Lecture 19: Eigenvalues and eigenvectors

19.1 Definition of Eigenvalues and Eigenvectors

19.1.1 Basic Concepts

#####Definition: Eigenvaluesandeigenvectors

description: Let $A$ be the $n$ order matrix $, \lambda$ be a number $, $ If there is a $n$ dimensional non-zero column vector $\xi, $ such that $: $

$$A\xi=\lambda\xi$$

is called:

1. $\lambda$ is the eigenvalue;

2. $\xi$ is the eigenvector of A corresponding to the eigenvalue $\lambda$;

Explanation

Features:
React to a property of matrix A;
Explanation: eigenvalue
For example, when A is a matrix, when A acts on a vector, its result can be represented by a number (scaling the vector with a number), and this number is the eigenvalue;
Spatially, it can reflect properties such as extremes and similarities within space;
Concept:
A matrix of n order will have n eigenvalues. By looking at this number, one can know certain characteristics of the A matrix;
Explanation: $\xi$
$\xi$ must not be zero;
Because if $\xi$ equals 0, it means the current matrix has only a unique zero solution, i.e., S=r(A). Acting A on $\xi$ does not reflect A's effect, because the matrix is still zero on the zero vector, so here $\xi$ is a nonzero vector;
Demonstrate the effect, so you can understand the role of eigenvalues;
Meaning:

Analysis: Analyze the $A\xi=\lambda\xi$

Steps:
-> $\lambda\xi-A\xi=0$ // $\xi$ is a nonzero vector
-> $(\lambda E-A)\xi=0$ // Extract the common factor
-> $(\lambda E-A)X=0$
// Set $\xi$ to be the solution to the current system of equations;
// Because X is $\xi$, $\xi$ is a nonzero vector, and the current expression equals 0 -> means the current equation is homogeneous;
// Because this homogeneous system has a nonzero solution, the columns of $\lambda E-A$ are linearly dependent <- Theorem 4 on linear dependence: Lecture 11: Linear Dependence of Vectors and Vector Sets;
// Therefore, $r(\lambda E-A)<n$: the rank of the matrix is less than $n$;
// For $n$ column vectors $a_1,a_2,\cdots,a_n$ in $\mathbb{R}^n$, linear dependence is equivalent to $\left|a_1,a_2,\cdots,a_n\right|=0$. Hence $|\lambda E-A|=0$
-> $|\lambda E-A|=0$
// Here you can find $\lambda_i(i=1,2,3...)$
Illustration:

#####Definition: Characteristicequations

description: $$\begin{vmatrix}\lambda E-A\end{vmatrix}=\begin{vmatrix}\lambda-a_{11}&-a_{12}&\cdots&-a_{1n}\\-a_{21}&\lambda-a_{22}&\cdots&-a_{2n}\\\vdots&\vdots&&\vdots\\-a_{n1}&-a_{n2}&\cdots&\lambda-a_{nn}\end{vmatrix}=0$$

Explanation

$|\lambda E-A|$: called the characteristic equation

19.1.2 Geometric Explanation

Concepts: Geometric explanation of eigenvalues and eigenvectors

Concept:
After the basis vector is transformed, some transformations correspond only to proportional transformations after and before the solution, while the angular position remains unchanged;
The value used to describe the transformation ratio of an eigenvector, called an eigenvalue;
Eigenvectors are those vectors that remain on the original line after transformation;
Illustration:

Concept: When the eigenvalue is negative

Illustration:

19.2 Calculation method

19.2.1 Methods for Finding Eigenvalues and Eigenvectors

Method: Find eigenvalues

Step 1: Write the characteristic equation $|\lambda E-A|$
$A=\begin{bmatrix}2&2&-2\\2&5&-4\\-2&-4&5\end{bmatrix}$
$|\lambda E-A|=\left|\begin{matrix}\lambda-2&-2&2\\-2&\lambda-5&4\\2&4&\lambda-5\end{matrix}\right|$
Step 2: Calculate the determinant of the characteristic equation to obtain the equation about $\lambda$;
In the end, you will definitely get an equation similar to $\lambda^3+a\lambda^2+b\lambda+c=0$, which can then be transformed into a form that finds roots;
Step 3: Find all eigenvalues (the roots of the characteristic polynomial);

Method: Find eigenvectors

Step 1: For each eigenvalue $\lambda$, substitute it into $(\lambda E-A)x=0$;
Step 2: Row-reduce $\lambda E-A$ and find a basis for its null space;
The eigenspace dimension is $S=n-r(\lambda E-A)$;
These S vectors solved form the space of understanding. All vectors on this two-dimensional plane are eigenvectors corresponding to the current substituted $\lambda_i$;
Except for the zero vector;
2.2 If simplification does not allow for a stepped matrix, you can also determine the current rank by using the determinant = 0 for the restriction on rank and the number of irrelevant term vectors in the current matrix $\lambda E-A$, to determine its current rank and obtain its maximally independent group;
Step 3: According to step two, when you get the current $\lambda=\text{ A certain value }$, you get $\xi=k_1\xi_1+k_2\xi_2$
Viewing k1 and k2 cannot be 0 at the same time;
Step 4: If there are other $\lambda_i$, continue using the above steps to find the eigenvector $\xi$;

19.2.2 Finding the Root

Supplement: Method to find roots

If there are no constant terms in $\lambda^3+a\lambda^2+b\lambda+c=0$, then $\lambda=0$ is definitely rooted;
If $\lambda^3+a\lambda^2+b\lambda+c=0$ among them, $a+b+c=0$, then $\lambda =1$ is definitely their root: $(\lambda -1)$
If the sum of the even-degree terms in $\lambda^3+a\lambda^2+b\lambda+c=0$ equals the sum of odd-degree terms, then $f(-1)=0$, so $-1$ is its root;

Method: Root testing method

Concept:
Using the root-finding method to obtain the sum of a given root, using polynomials to obtain codivision with co-division, and using a calculator for quadratic terms;
Example:

\begin{aligned} \lambda^{2}-2\lambda-8 \\ \begin{aligned}\lambda-1\sqrt{\lambda^{3}-3\lambda^{2}-6\lambda+8}\\\frac{\lambda^{3}-\lambda^{2}}{-2\lambda^{2}-6\lambda}\end{aligned} \text{.} \\ \frac{-2\lambda^{2}+2\lambda}{-8\lambda+8} \\ \frac{-8\lambda+8}{0} \end{aligned}

#####Theorem: PolynomialRoots

description: let the following formula be a polynomial with coefficients $a_i$ both integers: $$f\left(x\right)=1·x^{k}+a_{k-1}x^{k-1}+\cdots+a_{1}x+a_{0}$$

then the rational roots of $f(x)=0$ are integers, and $a_0$ factor of.

Explanation

The solution to the equation is a factor of $a_0$
Therefore, its factors can be framed first, and once the factors are known, their range is determined;

19.3 Significant Nature and Conclusion

19.3.1 Properties and Conclusions of Eigenvalues

Concept: Property one

$$\lambda_0\text{ is an eigenvalue of }A\Leftrightarrow|\lambda_0 E-A|=0.$$
Similarly, $$\lambda_0\text{ is not an eigenvalue of }A\Leftrightarrow|\lambda_0E-A|\neq0,$$ in which case $\lambda_0E-A$ is invertible and has full rank.
Supplement:
If $|aA+bE|=0$ (equivalently, $aA+bE$ is singular) and $a\neq0$, then $-\frac{b}{a}$ is an eigenvalue of $A$.

Concept: Property two

Nature:
If $\lambda_1,\lambda_2,\ldots,\lambda_n$ are the $n$ eigenvalues of $A$, counted with algebraic multiplicity, then:
$$\begin{cases}\left|A\right|=\lambda_{1}\lambda_{2}\cdots\lambda_{n},\\\mathrm{tr}\left(A\right)=\lambda_{1}+\lambda_{2}+\cdots+\lambda_{n}.\end{cases}$$

-Conclusion 1:

1. There is a matrix with eigenvalue 0, whose determinant must be 0
2. $\lambda_{1}+\lambda_{2}+\cdots+\lambda_{n}$ The sum of eigenvalues equals the sum of the main diagonal elements of the current matrix A tr(A)

-Conclusion Two: Based on Conclusions One and Proof

From equations (1) and (2), we get:
$$\begin{cases}a_{11}+a_{22}+a_{33}=\lambda_{1}+\lambda_{2}+\lambda_{3},\\A_{11}+A_{22}+A_{33}=\lambda_{2}\lambda_{3}+\lambda_{1}\lambda_{3}+\lambda_{1}\lambda_{2},\\\left|A\right|=\lambda_{1}\lambda_{2}\lambda_{3}.\end{cases}$$
More generally, the sum of the principal minors of order $k$ equals the $k$th elementary symmetric sum of the eigenvalues.
Proof for order $3$: expand the characteristic polynomial $|\lambda E-A|$ in two ways.
$$\text{Equation (1)}\quad|\lambda E-A|=\begin{vmatrix}\lambda-a_{11}&-a_{12}&-a_{13}\\-a_{21}&\lambda-a_{22}&-a_{23}\\-a_{31}&-a_{32}&\lambda-a_{33}\end{vmatrix}=\lambda^{3}-\left(a_{11}+a_{22}+a_{33}\right)\lambda^{2}+\left(A_{11}+A_{22}+A_{33}\right)\lambda-\left|A\right|$$
$$\text{Equation (2)}\quad\left|\lambda E-A\right|=\left(\lambda-\lambda_1\right)\left(\lambda-\lambda_2\right)\left(\lambda-\lambda_3\right)=\lambda^3-(\lambda_1+\lambda_2+\lambda_3)\lambda^2+(\lambda_1\lambda_2+\lambda_1\lambda_3+\lambda_2\lambda_3)\lambda-\lambda_1\lambda_2\lambda_3$$
Supplement: Principal minors
A principal minor uses the same index set for its selected rows and columns.
For a $3\times3$ matrix, the three principal minors of order $2$ are $\begin{vmatrix}a_{22}&a_{23}\\a_{32}&a_{33}\end{vmatrix}=A_{11},\begin{vmatrix}a_{11}&a_{13}\\a_{31}&a_{33}\end{vmatrix}=A_{22},\begin{vmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{vmatrix}=A_{33}$.

19.3.2 Properties and Conclusions of Eigenvectors

Concept: Property One

A nonzero vector $\xi$ is an eigenvector of $A$ associated with $\lambda_0$ if and only if $\xi$ is a nonzero solution of $(\lambda_0E-A)x=\mathbf{0}$;

Concept: Conclusion One

An eigenvalue $\lambda$ of algebraic multiplicity k has at most k linearly independent eigenvectors;

Concept: Conclusion two

If $\xi_1,\xi_2$ are eigenvectors of A associated with distinct eigenvalues $\lambda_1,\lambda_2$, then $\xi_1,\xi_2$ are linearly independent;

Summary: $\text{For matrix }A,\begin{cases}\lambda_{1}\neq\lambda_{2}\Rightarrow\xi_{1},\xi_{2}\text{ are linearly independent},\\\lambda_{1}=\lambda_{2}\Rightarrow\xi_{1},\xi_{2}\text{ may be linearly dependent or linearly independent}.\end{cases}$

Concept: Conclusion three

If $\xi_1,\xi_2$ are eigenvectors of $A$ associated with the same eigenvalue $\lambda$ and $k_1\xi_1+k_2\xi_2\neq0$, then $k_1\xi_1+k_2\xi_2$ is also an eigenvector associated with $\lambda$ (a common special case has one coefficient, such as $k_2$, equal to 0);

Concept: Conclusion Four

If $\xi_1,\xi_2$ are eigenvectors of $A$ associated with distinct eigenvalues $\lambda_1,\lambda_2$ and $k_1,k_2\neq0$, then $k_1\xi_1+k_2\xi_2$ is not an eigenvector of $A$ (a common special case is $k_1=k_2=1$);

Concept: Conclusion Five

Let $\lambda_1,\lambda_2$ be distinct eigenvalues of $A$. If $\xi$ is an eigenvector associated with $\lambda_1$, then it cannot also be an eigenvector associated with $\lambda_2$;

19.3.3 Common Eigenvalues and Eigenvectors of Matrices

Summary: Eigenvalues and eigenvectors of commonly used matrices

Illustration:

Concept: The matrix is $f(A)$

Important formula - When A is a polynomial equal to 0
Formula:
$$A\xi=\lambda \xi \rightarrow f(A)\xi=f(\lambda)\xi$$
Proof:
For example: $A^2\xi=\lambda A\xi=\lambda^2\xi$
Example:
When $A^2-2A+3E$ $\lambda$: $\lambda^2-2\lambda+3$
Note:
The equation obtained from the polynomial of $A$ about $\lambda$ only represents the possible values of $\lambda$ under the current polynomial conditions of A (what relation or range they satisfy), and does not represent the different $\lambda$ of the current characteristic equation;

Concept: The matrix is $A^*$

Formula:
When the matrix is $A^*$, its eigenvalue equals $\frac{|A|}{\lambda}$;
And since $|A|=\lambda_1\lambda_2\lambda_3$, so $\frac{|A|}{\lambda}=\lambda_1\lambda_2\text{ or }\lambda_1\lambda_3\text{ or }\lambda_2\lambda_3$

Supplement: $P^{-1}AP$, its eigenvalues do not change -> similar; but the eigenvectors change: $P^{-1}\xi$;

Concept: The conclusion of a rank 1 matrix

When r(A)=1, matrix $A_{n*n}$ can definitely be reduced to two nonzero determinants: $\alpha\beta^{T}$ product of the two;
And: $\lambda_1=\lambda_2=\lambda_3=...=\lambda_{n-1}=0$
And: $\lambda_n=tr(A)=\beta^T\alpha$