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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?

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

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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.

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  • $\begingroup$ 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. $\endgroup$
    – mathslover
    Apr 30, 2021 at 16:37
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    $\begingroup$ Yes. And that happens if and only if the dimension of the null space of the matrix is $1$. $\endgroup$ Apr 30, 2021 at 16:37
  • $\begingroup$ But can the dimension of the null-space be greater than 1? $\endgroup$
    – mathslover
    Apr 30, 2021 at 16:48
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    $\begingroup$ Yes. Take the null matrix, for instance. $\endgroup$ Apr 30, 2021 at 16:50
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No. All vectors (including non-parallel ones) are eigenvectors of the identity transformation with eigenvalue 1. So eigenvectors needn't be parallel.

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