$N>M$ where both are natural numbers. $\mathbf{E}\in\mathbb{C}^{N\times M}$ is another matrix with orthonormal columns, i.e. $\mathbf{E}^{\text{H}}\mathbf{E}=\mathbf{I}$.
$\mathbf{A}\in\mathbb{C}^{N\times N}$ is another Hermitian symmetric, positive definite matrix.
Now is it possible to simplify the expression $\mathbf{E}\left(\mathbf{E}^{\text{H}}\mathbf{A}\mathbf{E}\right)^{-1}\mathbf{E}^{\text{H}}$, at least to reduce the number of $\mathbf{E}$? Also, I have the hunch that this matrix is singular, but cannot really get around to prove it.
The rule for invertible matrices is that $(AB)^{-1}=B^{-1}A^{-1}$ then $(E^tAE)^{-1}=E^{-1}A^{-1}(E^t)^{-1}$ and $$E(E^tAE)^{-1}E^t=(EE^{-1})A^{-1}((E^t)^{-1}E^t)=A^{-1}$$ because the matrices product is associative.