Invertibility of Hermitian matrix Suppose $G$ is a Hermitian $n \times n$ matrix and $A$ is some $n \times n$ matrix over complex numbers. 
such that $G-A^H G A$ is positive-definite. 
Then can we show that $G$ is invertible?
Also, can we conclude anything about the eigenvalues of the matrix $A$ in terms of the eigenvalues of $G$ (for instance, relate the number of positive eigenvalues of $G$ with the eigenvalues of $A$ of norm less than 1)?
 A: Let $G$ be non invertible matrix. Then singular value decomposition of $G$, $G=USV^T$ will have at least one singular value $\sigma_n=0$.
Now consider the singluar value decomposition of $G-A^HGA=U^{'}S^{'}V^{'T}$ where we have $\sigma^{'}_i\neq 0$ for all $i$.
As $G$ has singularity, $A^HGA$ shouldnt such that $G-A^HGA$ is positive definite. From this, we can write the SVD of $A^HGA$ as 
$$A^HGA=(U^*S^*V^{*})^HUSV^TU^*S^*V^{*}=usv^T$$
where 
$s=S^*S^*S$
which has $\sigma_n=0$. Therefore $G$ can not be non-invertible.
A: Since $G-A^*GA$ is Hermitian and positive definite, we can find $m>0$ such that 
$$\forall x\in\Bbb C^n,\quad x^*Gx-(Ax)^*G(Ax)\geq m\lVert x\rVert^2.$$
This gives, by induction, that for all $x\in\Bbb C^n$,
$$(A^px)^*G(A^px)\leq x^*Gx-m\sum_{j=0}^{p-1}\lVert A^jx\rVert^2.$$
The initial assumption gives that $G-(A^p)^*GA^p\geq 0$ and that each eigenvalue of $A$ is of modulus $<1$. Therefore, using Jordan form of $A$, we get that $G\geq 0$. We get 
$$m\sum_{j=0}^{p-1}\lVert A^jx\rVert^2\leq x^*Gx,$$
and the wanted result.
A: $G-A^TGA$ is positive definite, so $$x^T(G-A^TGA)x>0$$ 
or $$x^TGx>x^TA^TGAx$$ for any $x \in \mathbb{C}^n$.
Then suppose $G$ is not invertible, then there exists $y \in \mathbb{C}^n$ such that $Gy=0$, then we have $$y^TA^TGAy<y^TGy=0$$
Since $A$ is arbitrary, the left side of the above inequality means $G$ is negative definite, which is contradict to the assumption that $G$ is not invertible.
So $G$ is invertible.
A: Another solution would be to show that $G$ does not admit the eigenvalue $0$. (Then G $is$ invertible because it is hermitien). This can be shown by contradiction: Let $x$ be the eigenvector to the eigenvalue $0$ of $G$. Then $x^t(G-A^tGA)x^t>0$ gives (since $Gx=0$):
$-(Ax)^tG(Ax)>0$.
Now you can argue in the same way as chaohuang.
