Eigenvalue Equation

Equivalent forms

(A - \lambda I)\mathbf{v} = 0

\det(A - \lambda I) = 0

Quantized energy levels, principal axes of inertia, and normal modes of vibration are all the same statement: A v = λ v.

Symbol	Name	SI unit	Dimension	Range
$a$	Matrix entry a Top-left of the 2×2 matrix	1	1	-2 – 3
$b$	Matrix entry b Off-diagonal coupling	1	1	-2 – 2
$\lambda_1$	Eigenvalue 1output First stretch factor	1	1	-4 – 4
$\lambda_2$	Eigenvalue 2output Second stretch factor	1	1	-4 – 4

Unit systems

Symbol	SI	CGS	Imperial
$a$	1	1	1

Where it holds

Every square matrix over ℂ has eigenvalues; real matrices may have complex ones. Defective matrices lack a full eigenbasis.

Dimensional analysis

A v

\lambda v

\lambda is

a pure scalar multiplier, so both sides carry the dimensions of v; A must be dimensionless for the equation to be homogeneous, as it is here.

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Discovery

Cauchy, Hilbert · 1826

Euler and Lagrange met eigenvalues in rotating rigid bodies; Cauchy proved real symmetric matrices have real eigenvalues (1820s); Hilbert coined 'eigen' (own/characteristic). In 1926 they became the backbone of quantum mechanics, where observables ARE eigenvalue problems.

Try this

Most vectors get knocked sideways when a matrix hits them. A special few just get longer or shorter, keeping their direction. Find those and you've found the natural axes of the whole system.

Spin an input vector around the circle; its image traces an ellipse. Watch the two directions where input and output line up perfectly — the eigenvectors.

Research status: stable

Real-world applications

Energy levels and observables in quantum mechanics
Normal modes of vibrating structures (bridges, molecules)
Principal component analysis in data science
PageRank as the dominant eigenvector of the link matrix

Common misconceptions

Eigenvectors are unique — any nonzero scalar multiple is also an eigenvector.
All matrices are diagonalizable — defective matrices (repeated $\lambda$ , too few eigenvectors) are not.
Real matrices have real eigenvalues — rotations have complex ones.

Experimental verification

Atomic spectral lines are differences of energy eigenvalues — measured to 12+ digits. Principal axes of inertia (eigenvectors of the inertia tensor) predict how a tossed object tumbles, seen in the 'tennis-racket' / Dzhanibekov effect.

Derivation

A

v = \lambda v

means (A-

\lambda I)v = 0

has a nonzero solution, which requires det(A-

\lambda I)=0

.

For

2\times 2 [[

a,b],[c,d]] this

is \lambda ^{2}

- (a+

d)\lambda + (ad-bc)=0

, whose roots are the eigenvalues; back-substitute to find each eigenvector.

Limiting cases

b = 0

⟶

\lambda = a

, dA diagonal matrix has its diagonal as eigenvalues and the axes as eigenvectors.

symmetric A⟶ real

\lambda

, orthogonal vGuaranteed by the spectral theorem.

det(A-

\lambda I)=0

⟶ characteristic equationEigenvalues are exactly the roots that make A-

\lambda I

singular.

What if…

What if two eigenvalues are equal?

The eigenspace may be 2D (still diagonalizable) or deficient (defective) — the geometric vs algebraic multiplicity decides.

What if A is a rotation?

No real eigenvectors — every direction turns — so the eigenvalues are complex, $e^{\pm i\theta }$ .

1

Eigenvalues of [[2,1],[1,2]]

Given ·

A:: [[2,1],[1,2]]

Find ·

\lambda

Steps

Trace $= 4$ , $det = 3$
$\lambda ^{2} - 4\lambda + 3 = 0$
$(\lambda -3)(\lambda -1)=0 \to 3$ , 1

Result ·

\lambda = 3

and 1

2

Eigenvector for λ = 3

Given ·

A:: [[2,1],[1,2]]
\lambda:: 3

Find · v

Steps

(A- $3I) = [[-1$ ,1],[1,-1]]
Row gives - $v_{1}+v_{2}=0$
$v = (1$ ,1)

Result ·

v \propto (1

,1)

Characteristic Polynomial→diagonalizationDeterminant→Rotation Matrices→