CBMS-2023-08

题目来源：Problem 8 日期：2024-07-09 题目主题：Math-线性代数-特征值与特征向量、正定矩阵、矩阵的非奇异性

具体题目

(1) Describe the eigenvalues and eigenvectors of the following matrix. ($\lambda$ is a real value.)

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

(2) What is the range of $\lambda$ such that ${\mathbf{A}}_{\lambda }$ is positive semidefinite.

(3) Consider an $n\times n$ symmetric matrix where all diagonal elements are $b$ and all non-diagonal elements are $a$. Show that this matrix is non-singular when $|b|>|(n-1)a|$.

正确解答

Let’s correct the simplification process by subtracting the second and third rows from the first row to get the required form. Here is the detailed solution:

正确解答

(1) Eigenvalues and Eigenvectors

To find the eigenvalues ${\mu }_i$ of the matrix ${\mathbf{A}}_{\lambda }$, we need to solve the characteristic equation $\det ({\mathbf{A}}_{\lambda }-\mu \mathbf{I})=0$.

$${\mathbf{A}}_{\lambda }-\mu \mathbf{I}=\left(\begin{array}{ccc}\lambda -\mu & 1& 1\\ 1& \lambda -\mu & 1\\ 1& 1& \lambda -\mu \end{array}\right)$$

Using the row addition method for simplification, we can add all rows to the first row:

$$\left|\begin{array}{ccc}\lambda -\mu & 1& 1\\ 1& \lambda -\mu & 1\\ 1& 1& \lambda -\mu \end{array}\right|=\left|\begin{array}{ccc}\lambda -\mu +1+1& 1+(\lambda -\mu )+1& 1+1+(\lambda -\mu )\\ 1& \lambda -\mu & 1\\ 1& 1& \lambda -\mu \end{array}\right|=\left|\begin{array}{ccc}\lambda -\mu +2& \lambda -\mu +2& \lambda -\mu +2\\ 1& \lambda -\mu & 1\\ 1& 1& \lambda -\mu \end{array}\right|$$

Now, we can absorb the common factor $(\lambda -\mu +2)$ from the first row:

$$=(\lambda -\mu +2)\left|\begin{array}{ccc}1& 1& 1\\ 1& \lambda -\mu & 1\\ 1& 1& \lambda -\mu \end{array}\right|=(\lambda -\mu +2)\left|\begin{array}{ccc}1& 1& 1\\ 0& \lambda -\mu -1& 0\\ 0& 0& \lambda -\mu -1\end{array}\right|=(\lambda -\mu +2)(\lambda -\mu -1{)}^2$$

Setting the determinant to zero:

So, the eigenvalues are:

$${\mu }_1=\lambda +2,{\mu }_2=\lambda -1\text{(with multiplicity 2)}$$

Eigenvectors

For ${\mu }_1=\lambda +2$:

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

Solving

$$\left(\begin{array}{ccc}-2& 1& 1\\ 1& -2& 1\\ 1& 1& -2\end{array}\right)\mathbf{v}=0$$

, we find:

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

For ${\mu }_2=\lambda -1$:

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

We need to solve ${\mathbf{A}}_{\lambda }\mathbf{v}=(\lambda -1)\mathbf{v}$, which gives the eigenvectors corresponding to ${\mu }_2$. This generally results in two linearly independent eigenvectors orthogonal to ${\mathbf{v}}_1$:

$${\mathbf{v}}_2=\left(\begin{array}{c}1\\ 0\\ -1\end{array}\right),{\mathbf{v}}_3=\left(\begin{array}{c}0\\ 1\\ -1\end{array}\right)$$

(2) Positive Semi-definiteness

A matrix is positive semidefinite if all its eigenvalues are non-negative. For ${\mathbf{A}}_{\lambda }$ to be positive semidefinite:

$${\mu }_1=\lambda +2\ge 0$$ $${\mu }_2=\lambda -1\ge 0$$

Solving these inequalities:

$$\lambda +2\ge 0⟹\lambda \ge -2$$ $$\lambda -1\ge 0⟹\lambda \ge 1$$

The most restrictive condition is $\lambda \ge 1$.

Thus, the range of $\lambda$ such that ${\mathbf{A}}_{\lambda }$ is positive semidefinite is:

$$\lambda \ge 1$$

(3) Non-Singularity of Symmetric Matrix

Consider an $n\times n$ symmetric matrix $\mathbf{B}$ where all diagonal elements are $b$ and all non-diagonal elements are $a$.

$$\mathbf{B}=\left(\begin{array}{ccccc}b& a& a& \cdots & a\\ a& b& a& \cdots & a\\ a& a& b& \cdots & a\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ a& a& a& \cdots & b\end{array}\right)$$

The matrix $\mathbf{B}$ can be written as:

$$\mathbf{B}=b\mathbf{I}+a\mathbf{J}$$

where $\mathbf{J}$ is the $n\times n$ matrix with all elements equal to 1.

The eigenvalues of $\mathbf{J}$ are $n$ (with multiplicity 1) and $0$ (with multiplicity $n-1$). Thus, the eigenvalues of $\mathbf{B}$ are:

$${\lambda }_1=b+(n-1)a\text{and}{\lambda }_2=b-a(\text{with multiplicity }n-1)$$

The matrix $\mathbf{B}$ is non-singular if all its eigenvalues are non-zero:

$$b+(n-1)a\ne 0$$ $$b-a\ne 0$$

This holds if:

$$|b|>|(n-1)a|$$

since $|b|>(n-1)|a|$ implies $|b|>|a|$ when $n\ge 2$.

Thus, the matrix $\mathbf{B}$ is non-singular when $|b|>|(n-1)a|$.

知识点

特征值特征向量正定矩阵

难点解题思路

1. 通过求解特征方程来找到特征值。
2. 根据特征值的符号判断矩阵的半正定性。
3. 使用矩阵特征值的性质判断矩阵的非奇异性。

解题技巧和信息

1. 计算特征方程时，使用行列式和代数余子式。
2. 确定半正定矩阵时，所有特征值必须为非负数。
3. 判断矩阵是否非奇异，可以通过特征值是否全非零来实现。

重点词汇

eigenvalue 特征值

eigenvector 特征向量

positive semidefinite 正半定

non-singular 非奇异

参考资料

1. Linear Algebra and Its Applications by Gilbert Strang, Chap. 6
2. Introduction to Linear Algebra by Gilbert Strang, Chap. 7