For each eigenvalue, the eigenvectors (together with $\mathbf{0}$) form a subspace. How big that subspace is decides a great deal, as the next topic will show. We find eigenvalues with the characteristic equation and then their eigenvectors by row reduction.
On an eigenspace the matrix acts very simply: every vector in $E_\lambda$ is just multiplied by $\lambda$. A line eigenspace is a line the matrix stretches; a plane eigenspace is a whole plane the matrix stretches uniformly.
The eigenspace of $A$ for $\lambda$ is $E_\lambda = \operatorname{Nul}(A - \lambda I)$: all eigenvectors for $\lambda$ plus the zero vector. Its dimension is the geometric multiplicity of $\lambda$. The geometric multiplicity is always at least $1$ and at most the algebraic multiplicity.
Worked example: An eigenspace that is only a line
- $B - 2I = \begin{bmatrix} 0 & 0 & 0 \\ 1 & 1 & -1 \\ 0 & 2 & -2 \end{bmatrix}$, which reduces to $\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{bmatrix}$.
- Two pivots, one free variable: $x_1 = 0$, $x_2 = x_3 = t$.
- $E_2 = \operatorname{Span}\left\{\begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}\right\}$. Check: $B\begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 2 \\ 2 \end{bmatrix}$. ✓
- So $\lambda = 2$ has algebraic multiplicity 2 but geometric multiplicity only 1.
Answer: $\left\{(0, 1, 1)\right\}$
Worked example: An eigenspace that is a plane
- $C - 3I = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 1 & 2 & -2 \end{bmatrix}$: one pivot, two free variables.
- $x_1 = -2x_2 + 2x_3$. With $x_2 = s$, $x_3 = t$: $\mathbf{x} = s\begin{bmatrix} -2 \\ 1 \\ 0 \end{bmatrix} + t\begin{bmatrix} 2 \\ 0 \\ 1 \end{bmatrix}$.
- $E_3$ is a plane with basis $\{(-2, 1, 0), (2, 0, 1)\}$: geometric multiplicity 2, equal to the algebraic multiplicity.
Answer: $\{(-2, 1, 0),\ (2, 0, 1)\}$
Why it works
| Matrix | Eigenvalues | Eigenspaces |
|---|---|---|
| $2I$ (scaling) | $2, 2$ | all of $\mathbb{R}^2$ (dimension 2) |
| $\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$ (shear) | $1, 1$ | one line (dimension 1) |
| $\begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix}$ | $5, 2$ | two lines |
| rotation by $90^\circ$ | $\pm i$ (not real) | no real eigenvectors |
A practical routine for a whole matrix: (1) find the eigenvalues from $\det(A - \lambda I) = 0$ and note their multiplicities; (2) for each eigenvalue, row reduce $A - \lambda I$; (3) count the free variables to get the dimension and read off one basis vector per free variable. If a row reduction of $A - \lambda I$ ever shows a pivot in every column, that $\lambda$ was not an eigenvalue after all: go back and check the characteristic polynomial.
Key idea