Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question

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 $W$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.