IS Math-2015-01

题目来源：Problem 1 日期：2024-08-11 题目主题：Math-Linear Algebra-Matrix Analysis and Quadratic Forms

解题思路

这道题涉及几个关键的线性代数概念：特征多项式、矩阵函数、二次型、矩阵对称化、特征值和特征向量、正定性以及二次函数的驻点。我们将逐步解决每个小问，并在需要时使用前面小问的结果。

Solution

1. Characteristic polynomial of $\mathbf{A}$

To find the characteristic polynomial of $\mathbf{A}$, we need to calculate $\det (\lambda \mathbf{I}-\mathbf{A})$.

$$\lambda \mathbf{I}-\mathbf{A}=\left(\begin{array}{ccc}\lambda +3& 0& 0\\ 2& \lambda +3& -1\\ -2& 3& \lambda +3\end{array}\right)$$

The characteristic polynomial is:

$$\begin{array}{rl}\det (\lambda \mathbf{I}-\mathbf{A})& =(\lambda +3)[(\lambda +3)(\lambda +3)-(-1)(3)]\\ & =(\lambda +3)[(\lambda +3{)}^2+3]\\ & =(\lambda +3{)}^3+3(\lambda +3)\\ & ={\lambda }^3+9{\lambda }^2+27\lambda +27+3\lambda +9\\ & ={\lambda }^3+9{\lambda }^2+30\lambda +36\end{array}$$

Therefore, the characteristic polynomial of $\mathbf{A}$ is ${\lambda }^3+9{\lambda }^2+30\lambda +36$.

2. Calculation of $\mathbf{C}$

Given $\mathbf{C}={\mathbf{A}}^5+9{\mathbf{A}}^4+30{\mathbf{A}}^3+36{\mathbf{A}}^2+30\mathbf{A}+9\mathbf{I}$, we can use the Cayley-Hamilton theorem, which states that every square matrix satisfies its own characteristic equation.

From the characteristic polynomial, we know that:

${\mathbf{A}}^3+9{\mathbf{A}}^2+30\mathbf{A}+36\mathbf{I}=0$

Multiplying both sides by ${\mathbf{A}}^2$:

${\mathbf{A}}^5+9{\mathbf{A}}^4+30{\mathbf{A}}^3+36{\mathbf{A}}^2=0$

Therefore:

$\mathbf{C}=0+30\mathbf{A}+9\mathbf{I}=30\mathbf{A}+9\mathbf{I}$

Substituting the value of $\mathbf{A}$:

$$\mathbf{C}=\left(\begin{array}{ccc}-81& 0& 0\\ -60& -81& 30\\ 60& -90& -81\end{array}\right)$$

3. Partial derivative of $x^T\mathbf{A}x$

Let $f(x)=x^T\mathbf{A}x$. We can expand this as:

$$f(x)=\left(\begin{array}{ccc}{x}_1& x_2& x_3\end{array}\right)\left(\begin{array}{ccc}-3& 0& 0\\ -2& -3& 1\\ 2& -3& -3\end{array}\right)\left(\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right)$$

Expanding further:

$f(x)=-3x_1^2+(-2x_1-3x_2+2x_3)x_2+(x_2-3x_3)x_3$

Now, we can calculate the partial derivatives:

$$\begin{array}{rl}\frac{\partial f}{\partial x_1}& =-6x_1-2x_2\\ \frac{\partial f}{\partial x_2}& =-2x_1-6x_2+x_3\\ \frac{\partial f}{\partial x_3}& =2x_2-6x_3\end{array}$$

Therefore:

$\frac{\partial }{\partial \mathbf{x}}(x^T\mathbf{A}x)=(-6x_1-2x_2,-2x_1-6x_2+x_3,2x_2-6x_3)$

Alternatively, we can express this as:

$\frac{\partial }{\partial \mathbf{x}}(x^T\mathbf{A}x)=(\mathbf{A}+{\mathbf{A}}^T)x$

4. Symmetric matrix $\tilde{\mathbf{A}}$, eigenvalues, and eigenvectors

For any vector $\mathbf{x}$:

$$\begin{array}{rl}{x}^T\mathbf{A}x& =\sum _{i,j}{a}_{ij}{x}_i{x}_j\\ & =\sum _{i,j}{a}_{ij}{x}_i{x}_j+\sum _{i,j}{a}_{ji}{x}_j{x}_i\\ & =\sum _{i,j}(a_{ij}+a_{ji})x_i{x}_j\\ & =x^T\left(\frac{1}{2}(\mathbf{A}+{\mathbf{A}}^T)\right)x\end{array}$$

This transformation preserves the value of the quadratic form for all vectors $\mathbf{x}$ while ensuring that the resulting matrix $\tilde{\mathbf{A}}$ is symmetric. The symmetry is crucial for guaranteeing real eigenvalues and orthogonal eigenvectors, which are essential for analyzing the properties of the quadratic form.

To find $\tilde{\mathbf{A}}$, we need to symmetrize $\mathbf{A}$:

$$\tilde{\mathbf{A}}=\frac{1}{2}(\mathbf{A}+{\mathbf{A}}^T)=\left(\begin{array}{ccc}-3& -1& 1\\ -1& -3& -1\\ 1& -1& -3\end{array}\right)$$

Now, we need to find the eigenvalues of $\tilde{\mathbf{A}}$. The characteristic polynomial is:

$\det (\lambda \mathbf{I}-\tilde{\mathbf{A}})=(\lambda +5)(\lambda +3)(\lambda +1)$

Therefore, the eigenvalues are:

${\lambda }_1=-1,{\lambda }_2=-3,{\lambda }_3=-5$

For the eigenvectors:

1. For ${\lambda }_1=-1$: $(\tilde{\mathbf{A}}+\mathbf{I}){\mathbf{v}}_1=0$ gives ${\mathbf{v}}_1=(1,-1,1{)}^T$

2. For ${\lambda }_2=-3$: $(\tilde{\mathbf{A}}+3\mathbf{I}){\mathbf{v}}_2=0$ gives ${\mathbf{v}}_2=(1,0,-1{)}^T$

3. For ${\lambda }_3=-5$: $(\tilde{\mathbf{A}}+5\mathbf{I}){\mathbf{v}}_3=0$ gives ${\mathbf{v}}_3=(1,2,1{)}^T$

Normalizing these vectors to make $\mathbf{V}$ orthogonal:

$$\mathbf{V}=\left(\begin{array}{ccc}\frac{1}{\sqrt{3}}& \frac{1}{\sqrt{2}}& \frac{1}{\sqrt{6}}\\ -\frac{1}{\sqrt{3}}& 0& \frac{2}{\sqrt{6}}\\ \frac{1}{\sqrt{3}}& -\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{6}}\end{array}\right)$$

5. Proving $x^T\mathbf{A}x\le 0$

We know that $x^T\mathbf{A}x=x^T\tilde{\mathbf{A}}x$. Since $\tilde{\mathbf{A}}$ is symmetric, we can diagonalize it:

$x^T\tilde{\mathbf{A}}x=x^T\mathbf{V\Lambda }{\mathbf{V}}^{T}x$

where $\mathbf{\Lambda }=\text{diag}(-1,-3,-5)$ and $\mathbf{V}$ is orthogonal.

Let $y={\mathbf{V}}^{T}x$. Then:

$x^T\tilde{\mathbf{A}}x=y^T\mathbf{\Lambda }y=-y_1^2-3y_2^2-5y_3^2\le 0$

This proves that $x^T\mathbf{A}x\le 0$ for any real vector $x$.

6. Stationary point of $g(x)=x^T\mathbf{A}x+2{\mathbf{b}}^{T}x$

To find the stationary point, we set the gradient to zero:

$\nabla g(x)=(\mathbf{A}+{\mathbf{A}}^T)x+2\mathbf{b}=0$

$2\tilde{\mathbf{A}}x+2\mathbf{b}=0$

$\tilde{\mathbf{A}}x=-\mathbf{b}$

Solving this linear system:

$$\left(\begin{array}{ccc}-3& -1& 1\\ -1& -3& -1\\ 1& -1& -3\end{array}\right)\left(\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right)=-\left(\begin{array}{c}1\\ 1\\ 0\end{array}\right)$$

The solution is:

$x=\left(\begin{array}{c}-\frac{1}{2}\\ -\frac{1}{2}\\ \frac{1}{2}\end{array}\right)$

Therefore, the stationary point of $g(x)$ is $(-\frac{1}{2},-\frac{1}{2},\frac{1}{2})$.

知识点

特征多项式Cayley-Hamilton定理矩阵函数二次型矩阵对称化特征值和特征向量正定矩阵

难点思路

本题的难点在于 Cayley-Hamilton 定理的应用和矩阵的对称化。Cayley-Hamilton 定理允许我们将高次矩阵表达式简化，而矩阵的对称化则是处理二次型问题的关键步骤。

解题技巧和信息

1. 在处理特征多项式时，可以利用矩阵的结构（如上三角）来简化计算。
2. 使用 Cayley-Hamilton 定理可以大大简化高次矩阵表达式的计算。
3. 在处理非对称矩阵的二次型时，将其对称化是一个重要步骤。
4. 对于二次型的非正定性证明，可以利用特征值的性质。
5. 在求解线性方程组时，可以利用矩阵的特殊结构（如对称性）来简化计算。

重点词汇

characteristic polynomial 特征多项式

Cayley-Hamilton theorem Cayley-Hamilton 定理

quadratic form 二次型

symmetric matrix 对称矩阵

eigenvalue 特征值

eigenvector 特征向量

positive definiteness 正定性

stationary point 驻点

参考资料

1. Gilbert Strang, “Linear Algebra and Its Applications”, 4th Edition, Chapter 6
2. Carl D. Meyer, “Matrix Analysis and Applied Linear Algebra”, Chapter 3 and 7
3. Roger A. Horn, Charles R. Johnson, “Matrix Analysis”, Chapter 2 and 4