What is meant by "find the eigenvectors of a matrix"

Question

Let's say I have a question asking me to find the eigenvectors associated with $\lambda = 2$ for $A$ (a $3 \times 3$ matrix).

I find that the eigenvectors associated with $\lambda = 2$ are all vectors spanned by $\mathbf{u}_1$ and $\mathbf{u}_2$ (a plane).

Can I call $\mathbf{u}_1$ and $\mathbf{u}_2$ the eigenvectors associated with $\lambda = 2$, even though there is a whole plane of them? I can find a basis for the space but I don't see how I can give a finite number of vectors. Is the correct way to word it that eigenvectors are $\operatorname{span} \{\mathbf{u}_1, \mathbf{u}_2 \}$?

Answer 1

As you've observed, sometimes the eigenspace $V_\lambda$ associated to an eigenvalue is more than $1$-dimensional. To find "the eigenvectors" is to find a linearly independent set of eigenvectors for each eigenvalue that span this subspace, i.e. an eigenbasis. (They might not span the whole space if $A$ is not diagonalizable.)

Edited to clarify: An eigenvector must be nonzero, so the eigenspace $V_\lambda$ is the subspace of all eigenvectors of the given eigenvalue $\lambda$, together with the zero vector $\mathbf{0}$: $$ V_\lambda = \{ v \in V \mid A\mathrm{v} = \lambda\mathrm{v} \}. $$ You are describing this vector subspace with a minimum of information, so just as you would with any subspace, describe a basis for it.

Answer 2

When you write that "$\mathbf{u}_1$ and $\mathbf{u}_2$ are the eigenvectors associated with $\lambda=2$", the usage of the definite article the definitely carries a taint of uniqueness, which is something you don't want.

You could simply say instead that "$\mathbf{u}_1$ and $\mathbf{u}_2$ span the eigenvectors associated with $\lambda=2$", or even better that they span the $\lambda=2$ eigenspace. Your last way of writing it, with $\text{span}\{\mathbf{u}_1,\mathbf{u}_2\}$, is also a good way to do it with fewer words and more symbols, if that is desirable.

Answer 3

Let‘s review the commonly used terms for eigenvalue problems:

The idea behind eigenvalue problems is that we would like to find the vectors $v \neq 0$ which are mapped to a multiple of themselves under an endomorphism, or, after choosing a basis, a matrix $A$. Meaning, $$A \cdot v = \lambda \cdot v,$$ where $\lambda$ is an element in the underlying field, for example the real numbers.

If there is such a non-trivial vector to a specific $\lambda$ such that this equation holds, $\lambda$ is called an eigenvalue. Any $v \neq 0$ which makes the equation true is then called eigenvector to the eigenvalue $\lambda$, so yes, $u_1$ and $u_2$ are most certainly called eigenvectors!

It is easy to see that if $v$ is an eigenvector, so is any non-zero multiple of $v$! Therefore, the space of all eigenvectors to an eigenvalue together with the zero vector, which is called the eigenspace $V_\lambda$ of the eigenvalue $\lambda$, is usually not finite (And never finite over the real or complex numbers, except you define eigenspaces even for $\lambda$‘s which are not eigenvalues, in which case the eigenspace is $\{0\}$). To go back to your question: You cannot give a finite number of eigenvectors as your answer as there are many more!

So, the correct term for $span(u_1,u_2)$ is eigenspace of the eigenvalue 2.

In case you are a physicist: If they talk about ”the eigenvectors“, they usually mean one should pick an orthonormal basis of the eigenspace, even though this of course does not yield unique results.