Eigenvalue and Eigenvectors(特征值和

作者: 默写年华Antifragile | 来源:发表于2018-11-30 20:32 被阅读9次

This chapter begins the "second half" of linear algebra. The first half was about $Ax=b$ . The new problem $Ax=λx$ will stil be solved by simplifying a matrix --- making it diagonal if possible. The basic step is no longer to subtract a multiple of one row from another: Elimination changes the eigenvalues, which we don't want.

Eigenvalue equation: $Ax=λx$

The number λ is an eigenvalue of the matrix A
and the vector x is the associated eigenvector

1. Introduction

$Ax=λx$ is a nonlinear equation
$Ax=λx$ $\rightarrow$ $(A-λI)x=0$
The vector x is in nullspace of $A-λI$
The number λ is chosen so that $A-λI$ has a nullspace
The number λ is an eigenvalue of A if and only if $A-λI$ is singular:
$det(A-λI)=0$
if $A-λI$ is invertible, then $(A-λI)^{-1}(A-λI)x=0$ , then x = 0; however, we need the nullspace of $A-λI$ contain vectors other than zero. So if and only if $A-λI$ is singular.
Eigenspace: all vectors in the nullspace of $A-λI$
Steps in solving $Ax=λx$ :
- Compute the determinant of $A-λI$ . With λ subtracted along the diagonal, this determinant is a polynomial of degree n. It starts with $(-λ)^{n}$
- Find the roots of this polynomial. The n roots are the eigenvalue of A
- For each eigenvalue solve the equation $(A-λI)x=0$ . Since the determinant is zero, there are solutions other than x=0. These are the eigenvectors.
A zero eigenvalue signals that A is singular(not invertible); its determinant is zero. Invertible matrices have all $λ\neq 0$
The eigenvalue are on the main diagonal when A is triangular; so we need to transform A into a diagonal or triangular matrix without changing its eigenvalues. However, Gaussian factorizstion A=LU is not suited to this purpose. The eigenvalue of U may be visible on the diagonal, but they are not the eigenvalues of A.
The sum of the n eigenvalues equals the sum of the n diagonal entries:
Trace of $A=λ_1+\cdots+λ_n=a_{11}+\cdots+a_{nn}$
The Product of the n eigenvalues equals the determinant of A.