Eigenvalues and Eigenvectors

Matrices are characterized by their eigenvalues and eigenvectors.

Introduction

The square matrix A is a linear transformation mapping vector a to vector b. There is at least one vector that maps to a linear combination of itself, i.e. multiplying by A works out to multiplying by some scalar λ.

Each of these vectors is an eigenvector. Each eigenvector has a corresponding scaling factor λ. These are eigenvalues.

Consider the matrix representing rotation in the y-axis:

┌                 ┐
│  cos(θ) 0 sin(θ)│
│       0 1      0│
│ -sin(θ) 0 cos(θ)|
└                 ┘

Given this as A, the vector [0 1 0]] (and any linear transformation of it) maps to itself. The members of this infinite set of vectors are all eigenvectors of A. (And because there is no scaling, their corresponding eigenvalues λ are all 1.)

Eigenvectors and eigenvalues often include complex numbers.

julia> using LinearAlgebra

julia> A = [0 0 1; 0 1 0; -1 0 0]
3×3 Matrix{Int64}:
  0  0  1
  0  1  0
 -1  0  0

julia> eigvals(A)
3-element Vector{ComplexF64}:
 0.0 - 1.0im
 0.0 + 1.0im
 1.0 + 0.0im

julia> eigvecs(A)
3×3 Matrix{ComplexF64}:
 0.707107-0.0im       0.707107+0.0im       0.0+0.0im
      0.0-0.0im            0.0+0.0im       1.0+0.0im
      0.0-0.707107im       0.0+0.707107im  0.0+0.0im

Note that in the above, eigenvectors are returned as an eigenvector matrix. This is usually notated as S.

Description

Eigenvalues and eigenvectors are the pairs of λ and x that satisfy Ax = λx and |A - λI| = 0.

Given a matrix of size n x n, either there are n unique pairs of eigenvectors and eigenvalues, or the matrix is defective.

Properties

Only square matrices have eigenvectors.

Adding nI to A does not change its eigenvectors and adds n to the eigenvalues.

The trace is the sum of eigenvalues. The determinant is the product the eigenvalues.

A diagonal matrix is a trivial case because...

• the columns are its eigenvectors
• the numbers in the diagonal are its eigenvalues

This also means that any diagonalizable matrix of size n x n must have n unique pairs of eigenvectors and eigenvalues, and cannot be defective.

Calculation

Conventional Method

Because eigenvalues are characterized by |A - λI| = 0, they can be solved for by:

• subtracting λ from each value on the diagonal

• formulating the determinant for this difference
• setting the formulation for 0

• solving for λ

In a simple 2 x 2 matrix, this looks like:

|    A   -  λI   | = 0

│ ┌    ┐  ┌    ┐ │
│ │ a b│ -│ λ 0│ │ = 0
│ │ c d│  │ 0 λ│ │
│ └    ┘  └    ┘ │

│ ┌        ┐ │
│ │ a-λ   b│ │ = 0
│ │   c d-λ│ │
│ └        ┘ │

(a-λ)(d-λ) - bc = 0

This leads to the characteristic polynomial of A; solving for the roots, as through either factorization or the quadratic formula, gives the eigenvalues.

Because eigenvectors are characterized by...

Ax = λx

Ax - λx = 0

(A - λI)x = 0

...eigenvectors can be solved for given eigenvalues using substitution.

In a simple 2 x 2 matrix, this looks like:

(   A    -  λI   )   x  =   0

/ ┌    ┐  ┌    ┐ \ ┌  ┐   ┌  ┐
│ │ a b│ -│ λ 0│ │ │ u│ = │ 0│
│ │ c d│  │ 0 λ│ │ │ v│   │ 0│
\ └    ┘  └    ┘ / └  ┘   └  ┘

┌        ┐ ┌  ┐   ┌  ┐
│ a-λ   b│ │ u│ = │ 0│
│   c d-λ│ │ v│   │ 0│
└        ┘ └  ┘   └  ┘

┌        ┐ ┌  ┐   ┌  ┐
│ a-λ   b│ │ u│ = │ 0│
│   c d-λ│ │ v│   │ 0│
└        ┘ └  ┘   └  ┘

(a-λ)u + (b)v = 0
(c)u + (d-λ)v = 0

Shortcut

This method is only applicable to 2 x 2 matrices.

• Because the trace is equal to the sum of eigenvalues, it follows that 1/2 of the trace is also the mean of the eigenvalues.

• Because the characteristic polynomial must be quadratic, the eigenvalues must be evenly spaced from the center (i.e., the mean). Given the above mean as m and an unknown distance as d, the eigenvalues must be (m-d) and (m+d).

• By definition of the determinant, |A| = (m-d)(m+d) = m² - d². This can be solved for d.

Altogether,

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