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Eigenvalue & Eigenvector Calculator

Enter a 2×2 or 3×3 matrix to find its eigenvalues and eigenvectors. The tool shows the characteristic polynomial det(A − λI) = 0, solves for real and complex eigenvalues, computes normalized eigenvectors, and builds the diagonalization A = PDP⁻¹ when possible.

Matrix Input

Matrix Size

Matrix A (2×2)

[a b][c d]

Presets

Determinant

8

Trace

6

Symmetric

Yes

Diagonalizable

Yes

Characteristic Polynomial

det(A − λI) = 0

λ2−6λ+8=0\lambda^2 - 6\lambda + 8 = 0

Eigenvalues & Eigenvectors

λ₁4
v₁ =[0.7071, 0.7071]
λ₂2
v₂ =[0.7071, -0.7071]

Diagonalization

A = P D P⁻¹

P (eigenvectors)

(0.70710.70710.7071−0.7071)\begin{pmatrix}0.7071 & 0.7071 \\ 0.7071 & -0.7071\end{pmatrix}

D (eigenvalues)

(4002)\begin{pmatrix}4 & 0 \\ 0 & 2\end{pmatrix}

Eigenvector Directions (2D)

Original basisTransformed basisEigenvector 1Eigenvector 2

Reference Guide

Eigenvalues

A scalar λ is an eigenvalue of matrix A if there exists a nonzero vector v such that Av = λv. Geometrically, eigenvectors are directions that only scale under the transformation — they do not rotate.

To find eigenvalues, set up the characteristic equation and solve for λ.

det⁡(A−λI)=0\det(A - \lambda I) = 0

A real n×n matrix always has exactly n eigenvalues (counting multiplicity) in ℂ. They can be real or come in complex conjugate pairs.

Eigenvectors

Once eigenvalue λ is known, find the eigenvector v by solving the homogeneous system:

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

The solution is the null space of (A − λI). Any nonzero vector in this null space is a valid eigenvector. This tool returns normalized eigenvectors (magnitude 1) for each real eigenvalue.

If λ is complex, the eigenvectors are also complex and are not shown on the real-plane plot.

Characteristic Polynomial

For a 2×2 matrix the characteristic polynomial is a quadratic with coefficients determined by the trace and determinant:

λ2−tr(A) λ+det⁡(A)=0\lambda^2 - \text{tr}(A)\,\lambda + \det(A) = 0

For a 3×3 matrix it is a cubic:

λ3−tr(A) λ2+(M11+M22+M33) λ−det⁡(A)=0\lambda^3 - \text{tr}(A)\,\lambda^2 + (M_{11}+M_{22}+M_{33})\,\lambda - \det(A) = 0

where M₁₁, M₂₂, M₃₃ are the 2×2 principal minors (upper-left, center, lower-right cofactors).

Diagonalization

A matrix A is diagonalizable if it can be written as A = PDP⁻¹ where D is diagonal and P contains eigenvectors as columns.

A=PDP−1,D=(λ100λ2)A = P D P^{-1}, \quad D = \begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{pmatrix}

Sufficient conditions for diagonalizability:

  • n distinct eigenvalues always give n independent eigenvectors.
  • Real symmetric matrices are always orthogonally diagonalizable (spectral theorem).
  • A repeated eigenvalue may or may not be diagonalizable — it fails only when the geometric multiplicity is less than the algebraic multiplicity.