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Let $A$ be real and let $\lambda = \alpha + i \beta$ be a complex eigenvalue of $A$ with eigenvector $x + iy$, show that the space spanned by $x$ and $y$ is an invariant subspace of $A$.

What I believe I need to show: I think I want to show $Av=xv$ and $Av=yv$ where $v$ is the eigenvector given above. Is this assumption correct? And if so, I wasn't having any luck proving this. If this isn't what I'm suppose to prove, could someone explain what I am suppose to try and prove for this problem. Thank you.

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1 Answer 1

Hint: You need to show that $Ax$ is a linear combination of $x$ and $y$, ditto for $Ay$. A natural thing to do is to exploit facts about complex conjugates. What does $A$ do to $x-iy$? Here the fact that $A$ is real is crucial.

It may be useful to note that $$Av=\lambda v=(\alpha + i\beta)(x+i y).$$ Expand the right-hand side.

Note on attempt: If you could prove what you tried to prove, that would indeed finish things. But what you tried to prove is not necessarily true. The space spanned by $x$ is not necessarily invariant, and neither is the space spanned by $y$. What you are asked to show is that if $W$ is the space spanned by $x$ and $y$, then $AW\subseteq W$. But $A$ may do quite a bit of scrambling of $W$.

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