For a matrix $A$, an eigenvector satisfies

$$A\mathbf v=\lambda\mathbf v.$$

Rearranging gives

$$(A-\lambda I)\mathbf v=\mathbf0.$$

A nonzero solution exists only when $A-\lambda I$ is not invertible. Therefore its determinant must vanish.

The characteristic equation

Eigenvalues are found from

$$\det(A-\lambda I)=0.$$

The expression

$$p_A(\lambda)=\det(A-\lambda I)$$

is the characteristic polynomial.

For example, if

$$A=\begin{pmatrix}2&0\0&3\end{pmatrix},$$

then

$$\det(A-\lambda I)=(2-\lambda)(3-\lambda),$$

so the eigenvalues are $2$ and $3$.

Finding eigenvectors

For each eigenvalue, solve

$$(A-\lambda I)\mathbf v=\mathbf0.$$

For $\lambda=2$ in the example,

$$A-2I=\begin{pmatrix}0&0\0&1\end{pmatrix},$$

so $v_2=0$ and $v_1$ is free. The eigenspace is therefore

$$\operatorname{span}((1,0)).$$

Similarly, the eigenspace for $\lambda=3$ is $\operatorname{span}((0,1)).$$

Multiplicity

An eigenvalue may occur more than once as a root of the characteristic polynomial. This is its algebraic multiplicity.

The dimension of its eigenspace is its geometric multiplicity. The geometric multiplicity is at least $1$ and cannot exceed the algebraic multiplicity.

Calculating eigenvectors therefore combines determinant equations with homogeneous linear systems.

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