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.