# Are eigenvectors corresponding to the same eigenvalue always parallel?

Suppose that $$A \underline{x}_1 = \lambda \underline{x}_1$$ and $$A\underline{x}_2=\lambda \underline{x}_2$$, where $$A \in \mathbb{R}^{n \times m}, \lambda \in \mathbb{C}, \ \mathrm{and} \ \underline{x}_1, \underline{x}_2 \in \mathbb{R}^m$$.

It is clear that there exists infinitely many eigenvectors, since for all $$\alpha \in \mathbb{C}$$, $$\alpha \underline{x}_1$$ is an eigenvector as well.

But is it generally true that all eigenvectors corresponding to the same eigenvalue $$\lambda$$ are parallel, i.e. $$\underline{x}_1=\alpha \underline{x}_2$$? How could it be proven?

• NB: All nonzero $\alpha \in \mathbb C$. Apr 30, 2021 at 16:26

Not, it is true. If $$A\in\Bbb R^{n\times n}$$ is the identity matrix, then all vectors of $$\Bbb R^n$$ are eigenvectors of $$A$$ with eigenvalue $$1$$. So, if $$n>1$$, they're not all parallel.
• What about if $\lambda=0$? I guess it is the same as asking if all vectors in the null-space of some matrix are parallel. Apr 30, 2021 at 16:37
• Yes. And that happens if and only if the dimension of the null space of the matrix is $1$. Apr 30, 2021 at 16:37