What eigenvalues and eigenvectors are
Most vectors change direction when a matrix acts on them. A few do not — they only get longer or shorter, or flipped. Those special directions are the eigenvectors, and the factor each one is scaled by is its eigenvalue. In symbols, Av = λv with v ≠ 0.
Finding them turns a matrix from an opaque grid of numbers into something you can describe in a sentence. A matrix with eigenvalues 3 and 0.5 stretches by 3 along one axis and squashes by half along another, and once you know those axes you know everything the matrix does. That is why eigenvalues appear as the answer to so many different questions: the natural frequencies of a vibrating structure, the principal components of a data set, the long-run distribution of a Markov chain, the stability of an equilibrium in a differential equation, and the ranking vector behind PageRank are all eigenvector problems.
Two identities give you an instant check on any answer. The eigenvalues sum to the trace and multiply to the determinant. For the default matrix, the eigenvalues 11, 2 and 1 sum to 14 and multiply to 22, which are exactly the trace and determinant reported above.
Why the determinant appears in the recipe
Start from Av = λv and move everything to one side: (A − λI)v = 0. You are asking for a non-zero vector that the matrix A − λI sends to zero. A matrix does that only if it is singular — an invertible matrix sends nothing but zero to zero. So λ is an eigenvalue precisely when det(A − λI) = 0.
Expanding that determinant gives a polynomial of degree n in λ, the characteristic polynomial. For a 2×2 it is λ² − (tr A)λ + det A, so the eigenvalues come straight from the quadratic formula. For a 3×3 it is λ³ − (tr A)λ² + (sum of the three principal 2×2 minors)λ − det A, a cubic that this calculator solves with the trigonometric form of Cardano's formula and then refines with a few Newton steps.
Once you have λ, the eigenvectors are the null space of A − λI, found by row reduction. There is always at least one, because the matrix is singular by construction; the dimension of that null space is the geometric multiplicity.
The other multiplicity comes from the polynomial: if (λ − 3)² divides the characteristic polynomial, then 3 has algebraic multiplicity 2. The two need not agree. Geometric multiplicity is always at least 1 and never exceeds the algebraic one, and the gap between them is what makes a matrix defective.
Worked example: the 3×3 matrix [[2,0,0],[0,3,4],[0,4,9]]
The first row and column isolate the entry 2, so the matrix splits into a 1×1 block and a 2×2 block, and the eigenvalues are the union of the blocks' eigenvalues.
- Coefficients of the characteristic polynomial. The trace is 2 + 3 + 9 = 14. The three principal 2×2 minors are det[[2,0],[0,3]] = 6, det[[2,0],[0,9]] = 18 and det[[3,4],[4,9]] = 27 − 16 = 11, summing to 35. The determinant is 2 × 11 = 22. So the polynomial is λ³ − 14λ² + 35λ − 22.
- Use the block structure to solve it. The 1×1 block gives λ = 2 immediately. The 2×2 block [[3,4],[4,9]] has trace 12 and determinant 11, so λ = (12 ± √(144 − 44))/2 = (12 ± 10)/2, giving 11 and 1.
- Check against the identities. 11 + 2 + 1 = 14 = trace ✓. 11 × 2 × 1 = 22 = determinant ✓.
- Eigenvector for λ = 11. A − 11I = [[−9,0,0],[0,−8,4],[0,4,−2]]. The first row forces x = 0. The second says −8y + 4z = 0, so z = 2y. Taking y = 1 gives (0, 1, 2), which the calculator reports scaled to (0, 0.5, 1).
- Eigenvector for λ = 2. A − 2I = [[0,0,0],[0,1,4],[0,4,7]]. The last two rows force y = z = 0 (their 2×2 determinant is 7 − 16 = −9, non-zero), leaving x free: (1, 0, 0).
- Eigenvector for λ = 1. A − I = [[1,0,0],[0,2,4],[0,4,8]]. The first row forces x = 0; the second gives 2y + 4z = 0, so y = −2z. Taking z = 1 gives (0, −2, 1), which the calculator reports as the proportional (0, 1, −0.5).
Verify one of them directly: A(0, 1, 2) = (0, 3(1) + 4(2), 4(1) + 9(2)) = (0, 11, 22) = 11 × (0, 1, 2) ✓.
Three distinct eigenvalues give three independent eigenvectors, so the matrix is diagonalisable: A = PDP⁻¹ with P holding those eigenvectors as columns and D = diag(11, 2, 1). Note also that the matrix is symmetric, and the three eigenvectors are mutually perpendicular — (0,1,2)·(1,0,0) = 0 and (0,1,2)·(0,−2,1) = −2 + 2 = 0 — exactly as the spectral theorem promises.
How to read the spectrum
The sign and size of each eigenvalue tell you what the map does along that axis. |λ| > 1 stretches, |λ| < 1 contracts, λ < 0 flips the direction, and λ = 0 collapses it entirely — which is why a zero eigenvalue is the same statement as a zero determinant and a singular matrix.
The spectral radius decides long-run behaviour. The powers Aⁿ tend to zero when max|λ| < 1 and blow up when it exceeds 1. That single number determines whether an iterative scheme converges, whether a discrete dynamical system settles, and whether a linear recurrence grows or decays.
Complex eigenvalues mean rotation. A real matrix can have complex conjugate pairs, and when it does the map has a rotational component in the plane spanned by the real and imaginary parts of the eigenvector. The 90° rotation matrix [[0,−1],[1,0]] is the pure case: eigenvalues ±i, modulus 1, no real eigenvector at all because no real direction survives a quarter turn.
Compare the two multiplicities. When they agree for every eigenvalue, there is a full basis of eigenvectors and the matrix diagonalises. When geometric falls short of algebraic, the matrix is defective: [[2,1],[0,2]] has λ = 2 with algebraic multiplicity 2 but only the one eigenvector direction (1, 0). Defective matrices cannot be diagonalised at all, and the nearest thing available is the Jordan form.
Small matrices and their spectra
| Matrix | Eigenvalues | Trace | Determinant | Diagonalisable? |
|---|---|---|---|---|
| [[4,1],[2,3]] | 5, 2 | 7 | 10 | Yes |
| [[2,0],[0,3]] | 2, 3 | 5 | 6 | Yes — already diagonal |
| [[2,1],[0,2]] | 2, 2 | 4 | 4 | No — defective |
| [[0,−1],[1,0]] | ±i | 0 | 1 | Not over the reals |
| [[1,2],[2,4]] | 5, 0 | 5 | 0 | Yes, but singular |
| [[3,1],[1,3]] | 4, 2 | 6 | 8 | Yes — symmetric |
| [[2,0,0],[0,3,4],[0,4,9]] | 11, 2, 1 | 14 | 22 | Yes — symmetric |
| I₃ | 1, 1, 1 | 3 | 1 | Yes — every vector is an eigenvector |
The identity and the shear both have a repeated eigenvalue, and they sit on opposite sides of the defective line: the identity has a three-dimensional eigenspace, the shear a one-dimensional one.
Mistakes to avoid
- Solving det(A − λI) = 0 with a sign slip. Subtract λ from the diagonal, not add it. The 2×2 characteristic polynomial is λ² − (tr A)λ + det A; the minus sign on the trace term is the one people drop.
- Assuming a repeated eigenvalue means the matrix is defective. It does not. The identity has an eigenvalue repeated n times and is perfectly diagonalisable. Only a shortfall in the geometric multiplicity makes a matrix defective.
- Expecting real eigenvalues from a real matrix. Rotations have none. If your matrix is symmetric, the spectral theorem guarantees real eigenvalues; otherwise there is no such promise.
- Normalising the eigenvector and thinking that made it unique. Any non-zero multiple of an eigenvector is an eigenvector, including the negative. Two correct answers can differ by sign or scale; this calculator scales the largest component to 1 for readability.
- Reporting the zero vector as an eigenvector. It satisfies Av = λv for every λ, which is exactly why the definition excludes it.
- Using the characteristic polynomial numerically for large matrices. Its roots are badly conditioned even when the eigenvalues are not, which is why real software uses the QR algorithm instead. At 2×2 and 3×3 the closed forms used here are exact enough that it does not matter.
Why diagonalisation is worth the trouble
If A = PDP⁻¹ then An = PDnP⁻¹, and raising a diagonal matrix to a power just raises each diagonal entry. A calculation whose cost would grow with n becomes constant. The same trick solves systems of linear differential equations — eAt = PeDtP⁻¹ decouples the system into independent scalar equations — and gives the closed form of any linear recurrence. Everything expensive about a matrix becomes cheap in its eigenbasis, which is why finding that basis is worth the effort.
Related tools
The characteristic polynomial is a determinant, so the determinant calculator is the natural companion — and its output is the product of the eigenvalues, which is the check this page reports. To find the eigenvectors by hand you row-reduce A − λI, which is exactly what the RREF calculator does; the eigenspace is the null space it reports, and its dimension is the nullity.
Verify any eigenvector by multiplying it back: Av should come out as a clean multiple of v, which the matrix multiplication calculator will show you. Building P from the eigenvectors requires inverting it for the diagonalisation, and the matrix inverse calculator also reports the condition number — which for a symmetric matrix is the ratio of the largest to the smallest absolute eigenvalue you found here.
Eigenvalues of a Hessian classify critical points of a function of several variables, so if you arrived here from a multivariable optimisation problem, build the Hessian with the partial derivative calculator and bring it back. And when a cubic characteristic polynomial has roots that are not nice numbers, the Newton-Raphson calculator will refine any of them from a starting guess.