The equivalence between the original LQT problem and the corresponding LQR problem with an augmented state space can be shown by using the block matrices composition rule
| \bm{M}=\footnotesize\begin{bmatrix}\bm{M}_{11}&\bm{M}_{12}\\ \bm{M}_{21}&\bm{M}_{22}\end{bmatrix}\quad\iff\quad\bm{M}^{-1}=\footnotesize\begin{bmatrix}\bm{I}&\bm{0}\\ -\bm{M}_{22}^{-1}\bm{M}_{21}&\bm{I}\end{bmatrix}\begin{bmatrix}\bm{S}^{-1}&\bm{0}\\ \bm{0}&\bm{M}_{22}^{-1}\end{bmatrix}\begin{bmatrix}\bm{I}&-\bm{M}_{12}\bm{M}_{22}^{-1}\\ \bm{0}&\bm{I}\end{bmatrix}, |
where \bm{M}_{11}\!\in\!\mathbb{R}^{d\times d} , \bm{M}_{12}\!\in\!\mathbb{R}^{d\times D} , \bm{M}_{21}\!\in\!\mathbb{R}^{D\times d} , \bm{M}_{22}\!\in\!\mathbb{R}^{D\times D} , and \bm{S}=\bm{M}_{11}-\bm{M}_{12}\bm{M}_{22}^{-1}\bm{M}_{21} the Schur complement of the matrix \bm{M} .
In our case, we have
| \begin{bmatrix}\bm{Q}^{-1}\!+\!\bm{\mu}\bm{\mu}^{\scriptscriptstyle\top}&\bm{\mu}\\ \bm{\mu}^{\scriptscriptstyle\top}&1\end{bmatrix}^{-1}=\begin{bmatrix}\bm{I}&\bm{0}\\ -\bm{\mu}^{\scriptscriptstyle\top}&1\end{bmatrix}\begin{bmatrix}\bm{Q}&\bm{0}\\ \bm{0}&1\end{bmatrix}\begin{bmatrix}\bm{I}&-\bm{\mu}\\ \bm{0}&1\end{bmatrix}, |
and it is easy to see that
| {(\bm{x}-\bm{\mu})}^{\scriptscriptstyle\top}\bm{Q}(\bm{x}-\bm{\mu})=\begin{bmatrix}\bm{x}\\ 1\end{bmatrix}^{\scriptscriptstyle\top}\begin{bmatrix}\bm{Q}^{-1}\!+\!\bm{\mu}\bm{\mu}^{\scriptscriptstyle\top}&\bm{\mu}\\ \bm{\mu}^{\scriptscriptstyle\top}&1\end{bmatrix}^{-1}\begin{bmatrix}\bm{x}\\ 1\end{bmatrix}-1. |