TITLE: From $Ax=\lambda x$, we have $Ax i = \lambda x i$ , where $i^2=-1$??
QUESTION [2 upvotes]: Actually I found this problem when I met a question, asking me to prove the eigenvector and eigenvalue of real symmetric matrices are all real.
I have already proved the eigenvalue part already, but for the eigenvector part, I found this problem.
Same to the title. Assume $\lambda$ is a real number, $A$ is a real matrix, $x$ is a real eigenvector, and we have
$$ Ax= \lambda x  $$
Then do we have 
$$Ax i =\lambda x i  $$
where $i^2=-1$?
This means that if it has an real eigenvector, then it has a complex eigenvector. I feel really strange of it.
Best

REPLY [2 votes]: The statement that eigenvalues and eigenvectors of symmetric matrices are real is for real matrices.  For symmetric matrices you can have a real eigenvector, but as you say if you are working over the complex numbers you can multiply any eigenvector by a complex constant and make its entries complex.  For asymmetric real matrices you can have a pair of complex conjugate eigenvalues.  You will not be able to have all the entries in the eigenvector real in those cases.  
When your matrices are over the complex numbers, the corresponding statement is that Hermitian matrices have real eigenvalues.  The eigenvectors need not be real, but ones corresponding to different eigenvalues are orthogonal.
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