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Under what conditions does a complex matrix with a real eigenvalue have a corresponding real eigenvector?

Say I have a complex matrix, $A$, with eigenvalue $\lambda = 1$, under what conditions, is the corresponding eigenvector strictly real?

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up vote 2 down vote accepted

$A$ has a real eigenvector corresponding to a real eigenvalue corresponding to a real eigenvalue if and only if an eigenvector of its real part (viewed as a matrix in the real matrix space) lies inside the null space of its imaginary part. Say $A=R+iC$ where $R$ and $C$ are real matrices. If $Av=\lambda v$ for some real scalar $\lambda$ and nonzero real vector $v$, then $Rv+iCv=\lambda v$. Hence $Rv=\lambda v$ and $Cv=0$.

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In particular, at minimum the imaginary part of the matrix must be singular... – Thomas Andrews Feb 20 '13 at 19:28

I'm not sure how much it helps, but here is a condition:

$Av=v$ for a nonzero real vector if and only if there is an invertible real matrix $P$ such that the first column of $P^{-1} A P$ is $(1,0,\ldots, 0)^t$.

Proof: Complete $v$ to a basis of $\mathbb{C}^n$ using real vectors and let $P$ be the corresponding transition matrix.

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