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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

Author

zephyr

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 \(A^k\).

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。

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, \(\mathbf{A}^k\) has the same eigenvectors as \(\mathbf{A}\), and the eigenvalues are the \(k\)-th powers of the eigenvalues of \(\mathbf{A}\). Therefore, the eigenvalue-eigenvector pairs for \(\mathbf{A}^k\) are:

  • Eigenvalue: \(\lambda_{ii}^k\)
  • Corresponding eigenvector: \(\mathbf{y}_i = \mathbf{P} \mathbf{e}_i\)

In summary, every eigenvalue \(\lambda_{ii}\) of \(\mathbf{A}\) raised to the power \(k\) is an eigenvalue of \(\mathbf{A}^k\), and the corresponding eigenvectors remain the same.

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