CBMS-2020-08

题目来源：[[做题/文字版题库/CBMS/2020#Question 8|2020#Question 8]] 日期：2024-07-23 题目主题：Math-线性代数-矩阵特征值与特征向量

解题思路

题目要求我们展示矩阵的逆矩阵、特征值和特征向量，并研究矩阵的幂次与特征值和特征向量的关系。我们可以使用相似变换的性质来简化这些问题。

Solution

Problem 1: Show the inverse matrix for each of $\mathbf{\Lambda }$ and $A$

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 $\mathbf{\Lambda }$:

$${\mathbf{\Lambda }}^{-1}=\left(\begin{array}{cccc}\frac{1}{{\lambda }_{11}}& 0& \cdots & 0\\ 0& \frac{1}{{\lambda }_{22}}& \cdots & 0\\ ⋮& ⋮& \ddots & ⋮\\ 0& 0& \cdots & \frac{1}{{\lambda }_{nn}}\end{array}\right)$$

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:

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

Taking the inverse of both sides, we get:

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

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:

$${\mathbf{A}}^{-1}=\mathbf{P}\left(\begin{array}{cccc}\frac{1}{{\lambda }_{11}}& 0& \cdots & 0\\ 0& \frac{1}{{\lambda }_{22}}& \cdots & 0\\ ⋮& ⋮& \ddots & ⋮\\ 0& 0& \cdots & \frac{1}{{\lambda }_{nn}}\end{array}\right){\mathbf{P}}^{-1}$$

Problem 2: Show that ${\lambda }_{ii}$ is one of the eigenvalues of $\mathbf{A}$, and show one of the corresponding eigenvectors of $\mathbf{A}$

Given the equation ${\mathbf{P}}^{-1}\mathbf{AP}=\mathbf{\Lambda }$, 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:

$$\mathbf{\Lambda }{\mathbf{e}}_i=\left(\begin{array}{cccc}{\lambda }_{11}& 0& \cdots & 0\\ 0& {\lambda }_{22}& \cdots & 0\\ ⋮& ⋮& \ddots & ⋮\\ 0& 0& \cdots & {\lambda }_{nn}\end{array}\right)\left(\begin{array}{c}0\\ ⋮\\ 1\\ ⋮\\ 0\end{array}\right)={\lambda }_{ii}{\mathbf{e}}_i$$

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 ${\lambda }_{ii}$.

Problem 3: Show every pair of eigenvalue and corresponding eigenvector of ${\mathbf{A}}^k$

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 $k$:

$${\mathbf{\Lambda }}^k=\left(\begin{array}{cccc}{\lambda }_{11}^k& 0& \cdots & 0\\ 0& {\lambda }_{22}^k& \cdots & 0\\ ⋮& ⋮& \ddots & ⋮\\ 0& 0& \cdots & {\lambda }_{nn}^k\end{array}\right)$$

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.

知识点

特征值特征向量相似变换

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