How can I prove this formula is nonsingular? I have a formula
$$A^{\mathrm T}+QA^{-1}G$$
where $A$ is nonsingular, $Q$ is positive semi-definite, $G$ is positive semi-definite, $(A,G)$ is controllable, and $(Q,A)$ is observable. My question is how to prove that this formula is nonsingular. Thanks in advance!
 A: This has nothing to do with controllability or observability. We only need nonsingularity of $A$ and positive semidefiniteness of $Q$ and $G$.
\begin{align}
&(A^T + QA^{-1}G)x = 0\tag{1}\\
\Rightarrow&x + A^{-T}QA^{-1}Gx=0\tag{2}\\
\Rightarrow&G^{1/2}x + G^{1/2}A^{-T}QA^{-1}Gx=0\\
\Rightarrow&(I + G^{1/2}A^{-T}QA^{-1}G^{1/2})G^{1/2}x=0\\
\Rightarrow&G^{1/2}x=0\quad (\because I + G^{1/2}A^{-T}QA^{-1}G^{1/2}\succ0)\\
\Rightarrow&Gx=0\\
\Rightarrow&x=0\quad (\text{by } (2)).
\end{align}
Therefore, equation $(1)$ has only the trivial solution, i.e. $A^T + QA^{-1}G$ is nonsingular.
Edit. Alternatively, $A^T + QA^{-1}G$ is invertible if and only if $I + A^{-T}QA^{-1}G$ is invertible. Using the fact that $XY$ and $YX$ have identical spectra for any two square matrices $X$ and $Y$, we see that $A^{-T}QA^{-1}G$ and $G^{1/2}A^{-T}QA^{-1}G^{1/2}$ have identical eigenvalues. Hence $I+A^{-T}QA^{-1}G$ and $I+G^{1/2}A^{-T}QA^{-1}G^{1/2}$ also have identical eigenvalues. Since the latter sum is positive definite, the former one is invertible.
