Solve the following matrix equation in $D$
$$ A=D^{T}(DVD^{T}+\alpha\lambda_{\max}(D^{T}D)I)^{-1}D$$
where $I$ is the identity matrix, $A$ and $V$ are known matrices, $\alpha$ is a known real positive constant, and $\lambda_{\max}$ denotes the maximum eigenvalue.
I tried to use SVD decomposition of $A$ in order to find the solution $D$, but it seems like the matrix $A$ has to respect a condition in order to have a solution $D$. Can anyone help me find this condition please?
Does anyone have any idea how to solve it in order to find $D$, please? Thanks, Gerard.
According to the first part of my comment above, we may assume that $\lambda_{max}(D^TD)=1/\alpha$. Then $A=D^T(I+DVD^T)^{-1}D$, that is, (if $A$ is invertible) $DA^{-1}D^T=I+DVD^T$ and finally,
$(*)$ $A^{-1}-V=D^{-1}D^{-T}=EE^T$ where $E=D^{-1}$.
Note that, necessarily, $D$ is invertible and overall,
cond 1. $A^{-1}-V$ is a symmetric $>0$ matrix.
A particular solution $E_0$ of $(*)$ is given by cholesky; the general solution is in the form $E=E_0U$ where $U$ is an orthogonal matrix; finally $D=U^TE_0^{-1}$. Note that $D^TD=(A^{-1}-V)^{-1}$ and it remains to consider the condition:
cond 2. $\lambda_{max}((A^{-1}-V)^{-1})=1/\alpha$. (the condition when $n=1$)