An eigenvector of A is a vector that, under A,

Keeps its direction and is only scaled. By definition A v = λ v: the direction is preserved and the vector is scaled by the eigenvalue λ.

The eigenvalues of A are found by solving:

det(A − λI) = 0. The characteristic equation det(A − λI) = 0 yields the eigenvalues.

Once you have an eigenvalue λ, its eigenvectors are:

The nullspace of (A − λI). Substitute λ and solve (A − λI)v = 0; the solutions are the eigenvectors.

For A = [[2, 1], [1, 2]], the eigenvalues are:

1 and 3. (2−λ)² − 1 = (λ−1)(λ−3) = 0, so λ = 1 and λ = 3 (they sum to the trace 4 and multiply to det 3).

Diagonalizing A = P D P⁻¹ makes which computation easy?

High powers Aⁿ. Aⁿ = P Dⁿ P⁻¹, and powers of a diagonal matrix are found entry by entry.

The eigenvalues of a matrix always multiply to give:

The determinant. The product of the eigenvalues equals the determinant; their sum equals the trace.

Eigenvalues & Eigenvectors

Directions the transformation leaves alone

Apply a matrix to most vectors and they swing to a new direction. But special directions survive: the vector comes out pointing the same way (or exactly reversed), merely stretched or shrunk. Such a vector is an eigenvector, and the factor it is scaled by is its eigenvalue λ.

The whole idea is one equation: A v = λ v. The matrix acting on v does nothing more interesting than multiplying it by a number.

Set a matrix and look for a direction whose arrow stays on the same line through the origin after transforming — only longer or shorter. That line is an eigenvector; how much it stretches is the eigenvalue. Most directions get knocked off their line; eigen-directions do not.

Why anyone cares

Eigenvectors are the natural axes of a transformation. Along them the matrix is just multiplication, so anything involving repeating the transformation becomes easy: raising A to a power, the long-run behaviour of a Markov process, the stability of a system of differential equations.

They are also the engine behind famous algorithms — Google's original PageRank is an eigenvector, and Principal Component Analysis (the backbone of a lot of data science) finds the eigenvectors of a covariance matrix to see which directions carry the most information.

Finding the eigenvalues

Rewrite A v = λ v as (A − λI) v = 0. For a non-zero v to be sent to zero, the matrix (A − λI) must be singular — so its determinant must vanish. That gives the characteristic equation:

det(A − λI) = 0solve this polynomial in λ to get the eigenvalues; a 2×2 gives a quadratic

Finding the eigenvectors, and diagonalization

For each eigenvalue λ, substitute it back and solve (A − λI) v = 0 — the eigenvectors are that nullspace.

When an n × n matrix has a full set of n independent eigenvectors, you can diagonalize it: A = P D P⁻¹, where the columns of P are the eigenvectors and D is the diagonal matrix of eigenvalues. This is the payoff — powers become trivial, since Aⁿ = P Dⁿ P⁻¹ and raising a diagonal matrix to a power just raises each entry.

Two shortcuts to check your work

The eigenvalues of a matrix always sum to its trace (the diagonal total) and multiply to its determinant. If your eigenvalues fail either test, you have made an arithmetic slip.

A worked example

Worked example

Find the eigenvalues and eigenvectors of A = [[2, 1], [1, 2]].

Characteristic equation: det[[2−λ, 1], [1, 2−λ]] = (2−λ)² − 1 = λ² − 4λ + 3 = (λ−1)(λ−3).Expand det(A − λI) and set it to zero.
Eigenvalues λ = 1 and λ = 3. (Check: sum 4 = trace, product 3 = det.)The roots of the characteristic polynomial, verified by the trace/determinant shortcuts.
For λ = 3: (A − 3I)v = [[−1, 1], [1, −1]]v = 0 → v = [1, 1].Solve the nullspace; the equations say the components are equal.
For λ = 1: (A − I)v = [[1, 1], [1, 1]]v = 0 → v = [1, −1].Here the components are opposite.

λ = 3 with eigenvector [1, 1], and λ = 1 with eigenvector [1, −1] — two perpendicular directions the matrix only stretches.