How to Find a Basis for Each Eigenspace

An eigenvalue can have many eigenvectors, so the goal is not usually to list them all. Instead, find a basis for its eigenspace: a linearly independent set of eigenvectors whose span contains every eigenvector for that eigenvalue. This gives a compact description of all the possibilities.

What is an eigenspace?

For a square matrix $A$ and eigenvalue $\lambda$, an eigenvector $v$ satisfies $Av=\lambda v$ and must be nonzero. Rearranging gives

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

The solution set of this homogeneous system is the nullspace of $A-\lambda I$, called the eigenspace for $\lambda$:

$$E_\lambda=\ker(A-\lambda I).$$

The eigenspace includes the zero vector because every nullspace does. But the zero vector is not an eigenvector. Any nonzero vector in the eigenspace is an eigenvector for $\lambda$.

How to find a basis

For each eigenvalue, follow these steps:

1. Form $A-\lambda I$.
2. Solve $(A-\lambda I)x=0$, usually by row-reducing the matrix.
3. Express the solution using free variables as a linear combination of vectors.
4. Use the vectors in that expression as a basis for the eigenspace.

The vectors must be linearly independent and span the full solution set. A basis is not unique: different choices of independent spanning vectors can describe the same eigenspace.

Worked example

Consider

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

Its characteristic polynomial is $(2-\lambda)^2(3-\lambda)$, so its eigenvalues are $2$ and $3$. Find a basis for each eigenspace.

Eigenvalue $2$

Subtract $2I$ and solve:

$$A-2I=\begin{pmatrix} 0&0&1\\ 0&0&0\\ 0&0&1 \end{pmatrix}, \qquad (A-2I)\begin{pmatrix}x\\y\\z\end{pmatrix}=0.$$

The equations require $z=0$; $x$ and $y$ are free. Thus

$$\begin{pmatrix}x\\y\\z\end{pmatrix} =x\begin{pmatrix}1\\0\\0\end{pmatrix} +y\begin{pmatrix}0\\1\\0\end{pmatrix}.$$

A basis is $\{(1,0,0),(0,1,0)\}$. The eigenspace is the plane of vectors $(x,y,0)$. For instance, $(3,-1,0)$ is an eigenvector for $2$, since it is a nonzero vector in this plane. The vector $(0,0,0)$ is in the eigenspace but is not an eigenvector.

Eigenvalue $3$

Now solve $(A-3I)v=0$:

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

The equations give $x=z$ and $y=0$, so every solution has the form

$$\begin{pmatrix}x\\y\\z\end{pmatrix} =z\begin{pmatrix}1\\0\\1\end{pmatrix}.$$

Therefore, a basis for the eigenspace of $3$ is $\{(1,0,1)\}$. Since this eigenspace is one-dimensional, every eigenvector for $3$ is a nonzero scalar multiple of $(1,0,1)$.

How many eigenvectors are there?

A basis lists only enough vectors to describe the eigenspace; it does not list every eigenvector. If $v_1,\ldots,v_k$ is a basis for $E_\lambda$, then every vector in the eigenspace is a combination

$$c_1v_1+\cdots+c_kv_k.$$

Every such combination is an eigenvector when it is nonzero. Over an infinite field, there are infinitely many eigenvectors whenever the eigenspace has positive dimension, because any nonzero scalar multiple of an eigenvector is another eigenvector.

The number of vectors in a basis for $E_\lambda$ is its dimension, also called the geometric multiplicity of $\lambda$. It can be smaller than the eigenvalue’s multiplicity as a root of the characteristic polynomial. In the example, $2$ is a repeated root, but its eigenspace has dimension two.

Common mistakes

• Including zero as an eigenvector: Zero belongs to the eigenspace, but the definition of eigenvector excludes it.
• Using the wrong equation: For eigenvalue $\lambda$, solve $(A-\lambda I)x=0$, not $Ax=0$ unless $\lambda=0$.
• Listing dependent vectors as a basis: A spanning set is not a basis unless its vectors are linearly independent. Remove redundant vectors.
• Confusing multiplicity with eigenspace dimension: A repeated eigenvalue does not automatically have an eigenspace of the same dimension. Compute the nullspace to determine its dimension.
• Thinking the basis is unique: Many different bases can span the same eigenspace; what matters is independence and spanning.