Basics of the eigenvalue problem固有値問題こゆうちもんだい

1Introduction

The eigenvalue problem固有値問題こゆうちもんだい asks for a scalar $\lambda$\lambda and a nonzero vector $v$v such that

Both $\lambda$\lambda and $v$v are unknown. The method is therefore two-stage: first find the values of $\lambda$\lambda for which $A-\lambda I$A-\lambda I loses rank, then find the kernel of $A-\lambda I$A-\lambda I.

2Terms and definitions

Let $A$A be an $n\times n$n\times n square matrix正方行列せいほうぎょうれつ. The standard eigenvalue problem標準固有値問題ひょうじゅんこゆうちもんだい is

The eigenspace固有空間こゆうくうかん for $\lambda$\lambda is

The characteristic polynomial特性多項式とくせいたこうしき is

3Plan

The solving order is:

1. Rewrite $Av=\lambda v$Av=\lambda v as $(A-\lambda I)v=0$(A-\lambda I)v=0.
2. Require a nonzero solution, so $\det (A-\lambda I)=0$\det(A-\lambda I)=0.
3. Solve the characteristic equation.
4. For each eigenvalue, solve $(A-\lambda I)v=0$(A-\lambda I)v=0 to obtain the eigenspace.

4Intuitive explanation

The eigenvalue problem looks for directions that $A$A does not turn. The matrix $A-\lambda I$A-\lambda I compares the action of $A$A with pure scaling by $\lambda$\lambda. A nonzero vector in the kernel of $A-\lambda I$A-\lambda I is a direction where these two actions agree.

5Precise explanation

5.11. Eigenvalues and rank deficiency

For $v\ne 0$v\ne0,

Thus $\lambda$\lambda is an eigenvalue exactly when the homogeneous system $(A-\lambda I)x=0$(A-\lambda I)x=0 has a nontrivial solution. Since $A-\lambda I$A-\lambda I is square,

This is why rank and determinants enter the eigenvalue problem.

5.22. Why the algorithm is valid

Solving $\det (A-\lambda I)=0$\det(A-\lambda I)=0 keeps exactly the $\lambda$\lambda values for which a nonzero kernel can exist. For every other $\lambda$\lambda, $A-\lambda I$A-\lambda I is invertible and the kernel is zero.

5.33. Generalized eigenvalue problem

The generalized eigenvalue problem一般化固有値問題いっぱんかこゆうちもんだい is

If $B$B is invertible, this reduces to $B^{-1}Av=\lambda v$B^{-1}Av=\lambda v. If $B$B is not invertible, one studies the matrix pencil $\det (A-\lambda B)=0$\det(A-\lambda B)=0 directly.

6Worked example

For

the characteristic equation is

Thus $\lambda =3,2$\lambda=3,2. For $\lambda =3$\lambda=3, the eigenspace is

For $\lambda =2$\lambda=2, the eigenspace is

7Connection to companion matrices

The eigenvalue problem also converts polynomial root problems into matrix problems. A companion matrix同伴行列どうはんぎょうれつ is built so that a given monic polynomial becomes its characteristic polynomial.

8How to recognize the method

For $Av=\lambda v$Av=\lambda v, first solve $\det (A-\lambda I)=0$\det(A-\lambda I)=0. Then find eigenvectors as the kernel of $A-\lambda I$A-\lambda I. For $Av=\lambda Bv$Av=\lambda Bv, first check whether $B$B is invertible.

9Scope

This lecture treats finite-dimensional square matrices. Real matrices may have complex eigenvalues, so extending scalars to the complex numbers is often necessary. In numerical computation, directly expanding the characteristic polynomial is often unstable; practical algorithms such as the QR method are used, but this lecture focuses on definitions and proofs.

10Final forms

11In one sentence

• The eigenvalue problem first finds where rank drops, then finds the kernel at that value.

12Exercise links

13Related links