IS_Math-2020-01

IS Math-2022-01

题目来源：Problem 1 日期：2024-07-05 题目主题：Math-线性代数-矩阵分析

具体题目

Square matrices $A$ and $B$ are given by

$$A=\left(\begin{array}{ccc}1& \sqrt{2}& 0\\ \sqrt{2}& 1& \sqrt{2}\\ 0& \sqrt{2}& 1\end{array}\right),B=\left(\begin{array}{ccc}0& -\frac{2}{3}& \frac{1}{3}\\ \frac{2}{3}& 0& -\frac{2}{3}\\ \frac{1}{3}& \frac{2}{3}& 0\end{array}\right).$$

For a real square matrix $X$, $\exp (X)$ is defined as follows:

$$\exp (X)=\sum _{k=0}^{\infty }\left(\frac{1}{k!}{X}^k\right)=I+X+\frac{1}{2!}{X}^2+\frac{1}{3!}{X}^3+\cdots ,$$

where $I$ is a unit matrix. Answer the following questions.

1. Calculate all eigenvalues of $A$ and the corresponding eigenvectors whose norm is one and whose first element is a nonnegative real.
2. Calculate $A^n$, where $n$ is a nonnegative integer.
3. Calculate $\exp (A)$.
4. Show that $\exp (\alpha B)$ is represented by the following equation:

$$\exp (\alpha B)=I+(\sin \alpha )B+(1-\cos \alpha )B^2,$$

where $\alpha$ is a real number. You may use the Cayley-Hamilton theorem.

5. A function $f$ of 3-dimensional real vector $x$ is given as follows:

$$f(x)=\sum _{k=1}^n{\Vert \exp \left(\frac{2\pi k}{n}B\right)a-x\Vert }^2,$$

where $a$ is a 3-dimensional real vector, and $n\ge 2$. Show that $f$ takes the minimum value at $x=(I+B^2)a$.

正确解答

1. Eigenvalues and Eigenvectors of $A$

To find the eigenvalues of $A$, we solve the characteristic polynomial $\det (A-\lambda I)=0$:

$$A-\lambda I=\left(\begin{array}{ccc}1-\lambda & \sqrt{2}& 0\\ \sqrt{2}& 1-\lambda & \sqrt{2}\\ 0& \sqrt{2}& 1-\lambda \end{array}\right)$$

The determinant is:

$$\left|\begin{array}{ccc}1-\lambda & \sqrt{2}& 0\\ \sqrt{2}& 1-\lambda & \sqrt{2}\\ 0& \sqrt{2}& 1-\lambda \end{array}\right|=(1-\lambda )\left[(1-\lambda {)}^2-2\right]-2(1-\lambda )=(1-\lambda )\left[{\lambda }^2-2\lambda -3\right]$$

Solving the polynomial:

$$(1-\lambda )(\lambda -3)(\lambda +1)=0$$

The eigenvalues are:

$${\lambda }_1=1,{\lambda }_2=3,{\lambda }_3=-1$$

Eigenvectors

1. For ${\lambda }_1=1$:

$$A-I=\left(\begin{array}{ccc}0& \sqrt{2}& 0\\ \sqrt{2}& 0& \sqrt{2}\\ 0& \sqrt{2}& 0\end{array}\right)\to \text{RREF}\to \left(\begin{array}{ccc}1& 0& -1\\ 0& 1& 0\\ 0& 0& 0\end{array}\right)$$

The eigenvector is:

$${\mathbf{v}}_1=\frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\ 0\\ -1\end{array}\right)$$

2. For ${\lambda }_2=3$:

$$A-3I=\left(\begin{array}{ccc}-2& \sqrt{2}& 0\\ \sqrt{2}& -2& \sqrt{2}\\ 0& \sqrt{2}& -2\end{array}\right)\to \text{RREF}\to \left(\begin{array}{ccc}1& \sqrt{2}& 1\\ 0& 1& \sqrt{2}\\ 0& 0& 0\end{array}\right)$$

The eigenvector is:

$${\mathbf{v}}_2=\frac{1}{2}\left(\begin{array}{c}1\\ \sqrt{2}\\ 1\end{array}\right)$$

3. For ${\lambda }_3=-1$:

$$A+I=\left(\begin{array}{ccc}2& \sqrt{2}& 0\\ \sqrt{2}& 2& \sqrt{2}\\ 0& \sqrt{2}& 2\end{array}\right)\to \text{RREF}\to \left(\begin{array}{ccc}1& \sqrt{2}& 1\\ 0& 1& \sqrt{2}\\ 0& 0& 0\end{array}\right)$$

The eigenvector is:

$${\mathbf{v}}_3=\frac{1}{2}\left(\begin{array}{c}1\\ -\sqrt{2}\\ 1\end{array}\right)$$

2. Calculating $A^n$

Using the spectral decomposition $A=PD{P}^{-1}$:

$$D=\left(\begin{array}{ccc}1& 0& 0\\ 0& 3& 0\\ 0& 0& -1\end{array}\right),P=\left(\begin{array}{ccc}\frac{1}{\sqrt{2}}& \frac{1}{2}& \frac{1}{2}\\ 0& \frac{\sqrt{2}}{2}& -\frac{\sqrt{2}}{2}\\ -\frac{1}{\sqrt{2}}& \frac{1}{2}& \frac{1}{2}\end{array}\right)$$

Thus:

$$A^n=P{D}^n{P}^{-1}=\left(\begin{array}{ccc}\frac{1}{\sqrt{2}}& \frac{1}{2}& \frac{1}{2}\\ 0& \frac{\sqrt{2}}{2}& -\frac{\sqrt{2}}{2}\\ -\frac{1}{\sqrt{2}}& \frac{1}{2}& \frac{1}{2}\end{array}\right)\left(\begin{array}{ccc}1& 0& 0\\ 0& 3^n& 0\\ 0& 0& (-1{)}^n\end{array}\right)\left(\begin{array}{ccc}\frac{1}{\sqrt{2}}& 0& -\frac{1}{\sqrt{2}}\\ \frac{1}{2}& \frac{\sqrt{2}}{2}& \frac{1}{2}\\ \frac{1}{2}& -\frac{\sqrt{2}}{2}& \frac{1}{2}\end{array}\right)$$

3. Calculate $\exp (A)$

To calculate $\exp (A)$, we use the definition of the matrix exponential:

$$\exp (A)=\sum _{k=0}^{\infty }\frac{1}{k!}{A}^k=I+A+\frac{1}{2!}{A}^2+\frac{1}{3!}{A}^3+\cdots$$

Given the eigenvalues and eigenvectors from the previous steps, we use the spectral decomposition of $A$:

$$A=PD{P}^{-1}$$

where

$$D=\left(\begin{array}{ccc}1& 0& 0\\ 0& 3& 0\\ 0& 0& -1\end{array}\right),P=\left(\begin{array}{ccc}\frac{1}{\sqrt{2}}& \frac{1}{2}& \frac{1}{2}\\ 0& \frac{\sqrt{2}}{2}& -\frac{\sqrt{2}}{2}\\ -\frac{1}{\sqrt{2}}& \frac{1}{2}& \frac{1}{2}\end{array}\right)$$

Therefore,

$$A^k=P{D}^k{P}^{-1}$$

Since $D$ is diagonal, $D^k$ is straightforward to calculate:

$$D^k=\left(\begin{array}{ccc}{1}^k& 0& 0\\ 0& 3^k& 0\\ 0& 0& (-1{)}^k\end{array}\right)$$

Now, compute $\exp (D)$:

$$\exp (D)=\sum _{k=0}^{\infty }\frac{1}{k!}{D}^k=\left(\begin{array}{ccc}\exp (1)& 0& 0\\ 0& \exp (3)& 0\\ 0& 0& \exp (-1)\end{array}\right)$$

Thus,

$$\exp (A)=P\exp (D)P^{-1}$$

Compute $\exp (D)$:

$$\exp (D)=\left(\begin{array}{ccc}e& 0& 0\\ 0& e^3& 0\\ 0& 0& \frac{1}{e}\end{array}\right)$$

Therefore,

$$\exp (A)=P\left(\begin{array}{ccc}e& 0& 0\\ 0& e^3& 0\\ 0& 0& \frac{1}{e}\end{array}\right)P^{-1}$$

We already have $P$ and $P^{-1}$. Calculate $P^{-1}$:

$$P^{-1}=\left(\begin{array}{ccc}\frac{1}{\sqrt{2}}& 0& -\frac{1}{\sqrt{2}}\\ \frac{1}{2}& \frac{\sqrt{2}}{2}& \frac{1}{2}\\ \frac{1}{2}& -\frac{\sqrt{2}}{2}& \frac{1}{2}\end{array}\right)$$

Multiply the matrices:

$$\exp (A)=\left(\begin{array}{ccc}\frac{1}{\sqrt{2}}& \frac{1}{2}& \frac{1}{2}\\ 0& \frac{\sqrt{2}}{2}& -\frac{\sqrt{2}}{2}\\ -\frac{1}{\sqrt{2}}& \frac{1}{2}& \frac{1}{2}\end{array}\right)\left(\begin{array}{ccc}e& 0& 0\\ 0& e^3& 0\\ 0& 0& \frac{1}{e}\end{array}\right)\left(\begin{array}{ccc}\frac{1}{\sqrt{2}}& 0& -\frac{1}{\sqrt{2}}\\ \frac{1}{2}& \frac{\sqrt{2}}{2}& \frac{1}{2}\\ \frac{1}{2}& -\frac{\sqrt{2}}{2}& \frac{1}{2}\end{array}\right)$$

4. Showing $\exp (\alpha B)$

Using the Cayley-Hamilton theorem, the characteristic polynomial of $B$ gives $B^3-(\text{tr}(B))B^2+(\text{sum of principal minors})B-\det (B)I=0$. Since $B$ is traceless and its determinant is zero, we simplify to $B^3=-B$.

The matrix exponential $\exp (\alpha B)$ is given by:

$$\exp (\alpha B)=\sum _{k=0}^{\infty }\frac{(\alpha B{)}^k}{k!}$$

Given the properties of $B$, we can write:

$$\exp (\alpha B)=I+\alpha B+\frac{(\alpha B{)}^2}{2!}+\frac{(\alpha B{)}^3}{3!}+\cdots$$

Since $B^3=-B$, higher powers of $B$ can be expressed in terms of $B$ and $B^2$. The series becomes:

$$\exp (\alpha B)=I+\left(\alpha -\frac{{\alpha }^3}{3!}+\frac{{\alpha }^5}{5!}-\cdots \right)B+\left(\frac{{\alpha }^2}{2!}-\frac{{\alpha }^4}{4!}+\cdots \right)B^2$$

Recognizing the Taylor series for $\sin (\alpha )$ and $1-\cos (\alpha )$:

$$\sin (\alpha )=\alpha -\frac{{\alpha }^3}{3!}+\frac{{\alpha }^5}{5!}-\cdots$$ $$1-\cos (\alpha )=\frac{{\alpha }^2}{2!}-\frac{{\alpha }^4}{4!}+\cdots$$

Thus,

$$\exp (\alpha B)=I+(\sin \alpha )B+(1-\cos \alpha )B^2$$

5. Minimizing $f(x)$

Given the function:

$$f(x)=\sum _{k=1}^n{\Vert \exp \left(\frac{2\pi k}{n}B\right)a-x\Vert }^2$$

where $a$ is a 3-dimensional real vector, and $n\ge 2$.

First, we express the exponential term $\exp \left(\frac{2\pi k}{n}B\right)$. From the previous question, we know:

$$\exp (\alpha B)=I+(\sin \alpha )B+(1-\cos \alpha )B^2$$

Therefore, for $\alpha =\frac{2\pi k}{n}$:

$$\exp \left(\frac{2\pi k}{n}B\right)=I+\sin \left(\frac{2\pi k}{n}\right)B+\left(1-\cos \left(\frac{2\pi k}{n}\right)\right)B^2$$

Substitute this into the function $f(x)$:

$$f(x)=\sum _{k=1}^n{\Vert \left(I+\sin \left(\frac{2\pi k}{n}\right)B+\left(1-\cos \left(\frac{2\pi k}{n}\right)\right)B^2\right)a-x\Vert }^2$$

Now, let’s denote $\exp \left(\frac{2\pi k}{n}B\right)a$ as $y_k$:

$$y_k=\left(I+\sin \left(\frac{2\pi k}{n}\right)B+\left(1-\cos \left(\frac{2\pi k}{n}\right)\right)B^2\right)a$$

Thus, $f(x)$ becomes:

$$f(x)=\sum _{k=1}^n{\Vert y_k-x\Vert }^2$$

We need to minimize $f(x)$ with respect to $x$. To do this, we find the critical point by setting the gradient $\nabla f(x)$ to zero.

The gradient of $f(x)$ with respect to $x$ is:

$$\nabla f(x)=-2\sum _{k=1}^n(y_k-x)$$

Set the gradient to zero:

$$-2\sum _{k=1}^n(y_k-x)=0$$ $$\sum _{k=1}^n(y_k-x)=0$$ $$\sum _{k=1}^n{y}_k=nx$$

Therefore, the minimizing $x$ is given by:

$$x=\frac{1}{n}\sum _{k=1}^n{y}_k$$

Now, we compute the sum ${\sum }_{k=1}^n{y}_k$:

$$\sum _{k=1}^n{y}_k=\sum _{k=1}^n\left(I+\sin \left(\frac{2\pi k}{n}\right)B+\left(1-\cos \left(\frac{2\pi k}{n}\right)\right)B^2\right)a$$

Notice that the identity matrix $I$ and the vector $a$ are constants, so we can factor them out:

$$\sum _{k=1}^n{y}_k=\left(\sum _{k=1}^{n}I\right)a+\left(\sum _{k=1}^n\sin \left(\frac{2\pi k}{n}\right)B\right)a+\left(\sum _{k=1}^n\left(1-\cos \left(\frac{2\pi k}{n}\right)\right)B^2\right)a$$

We simplify each term separately:

1. For the first term:

$$\sum _{k=1}^{n}I=nI$$

2. For the second term, consider the summation of sine terms:

$$\sum _{k=1}^n\sin \left(\frac{2\pi k}{n}\right)$$

Since $\sin \left(\frac{2\pi k}{n}\right)$ is symmetric around the unit circle, the sum of sines over one period $2\pi$ is zero:

$$\sum _{k=1}^n\sin \left(\frac{2\pi k}{n}\right)=0$$

3. For the third term, consider the summation of cosine terms:

$$\sum _{k=1}^n\left(1-\cos \left(\frac{2\pi k}{n}\right)\right)$$

Since $\cos \left(\frac{2\pi k}{n}\right)$ is also symmetric around the unit circle, the sum of cosines over one period $2\pi$ is zero, and thus:

$$\sum _{k=1}^n\left(1-\cos \left(\frac{2\pi k}{n}\right)\right)=\sum _{k=1}^{n}1-\sum _{k=1}^n\cos \left(\frac{2\pi k}{n}\right)=n-0=n$$

So, the third term simplifies to:

$$n{B}^2$$

Combining these results:

$$\sum _{k=1}^n{y}_k=nIa+n{B}^{2}a=n(I+B^2)a$$

Thus,

$$x=\frac{1}{n}\sum _{k=1}^n{y}_k=\frac{1}{n}n(I+B^2)a=(I+B^2)a$$

Therefore, the function $f(x)$ takes the minimum value at:

$$x=(I+B^2)a$$

Convexity of $f(x)$

To show that $f(x)$ is convex, we need to prove that its Hessian is positive semi-definite. Recall that:

$$f(x)=\sum _{k=1}^n{\Vert y_k-x\Vert }^2$$

We expand this:

$$f(x)=\sum _{k=1}^n(y_k-x{)}^T(y_k-x)=\sum _{k=1}^n(y_k^T{y}_k-2y_k^{T}x+x^{T}x)$$

Since $y_k^T{y}_k$ is a constant and ${\sum }_{k=1}^n{y}_k$ is a vector, the quadratic term $x^{T}x$ dominates the expression. The Hessian of $f(x)$ is computed as:

$$H={\nabla }^{2}f(x)=2nI$$

Since the Hessian is a scaled identity matrix, it is positive definite. Therefore, $f(x)$ is a convex function.

知识点

线性代数特征值和特征向量矩阵指数Cayley-Hamilton定理凸函数

• Cayley-Hamilton 定理 Cayley-Hamilton Theorem
• [[Linear Equations (Ax=0 & Ax=b)#齐次线性方程组-ax—0-homogeneous-system-ax—0|齐次线性方程组 $Ax=0$ Homogeneous System $Ax=0$]]
• 3. 计算方法 (Computation Methods)
• 泰勒展开 Taylor Expansion
• 凸函数优化

解题技巧和信息

• 使用特征值分解简化矩阵幂和指数的计算
• Cayley-Hamilton 定理用于简化矩阵多项式
• 凸函数的性质帮助确定最优解的存在性和唯一性

重点词汇

eigenvalue 特征值

eigenvector 特征向量

matrix exponential 矩阵指数

Cayley-Hamilton theorem Cayley-Hamilton 定理

convex function 凸函数