How to prove that $E(\frac{\hat{E}^T\hat{E}}{n-q}) = \Sigma$ So quick explanation:
This is a multivariate statistics problem (specifically for multivariate regression). n is, as always, the sample size. q is the number of independent variables. Sigma is the covariance matrix of $e_i$. E is the error matrix. The bold is for vectors. E.g. $\bf{1}$ is a vector of 1's.
Y is the $n$x$p$ matrix of the dependent variables for n observations and p dependent variables $y$.

My skill with this is lacking.
I can get this to,
$$E[Y^T (I-H) Y] = \Sigma$$ but have no clue how to introduce $\Sigma$
OK. But the book offers a hint.
It states that,
$$E(\hat{e}^T\hat{e}) = (n-k)\sigma^2$$
and suggest that this can be solved using another problem in the book so I'll show what was proven in that:
$$E(x^TAx) = E(tr(xx^TA)) = tr(\Sigma A) = \mu^TA\mu$$
$$\sigma^{-2}E(x^TAx) = tr(A) = p-1$$
given that $\mu = \mu \bf{1}$ $\Sigma = \sigma^2 I$ and $A = I -\bf{11^T} p^{-1}$

Let me know if you need anything else. I really want to figure this out but this is very difficult for me. Thank you in advance. 
 A: Let us define the model as $y = X\beta + u$ where $X$ is $n \times q$ and $u$ is the $n \times 1$ error vector. 
First let us consider $\hat{E} = y - \hat{y}$.
This is equal to $\hat{E} = y - \hat{y} = y - X(X^TX)^{-1}X^Ty = (I - P)y$
where $P = X(X^TX)^{-1}X^T$.
Then, we can show trivially that $P$ is a projection matrix i.e. $P^T = P$ and $PP = P$ and this is left to the reader to check.
Then as $y = X\beta + u$ we can say that $\hat{E} = (I-P)(X\beta + u) = X\beta -X(X^TX)X^TX\beta + u - Pu = X\beta - X\beta + (I-P)u = (I-P)u$.
Then, as $P$ is a projection matrix, so is $(I-P)$, hence $(I-P)^T(I-P) = (I-P)(I-P) = (I-P)$.
Then if we consider $\hat{E}^T\hat{E} = u^T(I-P)^T(I-P)u = u^T(I-P)u$ by the above.
Then as $u$ is $n \times 1$, $u^T$ is $1 \times n$ and so $\hat{E}^T\hat{E}$ is a $1 \times 1$ scalar. 
Therefore $\hat{E}^T\hat{E} = tr\{\hat{E}^T\hat{E}\} = tr\{u^T(I-P)u\} = tr\{(I-P)uu^T\}.$
Then taking expectations of both sides we have that $E[\hat{E}^T\hat{E}] = E[tr\{(I-P)uu^T\}] = tr\{(I-P)E[uu^T]\}$ as $I-P$ is a constant.
Then as we assume $E[u] = 0$ we know that $E[uu^T] = var[u]$ and as we assume $Var[u] = \sigma_0^2$ we have that $E[uu^T] = \sigma_0^2$.
Hence $E[\hat{E}^T\hat{E}] = tr\{(I-P)E[uu^T]\} = tr\{\sigma_0^2(I-P)\} = \sigma_0^2 tr\{(I-P)\}$.
Then $tr\{(I-P)\} = tr\{I\} - tr\{P\} = n - tr\{X(X^TX)^{-1}X^T\} = n - tr\{(X^TX)^{-1}X^TX\} = n - tr\{I_q\} = n - q$.
Therefore $E[\hat{E}^T\hat{E}] = \sigma_0^2 (n - q)$.
Then $E[\frac{\hat{E}^T\hat{E}}{n-q}] = \frac{\sigma_0^2 (n - q)}{(n-q)} = \sigma_0^2$ 
