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 2019年8月実施 問題8

标签：

• Tokyo-University
• Matrix-Diagonalization
• Matrix-Inverse
• Eigenvalues-and-Eigenvectors
• Matrix-Power

Author

zephyr, 祭音Myyura

Description

Assume that the following equation holds for $n \times n$ square matrices, $A = \{a_{ij}\}$, $P = \{p_{ij}\}$ and $\mathbf{\Lambda} = \{\lambda_{ij}\}$:

$$P^{-1}AP = \mathbf{\Lambda}$$

Assume that $n \geq 2$, $P^{-1}$ is the inverse matrix of $P$, $\lambda_{ii} \neq 0$, and $\lambda_{ij} = 0$ if $i \neq j$.

Solve the following problems.

(1) Show the inverse matrix for each of $\mathbf{\Lambda}$ and $A$.

(2) Show that $\lambda_{ii}$ is one of the eigenvalues of $A$, and show one of the corresponding eigenvectors of $A$.

(3) Suppose that $k$ is a positive integer. Show every pair of eigenvalue and corresponding eigenvector of $A^k$.

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假设对于 $n \times n$ 的方阵 $A = \{a_{ij}\}$、$P = \{p_{ij}\}$ 和 $\mathbf{\Lambda} = \{\lambda_{ij}\}$，以下等式成立：

$$P^{-1}AP = \mathbf{\Lambda}$$

假设 $n \geq 2$，$P^{-1}$ 是 $P$ 的逆矩阵，$\lambda_{ii} \neq 0$，且当 $i \neq j$ 时 $\lambda_{ij} = 0$。

解决以下问题。

(1) 展示 $\mathbf{\Lambda}$ 和 $A$ 的逆矩阵。

(2) 证明 $\lambda_{ii}$ 是 $A$ 的一个特征值，并展示 $A$ 的一个对应特征向量。

(3) 假设 $k$ 是一个正整数。展示 $A^k$ 的每对特征值和对应特征向量。

题目描述

设 $A=(a_{ij})$、$P=(p_{ij})$、$\mathbf\Lambda=(\lambda_{ij})$ 均为 $n\times n$ 方阵，$n\ge2$，且

$$P^{-1}AP=\mathbf\Lambda.$$

其中 $P$ 可逆，$\mathbf\Lambda$ 为对角矩阵：

$$\lambda_{ii}\ne0,\qquad \lambda_{ij}=0\quad(i\ne j).$$

回答下列问题：

1. 分别写出 $\mathbf\Lambda^{-1}$ 与 $A^{-1}$。
2. 对每个对角元 $\lambda_{ii}$，证明它是 $A$ 的一个特征值，并用 $P$ 的元素或列向量写出一个对应特征向量。
3. 对任意正整数 $k$，列出 $A^k$ 的全部“特征值—对应特征向量”配对。

Kai

(1)

Since $\mathbf{\Lambda}$ is a diagonal matrix with diagonal entries $\lambda_{ii}$, its inverse, denoted as $\mathbf{\Lambda}^{-1}$, is also a diagonal matrix. The diagonal entries of $\mathbf{\Lambda}^{-1}$ are the reciprocals of the diagonal entries of $\mathbf{\Lambda}$:

$$\mathbf{\Lambda}^{-1} = \begin{pmatrix} \frac{1}{\lambda_{11}} & 0 & \cdots & 0 \\ 0 & \frac{1}{\lambda_{22}} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \frac{1}{\lambda_{nn}} \end{pmatrix}$$

To find the inverse matrix of $A$, use the given equation $P^{-1}AP = \mathbf{\Lambda}$. Multiply both sides by $P$ and $P^{-1}$ appropriately:

$$\mathbf{A} = \mathbf{P} \mathbf{\Lambda} \mathbf{P}^{-1}$$

Taking the inverse of both sides, we get:

$$\mathbf{A}^{-1} = (\mathbf{P} \mathbf{\Lambda} \mathbf{P}^{-1})^{-1}$$

Using the property of inverses for matrix products:

$$\mathbf{A}^{-1} = \mathbf{P} (\mathbf{\Lambda}^{-1}) \mathbf{P}^{-1}$$

Thus, $\mathbf{A}^{-1}$ is given by:

$$\mathbf{A}^{-1} = \mathbf{P} \begin{pmatrix} \frac{1}{\lambda_{11}} & 0 & \cdots & 0 \\ 0 & \frac{1}{\lambda_{22}} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \frac{1}{\lambda_{nn}} \end{pmatrix} \mathbf{P}^{-1}$$

(2)

Given the equation $\mathbf{P}^{-1}\mathbf{A}\mathbf{P} = \mathbf{\Lambda}$, this implies that $\mathbf{\Lambda}$ is the diagonal form of $\mathbf{A}$ under the similarity transformation by $P$. The diagonal entries of $\mathbf{\Lambda}$, denoted as $\lambda_{ii}$, are the eigenvalues of $A$.

To show this formally, consider $\mathbf{\Lambda} \mathbf{e}_i = \lambda_{ii} \mathbf{e}_i$, where $\mathbf{e}_i$ is the $i$-th standard basis vector. We have:

$$\mathbf{\Lambda} \mathbf{e}_i = \begin{pmatrix} \lambda_{11} & 0 & \cdots & 0 \\ 0 & \lambda_{22} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \lambda_{nn} \end{pmatrix} \begin{pmatrix} 0 \\ \vdots \\ 1 \\ \vdots \\ 0 \end{pmatrix} = \lambda_{ii} \mathbf{e}_i$$

Since $\mathbf{P}^{-1}\mathbf{AP} = \mathbf{\Lambda}$, let $\mathbf{y}_i = \mathbf{P} \mathbf{e}_i$. Then:

$$\mathbf{A} \mathbf{y}_i = \mathbf{A} (\mathbf{P} \mathbf{e}_i) = \mathbf{P} (\mathbf{\Lambda} \mathbf{e}_i) = \mathbf{P} (\lambda_{ii} \mathbf{e}_i) = \lambda_{ii} (\mathbf{P} \mathbf{e}_i) = \lambda_{ii} \mathbf{y}_i$$

Hence, $\mathbf{y}_i = \mathbf{P} \mathbf{e}_i$ is an eigenvector of $\mathbf{A}$ corresponding to the eigenvalue $\lambda_{ii}$.

(3)

From the similarity transformation $\mathbf{P}^{-1}\mathbf{AP} = \mathbf{\Lambda}$, raising both sides to the power $k$ gives:

$$(\mathbf{P}^{-1}\mathbf{AP})^k = \mathbf{\Lambda}^k$$

Since $\mathbf{P}^{-1}\mathbf{AP} = \mathbf{\Lambda}$:

$$\mathbf{P}^{-1}\mathbf{A}^k\mathbf{P} = \mathbf{\Lambda}^k$$

Because $\mathbf{\Lambda}$ is diagonal, $\mathbf{\Lambda}^k$ is also diagonal, with each diagonal element being raised to the power $k$:

$$\mathbf{\Lambda}^k = \begin{pmatrix} \lambda_{11}^k & 0 & \cdots & 0 \\ 0 & \lambda_{22}^k & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \lambda_{nn}^k \end{pmatrix}$$

Thus, each column $\mathbf y_i=\mathbf P\mathbf e_i$ is an eigenvector of $\mathbf A^k$ with eigenvalue $\lambda_{ii}^k$. More generally, for each distinct value $\mu$ among $\{\lambda_{11}^k,\ldots,\lambda_{nn}^k\}$, the full eigenspace is

$$\ker(\mathbf A^k-\mu\mathbf I) =\operatorname{span}\{\mathbf y_i:\lambda_{ii}^k=\mu\}.$$

Hence every eigenvector of $\mathbf A^k$ is a nonzero vector in one of these spaces.

Knowledge

特征值和特征向量 相似变换

重点词汇

• Matrix: 矩阵
• Eigenvalue: 特征值
• Eigenvector: 特征向量
• Inverse matrix: 逆矩阵
• Diagonal matrix: 对角矩阵
• Similarity transformation: 相似变换

参考资料

1. Axler, S. (2015). Linear Algebra Done Right. Springer. Chap. 8
2. Strang, G. (2009). Introduction to Linear Algebra. Wellesley-Cambridge Press. Chap. 5