How to determine the following matrix decomposition?

Question

Let $A$ and $V$ be matrices, with $V$ being semidefinite positive. We would like to decompose $AVA^{T}$ as a product of three matrices such that

\begin{equation} AVA^{T}=\begin{bmatrix}I & A\end{bmatrix}.X.\begin{bmatrix}I\\ A^{T} \end{bmatrix}, \end{equation} with $I$ being the identity matrix. So we would like to determine $X$. The matrix $X$ has to respect the following constraint: $X$ is symmetric, invertible and depends only on $V$. One can think for example that \begin{equation} X=\begin{bmatrix}0 & 0\\ 0 & V \end{bmatrix}, \end{equation} but it is not possible since $\begin{bmatrix}0 & 0\\ 0 & V \end{bmatrix}$ is not invertible. Can anyone help me determining $X$ please? Thanks.

Answer 1

I think you can try to write down $ X $ (using its symmetry) in $ \begin{bmatrix} X_1 & X_2 \\ X_2 & X_3 \end{bmatrix} $ (like in your example) and you will obtain (from the identity that you would have): $$ A (V-X_3) A^T = X_1 + 2 sym(AX_2) $$ (with $ sym(K) $ the symmetric part of $ K $). It is not clear for me if there always exists a solution of your problem. Maybe you can try to look for a counterexample or to do a particular assignment to some $ X_i, \, (i=1,2,3) $, and find the other matrices that form $ X $.

Answer 2

If you choose $X=\begin{bmatrix} X_{11}& X_{12} \\ X_{21} &X_{22}\end{bmatrix}$ you get (by multiplying your rhs) to $$ X_{11}+AX_{21}+(AX_{12})^T + AX_{22}A^T = AVA^T $$ By choosing $X_{11}=0$ and a skew-symmetric $X_{21}$ and $X_{22}=V$ you will get a valid matrix $X$ to fulfill your equation. I'm not sure about the regularity though.