Can One Eigenvalue Have Two Different Eigenvectors?

Yes. A single eigenvalue can have many different eigenvectors. The key is that eigenvectors for an eigenvalue are all the nonzero solutions of one linear equation, and there can be many such solutions.

What counts as an eigenvector?

For a matrix $A$, a scalar $\lambda$ is an eigenvalue if there is a nonzero vector $v$ such that

$$Av=\lambda v.$$

Rearranging gives

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

where $I$ is the identity matrix. Thus the eigenvectors for $\lambda$ are exactly the nonzero vectors in the null space of $A-\lambda I$. That null space is called the eigenspace for $\lambda$.

The zero vector is excluded: it satisfies $A0=\lambda 0$, but the definition of an eigenvector requires a nonzero vector.

Different eigenvectors can point in the same direction

If $v$ is an eigenvector for $\lambda$ and $c$ is any nonzero scalar, then

$$A(cv)=cAv=c\lambda v=\lambda(cv).$$

So $cv$ is also an eigenvector for the same eigenvalue. For example, if $v$ is an eigenvector, then $2v$ and $-v$ are different vectors but are still eigenvectors for the same eigenvalue. They are scalar multiples, so they are not linearly independent.

Over the usual fields of real or complex numbers, this means even a one-dimensional eigenspace contains infinitely many distinct eigenvectors. Those vectors all lie on the same line through the origin.

Worked example: two independent eigenvectors

Consider

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

For $\lambda=2$, subtract $2I$:

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

A vector $v=(x,y,z)$ is in the null space of this matrix exactly when $z=0$. Therefore the eigenspace for $2$ is

$$\{(x,y,0):x,y\in\mathbb{R}\}.$$

For instance, $(1,0,0)$ and $(0,1,0)$ are both eigenvectors for $2$:

$$A\begin{pmatrix}1\\0\\0\end{pmatrix}=2\begin{pmatrix}1\\0\\0\end{pmatrix}, \qquad A\begin{pmatrix}0\\1\\0\end{pmatrix}=2\begin{pmatrix}0\\1\\0\end{pmatrix}.$$

They are not scalar multiples, so they are linearly independent. Their sum, $(1,1,0)$, is another eigenvector for $2$. In fact, every nonzero vector of the form $(x,y,0)$ is one.

Does a repeated eigenvalue guarantee independent eigenvectors?

No. A repeated eigenvalue always has nonzero eigenvectors, and those include scalar multiples of one another. But it need not have two linearly independent eigenvectors. The number of independent eigenvectors for $\lambda$ is the dimension of its eigenspace, also called its geometric multiplicity.

For example, take

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

The only eigenvalue is $2$, and

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

Solving $(B-2I)(x,y)=0$ gives $y=0$. So the eigenspace is the line of vectors $(x,0)$, and it has dimension one. There are many distinct eigenvectors, such as $(1,0)$ and $(3,0)$, but they are scalar multiples; there are no two linearly independent eigenvectors for this eigenvalue.

This distinction matters when you are checking whether a matrix can be diagonalized. A matrix needs enough linearly independent eigenvectors to form a basis—not merely several different eigenvectors.

Common mistakes

• Calling the zero vector an eigenvector: it is always excluded, even though it satisfies the equation formally.
• Treating different as independent: two vectors can be distinct and still be scalar multiples.
• Assuming a repeated eigenvalue supplies multiple independent eigenvectors: solve $(A-\lambda I)v=0$ to find the eigenspace and its dimension.
• Confusing an eigenvalue with its eigenvectors: the eigenvalue is a scalar; its eigenvectors are nonzero vectors in the associated eigenspace.