Are all matrices similar to a real diagonal matrix, real symmetric?

First off, the question: $A=QLQ'$ given that $Q$ is orthogonal, and $L$ is a real diagonal matrix, is the following true: $A$ is a real symmetric matrix.

My attempts

The easy part is to recognize $Q' = Q^T$. Therefore we have a similarity equation between A and L. This question basically boils down to whether all similar matrices to L is real symmetric. I know that L has all real eigenvalues, and so does A. However, could there exist not-real-symmetric matrices that also have all real eigenvalues?

So first I tried to think of easy contradictory examples but they all came out to be real symmetric. Take this 1 for example:

Take L to be $\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$, I know that the characteristic polynomial must be the same, the rank must be the same, and the geometric multiplicities of its eigenvalues(consequently) is also the same.

The charpoly(L) is $(1-\lambda)(-1-\lambda)$, which is $\lambda^2-1$.

Take A to be $\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$. The charpoly(A) is $\lambda^2 - 1$ as well. Its rank is also 2.

But argh, this doesn't help! This is a real symmetric!

So I turn to math SO for help - am I missing an obvious example here? If this is actually true, then what's the proof behind it? Thanks!

• Hint: use the property of transpose: $(AB)^T = B^T A^T$. It is much simpler than you think. Sep 30, 2016 at 7:21
• Let me make an educated guess here: Since $L^T = L$, we can transpose both sides of the equation, giving $A^T = (QLQ^{-1})^T$, which gives us $A^T = Q^TLQ$. We know that $Q^-1=Q^T$, then we can see that $QA^T = LQ$. We can see that $A^T$ is also similar to L, which we can now say $A^T$ is similar to $A$, which... makes it real symmetric? I think somewhere in there I went wrong. Sep 30, 2016 at 7:25
• Oh wait I am stupid. Sep 30, 2016 at 7:31

$A = QLQ^T$
$A^T = (QLQ^T)^T = (Q^T)^TL^TQ^T = QL^TQ^T$. We know $L^T=L$, therefore $A^T=QLQ^T$. Therefore $A^T=A$