Let $A \in \mathcal{M}_{12 \times 12}$ and let $v_1,v_2$ be eigenvectors of $A$ such that $Av_1 = v_1$ and $Av_2 = 2v_2$. Is it true that vector $v_1 - v_2$ is not eigenvector of $A$?

My answer:

We have $A(v_1 - v_2) = Av_1 - Av_2 = v_1 - 2v_2$. Let suppose that $v_1-v_2$ is eigenvector of $A$. So exists $\lambda \neq 0 \in \mathbb{R} $ such that $v_1 - 2v_2 = \lambda (v_1 - v_2)$ hence $(1- \lambda)v_1 + (\lambda - 2)v_2 = 0$. Of course, $v_1,v_2$ are linear independenc so $\lambda =1$ and $\lambda=2$. We have conflict, so the answer is FALSE.

But answer in my book is TRUE. Could you tell me why?

  • 7
    $\begingroup$ Maybe it's a parsing problem. It is true that $v_1 - v_2$ is not an eigenvector. $\endgroup$ Sep 7, 2013 at 13:14
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    $\begingroup$ You stated the problem as "Is it true that vector $v_1-v_2$ is not an eigenvector of $A$?" I would say that's true -- it is not an eigenvector $A$. $\endgroup$ Sep 7, 2013 at 13:14
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    $\begingroup$ It's my mistake, because the ask in book is: vector $v_1-v_2$ is not eigenvector of $A$. Probably I have read it as $v_1-v_2$ is eigenvector of $A$. Sorry for performance. $\endgroup$
    – Thomas
    Sep 7, 2013 at 13:30

2 Answers 2


Your maths logic is correct, eigenvectors do not generally combine to form new eigenvectors.

So the answer to your title question is No, it is not true that $v_1-v_2$ is an eigenvector.

The answer to your book's question is Yes, it is true that $v_1-v_2$ is not an eigenvector.


A linear combination of two eigenvectors is only an eigenvector when the two eigenvectors possess the same eigenvalue.

  • $\begingroup$ I know what you're getting at, but this isn't true in general. As the simplest possible example, if you've got two eigenvectors $v_1,v_2$ with distinct eigenvalues, then $0v_1+v_2=v_2$ is a linear combination of two eigenvectors with different eigenvalues, but it's an eigenvalue. I think you should edit your post because what you've written probably isn't what you were trying to say. $\endgroup$ Oct 3, 2013 at 16:20
  • $\begingroup$ Since nonzero multiples of eigenvectors are eigenvectors, this can be corrected by avoiding linear combinations altogether: a sum of eigenvectors is a eigenvector only if the corresponding eigenvalues are equal. In the question apply this to the eigenvectors $v_1$ and $-v_2$. $\endgroup$ Oct 3, 2013 at 16:57

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