$Ax = kx$ for complex scalar $k$ and complex vector $x$, and symmetric matrix $A$ Prove that if $Ax = kx$ for some non-zero, possibly complex valued scalar $k$ and non-zero complex valued vector $x$, and real symmetric $n$x$n$ matrix $A$, then $k$ must be real valued.
A hint was given saying to consider the complex conjugates. I'm unsure how to go about this proof. I tried splitting $k$ and $x$ into complex form, $a + bi$, and $c + di$ respectively (note that c and d are $n$x$1$ column vectors and a and b are real valued scalars). So $Ax =  Ac + Adi = ac - bd + adi + bci$. We want to prove that $b = 0$. So it may be helpful to isolate that term better. $Ax = a(c + di) + b(ci - d)$. However, I don't know where to go from here or where the complex conjugate becomes relevant, nor how the fact that A is symmetric is useful. Any help is appreciated. Thanks!
 A: I believe that $A$ is a real matrix; denote by $H$ the conjugate transpose: then $A^H=A^T=A$. Then
$$
k(x^Hx)=x^H(kx)=x^HAx=x^HA^Hx=(Ax)^Hx=(kx)^Hx=\bar{k}(x^Hx)
$$
Since $x\ne0$, also $x^Hx\ne0$ and so $k=\bar{k}$.
Note that the proof just requires that $A=A^H$, so it can even have complex coefficients (it's called Hermitian, in this case).

Without the conjugate transpose, let $\bar{x}$ be the vector you get from $x$ by conjugating all entries. Then do the same:
$$
k(\bar{x}^Tx)=\bar{x}^T(kx)=\bar{x}^TAx=\bar{x}^TA^Tx=
(A\bar{x})^Tx=(\overline{Ax})^Tx=(\bar{k}\bar{x})^Tx=\bar{k}(\bar{x}^Tx)
$$
A: Your statement is not true in general. Take $A$ to be the following :
$$
A=\left[\begin{matrix}
1&i\\
i&1
\end{matrix}\right].
$$
then $k=1+i$ and $x=[1, 1]^T$ stasify $Ax=kx$. 
Now if $A$ is Hermitian then this is true and the proof is as follows. Take  $x^*Ax$, we can see $x^*Ax=(Ax)^*x=k^*x^*x$ and $x^*Ax=x^*kx=kx^*x$ and therefore $k=k^*$ which means that $k$ is real. ($x^*$ is the conjugate transpose.) If $A$ is real matrix then hermitian boils down to symmetric.
