How to Prove Variance using Matrices $\operatorname{Var}(C\hat\beta) = \sigma^2C(X'X)^{-1}C'$? Assuming a Gauss-Markov model such that $H_0 : C\beta = d$, (I'm posting here as suggested by the SE Statistics)
(a) how do I prove that the variance of $C\hat\beta \sim N(C\beta, \sigma^2C(X'X)^{-1}C')$?
(b) AND then, what if $C\beta$ is NOT estimable. What is the simplified expression for the Variance $\operatorname{Var}(C(X'X)^{-1}X'y)$ if not??
My Work...Which I Know is Not Correct, when assuming it is testable:
Prove $\operatorname{Var}(C\hat\beta) = \sigma^2C(X'X)^{-1}C'$.
Let $\hat\beta = (X'X)^{-1}X'y$   By definition.
$C = AX$   Given $C\beta$ is estimable, for some matrix $A$.
$=\sigma^2AX(X'X)^{-1}AX'y$   Substitution.
$=\sigma^2AX(X'X)^{-1}X'AA'$
$\operatorname{Var}(y)$ in linear transformation $= A\operatorname{Var}(y)A'$.
$Px=X(X'X)^{-1}X'$   Def. of projection operator.
$=\sigma^2APxAA'$    Substitution.
....don't know how to finish.
 A: The design matrix $X \in\mathbb R^{n\times p}$ typically has many more rows than columns.  If the columns of $X$ are linearly independent, then $X'X$ is invertible.  In that case we have
$$\hat \beta = (X'X)^{-1} X' y.
$$
Therefore
\begin{align}
\operatorname{var}(\hat\beta) & = \Big( (X'X)^{-1} X'\Big)\Big( \operatorname{var}(y)\Big)\Big( (X'X)^{-1} X'\Big)' \\[8pt]
& = \Big( (X'X)^{-1} X'\Big)\Big( \sigma^2 I_n \Big)\Big( (X'X)^{-1} X'\Big)' \\[8pt]
& = \sigma^2 \Big( (X'X)^{-1} X' \Big) \Big( X'(X'X)^{-1}\Big) = \sigma^2 (X'X)^{-1}.
\end{align}
Hence
\begin{align}
\operatorname{var}(C\hat\beta) = C\Big( \operatorname{var}(\hat\beta)\Big) C'.
\end{align}
The expected value is simpler:
$$
\operatorname{E}(C\hat\beta) = C \operatorname{E}(\hat\beta).
$$
Generally if $W\in\mathbb R^k$ is normally distributed and $M\in\mathbb R^{\ell\times k}$ is a constant (i.e. non-random) matrix then $MW\in\mathbb R^\ell$ is normally distributed.
No linear combination of components of $\beta$ can fail to be estimable unless the columns of $X$ are linearly dependent, and that's a separate case I will deal with here later${}.$
