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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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    Tha$n$ks both of you! That's a much easier way to see it!2011-11-08

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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 ago2011-11-08