Characteristic Polynomial

Equivalent forms

\lambda^n - (\mathrm{tr}\,A)\lambda^{n-1}+\dots+(-1)^n\det A=0

One polynomial packages every eigenvalue, and its coefficients are exactly the rotation-invariant quantities (trace, determinant) of the matrix.

Symbol	Name	SI unit	Dimension	Range
$\lambda$	Eigenvalueoutput Root of the characteristic polynomial	1	1	—
$A$	Matrix The n×n operator being analyzed	1	1	—
$\det$	Determinant Determinant of A−λI	1	1	—
$n$	Dimension Size of the square matrix	1	1	2 – 5

Where it holds

Any square matrix over a field. Real matrices may have complex-conjugate eigenvalue pairs; symmetric/Hermitian matrices always have real eigenvalues.

Dimensional analysis

p(\lambda )=det

(A-\lambda I)

→ dimensionless for a numeric matrix

0

\lambda

carries the units of A's entries; for the schematic numeric matrix here it is a pure number.

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Discovery

Cauchy / Lagrange (secular perturbations) · 1829

The 'secular equation' appeared in 18th-century celestial mechanics (Lagrange, Laplace) analyzing long-term orbital stability. Cauchy (1829) proved real symmetric matrices have real eigenvalues. The Cayley–Hamilton theorem (1858) later showed every matrix satisfies its own characteristic polynomial.

Research status: stable

Real-world applications

Normal modes and resonant frequencies of coupled oscillators
Quantum energy levels from a Hamiltonian matrix
Stability analysis of dynamical systems (eigenvalue signs)

Common misconceptions

Coefficients alternate with the elementary symmetric functions of eigenvalues, not arbitrary numbers
Algebraic multiplicity (root order) can exceed geometric multiplicity (eigenvectors)
Computing it via cofactor expansion is numerically unstable — use QR iteration in practice

Experimental verification

Underlies every measured normal-mode frequency (masses on springs, molecular vibrations) and quantum energy level (Hamiltonian eigenvalues), matching spectra to high precision.

Derivation

An eigenvalue

\lambda

satisfies

Av=\lambda v

for some nonzero v, i.e.

(A-\lambda I)v=0

has a nontrivial solution.

A homogeneous system has nonzero solutions exactly when its matrix is singular,

det(A-\lambda I)=0

.

Expanding the determinant gives a degree-n polynomial

p(\lambda )=(-1)^{n}(\lambda ^{n} - (tr

A

)\lambda ^{n-1}

+ … +

(-1)^{n} det

A).

Its n roots (with multiplicity, over ℂ) are the eigenvalues.

What if…

What if A is symmetric?

All roots are real and eigenvectors are orthogonal — the spectral theorem, why observables in QM are Hermitian.

What if two roots coincide (degeneracy)?

The eigenvalue is repeated; there may be fewer independent eigenvectors, signalling a defective (non-diagonalizable) matrix.

1

2×2 eigenvalues

Given ·

A = [[2

,1],[1,2]]

Find · eigenvalues

Steps

$det(A-\lambda I) = (2-\lambda )^{2}-1 = 0$
$\lambda ^{2}-4\lambda +3 = 0$
$\lambda = 1$ , 3 (trace $4 = 1+3$ , $det 3 = 1\times 3)$

Result ·

\lambda = 1

and 3

Eigenvalue Equation→cayley hamilton theoremdeterminant