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“矩阵代数初步”(Introduction to MATRIX ALGEBRA)课程由Prof. A.K.Kaw(University of South Florida)设计并讲授。
PDF格式学习笔记下载（Academia.edu）
第10章课程讲义下载（PDF）
Summary
A = matrix(c(1.5, -0.5, -0.5, 0, 0.5, 0, 1, -0.5, 0), ncol = 3)
PowerEigen(A, x0 = c(1, 1, 1))
Converged after 23 iterations
$value
[1] 1
$vector
[,1]
[1,] 1.0
[2,] -0.5
[3,] -0.5
Selected Problems
1. The eigenvalues $\lambda$ of matrix $[A]$ are found by solving the equation ( ).
Solution: $$|A-\lambda I| = 0$$
2. Find the eigenvalues and eigenvectors of $$[A] = \begin{bmatrix} 10& 9\\ 2& 3\end{bmatrix}$$ using the determinant method.
Solution: $$|A-\lambda I| = 0$$ $$\Rightarrow \det\left(\begin{bmatrix} 10-\lambda & 9\\ 2 & 3-\lambda\end{bmatrix}\right) = 0$$ $$\Rightarrow (10-\lambda)(3-\lambda) - 18=0$$ $$\Rightarrow \lambda^2 - 13\lambda +12 =0$$ $$\Rightarrow \lambda_1=1,\ \lambda_2=12$$ For $\lambda_1=1$, we have $$\begin{bmatrix} 10-\lambda & 9\\ 2 & 3-\lambda\end{bmatrix} \begin{bmatrix}x_1 \\ x_2\end{bmatrix} =\begin{bmatrix} 0\\ 0\end{bmatrix} $$ $$\Rightarrow \begin{bmatrix} 9 & 9\\ 2 & 2\end{bmatrix}\begin{bmatrix}x_1 \\ x_2\end{bmatrix} = \begin{bmatrix} 0\\ 0 \end{bmatrix}$$ $$\Rightarrow x_2 =-x_1$$ $$\Rightarrow X=\begin{bmatrix}x_1\\ -x_1 \end{bmatrix} = x_1\begin{bmatrix} 1\\ -1\end{bmatrix}$$ Thus the eigenvector corresponding to $\lambda_1=1$ is $\begin{bmatrix} 1\\ -1\end{bmatrix}$. Similarly, we can find the second eigenvector corresponding to $\lambda_2=12$: $$\begin{bmatrix} 10-\lambda & 9\\ 2 & 3-\lambda\end{bmatrix} \begin{bmatrix}x_1 \\ x_2\end{bmatrix} =\begin{bmatrix} 0\\ 0\end{bmatrix} $$ $$\Rightarrow \begin{bmatrix} -2 & 9\\ 2 & -9\end{bmatrix}\begin{bmatrix}x_1 \\ x_2\end{bmatrix} = \begin{bmatrix} 0\\ 0 \end{bmatrix}$$ $$\Rightarrow -2x_1+9x_2 = 0 \Rightarrow x_2 = {2\over9}x_1$$ $$\Rightarrow X=\begin{bmatrix}x_1\\ {2\over9}x_1 \end{bmatrix} = x_1\begin{bmatrix} 1\\ {2\over9}\end{bmatrix} \Rightarrow \begin{bmatrix} 9\\ 2\end{bmatrix}$$ Thus the eigenvector corresponding to $\lambda_2=12$ is $\begin{bmatrix} 9\\ 2\end{bmatrix}$.
3. Find the eigenvalues and eigenvectors of $$[A] = \begin{bmatrix}4& 0& 1\\ -2& 0& 1\\ 2& 0& 1\end{bmatrix}$$ using the determinant method.
Solution:
First of all, we can read off that $\lambda = 0$ is an eigenvalue of this matrix since it is singular. Then from the definition we have $$|A-\lambda I| = 0$$ $$\Rightarrow \det\left( \begin{bmatrix} 4-\lambda & 0 & 1\\ -2 & -\lambda & 1\\ 2 & 0& 1-\lambda \end{bmatrix}\right) = 0$$ $$\Rightarrow (4-\lambda)\left[(-\lambda)(1-\lambda) - 0\right]+\left[1\cdot(0+2\lambda)\right] =0$$ $$\Rightarrow (4-\lambda)(\lambda^2-\lambda) +2\lambda= 0$$ $$\Rightarrow \lambda(-\lambda^2+5\lambda-4+2) =0$$ $$\Rightarrow \lambda(\lambda^2-5\lambda+2) =0$$ $$\Rightarrow \lambda_1=0,\ \lambda_2 = 4.561553,\ \lambda_3=0.4384472.$$ For $\lambda_1 =0$, we have $$\begin{bmatrix} 4-\lambda & 0 & 1\\ -2 & -\lambda & 1\\ 2 & 0& 1-\lambda \end{bmatrix}\begin{bmatrix}x_1\\ x_2\\ x_3 \end{bmatrix}=\begin{bmatrix}0\\ 0\\ 0 \end{bmatrix}$$ $$\Rightarrow \begin{bmatrix} 4 & 0 & 1\\ -2 & 0 & 1\\ 2 & 0& 1 \end{bmatrix}\begin{bmatrix}x_1\\ x_2\\ x_3 \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}$$ The coefficient matrix is $$\begin{bmatrix} 4& 0& 1\\ -2& 0& 1\\ 2& 0& 1 \end{bmatrix} \Rightarrow \begin{bmatrix} 0& 0& 3\\ -2& 0& 1\\ 0& 0& 2 \end{bmatrix} \Rightarrow \begin{bmatrix} 0& 0& 0\\ -2& 0& 0\\ 0& 0& 2 \end{bmatrix}$$ that is, $x_1=x_3=0$ and $x_2$ is arbitrary. Hence the eigenvector corresponding to $\lambda_1=0$ is $\begin{bmatrix}0 \\ 1\\ 0 \end{bmatrix}$. For $\lambda_2= 4.561553$, we have $$\begin{bmatrix} 4-\lambda & 0 & 1\\ -2 & -\lambda & 1\\ 2 & 0& 1-\lambda \end{bmatrix} \begin{bmatrix}x_1\\ x_2\\ x_3 \end{bmatrix}=\begin{bmatrix}0\\ 0\\ 0 \end{bmatrix}$$ $$\Rightarrow \begin{bmatrix} -0.561553 & 0 & 1\\ -2 & -4.561553 & 1\\ 2 & 0& -3.561553 \end{bmatrix}\begin{bmatrix}x_1\\ x_2\\ x_3 \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}$$ The coefficient matrix is $$\begin{bmatrix} -0.561553 & 0 & 1\\ -2 & -4.561553 & 1\\ 2 & 0& -3.561553 \end{bmatrix} \Rightarrow \begin{bmatrix} -0.561553 & 0 & 1\\ 0 & -4.561553 & -2.561553\\ 0 & 0& 0 \end{bmatrix}$$ $$\Rightarrow \begin{cases}x_1= 1.780776x_3\\ x_2 = -0.5615528x_3\end{cases}$$ where $x_3$ is arbitrary. Thus the eigenvector corresponding to $\lambda_2=4.561553$ is $\begin{bmatrix}1.780776\\ -0.5615528\\ 1 \end{bmatrix}$. For $\lambda_3= 0.4384472$, we have $$\begin{bmatrix} 4-\lambda & 0 & 1\\ -2 & -\lambda & 1\\ 2 & 0& 1-\lambda \end{bmatrix} \begin{bmatrix}x_1\\ x_2\\ x_3 \end{bmatrix}=\begin{bmatrix}0\\ 0\\ 0 \end{bmatrix}$$ $$\Rightarrow \begin{bmatrix} 3.561553 & 0 & 1\\ -2 & -0.4384472 & 1\\ 2 & 0& 0.5615528 \end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}$$ The coefficient matrix is $$\begin{bmatrix} 3.561553 & 0 & 1\\ -2 & -0.4384472 & 1\\ 2 & 0& 0.5615528 \end{bmatrix} \Rightarrow \begin{bmatrix} 3.561553 & 0 & 1\\ 0 & -0.4384472 & 1.561553\\ 0 & 0& 0 \end{bmatrix}$$ $$\Rightarrow \begin{cases}x_1= -0.2807764x_3\\ x_2 = 3.561553x_3\end{cases}$$ where $x_3$ is arbitrary. Thus the eigenvector corresponding to $\lambda_3= 0.4384472$ is $\begin{bmatrix}-0.2807764\\ 3.561553 \\ 1 \end{bmatrix}$.
4. Find the eigenvalues of these matrices by inspection: (A) $\begin{bmatrix}2& 0& 0\\ 0& -3& 0\\ 0& 0& 6\end{bmatrix}$; (B) $\begin{bmatrix}3& 5& 7\\ 0& -2& 1\\ 0& 0& 0\end{bmatrix}$; (C) $\begin{bmatrix}2& 0& 0\\ 3& 5& 0\\ 2& 1& 6\end{bmatrix}$.
Solution:
The eigenvalues of a triangular matrix are the diagonal entries of the matrix. Thus, (A) $\lambda_1=2,\ \lambda_2=-3,\ \lambda_3=6$. (B) $\lambda_1=3,\ \lambda_2=-2,\ \lambda_3=0$. (C) $\lambda_1=2,\ \lambda_2=5,\ \lambda_3=6$.
5. Find the largest eigenvalue in magnitude and its corresponding vector by using the power method $$[A] = \begin{bmatrix}4& 0& 1\\ -2& 0& 1\\ 2& 0& 1 \end{bmatrix}$$ Start with an initial guess of the eigenvector as $\begin{bmatrix}1\\ -0.5\\ 0.5 \end{bmatrix}$.
Solution:
We will use the R script directly,
A = matrix(c(4, -2, 2, 0, 0, 0, 1, 1, 1), ncol = 3)
PowerEigen(A, x0 = c(1, -0.5, -0.5))
Converged after 9 iterations
$value
[1] 4.561553
$vector
[,1]
[1,] 1.0000000
[2,] -0.3153416
[3,] 0.5615528
6. Prove if $\lambda$ is an eigenvalue of $[A]$, then ${1\over\lambda}$ is an eigenvalue of $[A]^{-1}$.
Solution:
We hope to prove that $A^{-1}X={1\over\lambda}X$ where $AX=\lambda X$. $$A^{-1}X=A^{-1}(\lambda \cdot {1\over\lambda}) X = {1\over\lambda} A^{-1}\lambda X = {1\over\lambda} A^{-1}A X = {1\over\lambda}X$$
7. Prove that square matrices $[A]$ and $[A]^{T}$ have the same eigenvalues.
Solution:
We hope to prove that $\det(A-\lambda I) = \det(A^{T}-\lambda I)$, and an important result is $\det(A) = \det\left(A^{T}\right)$ for $A$ is a square matrix. $$\det(A-\lambda I) = \det\left((A-\lambda I)^{T}\right)$$ $$=\det\left(A^{T}-(\lambda I)^{T}\right)$$ $$=\det\left(A^{T}-\lambda I\right)$$
8. Show that $|\det(A)|$ is the product of the absolute values of the eigenvalues of $[A]$.
Solution:
We hope to prove that $$|\det(A)| =\prod_{i=1}^{n}|\lambda_i|$$ where $\lambda_i$ is the eigenvalues of matrix $A$. By the definition we have $$|\det(A-\lambda I)| = |f(\lambda)| =|(\lambda_1-\lambda)(\lambda_2-\lambda)\cdots(\lambda_n-\lambda)| $$ Set $\lambda=0$ (since it is a variable), we have $$|\det(A)| = |\lambda_1\lambda_2\cdots\lambda_n|= \prod_{i=1}^{n}|\lambda_i|$$
9. What are the eigenvalues of the following matrix? $$\begin{bmatrix}5& 6& 17\\ 0& -19& 23\\ 0& 0& 37 \end{bmatrix}$$
Solution:
This is an upper triangular matrix, hence its eigenvalues are the diagonal elements, that is, 5, -19, and 37.
10. If $\begin{bmatrix}-4.5\\ -4\\ 1 \end{bmatrix}$ is an eigenvector of $\begin{bmatrix}8& -4& 2\\ 4& 0& 2\\ 0& -2& -4 \end{bmatrix}$, what is the eigenvalue corresponding to the eigenvector?
Solution:
From the definition we have $AX=\lambda X$, that is $$\begin{bmatrix}8& -4& 2\\ 4& 0& 2\\ 0& -2& -4 \end{bmatrix} \begin{bmatrix}-4.5\\ -4\\ 1 \end{bmatrix} =\lambda \begin{bmatrix}-4.5\\ -4\\ 1 \end{bmatrix}$$ $$\Rightarrow \begin{bmatrix}-18\\ -16\\ 4 \end{bmatrix} = \lambda \begin{bmatrix}-4.5\\ -4\\ 1 \end{bmatrix}$$ Hence $\lambda = 4$.
11. The eigenvalues of the following matrix $$\begin{bmatrix}3& 2& 9\\ 7& 5& 13\\ 6& 17& 19\end{bmatrix}$$ are given by solving the cubic equation ( ).
Solution: $$|A-\lambda I| =\det\left( \begin{bmatrix}3-\lambda& 2& 9\\ 7& 5-\lambda& 13\\ 6& 17& 19-\lambda\end{bmatrix}\right)$$ $$= (3-\lambda)\begin{vmatrix}5-\lambda & 13\\ 17 & 19-\lambda\end{vmatrix} - 2\begin{vmatrix}7 & 13\\ 6 & 19-\lambda\end{vmatrix} + 9\begin{vmatrix}7 & 5-\lambda\\ 6 & 17\end{vmatrix}$$ $$= (3-\lambda)\left((5-\lambda)(19-\lambda) - 13\times17\right) - 2\times \left(7(19 - \lambda) - 6 \times 13 \right) + 9 \left(7\times17-6(5-\lambda)\right)$$ $$=\lambda^3 - 27\lambda^2 -122\lambda -313$$
12. The eigenvalues of a $4\times4$ matrix $[A]$ are given as 2, -3, 13, and 7. What is the $|\det(A)|$?
Solution:
Since for a $n\times n$ matrix $$|\det(A)| = \prod_{i=1}^{n}|\lambda_i|$$ Hence we have $$|\det(A)| = |2\times(-3)\times13\times7| = 546$$
13. If one of the eigenvalues of $[A]_{n\times n}$ is zero, it implies ( ).
Solution:
If an eigenvalue is zero, then its determinant must be zero. Furthermore, this means it is a singular matrix (i.e. non-invertible).
14. Given that matrix $$[A] = \begin{bmatrix}8& -4& 2\\ 4& 0& 2\\ 0& -2& -3 \end{bmatrix}$$ has an eigenvalue value of 4 with the corresponding eigenvectors of $[x]=\begin{bmatrix}-4.5\\ -4\\ 1\end{bmatrix}$, then what is the value of $[A]^{5}[X]$?
Solution:
Firstly, we show that $A^{m}X=\lambda^{m}X$, where $\lambda$ is an eigenvalue of $[A]$. By Mathematical Induction, we can read off that $n=1$ is correct.\\ Then suppose that $n=m-1$ is correct, that is, $A^{m-1}X = \lambda^{m-1}X$ holds. For $n=m$, we have $$A^{m}X = AA^{m-1}X = A\lambda^{m-1}X =\lambda^{m-1}AX = \lambda^{m-1}\lambda X =\lambda^{m}X$$ as desired. From this result, we have $$A^5X=\lambda^{5}X = 4^5\begin{bmatrix}-4.5\\ -4\\ 1\end{bmatrix} = \begin{bmatrix}-4608\\ -4096\\ 1024\end{bmatrix}$$
A.Kaw矩阵代数初步学习笔记 10. Eigenvalues and Eigenvectors
标签：des style http ar io color os sp for
原文地址：http://www.cnblogs.com/zhaoyin/p/4170081.html
