IS Math-2017-01

题目来源：Problem 1 日期：2024-07-28 题目主题：Math-线性代数-特征值和特征向量

解题思路

这道题主要考察线性代数中矩阵特征值和特征向量的知识。我们需要通过矩阵 $A$ 的特征值分解，来逐步解决各个问题。以下是每个问题的解决思路：

1. 计算 $x_n+y_n+z_n$，需要利用矩阵 $A$ 的性质，特别是行和的性质。
2. 通过计算特征多项式来求解特征值，并利用特征值求解特征向量。
3. 使用特征值和特征向量来表达矩阵 $A$。
4. 使用特征值分解来表达向量 $\left(\begin{array}{c}{x}_n\\ y_n\\ z_n\end{array}\right)$。
5. 计算 $n\to \infty$ 时，向量 $\left(\begin{array}{c}{x}_n\\ y_n\\ z_n\end{array}\right)$ 的极限。
6. 计算函数 $f(x_0,y_0,z_0)$ 的最大值和最小值，通过矩阵 $A$ 的特征值来分析。

Solution

Question 1: Express $x_n+y_n+z_n$ using $x_0,y_0$ and $z_0$

To find the expression for $x_n+y_n+z_n$, we first note that:

$$\left(\begin{array}{c}{x}_{n+1}\\ y_{n+1}\\ z_{n+1}\end{array}\right)=A\left(\begin{array}{c}{x}_n\\ y_n\\ z_n\end{array}\right)$$

The sum $x_n+y_n+z_n$ can be written as:

$$x_n+y_n+z_n=1^{\top }\left(\begin{array}{c}{x}_n\\ y_n\\ z_n\end{array}\right),$$

where $1^{\top }=\left(\begin{array}{ccc}1& 1& 1\end{array}\right)$.

Next, consider $1^{\top }A$:

$$1^{\top }A=\left(\begin{array}{ccc}1& 1& 1\end{array}\right)\left(\begin{array}{ccc}1-2\alpha & \alpha & \alpha \\ \alpha & 1-\alpha & 0\\ \alpha & 0& 1-\alpha \end{array}\right)=\left(\begin{array}{ccc}1& 1& 1\end{array}\right)$$

Therefore, we have:

$$x_{n+1}+y_{n+1}+z_{n+1}=x_n+y_n+z_n$$

Since this relation holds for all $n$, it implies that $x_n+y_n+z_n$ is constant for all $n$. Thus,

$$x_n+y_n+z_n=x_0+y_0+z_0$$

Question 2: Obtain the eigenvalues ${\lambda }_1,$\lambda_2, and ${\lambda }_3$, and their corresponding eigenvectors ${\mathbf{v}}_1,{\mathbf{v}}_2$, and ${\mathbf{v}}_3$ of the matrix $A$

To find the eigenvalues of $A$, we solve the characteristic equation $\det (A-\lambda I)=0$.

$$A-\lambda I=\left(\begin{array}{ccc}1-2\alpha -\lambda & \alpha & \alpha \\ \alpha & 1-\alpha -\lambda & 0\\ \alpha & 0& 1-\alpha -\lambda \end{array}\right)$$

Simplifying the determinant, we get:

$$(1-\lambda )(1-3\alpha -\lambda )(1-\alpha -\lambda )=0$$

Setting each factor to zero, we find the eigenvalues:

1. ${\lambda }_1=1$.
2. ${\lambda }_2=1-3\alpha$.
3. ${\lambda }_3=1-\alpha$.

To find the corresponding eigenvectors:

For ${\lambda }_1=1$:

$$(A-I){\mathbf{v}}_1=0$$ $$\left(\begin{array}{ccc}-2\alpha & \alpha & \alpha \\ \alpha & -\alpha & 0\\ \alpha & 0& -\alpha \end{array}\right){\mathbf{v}}_1=0$$

Solving this system, one solution is ${\mathbf{v}}_1=\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right)$.

For ${\lambda }_2=1-3\alpha$:

$$(A-(1-3\alpha )I){\mathbf{v}}_2=0$$ $$\left(\begin{array}{ccc}\alpha & \alpha & \alpha \\ \alpha & 2\alpha & 0\\ \alpha & 0& 2\alpha \end{array}\right){\mathbf{v}}_2=0$$

Solving this system, one solution is ${\mathbf{v}}_2=\left(\begin{array}{c}2\\ -1\\ -1\end{array}\right)$.

For ${\lambda }_3=1-\alpha$:

$$(A-(1-\alpha )I){\mathbf{v}}_3=0$$ $$\left(\begin{array}{ccc}1-3\alpha & \alpha & \alpha \\ \alpha & -\alpha & 0\\ \alpha & 0& -\alpha \end{array}\right){\mathbf{v}}_3=0$$

Solving this system, one solution is ${\mathbf{v}}_3=\left(\begin{array}{c}0\\ 1\\ -1\end{array}\right)$.

Question 3: Express the matrix $A$ using ${\lambda }_1,{\lambda }_2,{\lambda }_3,{\mathbf{v}}_1,{\mathbf{v}}_2$ and ${\mathbf{v}}_3$

The matrix $A$ can be expressed as:

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

where $P$ is the matrix of eigenvectors and $\Lambda$ is the diagonal matrix of eigenvalues.

$$P=\left(\begin{array}{ccc}1& 2& 0\\ 1& -1& 1\\ 1& -1& -1\end{array}\right),\Lambda =\left(\begin{array}{ccc}1& 0& 0\\ 0& 1-3\alpha & 0\\ 0& 0& 1-\alpha \end{array}\right)$$

Now let’s do the row operations to get $P^{-1}$.

Step-by-Step Row Operations

Starting with the augmented matrix $\left(P\mid I\right)$:

$$\left(P\mid I\right)=\left(\begin{array}{cccccc}1& 2& 0& 1& 0& 0\\ 1& -1& 1& 0& 1& 0\\ 1& -1& -1& 0& 0& 1\end{array}\right)$$

1. Step 1: Subtract the first row from the second row:

$$R2\leftarrow R2-R1\Rightarrow \left(\begin{array}{cccccc}1& 2& 0& 1& 0& 0\\ 0& -3& 1& -1& 1& 0\\ 1& -1& -1& 0& 0& 1\end{array}\right)$$

1. Step 2: Subtract the first row from the third row:

$$R3\leftarrow R3-R1\Rightarrow \left(\begin{array}{cccccc}1& 2& 0& 1& 0& 0\\ 0& -3& 1& -1& 1& 0\\ 0& -3& -1& -1& 0& 1\end{array}\right)$$

1. Step 3: Subtract the second row from the third row:

$$R3\leftarrow R3-R2\Rightarrow \left(\begin{array}{cccccc}1& 2& 0& 1& 0& 0\\ 0& -3& 1& -1& 1& 0\\ 0& 0& -2& 0& -1& 1\end{array}\right)$$

1. Step 4: Multiply the second row by $-\frac{1}{3}$:

$$R2\leftarrow -\frac{1}{3}R2\Rightarrow \left(\begin{array}{cccccc}1& 2& 0& 1& 0& 0\\ 0& 1& -\frac{1}{3}& \frac{1}{3}& -\frac{1}{3}& 0\\ 0& 0& -2& 0& -1& 1\end{array}\right)$$

1. Step 5: Multiply the third row by $-\frac{1}{2}$:

$$R3\leftarrow -\frac{1}{2}R3\Rightarrow \left(\begin{array}{cccccc}1& 2& 0& 1& 0& 0\\ 0& 1& -\frac{1}{3}& \frac{1}{3}& -\frac{1}{3}& 0\\ 0& 0& 1& 0& \frac{1}{2}& -\frac{1}{2}\end{array}\right)$$

1. Step 6: Subtract 2 times the second row from the first row:

$$R1\leftarrow R1-2R2\Rightarrow \left(\begin{array}{cccccc}1& 0& \frac{2}{3}& \frac{1}{3}& \frac{2}{3}& 0\\ 0& 1& -\frac{1}{3}& \frac{1}{3}& -\frac{1}{3}& 0\\ 0& 0& 1& 0& \frac{1}{2}& -\frac{1}{2}\end{array}\right)$$

1. Step 7: Add $\frac{1}{3}$ times the third row to the second row:

$$R2\leftarrow R2+\frac{1}{3}R3\Rightarrow \left(\begin{array}{cccccc}1& 0& \frac{2}{3}& \frac{1}{3}& \frac{2}{3}& 0\\ 0& 1& 0& \frac{1}{3}& -\frac{1}{6}& -\frac{1}{6}\\ 0& 0& 1& 0& \frac{1}{2}& -\frac{1}{2}\end{array}\right)$$

1. Step 8: Subtract $\frac{2}{3}$ times the third row from the first row:

$$R1\leftarrow R1-\frac{2}{3}R3\Rightarrow \left(\begin{array}{cccccc}1& 0& 0& \frac{1}{3}& \frac{1}{3}& \frac{1}{3}\\ 0& 1& 0& \frac{1}{3}& -\frac{1}{6}& -\frac{1}{6}\\ 0& 0& 1& 0& \frac{1}{2}& -\frac{1}{2}\end{array}\right)$$

Thus, the inverse matrix $P^{-1}$ is:

$$P^{-1}=\left(\begin{array}{ccc}\frac{1}{3}& \frac{1}{3}& \frac{1}{3}\\ \frac{1}{3}& -\frac{1}{6}& -\frac{1}{6}\\ 0& \frac{1}{2}& -\frac{1}{2}\end{array}\right)$$

Question 4: Express $\left(\begin{array}{c}{x}_n\\ y_n\\ z_n\end{array}\right)$ using $x_0,y_0,z_0$ and $\alpha$

To express $\left(\begin{array}{c}{x}_n\\ y_n\\ z_n\end{array}\right)$ using $x_0,y_0,z_0$ and $\alpha$, we need to compute the product $P{\Lambda }^n{P}^{-1}$.

Given:

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

where

$$P=\left(\begin{array}{ccc}1& 2& 0\\ 1& -1& 1\\ 1& -1& -1\end{array}\right),\Lambda =\left(\begin{array}{ccc}1& 0& 0\\ 0& (1-3\alpha {)}^n& 0\\ 0& 0& (1-\alpha {)}^n\end{array}\right),P^{-1}=\left(\begin{array}{ccc}\frac{1}{3}& \frac{1}{3}& \frac{1}{3}\\ \frac{1}{3}& -\frac{1}{6}& -\frac{1}{6}\\ 0& \frac{1}{2}& -\frac{1}{2}\end{array}\right)$$

We need to compute:

$$P{\Lambda }^n{P}^{-1}$$

First, let’s compute ${\Lambda }^n$:

$${\Lambda }^n=\left(\begin{array}{ccc}1& 0& 0\\ 0& (1-3\alpha {)}^n& 0\\ 0& 0& (1-\alpha {)}^n\end{array}\right)$$

Next, we compute $P{\Lambda }^n$:

$$P{\Lambda }^n=\left(\begin{array}{ccc}1& 2& 0\\ 1& -1& 1\\ 1& -1& -1\end{array}\right)\left(\begin{array}{ccc}1& 0& 0\\ 0& (1-3\alpha {)}^n& 0\\ 0& 0& (1-\alpha {)}^n\end{array}\right)$$

Performing the multiplication:

$$P{\Lambda }^n=\left(\begin{array}{ccc}1& 2(1-3\alpha {)}^n& 0\\ 1& -(1-3\alpha {)}^n& (1-\alpha {)}^n\\ 1& -(1-3\alpha {)}^n& -(1-\alpha {)}^n\end{array}\right)$$

Finally, we compute $P{\Lambda }^n{P}^{-1}$:

$$P{\Lambda }^n{P}^{-1}=\left(\begin{array}{ccc}1& 2(1-3\alpha {)}^n& 0\\ 1& -(1-3\alpha {)}^n& (1-\alpha {)}^n\\ 1& -(1-3\alpha {)}^n& -(1-\alpha {)}^n\end{array}\right)\left(\begin{array}{ccc}\frac{1}{3}& \frac{1}{3}& \frac{1}{3}\\ \frac{1}{3}& -\frac{1}{6}& -\frac{1}{6}\\ 0& \frac{1}{2}& -\frac{1}{2}\end{array}\right)$$

Performing this multiplication step by step:

1. Compute the first row of the product:

$$\left(\begin{array}{ccc}1& 2(1-3\alpha {)}^n& 0\end{array}\right)\left(\begin{array}{ccc}\frac{1}{3}& \frac{1}{3}& \frac{1}{3}\\ \frac{1}{3}& -\frac{1}{6}& -\frac{1}{6}\\ 0& \frac{1}{2}& -\frac{1}{2}\end{array}\right)=\left(\begin{array}{ccc}\frac{1}{3}+\frac{2}{3}(1-3\alpha {)}^n& \frac{1}{3}+-\frac{1}{3}(1-3\alpha {)}^n& \frac{1}{3}+-\frac{1}{3}(1-3\alpha {)}^n\end{array}\right)$$

1. Compute the second row of the product:

$$\left(\begin{array}{ccc}1& -(1-3\alpha {)}^n& (1-\alpha {)}^n\end{array}\right)\left(\begin{array}{ccc}\frac{1}{3}& \frac{1}{3}& \frac{1}{3}\\ \frac{1}{3}& -\frac{1}{6}& -\frac{1}{6}\\ 0& \frac{1}{2}& -\frac{1}{2}\end{array}\right)=\left(\begin{array}{ccc}\frac{1}{3}-\frac{1}{3}(1-3\alpha {)}^n& \frac{1}{3}-\frac{1}{3}(1-3\alpha {)}^n+\frac{1}{2}(1-\alpha {)}^n& \frac{1}{3}-\frac{1}{3}(1-3\alpha {)}^n-\frac{1}{2}(1-\alpha {)}^n\end{array}\right)$$

1. Compute the third row of the product:

$$\left(\begin{array}{ccc}1& -(1-3\alpha {)}^n& -(1-\alpha {)}^n\end{array}\right)\left(\begin{array}{ccc}\frac{1}{3}& \frac{1}{3}& \frac{1}{3}\\ \frac{1}{3}& -\frac{1}{6}& -\frac{1}{6}\\ 0& \frac{1}{2}& -\frac{1}{2}\end{array}\right)=\left(\begin{array}{ccc}\frac{1}{3}-\frac{1}{3}(1-3\alpha {)}^n& \frac{1}{3}-\frac{1}{3}(1-3\alpha {)}^n-\frac{1}{2}(1-\alpha {)}^n& \frac{1}{3}-\frac{1}{3}(1-3\alpha {)}^n+\frac{1}{2}(1-\alpha {)}^n\end{array}\right)$$

Putting it all together, we have:

$$P{\Lambda }^n{P}^{-1}=\left(\begin{array}{ccc}\frac{1}{3}+\frac{2}{3}(1-3\alpha {)}^n& \frac{1}{3}-\frac{1}{3}(1-3\alpha {)}^n& \frac{1}{3}-\frac{1}{3}(1-3\alpha {)}^n\\ \frac{1}{3}-\frac{1}{3}(1-3\alpha {)}^n& \frac{1}{3}-\frac{1}{3}(1-3\alpha {)}^n+\frac{1}{2}(1-\alpha {)}^n& \frac{1}{3}-\frac{1}{3}(1-3\alpha {)}^n-\frac{1}{2}(1-\alpha {)}^n\\ \frac{1}{3}-\frac{1}{3}(1-3\alpha {)}^n& \frac{1}{3}-\frac{1}{3}(1-3\alpha {)}^n-\frac{1}{2}(1-\alpha {)}^n& \frac{1}{3}-\frac{1}{3}(1-3\alpha {)}^n+\frac{1}{2}(1-\alpha {)}^n\end{array}\right)$$

This matrix represents $A^n$. To get ${\mathbf{v}}_n$, we multiply this result by ${\mathbf{v}}_0$:

$${\mathbf{v}}_n=\left(P{\Lambda }^n{P}^{-1}\right){\mathbf{v}}_0$$

Thus, the final result is:

$$\left(\begin{array}{c}{x}_n\\ y_n\\ z_n\end{array}\right)=\left(\begin{array}{ccc}\frac{1}{3}+\frac{2}{3}(1-3\alpha {)}^n& \frac{1}{3}-\frac{1}{3}(1-3\alpha {)}^n& \frac{1}{3}-\frac{1}{3}(1-3\alpha {)}^n\\ \frac{1}{3}-\frac{1}{3}(1-3\alpha {)}^n& \frac{1}{3}-\frac{1}{3}(1-3\alpha {)}^n+\frac{1}{2}(1-\alpha {)}^n& \frac{1}{3}-\frac{1}{3}(1-3\alpha {)}^n-\frac{1}{2}(1-\alpha {)}^n\\ \frac{1}{3}-\frac{1}{3}(1-3\alpha {)}^n& \frac{1}{3}-\frac{1}{3}(1-3\alpha {)}^n-\frac{1}{2}(1-\alpha {)}^n& \frac{1}{3}-\frac{1}{3}(1-3\alpha {)}^n+\frac{1}{2}(1-\alpha {)}^n\end{array}\right)\left(\begin{array}{c}{x}_0\\ y_0\\ z_0\end{array}\right)$$

Question 5: Obtain ${\lim }_{n\to \infty }\left(\begin{array}{c}{x}_n\\ y_n\\ z_n\end{array}\right)$

Given the result from Question 4, we have:

$$\left(\begin{array}{c}{x}_n\\ y_n\\ z_n\end{array}\right)=\left(\begin{array}{ccc}\frac{1}{3}+\frac{2}{3}(1-3\alpha {)}^n& \frac{1}{3}-\frac{1}{3}(1-3\alpha {)}^n& \frac{1}{3}-\frac{1}{3}(1-3\alpha {)}^n\\ \frac{1}{3}-\frac{1}{3}(1-3\alpha {)}^n& \frac{1}{3}-\frac{1}{3}(1-3\alpha {)}^n+\frac{1}{2}(1-\alpha {)}^n& \frac{1}{3}-\frac{1}{3}(1-3\alpha {)}^n-\frac{1}{2}(1-\alpha {)}^n\\ \frac{1}{3}-\frac{1}{3}(1-3\alpha {)}^n& \frac{1}{3}-\frac{1}{3}(1-3\alpha {)}^n-\frac{1}{2}(1-\alpha {)}^n& \frac{1}{3}-\frac{1}{3}(1-3\alpha {)}^n+\frac{1}{2}(1-\alpha {)}^n\end{array}\right)\left(\begin{array}{c}{x}_0\\ y_0\\ z_0\end{array}\right)$$

To find the limit as $n\to \infty$, we need to consider the behavior of the terms $(1-3\alpha {)}^n$ and $(1-\alpha {)}^n$.

Given that $0<\alpha <\frac{1}{3}$:

• $|1-3\alpha |<1$
• $|1-\alpha |<1$

As $n\to \infty$:

• $(1-3\alpha {)}^n\to 0$
• $(1-\alpha {)}^n\to 0$

Thus, the terms involving $(1-3\alpha {)}^n$ and $(1-\alpha {)}^n$ will approach zero.

Therefore, the limit of the matrix as $n\to \infty$ is:

$$\underset{n\to \infty }{\lim }P{\Lambda }^n{P}^{-1}=\left(\begin{array}{ccc}\frac{1}{3}& \frac{1}{3}& \frac{1}{3}\\ \frac{1}{3}& \frac{1}{3}& \frac{1}{3}\\ \frac{1}{3}& \frac{1}{3}& \frac{1}{3}\end{array}\right)$$

So, the limit of the vector $\left(\begin{array}{c}{x}_n\\ y_n\\ z_n\end{array}\right)$ is:

$$\underset{n\to \infty }{\lim }\left(\begin{array}{c}{x}_n\\ y_n\\ z_n\end{array}\right)=\left(\begin{array}{ccc}\frac{1}{3}& \frac{1}{3}& \frac{1}{3}\\ \frac{1}{3}& \frac{1}{3}& \frac{1}{3}\\ \frac{1}{3}& \frac{1}{3}& \frac{1}{3}\end{array}\right)\left(\begin{array}{c}{x}_0\\ y_0\\ z_0\end{array}\right)$$

Multiplying this matrix by the vector $\left(\begin{array}{c}{x}_0\\ y_0\\ z_0\end{array}\right)$:

$$\underset{n\to \infty }{\lim }\left(\begin{array}{c}{x}_n\\ y_n\\ z_n\end{array}\right)=\left(\begin{array}{c}\frac{1}{3}(x_0+y_0+z_0)\\ \frac{1}{3}(x_0+y_0+z_0)\\ \frac{1}{3}(x_0+y_0+z_0)\end{array}\right)$$

Explanation using the largest eigenvalue

The behavior of the system as $n\to \infty$ can also be explained using the largest eigenvalue. The eigenvalues of $A$ are ${\lambda }_1=1$, ${\lambda }_2=1-3\alpha$, and ${\lambda }_3=1-\alpha$.

Since $0<\alpha <\frac{1}{3}$, we have:

• $|{\lambda }_2|=|1-3\alpha |<1$
• $|{\lambda }_3|=|1-\alpha |<1$

The largest eigenvalue ${\lambda }_1=1$ dominates the behavior of the matrix as $n\to \infty$. The terms involving ${\lambda }_2$ and ${\lambda }_3$ will decay to zero because their absolute values are less than 1.

Therefore, the dominant term is associated with the eigenvalue ${\lambda }_1=1$, and the corresponding eigenvector is $\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right)$. This explains why the limit is:

$$\underset{n\to \infty }{\lim }\left(\begin{array}{c}{x}_n\\ y_n\\ z_n\end{array}\right)=c\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right)$$

where $c=\frac{1}{3}(x_0+y_0+z_0)$.

Thus, the final result is:

$$\underset{n\to \infty }{\lim }\left(\begin{array}{c}{x}_n\\ y_n\\ z_n\end{array}\right)=\left(\begin{array}{c}\frac{1}{3}(x_0+y_0+z_0)\\ \frac{1}{3}(x_0+y_0+z_0)\\ \frac{1}{3}(x_0+y_0+z_0)\end{array}\right)$$

Question 6: Obtain the maximum and the minimum values of $f(x_0,y_0,z_0)$

Given that the matrix $A$ is symmetric, we can use its orthogonal decomposition. Specifically, we can decompose $A$ as:

$$A=Q\Lambda Q^{\top }$$

where $Q$ is an orthogonal matrix (i.e., $Q{Q}^{\top }=I$), and $\Lambda$ is a diagonal matrix containing the eigenvalues of $A$.

The eigenvalues of $A$ are:

1. ${\lambda }_1=1$
2. ${\lambda }_2=1-3\alpha$
3. ${\lambda }_3=1-\alpha$

The function $f(x_0,y_0,z_0)$ is given by:

$$f(x_0,y_0,z_0)=\frac{\left(\left(\begin{array}{c}{x}_n\\ y_n\\ z_n\end{array}\right),\left(\begin{array}{c}{x}_{n+1}\\ y_{n+1}\\ z_{n+1}\end{array}\right)\right)}{\left(\left(\begin{array}{c}{x}_n\\ y_n\\ z_n\end{array}\right),\left(\begin{array}{c}{x}_n\\ y_n\\ z_n\end{array}\right)\right)}$$

Using the orthogonal decomposition $A=Q\Lambda Q^{\top }$, we have:

$$\left(\begin{array}{c}{x}_{n+1}\\ y_{n+1}\\ z_{n+1}\end{array}\right)=A\left(\begin{array}{c}{x}_n\\ y_n\\ z_n\end{array}\right)$$

Given:

$$\left(\begin{array}{c}{x}_n\\ y_n\\ z_n\end{array}\right)=Q{\Lambda }^n{Q}^{\top }\left(\begin{array}{c}{x}_0\\ y_0\\ z_0\end{array}\right)$$

We can express ${\mathbf{v}}_n$ as:

$${\mathbf{v}}_n=Q{\Lambda }^n{Q}^{\top }{\mathbf{v}}_0$$

Therefore:

$${\mathbf{v}}_{n+1}=Q\Lambda Q^{\top }Q{\Lambda }^n{Q}^{\top }{\mathbf{v}}_0=Q{\Lambda }^{n+1}{Q}^{\top }{\mathbf{v}}_0$$

To compute $f(x_0,y_0,z_0)$:

$$f(x_0,y_0,z_0)=\frac{(Q{\Lambda }^n{Q}^{\top }{\mathbf{v}}_0{)}^{\top }(Q\Lambda Q^{\top }Q{\Lambda }^n{Q}^{\top }{\mathbf{v}}_0)}{(Q{\Lambda }^n{Q}^{\top }{\mathbf{v}}_0{)}^{\top }(Q{\Lambda }^n{Q}^{\top }{\mathbf{v}}_0)}$$

Using the fact that $Q$ is orthogonal ($Q^{\top }Q=I$):

$$f(x_0,y_0,z_0)=\frac{({\Lambda }^n{Q}^{\top }{\mathbf{v}}_0{)}^{\top }(\Lambda {\Lambda }^n{Q}^{\top }{\mathbf{v}}_0)}{({\Lambda }^n{Q}^{\top }{\mathbf{v}}_0{)}^{\top }({\Lambda }^n{Q}^{\top }{\mathbf{v}}_0)}$$

Simplifying further:

$$f(x_0,y_0,z_0)=\frac{(Q^{\top }{\mathbf{v}}_0{)}^{\top }{\Lambda }^n\Lambda {\Lambda }^n(Q^{\top }{\mathbf{v}}_0)}{(Q^{\top }{\mathbf{v}}_0{)}^{\top }{\Lambda }^{2n}(Q^{\top }{\mathbf{v}}_0)}$$

Let $\mathbf{w}=Q^{\top }{\mathbf{v}}_0$, so:

$$f(x_0,y_0,z_0)=\frac{{\mathbf{w}}^{\top }{\Lambda }^{n+1}\mathbf{w}}{{\mathbf{w}}^{\top }{\Lambda }^{2n}\mathbf{w}}$$

The matrix $\Lambda$ is diagonal with eigenvalues ${\lambda }_1=1$, ${\lambda }_2=1-3\alpha$, and ${\lambda }_3=1-\alpha$, so:

$${\Lambda }^n=\left(\begin{array}{ccc}{1}^n& 0& 0\\ 0& (1-3\alpha {)}^n& 0\\ 0& 0& (1-\alpha {)}^n\end{array}\right)$$

Thus,

$${\Lambda }^{n+1}=\left(\begin{array}{ccc}{1}^{n+1}& 0& 0\\ 0& (1-3\alpha {)}^{n+1}& 0\\ 0& 0& (1-\alpha {)}^{n+1}\end{array}\right),{\Lambda }^{2n}=\left(\begin{array}{ccc}{1}^{2n}& 0& 0\\ 0& (1-3\alpha {)}^{2n}& 0\\ 0& 0& (1-\alpha {)}^{2n}\end{array}\right)$$

Therefore:

$$f(x_0,y_0,z_0)=\frac{w_1^2+w_2^2(1-3\alpha {)}^{2n+1}+w_3^2(1-\alpha {)}^{2n+1}}{w_1^2+w_2^2(1-3\alpha {)}^{2n}+w_3^2(1-\alpha {)}^{2n}}$$

Analysis

The function $f(x_0,y_0,z_0)$ depends on the components $w_1,w_2,w_3$, which are the coordinates of the vector $\mathbf{w}=Q^{\top }{\mathbf{v}}_0$. The eigenvalues ${\lambda }_1=1$, ${\lambda }_2=1-3\alpha$, and ${\lambda }_3=1-\alpha$ influence the function as constants.

Finding Maximum and Minimum Values

Maximum Value

To maximize $f(x_0,y_0,z_0)$, consider the scenario where $\mathbf{w}$ aligns with the eigenvector corresponding to the largest eigenvalue, which is ${\lambda }_1=1$:

If $\mathbf{w}$ is aligned with $\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)$:

$$w_1=\Vert \mathbf{w}\Vert ,w_2=0,w_3=0$$

Substituting into the function:

$$f(x_0,y_0,z_0)=\frac{w_1^2\cdot 1}{w_1^2\cdot 1}=1$$

Thus, the maximum value of $f(x_0,y_0,z_0)$ is:

$$\max f(x_0,y_0,z_0)=1$$

Minimum Value

To minimize $f(x_0,y_0,z_0)$, consider the scenario where $\mathbf{w}$ aligns with the eigenvector corresponding to the smallest eigenvalue, which is ${\lambda }_2=1-3\alpha$:

If $\mathbf{w}$ is aligned with $\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right)$:

$$w_1=0,w_2=\Vert \mathbf{w}\Vert ,w_3=0$$

Substituting into the function:

$$f(x_0,y_0,z_0)=\frac{w_2^2(1-3\alpha )}{w_2^2}=1-3\alpha$$

Thus, the minimum value of $f(x_0,y_0,z_0)$ is:

$$\min f(x_0,y_0,z_0)=1-3\alpha$$

Summary

The maximum and minimum values of the function $f(x_0,y_0,z_0)$ are determined by the largest and smallest eigenvalues of the matrix $A$, respectively:

$$\max f(x_0,y_0,z_0)=1,\min f(x_0,y_0,z_0)=1-3\alpha$$

知识点

线性代数特征值和特征向量正交分解Rayleigh商

难点思路

在本题中，我们需要通过矩阵分解和特征值的性质来计算函数 $f(x_0,y_0,z_0)$ 的最大值和最小值。理解和应用 Rayleigh 商是解决此类问题的关键。

解题技巧和信息

1. 矩阵的正交分解：利用矩阵 $A$ 的正交分解 $A=Q\Lambda Q^{\top }$，可以简化特征值问题的计算。
2. Rayleigh 商：Rayleigh 商在特征值分析中用于找到最大和最小特征值，这对确定函数 $f(x_0,y_0,z_0)$ 的范围非常有用。
3. 特征值和特征向量的对齐：Rayleigh 商最大值和最小值分别对应于二次型向量沿最大和最小特征值方向的值。

重点词汇

• eigenvalue 特征值
• eigenvector 特征向量
• matrix decomposition 矩阵分解
• characteristic polynomial 特征多项式

参考资料

1. Linear Algebra by Gilbert Strang, Chap. 5
2. Introduction to Linear Algebra by Serge Lang, Chap. 6