IS Math-2014-01

题目来源：Problem 1 日期：2024-08-10 题目主题：Math-线性代数-对称矩阵、特征值与二次型

解题思路

这道题目主要考察对称矩阵的特征值与特征向量的性质，以及如何利用这些性质进行函数的分析和不等式的证明。具体问题涉及以下几个核心知识点：

1. 对称矩阵的特征值和特征向量：包括特征值的实数性及其几何解释。
2. 二次型与线性变换：通过特征值分解，将二次型函数化简为标准形式，从而分析其极值问题。
3. Rayleigh 商：作为二次型与特征值之间的联系，Rayleigh 商在评估函数取值范围时起到了关键作用。
4. 函数关系的推导与不等式证明：结合对称矩阵的特征值性质，分析特定形式函数的取值范围，并通过极值点的分析证明不等式。

Solution

Question 1: Eigenvalues and Eigenvectors of Matrix $\mathbf{A}$

Given matrix:

$$\mathbf{A}=\left(\begin{array}{ccc}1& -1& 0\\ -1& 1& -1\\ 0& -1& 1\end{array}\right)$$

To find the eigenvalues and eigenvectors, we solve the characteristic equation:

$$\det (\mathbf{A}-\lambda \mathbf{I})=0$$

Expanding this determinant, we obtain the eigenvalues:

$${\lambda }_1=1-\sqrt{2},{\lambda }_2=1,{\lambda }_3=1+\sqrt{2}$$

The corresponding eigenvectors are found by solving $(\mathbf{A}-\lambda \mathbf{I})\mathbf{v}=0$ for each eigenvalue $\lambda$:

• For ${\lambda }_1=1-\sqrt{2}$:

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

• For ${\lambda }_2=1$:

  $${\mathbf{v}}_2=\left(\begin{array}{c}1\\ 0\\ -1\end{array}\right)$$

• For ${\lambda }_3=1+\sqrt{2}$:

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

Question 2: Prove that all eigenvalues of a symmetric matrix are real

Let $\mathbf{M}$ be a symmetric matrix, and consider the eigenvalue equation:

$$\mathbf{Mv}=\lambda \mathbf{v}$$

where $\mathbf{v}$ is an eigenvector and $\lambda$ is the corresponding eigenvalue.

To prove that $\lambda$ is real, multiply both sides of the equation by ${\mathbf{v}}^{\mathbf{T}}$ on the left:

$${\mathbf{v}}^{\mathbf{T}}\mathbf{Mv}=\lambda {\mathbf{v}}^{\mathbf{T}}\mathbf{v}$$

Since $\mathbf{M}$ is symmetric, ${\mathbf{v}}^{\mathbf{T}}\mathbf{Mv}$ is a real number because it represents a quadratic form. Additionally, ${\mathbf{v}}^{\mathbf{T}}\mathbf{v}$ is also a real and non-negative number (as it is the dot product of a vector with itself).

For the equation to hold, $\lambda$ must be real because it is the only scalar multiplying the real, non-negative value ${\mathbf{v}}^{\mathbf{T}}\mathbf{v}$. Therefore, all eigenvalues of a symmetric matrix are real.

Question 3: Example of a Real Matrix with Real Eigenvalues That Is Not Symmetric

Consider the matrix:

$$\mathbf{M}=\left(\begin{array}{cc}1& 1\\ 0& 2\end{array}\right)$$

The eigenvalues of $\mathbf{M}$ are ${\lambda }_1=1$ and ${\lambda }_2=2$, both real. However, $\mathbf{M}$ is not symmetric because:

$${\mathbf{M}}^{\mathbf{T}}=\left(\begin{array}{cc}1& 0\\ 1& 2\end{array}\right)\ne \mathbf{M}$$

Question 4: Show that $f(x,y,z)=1-2g(x,y,z)$

Given functions:

$$f(x,y,z)=\frac{{\mathbf{u}}^{\mathbf{T}}\mathbf{Au}}{{\mathbf{u}}^{\mathbf{T}}\mathbf{u}},g(x,y,z)=\frac{xy+yz}{x^2+y^2+z^2}$$

Expanding ${\mathbf{u}}^{\mathbf{T}}\mathbf{Au}$:

$${\mathbf{u}}^{\mathbf{T}}\mathbf{Au}=x^2+y^2+z^2-2xy-2yz$$

So:

$$f(x,y,z)=\frac{x^2+y^2+z^2-2xy-2yz}{x^2+y^2+z^2}$$

Simplifying, we get:

$$f(x,y,z)=1-2\frac{xy+yz}{x^2+y^2+z^2}=1-2g(x,y,z)$$

Question 5: Inequality for $g(x,y,z)$

Given the symmetric matrix $\mathbf{A}$, we can perform an eigenvalue decomposition:

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

where $\mathbf{V}$ is an orthogonal matrix whose columns are the eigenvectors of $\mathbf{A}$, and $\mathbf{\Lambda }$ is the diagonal matrix of eigenvalues ${\lambda }_1,{\lambda }_2,{\lambda }_3$.

Step 1: Expressing $f(x,y,z)$ Using Rayleigh Quotient

The function $f(x,y,z)$ can be expressed as the Rayleigh quotient:

$$f(x,y,z)=\frac{{\mathbf{u}}^{\mathbf{T}}\mathbf{Au}}{{\mathbf{u}}^{\mathbf{T}}\mathbf{u}}$$

Using the eigenvalue decomposition of $\mathbf{A}$:

$$f(x,y,z)=\frac{{\mathbf{u}}^{\mathbf{T}}\mathbf{V\Lambda }{\mathbf{V}}^{\mathbf{T}}\mathbf{u}}{{\mathbf{u}}^{\mathbf{T}}\mathbf{u}}$$

Let $\mathbf{w}={\mathbf{V}}^{\mathbf{T}}\mathbf{u}$, which is the vector $\mathbf{u}$ expressed in the basis of the eigenvectors of $\mathbf{A}$. Then we have:

$$f(x,y,z)=\frac{{\mathbf{w}}^{\mathbf{T}}\mathbf{\Lambda w}}{{\mathbf{w}}^{\mathbf{T}}\mathbf{w}}=\frac{{\lambda }_1{w}_1^2+{\lambda }_2{w}_2^2+{\lambda }_3{w}_3^2}{w_1^2+w_2^2+w_3^2}$$

This expression for $f(x,y,z)$ is a weighted average of the eigenvalues ${\lambda }_i$, with the weights given by the components $w_i^2$ of $\mathbf{w}$.

Step 2: Relation to $g(x,y,z)$ and Establishing the Inequality

From the earlier question, we know:

$$f(x,y,z)=1-2g(x,y,z)$$

Substituting this into our expression for $f(x,y,z)$ gives:

$$1-2g(x,y,z)=\frac{{\lambda }_1{w}_1^2+{\lambda }_2{w}_2^2+{\lambda }_3{w}_3^2}{w_1^2+w_2^2+w_3^2}$$

We already have the eigenvalues:

$${\lambda }_1=1-\sqrt{2},{\lambda }_2=1,{\lambda }_3=1+\sqrt{2}$$

Thus, the minimum and maximum values of the weighted sum on the right-hand side occur when all the weight is on the smallest or largest eigenvalue, respectively. This gives:

$$1-2g(x,y,z)\in [1-\sqrt{2},1+\sqrt{2}]$$

Solving for $g(x,y,z)$:

$$-\frac{1}{\sqrt{2}}\le g(x,y,z)\le \frac{1}{\sqrt{2}}$$

知识点

对称矩阵特征值和特征向量 Rayleigh商二次型

难点思路

本题的难点在于最后一步中对 $g(x,y,z)$ 的不等式证明，需要通过对称矩阵的特征值分解结合函数极值进行推导。这涉及到对矩阵分解和二次型函数的理解。

解题技巧和信息

1. 对称矩阵的特征值总是实数，这是因为特征向量对应的二次型形式具有实数解。
2. 不对称矩阵即使有实数特征值，也不一定是对称的。
3. 利用矩阵特征值分解可以将复杂的函数问题转化为标准形式，有助于分析函数的极值和不等式。

重点词汇

• Symmetric matrix 对称矩阵
• Eigenvalue 特征值
• Eigenvector 特征向量
• Eigenvalue decomposition 特征值分解
• Quadratic form 二次型

参考资料

1. Gilbert Strang, Linear Algebra and Its Applications, Chap. 5 (关于对称矩阵的特征值与特征向量)
2. Axler, Linear Algebra Done Right, Chap. 7 (对称矩阵的性质)