Inverse of $A^T Y A$

Question

Is there a formula for the inverse of a matrix $$ X = A^\top Y A $$ in terms of $A$ and $Y$, given that $A \in F^{m \times n}$ is full rank with $m >n$, and $Y$ is positive definite?

Answer 1

Assuming $A$ is a $n \times n$ matrix. Given that $A$ is full rank means its invertible. Since $A$ is full rank, $A^T$ is also full rank (column rank = row rank). Therefore, $A^T$ is also full rank and hence invertible. Using the following result about the inverse of the product of two invertible (full rank) matrices $A$ and $B$:

$$(AB)^{-1} = B^{-1} A^{-1}$$ and the fact that matrix multiplication is associative, se have that $X^{-1}$ for $X = A^T Y A$ is :

$$X^{-1} = A^{-1} Y^{-1} (A^{T})^{-1}$$

Also note that:

$$(A^{-1})^{T} = (A^T)^{-1} $$

Answer 2

Even with the restriction $m>n$, it need not be the case that $X=A^\top Y A$ be invertible. As a counterexample, suppose $m=2,n=1$. Then the choice $$Y=\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix},\; A=\begin{pmatrix} 1 \\ 1\end{pmatrix}$$ yields $$X = A^\top Y A = \begin{pmatrix} 1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} 1 \\ 1 \end{pmatrix}=\begin{pmatrix} 1 & -1 \end{pmatrix} \begin{pmatrix} 1 \\ 1 \end{pmatrix}=0$$ which is plainly not invertible.