How to Calculate an Eigenspace

An eigenspace collects all vectors that a matrix sends to scalar multiples of themselves for one fixed eigenvalue. To calculate it, turn the eigenvector equation into a homogeneous linear system, solve that system, and describe its full solution set.

The definition

Let $A$ be an $n\times n$ matrix over a field $F$ (typically $\mathbb{R}$ or $\mathbb{C}$), and let $\lambda\in F$ be an eigenvalue of $A$. A vector $x\in F^n$ is an eigenvector for $\lambda$ when $x\ne 0$ and

$$Ax=\lambda x.$$

Rearranging gives

$$(A-\lambda I)x=0,$$

where $I$ is the $n\times n$ identity matrix. Therefore the eigenspace is

$$E_\lambda=\operatorname{Null}(A-\lambda I)=\{x\in F^n:(A-\lambda I)x=0\}.$$

The eigenspace includes the zero vector because it is a null space. The eigenvectors themselves are its nonzero vectors.

How to find an eigenspace

For each eigenvalue you want to study:

1. Calculate $A-\lambda I$ by subtracting $\lambda$ from each diagonal entry of $A$.
2. Solve $(A-\lambda I)x=0$, usually by row-reducing the matrix.
3. Identify the free variables and express each variable in terms of them.
4. Write the general solution as a linear combination of vectors. Those vectors form a basis for the eigenspace if they are linearly independent.

If the eigenvalue is not supplied, first find it by solving the characteristic equation $\det(A-\lambda I)=0$. Either $\det(A-\lambda I)$ or $\det(\lambda I-A)$ is used in textbooks; the roots are the same.

Worked example

Consider

$$A=\begin{pmatrix}2&1&0\\0&3&0\\0&0&2\end{pmatrix}.$$

First find the eigenvalues. Since $A$ is upper triangular, its eigenvalues are its diagonal entries: $2$ and $3$.

For $\lambda=2$,

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

Writing $x=(x_1,x_2,x_3)^T$, the system says $x_2=0$. The variables $x_1$ and $x_3$ are free, so

$$x=\begin{pmatrix}x_1\\0\\x_3\end{pmatrix} =x_1\begin{pmatrix}1\\0\\0\end{pmatrix} +x_3\begin{pmatrix}0\\0\\1\end{pmatrix}.$$

Thus

$$E_2=\operatorname{span}\left\{\begin{pmatrix}1\\0\\0\end{pmatrix},\begin{pmatrix}0\\0\\1\end{pmatrix}\right\}.$$

For $\lambda=3$,

$$A-3I=\begin{pmatrix}-1&1&0\\0&0&0\\0&0&-1\end{pmatrix}.$$

The equations are $-x_1+x_2=0$ and $-x_3=0$. Therefore $x_2=x_1$ and $x_3=0$, giving

$$x=x_1\begin{pmatrix}1\\1\\0\end{pmatrix},\qquad E_3=\operatorname{span}\left\{\begin{pmatrix}1\\1\\0\end{pmatrix}\right\}.$$

You can check the basis vectors by multiplying by $A$: the first two are sent to twice themselves, while $(1,1,0)^T$ is sent to three times itself.

Interpreting the result

The dimension of an eigenspace is called the geometric multiplicity of its eigenvalue. In the example, $\dim(E_2)=2$ and $\dim(E_3)=1$. If an eigenvalue has algebraic multiplicity greater than one, its eigenspace may still have dimension smaller than that multiplicity. The null-space calculation determines the geometric multiplicity directly.

A basis is one convenient way to describe the whole eigenspace. Any linear combination of its basis vectors belongs to the eigenspace, and every vector in the eigenspace can be written that way. A single eigenvector is not usually a complete description if the eigenspace has dimension greater than one.

Common mistakes

• Using $A$ instead of $A-\lambda I$. The eigenvalue equation becomes a null-space problem only after subtracting $\lambda I$.
• Treating the zero vector as an eigenvector. It belongs to the eigenspace but is excluded from the definition of eigenvector.
• Stopping at one solution. You need the general solution, including all free variables, to describe the entire eigenspace.
• Confusing a basis with the eigenspace. The basis is a set of vectors; the eigenspace is their full span.
• Assuming every eigenvalue gives a one-dimensional space. A repeated eigenvalue can have an eigenspace of dimension greater than one.

If the matrix has real entries but an eigenvalue is complex, the corresponding eigenspace is naturally computed over $\mathbb{C}^n$. Over $\mathbb{R}$, that complex eigenvalue is not an eigenvalue in the first place.