Can $A^{T}(AA^{T})^{-1}A$ be simplified?

Question

Let $A$ is an $m\times n$ ($m<n$) real matrix with full positive entries and $\text{Rank}(A)=m$. Thus $(AA^{T})^{-1}$ is an $m\times m$ symmetric $M$-matrix since $AA^{T}$ is nonnegtive and positive definited.

I think $A^{T}(AA^{T})^{-1}A$ can be simplified to a simple form due to its special structure but I don't know how to do it. I tried using a pseudo inverse and the Cholesky decomposition but they made it more complex.

Thanks.

Answer 1

No you cannot simplify this, as far as I know. (I'm working with these things daily)

The matrix $P = A^T(AA^T)^{-1}A$ is a projector, that projects onto the image of $A^T$, because it holds \begin{align*} P^2 &= P \\ PA^T &= A^T. \end{align*}

Sorry.