variance of multiple regression coefficients If I consider universal kriging (or multiple spatial regression) in matrix form as:
${\bf{V = XA  + R }}$
where $\bf{R}$ is the residual and $\bf{A}$ are the trend coefficients, then the estimate of ${\bf{\hat A}}$ is:
${\bf{\hat A}}=(\bf{X^{T}C^{-1}X)^{-1}X^{T}C^{-1}V}$
(as I understand it), where $\bf{C}$ is the covariance matrix, if it is known. Then, the variance of the coefficients is:
$\text{VAR}({\bf{\hat A}})=(\bf{X^{T}C^{-1}X)^{-1}}$???
I am getting this from here.
How does one get from the estimate of ${\bf{\hat A}}$, to its variance? i.e. how can I derive that variance?
 A: $$\newcommand{\var}{\operatorname{var}}$$
First, recall that
$$
\var(MV) = M\Big(\var(V)\Big)M^T.
$$
so
$$
\begin{align}
& \var((X^T C^{-1}X)^{-1} X^T C^{-1}V) \\[10pt]
& = (X^T C^{-1}X)^{-1} X^T C^{-1}\Big(\var{V}\Big)\Big( (X^T C^{-1}X)^{-1} X^T C^{-1} \Big)^T. \tag{1}
\end{align}
$$
Then, recall that $(AB)^T$ (with $A$ to the left of $B$) is equal to $B^T A^T$ (with $A$ to the right of $B$).  With $X^T C^{-1} X$, one cannot invert all three matrices and multiply in the opposite order, since $X$ is not a square matrix.  But that matrix is symmetric, i.e. it is its own transpose.  And $C$ is also symmetric, and so is $C^{-1}$.  So we get:
$$
\Big( (X^T C^{-1}X)^{-1} X^T C^{-1} \Big)^T = C^{-1}X(X^TC^{-1}X)^{-1}.
$$
Then $(1)$ becomes
$$
\begin{align}
& (X^T C^{-1}X)^{-1} X^T C^{-1}\Big(\var{V}\Big) C^{-1}X(X^TC^{-1}X)^{-1} \\[10pt]
& = (X^T C^{-1}X)^{-1} X^T C^{-1}\Big( C \Big) C^{-1}X(X^TC^{-1}X)^{-1} \\[10pt]
& = (X^T C^{-1}X)^{-1} X^T C^{-1} X(X^TC^{-1}X)^{-1} \\[10pt]
& = (X^T C^{-1}X)^{-1}.
\end{align}
$$
