Q:

From $Ax=\lambda x$, we have $Ax i = \lambda x i$ , where $i^2=-1$??

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

A:

Usually, you'd write $\mu x$ for a scalar $\mu$ and a vector $x$. Even if $x$ is a vector consisting only of real entries, you can view it as a complex vector in $\mathbb C ^n$ (with the imaginary part of each of its entries being zero).
For such a vector in $\mathbb C^n$, the product with a complex number such as $\mu = i$ is well defined.
From $Ax = \lambda x$, we can deduce $$A(\mu x) = \mu A x = \mu \lambda x = \lambda (\mu x).$$
So viewed a linar map on a vector space which has $\mathbb C$ as its field, yes, then $ix$ is a complex eigenvector of $A$ (which not linearly independent from $x$).
Edit: Responding to your question that you'd need to show that a symmetric real matrix has only real eigenvalues and you can always find a real eigenvector to it: 
Let $z = (x_1 + iy_1,x_2+ iy_2, \dots, x_n+ iy_n)$, $x_j,y_j \in \mathbb R$ be a complex eigenvector of $A$, so that $$Az = \lambda z.$$ $\lambda$ is real (you said in the comment that  you've shown this already). Taking the componentwise complex conjugate of this equation, we end up with:
$$A \overline z = \lambda \overline z,$$
where $\overline z = (x_1- iy_1,x_2- iy_2, \dots, x_n- iy_n)$. (Why does the complex conjugate only apply to $z$?)
What's $A(z + \overline z)$? What's the imaginary part of each of the components?

