\ -\frac 1 3 &1 \end{matrix} \right]\\ q_1^T&= \left[ \begin{matrix} \frac{\sqrt{10}}{3} & 0 \end{matrix} \right]\\ q_2^T&=\left[ \begin{matrix} -\frac 1 3 &1 \end{matrix} \right] \end{aligned}$$

因此

$$\begin{aligned} R_1&=p_1 q_1^T\\ &=\frac 1 {\sqrt{10}}\left[ \begin{matrix} 3\\ 1 \end{matrix} \right]\left[ \begin{matrix} \frac{\sqrt{10}}{3} & 0 \end{matrix} \right]\\ &=\left[ \begin{matrix} 1 & 0\\ \frac 1 3 & 0 \end{matrix} \right]\\ R_2&=p_2 q_2^T\\ &=\left[ \begin{matrix} 0\\ 1 \end{matrix} \right]\left[ \begin{matrix} -\frac 1 3 &1 \end{matrix} \right]\\ &=\left[ \begin{matrix} 0 & 0\\ -\frac 1 3 &1 \end{matrix} \right] \end{aligned}$$

10.19

(a)注意到我们有

$$\begin{aligned} y(t) &=C e^{t A} x(0)\\ &=C e^{t A}\left(x(0) -x_0+x_0\right)\\ &=C e^{t A}\left(x(0) -x_0\right) + C e^{t A}x_0 \\ &=C e^{t A}\left(x(0) -x_0\right)+y_{\operatorname{nom}}(t)\\ \end{aligned}$$

因为

$$\left\|x(0)-x_{0}\right\| \leq r$$

所以

$$-\|C e^{t A} \| r+y_{\operatorname{nom}}(t)\le y(t) \le \|C e^{t A} \|r+y_{\operatorname{nom}}(t)$$

即

$$\begin{aligned} \overline{y}(t) &=\| C e^{t A} \|r+y_{\operatorname{nom}}(t) \\ \underline{y}(t) &=-\|C e^{t A} \|r+y_{\operatorname{nom}}(t) \end{aligned}$$

(b)

n = 6 ; 
A =[-0.2880 -1.0174 0.4625 1.2678 -1.6342 0.8384 ;     1.5861 0.3352 2.1051 0.2998 0.3260 0.8293 ;     0.2411 -2.3091 -0.0736 -0.6288 0.1439 0.5105 ;     -1.2803 0.4842 0.7187 -0.8074 0.0901 1.3939 ;     1.2931 1.0224 -0.7501 0.0724 0.0088 1.7703 ;     0.5874 -0.4287 0.5852 -1.4978 -1.9009 -0.1749 ];
C =[-10.3166  3.4759  -0.8583  -2.5407  -3.4990  8.0032];
x_0 =[-1.2413;       0.5541;       -0.3143;       1.0052;       -0.0480;       -0.2018];
r = 0.5000;

%(b)
N = 1000;
t = linspace(0, 10, N);
y_max = zeros(1, N);
y_min = zeros(1, N);
y_nom = zeros(1, N);

for i = 1:N
    ynom = C * expm(A * t(i)) * x_0;
    res = norm(C * expm(A * t(i)) * r);
    y_nom(i) = ynom;
    y_max(i) = ynom + res;
    y_min(i) = ynom - res;
end

plot(t, y_nom, 'b-', t, y_min, 'r--', t, y_max,'r--')

11.13

实正规矩阵的形式为

$$S^{-1} A S=\operatorname{diag}\left(\Lambda_{r},\left[\begin{array}{cc}{\sigma_{r+1}} & {\omega_{r+1}} \\ {-\omega_{r+1}} & {\sigma_{r+1}}\end{array}\right], \ldots,\left[\begin{array}{cc}{\sigma_{n}} & {\omega_{n}} \\ {-\omega_{n}} & {\sigma_{n}}\end{array}\right]\right)$$

其中

$$\lambda_{i}=\sigma_{i}+j \omega_{i}, \quad i=r+1, \ldots, n$$

下面讨论如何得到$S$，假设

$$S=\left[ \begin{matrix} s_1 &\ldots &s_n \end{matrix} \right]$$

对于$ 1\le i \le r$，取$s_i$为对应的特征向量即可。

对于$i\ge r+1$，假设$\lambda_i$对应的复特征向量为

$$p_i +jq_i$$

那么

$$\begin{aligned} A(p_i+jq_i) &=(\sigma_{i}+j \omega_{i})(p_i+jq_i)\\ &=(\sigma_i p_i -\omega_{i} q_i) +j(\omega_{i}p_i +\sigma_{i} q_i) \end{aligned}$$

对比实部虚部得到

$$\begin{aligned} Ap_i &=(\sigma_i p_i -\omega_{i} q_i)\\ Aq_i &=(\omega_{i}p_i +\sigma_{i} q_i) \end{aligned}$$

所以

$$A \left[ \begin{matrix} p_i &q_i \end{matrix} \right]=\left[ \begin{matrix} p_i &q_i \end{matrix} \right] \left[\begin{array}{cc}{\sigma_{i}} & {\omega_{i}} \\ {-\omega_{i}} & {\sigma_{i}}\end{array}\right]$$

利用上式计算即可：

N = 10;

while 1
    A = randn(N);
    %特征值分解
    [S0, Lambda] = eig(A);
    Lambda = diag(Lambda);
    %计算复特征值的数量
    index = imag(Lambda) ~= 0;
    if sum(index) > 0
        break
    end
end

S = zeros(N);
i = 1;
while i <= N
    if index(i)
        S(:, i) = real(S0(:, i));
        S(:, i + 1) = imag(S0(:, i));
        i = i + 2;
    else
        S(:, i) = S0(:, i);
        i = i + 1;
    end
end

S \ A * S

ans =
    3.3818    0.0000   -0.0000   -0.0000   -0.0000   -0.0000    0.0000   -0.0000    0.0000    0.0000
    0.0000    1.2037    2.1094   -0.0000   -0.0000   -0.0000    0.0000    0.0000    0.0000    0.0000
    0.0000   -2.1094    1.2037   -0.0000   -0.0000   -0.0000    0.0000   -0.0000   -0.0000    0.0000
    0.0000   -0.0000    0.0000   -0.4661    2.4649    0.0000   -0.0000    0.0000   -0.0000   -0.0000
   -0.0000    0.0000    0.0000   -2.4649   -0.4661   -0.0000   -0.0000    0.0000   -0.0000    0.0000
   -0.0000   -0.0000    0.0000   -0.0000    0.0000   -2.2108    0.9574   -0.0000    0.0000    0.0000
    0.0000   -0.0000    0.0000    0.0000   -0.0000   -0.9574   -2.2108    0.0000    0.0000   -0.0000
   -0.0000   -0.0000    0.0000    0.0000   -0.0000    0.0000   -0.0000   -1.3388    0.0000    0.0000
    0.0000    0.0000    0.0000    0.0000    0.0000   -0.0000    0.0000    0.0000    1.0193    0.4240
   -0.0000   -0.0000    0.0000    0.0000    0.0000   -0.0000    0.0000   -0.0000   -0.4240    1.0193

12.1

(a)

设

$$A=\left[ \begin{matrix} a_1^T\\ \vdots \\ a_m^T \end{matrix} \right],B=\left[ \begin{matrix} b_1 &\ldots & b_p \end{matrix} \right]$$

那么

$$\begin{aligned} {[AB]}_{ij}&=a_i^T b_j =0\\ AB&=\left[ \begin{matrix} a_1^T B\\ \vdots \\ a_m^TB \end{matrix} \right] =0\\ AB&=\left[ \begin{matrix} Ab_1 &\ldots &A b_p \end{matrix} \right] =0 \end{aligned}$$

如果$A$列满秩，那么$Ax=0$不存在非零解，从而$b_i =0$，因此$B=0$，这就与题设矛盾，所以$A$行满秩，即

$$\text{rank}(A)= m \le n$$

因为

$$a_i^T B=0 \Leftrightarrow B^T a_i =0$$

所以$B^T x=0$有$m$个线性无关的解，因此

$$n-p \ge m\Leftrightarrow n\ge m+p$$

(b)正确，实际上，对于$A \in \mathbb{R}^{(2k+1) \times (2k+1)}$，该结论都成立。

首先由条件可得

$$A=-A^T$$

取行列式可得

$$\det (A)= (-1)^{2k+1} \det (A) =-\det (A)$$

那么

$$\det (A) =0$$

(c)正确

$$(I-A)\left(\sum_{i=0}^{k-1} A^{i}\right)=I$$

(d)不正确，反例如下：

$$A= \left[ \begin{matrix} 1 &1\\ 0 &2 \end{matrix} \right],B=\left[ \begin{matrix} 2 &0\\ 0 &1 \end{matrix} \right],AB=\left[ \begin{matrix} 2 &1\\ 0 &2 \end{matrix} \right]$$

(e)正确，证明如下：

假设$AB$的$\lambda_i$的特征值对应的特征向量为$q_i$，那么

$$ABq_i =\lambda_i q_i\Rightarrow BA (Bq_i)=\lambda_i(Bq_i)$$

所以$\lambda_i $也是$BA$的特征值，反之同理。

(f)不正确，从上题中即可看出

(g)正确，利用反证法，假设$A$不可对角化，那么$A$存在阶数大于$1$的约当块$J$，不难看出$J^2$无法对角化，与题设矛盾。

13.1

首先

$$\begin{aligned} \left[\begin{matrix}{\dot x} \\ {\dot z}\end{matrix}\right] &=\left[\begin{matrix}A x+B_{1} u+B_{2} w_{1}\\ F z+G_{1} v+G_{2} w_{2} \end{matrix}\right]\\ &= \left[\begin{matrix}A x+B_{1} u+B_{2} H_{1} z\\ F z+G_{1} v+G_{2} \left(C x+D_{1} u+D_{2} w_{1}\right) \end{matrix}\right]\\ &=\left[\begin{matrix}A x+B_{1} u+B_{2} H_{1} z\\ F z+G_{1} v+G_{2} \left(C x+D_{1} u+D_{2} H_{1} z\right) \end{matrix}\right]\\ &=\left[\begin{matrix} A x+B_{2} H_{1} z\\ F z+G_{2} C x+G_{2} D_{2} H_{1} z \end{matrix}\right] + \left[\begin{matrix} B_1 u\\ G_{2} D_{1} u +G_1 v \end{matrix}\right] \\ &=\left[\begin{matrix} A & B_2 H_1 \\ G_2 C & F+G_2 D_2 H_1 \end{matrix}\right] \left[\begin{matrix}x \\ z\end{matrix}\right] +\left[\begin{matrix} B_1 & 0\\ G_2 D_1 &G_1 \end{matrix}\right] \left[\begin{matrix}{u} \\ {v}\end{matrix}\right] \end{aligned}$$

其次

$$\begin{aligned} y&=H_{2} z+J w_{2}\\ &=H_2 z +J\left(C x+D_{1} u+D_{2} w_{1}\right) \\ &=H_2 z +JCx +JD_1 u+JD_2 H_{1} z \\ &=\left[\begin{matrix} JC & H_2 + JD_2H_1 \end{matrix}\right] \left[\begin{matrix}x \\ z\end{matrix}\right] +\left[\begin{matrix} JD_1 & 0 \end{matrix}\right] \left[\begin{matrix}{u} \\ {v}\end{matrix}\right] \end{aligned}$$

补充题

1

因为

$$Av= \lambda v$$

所以

$$\begin{aligned} f(A) v &=\sum_{i=0}^{\infty} a_i A^i v\\ &=\left(\sum_{i=0}^{\infty} a_i \lambda^i\right) v\\ &= f(\lambda) v \end{aligned}$$

代码如下：

n = 3;
A = randn(n);
[S1, Lambda1] = eig(A);
Lambda1 = diag(Lambda1);

% method 1
B = (eye(n) + A) / (eye(n) - A);
[S2, Lambda2] = eig(B);
Lambda2 = diag(Lambda2);

% method 2
Lambda3 = (1 + Lambda1) ./ (1 - Lambda1);

Lambda2
Lambda3

Lambda2 =
   0.1312 + 1.4362i
   0.1312 - 1.4362i
   0.3422 + 0.0000i
Lambda3 =
   0.3422 + 0.0000i
   0.1312 + 1.4362i
   0.1312 - 1.4362i