Proof that Regression Sum of Squares and Residual Sum of Squares are independent random variables Having consulted a number of sources, I still can't find a complete proof that Regression Sum of Squares ($SS_{regression}$) and ($SS_{residual}$) are independent random variables. I'll be doubly pleased with a proof in matrix form. If it is too involved to be typed up here, I am happy to check it out myself if people can direct me to a book that contains the proof.
 A: Assume $y \sim \operatorname{Normal}(\beta X, \Sigma)$ with constant diagonal covariance $\operatorname{Cov}(y)=\Sigma=\sigma^2 \mathbb I$ and mean $\bar y=\mu$.
Using the hat matrix $\mathbb H$ we have that:
$$\hat y =\mathbb H y$$
And
$$\epsilon=y-\hat y=(\mathbb I -\mathbb H)y$$
Then (because the hat matrix is idempotent $\mathbb H^2 = \mathbb H$)
$$\begin{cases}
SSR = (\hat y - \mu)^T(\hat y - \mu)= y^T \mathbb Hy \color{red}{+n\mu^2-2\mu\mathbf 1^T\hat y}= y^T \mathbb Hy \color{red}{-n\mu^2}\\
SSE = \epsilon ^T \epsilon = y^T (\mathbb I -\mathbb H)y
\end{cases}$$
$$\operatorname{Cov}(SSR,SSE) = \operatorname{Cov}(y^T \mathbb Hy,y^T (\mathbb I -\mathbb H)y)$$
Using that $\operatorname{Cov}(x^TAx,x^TBx)=4\mu^TA\Sigma B\mu + 2\operatorname{tr}(A\Sigma B\Sigma)$ (see Prove that $\mathrm{Cov}(x^TAx,x^TBx) = 2 \mathrm{Tr}(A \Sigma B \Sigma) + 4 \mu^TA \Sigma B \mu$)
$$\operatorname{Cov}(SSR,SSE)
=4\mu^T\mathbb H\Sigma (\mathbb I -\mathbb H)\mu + 2\operatorname{tr}(\mathbb H\Sigma (\mathbb I -\mathbb H)\Sigma)\\
$$
Since $\Sigma=\sigma^2 \mathbb I$ is constant diagonal:
$$\operatorname{Cov}(SSR,SSE)
=4\sigma^2\mu^T\mathbb H (\mathbb I -\mathbb H)\mu + 2\sigma^4\operatorname{tr}(\mathbb H (\mathbb I -\mathbb H))\\
=4\sigma^2\mu^T (\mathbb H -\mathbb H^2)\mu +2\sigma^4\operatorname{tr}(\mathbb H -\mathbb H^2)\\
=0$$
