東京大学 新領域創成科学研究科 メディカル情報生命専攻 2022年8月実施 問題8

Author

zephyr

Description

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

\[ \mathbf{A}_\lambda = \begin{pmatrix} \lambda & 1 & 1 \\ 1 & \lambda & 1 \\ 1 & 1 & \lambda \end{pmatrix} \]

(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 \(|b| > |(n - 1)a|\).

Kai

(1) Eigenvalues and Eigenvectors

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

\[ \mathbf{A}_\lambda - \mu \mathbf{I} = \begin{pmatrix} \lambda - \mu & 1 & 1 \\ 1 & \lambda - \mu & 1 \\ 1 & 1 & \lambda - \mu \end{pmatrix} \]

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

\[ \begin{vmatrix} \lambda - \mu & 1 & 1 \\ 1 & \lambda - \mu & 1 \\ 1 & 1 & \lambda - \mu \end{vmatrix} = \begin{vmatrix} \lambda - \mu + 1 + 1 & 1 + (\lambda - \mu) + 1 & 1 + 1 + (\lambda - \mu) \\ 1 & \lambda - \mu & 1 \\ 1 & 1 & \lambda - \mu \end{vmatrix} = \begin{vmatrix} \lambda - \mu + 2 & \lambda - \mu + 2 & \lambda - \mu + 2 \\ 1 & \lambda - \mu & 1 \\ 1 & 1 & \lambda - \mu \end{vmatrix} \]

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

\[ = (\lambda - \mu + 2) \begin{vmatrix} 1 & 1 & 1 \\ 1 & \lambda - \mu & 1 \\ 1 & 1 & \lambda - \mu \end{vmatrix} = (\lambda - \mu + 2) \begin{vmatrix} 1 & 1 & 1 \\ 0 & \lambda - \mu - 1 & 0 \\ 0 & 0 & \lambda - \mu - 1 \end{vmatrix} = (\lambda - \mu + 2)(\lambda - \mu - 1)^2 \]

Setting the determinant to zero:

So, the eigenvalues are:

\[ \mu_1 = \lambda + 2, \quad \mu_2 = \lambda - 1 \quad \text{(with multiplicity 2)} \]

Eigenvectors

For \(\mu_1 = \lambda + 2\):

\[ \mathbf{A}_\lambda - (\lambda + 2) \mathbf{I} = \begin{pmatrix} -2 & 1 & 1 \\ 1 & -2 & 1 \\ 1 & 1 & -2 \end{pmatrix} \]

Solving

\[ \begin{pmatrix} -2 & 1 & 1 \\ 1 & -2 & 1 \\ 1 & 1 & -2 \end{pmatrix} \mathbf{v} = \mathbf{0} \]

, we find:

$$ \mathbf{v}_1 = \begin{pmatrix}

1 \

1 \

1

\end{pmatrix} $$

For \(\mu_2 = \lambda - 1\):

$$ \mathbf{A}_\lambda - (\lambda - 1) \mathbf{I} = \begin{pmatrix}

1 & 1 & 1 \

1 & 1 & 1 \

1 & 1 & 1

\end{pmatrix} $$

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 = \begin{pmatrix}

1 \

0 \

-1

\end{pmatrix}, \quad \mathbf{v}_3 = \begin{pmatrix}

0 \

1 \

-1

\end{pmatrix} $$

(2) Positive Semi-definiteness

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

\[ \mu_1 = \lambda + 2 \geq 0 \]

\[ \mu_2 = \lambda - 1 \geq 0 \]

Solving these inequalities:

\[ \lambda + 2 \geq 0 \implies \lambda \geq -2 \]

\[ \lambda - 1 \geq 0 \implies \lambda \geq 1 \]

The most restrictive condition is \(\lambda \geq 1\).

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

\[ \lambda \geq 1 \]

(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} = \begin{pmatrix}

b & a & a & \cdots & a \

a & b & a & \cdots & a \

a & a & b & \cdots & a \

\vdots & \vdots & \vdots & \ddots & \vdots \

a & a & a & \cdots & b

\end{pmatrix} $$

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:

\[ \lambda_1 = b + (n-1)a \quad \text{and} \quad \lambda_2 = b - a \quad (\text{with multiplicity } n-1) \]

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

\[ b + (n-1)a \neq 0 \]

\[ b - a \neq 0 \]

This holds if:

\[ |b| > |(n-1)a| \]

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

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

Knowledge

特征值和特征向量 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