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'''''A'''x'' is a linear transformation: '''''A''''' maps the input vector ''x'' to the output vector ''b''. The square matrix '''''A''''' is a linear transformation mapping vector ''a'' to vector ''b''. There are certain vectors that map to a linear combination of themself, i.e. multiplying by '''''A''''' works out to multiplying by some scalar ''λ''.
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For some linear transformations (i.e. some, but not all, matrices '''''A'''''), there are '''certain input vectors''' ''x'' where the output vector ''b'' is also just a linear transformation of the input vector ''x''. In other words, '''''A''''' mapped ''x'' to a scaled version of ''x''. That '''scaling factor''' is notated ''λ''. Another way to think of this is that there are certain vectors which maintain their direction through a transformation. For vectors in these directions, the transformation only involves some stretching factor ''λ''.
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For example, rotation around the ''y'' axis in 3 dimensions by ''θ'' degrees is calculated with: These certain vector are called '''eigenvectors'''. Each eigenvector has a corresponding scaling factor ''λ'', called '''eigenvalues'''.

Consider the [[LinearAlgebra/RotationMatrix|matrix representing rotation in the y-axis]]:
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The the vector ''[0 1 0]]'', and importantly any linear transformation of that unit vector, will not be transformed by this '''''A'''''. (And because the linear transformation involves no scaling, ''λ = 1''.) 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.)
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The '''certain input vectors''' are the '''eigenvectors''' of '''''A'''''. The '''scaling factors''' are the '''eigenvalues''' of '''''A'''''. Eigenvectors and eigenvalues often include complex numbers.
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Note that the other eigenvectors and eigenvalues are complex. Note also that the eigenvectors are returned as the eigenvector matrix, usually notated as '''''S'''''. Note that in the above, eigenvectors are returned as an eigenvector matrix. This is usually notated as '''''S'''''.
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== Definition == == Description ==
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Unless '''''A''''' is '''defective''', there should be ''n'' unique pairs of eigenvectors and eigenvalues. If there is a repeated eigenvalue, there may not be ''n'' independent eigenvectors. Given a matrix of size ''n x n'', either there are ''n'' unique pairs of eigenvectors and eigenvalues, or the matrix is '''defective'''.
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Only square matrices have eigenvectors.
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The sum of the eigenvalues is the trace (sum of diagonal). The product of the eigenvalues is the [[LinearAlgebra/Determinant|determinant]]. The [[LinearAlgebra/Trace|trace]] is the sum of eigenvalues. The [[LinearAlgebra/Determinant|determinant]] is the product the eigenvalues.

A [[LinearAlgebra/Diagonalization|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.
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== Solution == == Calculation ==
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=== Special Cases ===

A [[LinearAlgebra/Diagonalization|diagonal matrix]] is a special and trivial case for finding eigenvalues and eigenvectors.



=== Finding Eigenvalues ===
=== Conventional Method ===
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In a simple 2 x 2 matrix, this looks like: In a simple ''2 x 2'' matrix, this looks like:
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This leads to the '''characteristic polynomial''' of '''''A'''''; solving for the roots, as through either factorization or the quadratic formula, gives the eigenvalues.
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Because eigenvectors are characterized by...
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=== Finding Eigenvectors === '''''A'''x = λx''
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Because eigenvectors are characterered by: '''''A'''x - λx = 0''
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 * '''''A'''x = λx''
 * which can be rewritten as '''''A'''x - λx = 0''
 * which can be refactored as ''('''A''' - λ'''I''')x = 0''
''('''A''' - λ'''I''')x = 0''
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Eigenvectors can be solved for given eigenvalues. For each given ''λ,,i,,'', substitute it into that final equation and solve for the corresponding ''x,,i,,''. ...eigenvectors can be solved for given eigenvalues using substitution.
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In a simple 2 x 2 matrix, this looks like: In a simple ''2 x 2'' matrix, this looks like:
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=== 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) = m^2^ - d^2^''. This can be solved for ''d''.

Altogether,

{{attachment:shortcut.svg}}


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 are certain vectors that map to a linear combination of themself, i.e. multiplying by A works out to multiplying by some scalar λ.

Another way to think of this is that there are certain vectors which maintain their direction through a transformation. For vectors in these directions, the transformation only involves some stretching factor λ.

These certain vector are called eigenvectors. Each eigenvector has a corresponding scaling factor λ, called 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,

shortcut.svg


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LinearAlgebra/EigenvaluesAndEigenvectors (last edited 2026-02-03 23:52:07 by DominicRicottone)