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?
Best Answer
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.