$$\alpha=Q'\beta=\begin{pmatrix}\alpha_1\\\vdots\\ \alpha_p\end{pmatrix}, f=\begin{pmatrix}\delta_1\alpha_1\\\vdots\\\delta_p\alpha_p\end{pmatrix},F=\begin{pmatrix}\delta1\alpha_1& & \\ &\ddots & \\ &&\alpha_p \end{pmatrix},$$
$$\theta=\sum_{j=1}^p\delta_j\alpha_j=1_p'f,c=p^2\sigma^2+p\theta^2 , b=\begin{pmatrix}\delta_1\alpha_1^2\\\vdots\\\delta_p\alpha_p^2\end{pmatrix}=F\alpha,k=\begin{pmatrix}k_1\\ \vdots\\ k_p\end{pmatrix}$$
$$A=\begin{pmatrix}\sigma^2+\delta_1^2\alpha_1^2 \\&\ddots\\&&\ \sigma^2+\delta_p\alpha_p^2\end{pmatrix}=\sigma^2I_p+ F^2$$
and $G=\sigma^2I_p+\theta F$
Given the above the problem is to solve
$$\begin{pmatrix}A &&G\\ G&& (\frac{c}{p})I_p\end{pmatrix}\begin{pmatrix}k\\\varrho\end{pmatrix}=\begin{pmatrix}b\\ \theta \alpha\end{pmatrix},$$
The answer is (in algebraic notation):
$k_j=\frac{\sigma^2(p \delta_j \alpha_j^2-\theta \alpha_j)}{(p\sigma^2+\theta^2)(\sigma^2+\delta_j^2 \alpha_j^2)-(\sigma^2+\theta \delta_j \alpha_j)^2}$
$\varrho_j=\frac{\sigma^2(\theta \alpha_j-\delta_j\alpha_j^2)}{(p\sigma^2+\theta^2)(\sigma^2+\delta_j^2 \alpha_j^2)-(\sigma^2+\theta \delta_j \alpha_j)^2}$
I know they use the formula for inverting a partitioned matrix:
$$\begin{pmatrix}A_{11} && A_{12}\\ A_{21}&& A_{22}\end{pmatrix}^{-1}=\begin{pmatrix}A_{11}^{-1} + A_{11}^{-1}A_{12}A_{22.1}A_{21}A_{11}^{-1}&-A_{11}^{-1}A_{12}A_{22.1}^{-1}\\ -A_{22.1}^{-1}A_{21}A_{11}^{-1} &A_{22.1}^{-1}\end{pmatrix},$$
where $A_{22.1}=A_{22}-A_{21}A_{11}^{-1}A_{11}^{-1}$
and I think they take out the common factor so that
$$\begin{pmatrix}A_{11} && A_{12}\\ A_{21}&& A_{22}\end{pmatrix}^{-1}=\frac{1}{A_{22}-A_{12}A_{11}^{-1}A_{12}}\begin{pmatrix}(A_{22}-A_{12}A_{11}^{-1}A_{12})A_{11}^{-1} + A_{11}^{-1}A_{12}A_{21}A_{11}^{-1}&-A_{11}^{-1}A_{12}\\ A_{21}A_{11}^{-1} & 1\end{pmatrix},$$
Using this notation it is easy to find that the denominator shown for $k_j$ and $\varrho_j$ it is simply equal to the common factor. Similarly by subbing back into the original problem I am able to find the numerator for $\varrho$ , but when I try and do the same for k I find I can't get the correct result!
So first expanding out $A_{22.1}=\frac{c}{p}I_p-\frac{(\sigma^2+\theta \delta_j \alpha_j)^2}{(\sigma^2+\delta_j^2 \alpha_j^2)}$
$=\frac{\frac{c}{p}(\sigma^2+\delta_j^2\alpha_j^2)-(\sigma^2+\theta \delta_j \alpha_j)^2}{(\sigma^2+\delta_j^2 \alpha_j^2)}$
and hence $A_{22.1}^{-1}$
$=\frac{(\sigma^2+\delta_j^2 \alpha_j^2)}{\frac{c}{p}(\sigma^2+\delta_j^2\alpha_j^2)-(\sigma^2+\theta \delta_j \alpha_j)^2}$
Thus the first term (top right corner) of the inverted partitioned matrix is:
$=\frac{\frac{c}{p}(\sigma^2+\delta_j^2\alpha_j^2)-(\sigma^2+\theta \delta_j \alpha_j)^2}{(\sigma^2+\delta_j^2 \alpha_j^2)}\frac{1}{(\sigma^2+\delta_j^2 \alpha_j^2)}+\frac{(\sigma^2+\theta \delta_j \alpha_j)^2}{(\sigma^2+\delta_j^2 \alpha_j^2)^2 }$
$=\frac{c}{p}$
The top left element is
$\frac{(\sigma^2+\theta \delta_j \alpha_j)}{(\sigma^2+\delta_j^2 \alpha_j^2)}$
multiplying by $b$ and $\varrho$ respectively, and subbing in for c
I get
$=(p\sigma^2+\theta^2)\delta_j\alpha_j^2-\theta \alpha_j\frac{(\sigma^2+\theta \delta_j \alpha_j)}{(\sigma^2+\delta_j^2 \alpha_j^2)}$
Now if one forgets about the different denominators then one can simplify the two numerators to get the correct answer:
$=\sigma^2(p\delta_j\alpha_j^2-\theta \alpha_j)$
Trouble is I cannot just forget about the denominator?
1.) Have I made a mistake in my derivation?
2.) Is there some simplification I have missed?
Ah got it!
The $\frac{c}{p}$ term should be $\frac{c}{p(\sigma^2 + \delta^2 \alpha^2)}$ this then cancels with the numerator from the denominator term.