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If $\lambda$ is a real eigenvalue of a real matrix $M$, does there necessarily exist a real eigenvector of $M$ corresponding to $\lambda$?

Edit: Never mind. I figured it out.. If $\lambda$ is a real eigenvalue, then $\det(\lambda I-M)=0$, which means there exists a real vector $x$ such that $Mx=\lambda x$. Right?

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If $\mathbf A(\mathbf u+i\mathbf v)=\lambda(\mathbf u+i\mathbf v)$, then $\mathbf A\mathbf u=\lambda \mathbf u$ and $i\mathbf A\mathbf v=i\lambda\mathbf v$... make of that what you will. – J. M. Nov 8 '11 at 2:40
Hint: real linear combinations respect complex conjugation; that is, $\overline{(a_0z+a_1w+\ldots)} = a_0\bar{z}+a_1\bar{w}+\ldots$. If $v$ is an eigenvector, can you show that $\bar{v}$ is an eigenvector? If so, what about their sum? – Steven Stadnicki Nov 8 '11 at 2:40
Thanks both of you! That's a much easier way to see it! – user19078 Nov 8 '11 at 2:44

Yes. If there exists some ${\bf v} \not= 0$ (${\bf v}$ complex or whatever) such that $M{\bf v}=\lambda {\bf v}$, then $(M-\lambda I){\bf v}=0$ so $M-\lambda I$ has a non-trivial solution. Therefore, $\mathrm{det}(M-\lambda I)=0$ and so $M-\lambda I$ is a real singular matrix and thus has a real non-trivial solution. This is your desired eigenvector.

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Thanks.. I figured it out just a few seconds ago – user19078 Nov 8 '11 at 2:39

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