I am looking to prove the following
Let $z$ be an $m\times$ 1 random vector with $E(z)=\mu$ and $\operatorname{Cov}(z)=V$ and let $A$ be an $m\times m$ non-stochastic matrix. Then the following identity holds true:
$$E(z'Az) = \operatorname{trace}(AV) + \mu'A\mu$$
I am failing to find a straight forward proof to it.
p.s: This is not a homework assignment (although something in my homework DOES rely on this theorem, and I would love to know why it holds).
I reduced the problem to the following: If we call z'=X and z'A'=Y then we already know that: $\operatorname{cov}(X,Y) = E(XY')-E(X)E(Y)' \rightarrow E(XY')=\operatorname{cov}(X,Y)+E(X)E(Y)' \rightarrow $ $E(z'Az)=\operatorname{cov}(z',z'A')+E(z')E(z'A')'=\operatorname{cov}(z',z'A')+\mu'(\mu'A')'$ $=\operatorname{cov}(z',z'A')+\mu'A\mu$
Which leaves us to show that $\operatorname{cov}(z',z'A')=\operatorname{trace}(AV)$
Can someone please help by showing how this is true?
Is there another direction to this proof that I am missing?
Thanks.
It follows from the fact that $\operatorname{trace}(AB)=\operatorname{trace}(BA)$, i.e. one can cyclically permute matrices when taking a trace, it doesn't change the end result. Now, since $\operatorname{cov}(z',z'A')=(z-\mu)'(Az-A\mu)=(z-\mu)'A(z-\mu)$ is a number, $(z-\mu)'A(z-\mu)=\operatorname{trace}((z-\mu)'A(z-\mu))$ trivially. Then permute the $(z-\mu)'$ to the back to get $\operatorname{trace}(A(z-\mu)(z-\mu)')$. But the last two factors form just the covariance matrix of $z$ after taking the expectation of course.