# $v_1,v_2$ are eigenvectors of $A$. Is it true that $v_1-v_2$ is eigenvector of $A$?

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 $A$?

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?

• Maybe it's a parsing problem. It is true that $v_1 - v_2$ is not an eigenvector. Commented Sep 7, 2013 at 13:14
• 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 $A$. Commented Sep 7, 2013 at 13:14
• 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. Commented Sep 7, 2013 at 13:30

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.
• 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. Commented Oct 3, 2013 at 16:20
• 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 $-v_2$. Commented Oct 3, 2013 at 16:57