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In this paper on page 1924 it is stated that

\begin{equation} \text{Var}(u \mid y) = \sigma^2[G - GZ^\top H^{-1}ZG] \end{equation}

can be written as

\begin{equation} \text{Var}(u \mid y) = \sigma^2[G - (Z^\top R^{-1}Z + G^{-1})^{-1}Z^\top R^{-1}ZG], \end{equation}

where $H = ZGZ^\top + R$. Can someone show how this step is achieved?

Note that $H, G,$ and $R$ are symmetric matrices.

Edit: How does

\begin{equation} \begin{aligned} (Z^\top R^{-1}Z + G^{-1})GZ^\top &= Z^\top R^{-1}ZGZ^\top + Z^\top \\ &= Z^\top R^{-1}(ZGZ^\top + R) \\ &=Z^\top R^{-1}H. \end{aligned} \end{equation}

Imply

\begin{equation} (Z^\top R^{-1}Z + G^{-1})^{-1}Z^\top R^{-1} = GZ^\top H^{-1}. \end{equation}

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Noting that $(A + BCD)^{-1} = A^{-1} - A^{-1}B(C^{-1} + DA^{-1}B)^{-1}DA^{-1}$ we can write

\begin{equation} H^{-1} = (ZGZ^\top + R) = R^{-1} - R^{-1}Z(G^{-1} + Z^\top R^{-1}Z)^{-1}Z^\top R^{-1}. \end{equation}

Using $H^{-1}$ we can write

\begin{equation} \begin{aligned} GZ^\top H^{-1}ZG &= GZ^\top R^{-1}ZG - GZ^\top R^{-1}Z(G^{-1} + Z^\top R^{-1}Z)^{-1} Z^\top R^{-1} ZG \\ & = \Big[ G - GZ^\top R^{-1}Z(G^{-1} + Z^\top R^{-1}Z)^{-1}\Big]Z^\top R^{-1}ZG \\ & = \Big[G(G^{-1} + Z^\top R^{-1}Z) - GZ^\top R^{-1}Z\Big](G^{-1} + Z^\top R^{-1}Z)^{-1}Z^\top R^{-1}ZG \\ & = \Big[(I + GZ^\top R^{-1}Z) - GZ^\top R^{-1}Z\Big](G^{-1} + Z^\top R^{-1}Z)^{-1}Z^\top R^{-1}ZG \\ & = (G^{-1} + Z^\top R^{-1}Z)^{-1}Z^\top R^{-1}ZG. \end{aligned} \end{equation}

Thus $\text{Var}(u \mid y)$ can be written as

\begin{equation} \begin{aligned} \text{Var}(u \mid y) & = \sigma^2(G - GZ^\top H^{-1}ZG) \\ &= \sigma^2(G - [G^{-1} + Z^\top R^{-1}Z]^{-1}Z^\top R^{-1}ZG) \\ &= \sigma^2(G - [G^{-1} + Z^\top R^{-1}Z]^{-1}[Z^\top R^{-1}Z + G^{-1} - G^{-1}]G) \\ & = \sigma^2(G^{-1} + Z^\top R^{-1}Z)^{-1}. \end{aligned} \end{equation}

Edit: Noting that

\begin{equation} (Z^\top R^{-1}Z + G^{-1})GZ^\top = Z^\top R^{-1}H. \end{equation}

Multiplying on the left by $(Z^\top R^{-1}Z + G^{-1})^{-1}$ and on the right by $H^{-1}$ we obtain

\begin{equation} \begin{aligned} (Z^\top R^{-1}Z + G^{-1})^{-1}(Z^\top R^{-1}Z + G^{-1})GZ^\top H^{-1} & = (Z^\top R^{-1}Z + G^{-1})^{-1}Z^\top R^{-1}HH^{-1} \\ \implies GZ^\top H^{-1} & = (Z^\top R^{-1}Z + G^{-1})^{-1}Z^\top R^{-1}. \end{aligned} \end{equation}

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