I try to solve the following problem for a personal project :
$M$ is a wide matrix, $M \in \mathbb{R}^{m\times n}$, with $m << n$, such that $MM^t = D$, a diagonal matrix, and $A$ is a square, sparse, PSD matrix $A \in S^n_{++}$.
I know the pseudo inverse of $M$, $M^+$.
You might see $M$ as a way to project the variables on the set of constrained degrees of freedom I think are relevant to the problem.
I try to compute an approximation of $x^*$, which is the solution of $Ax^*=b$, by solving a much smaller problem. To bring back the solutions from $\mathbb{R}^m$ to $\mathbb{R}^n$ I use $M^+x$.
What I aim to compute is $\tilde{x} = M A^{-1} b$ because, when I solve the problem $AM^+ \hat{x} = b$ by least squares, which is easy because I just have to solve a $m\times m$ system, I get a solution that satisfies :
$||Ax^* - b|| < ||AM^+\hat{x} - b|| < ||AM^+\tilde{x}-b||$, but $||M^+\tilde{x} - x^*|| < ||M^+\hat{x} - x^*||$.
I care more about the second inequality that the first one.
So my question is : can I compute $\tilde{x}$ simply be solving an $m\times m$ system instead of an $n\times n$ system (which I have to do for the moment when I compute $A^{-1}b$).
I guess that it should be possible because $\tilde{x}$ has only $m$ degrees of freedom. I tried to play with pseudo-inverse properties such as $(AB)^+ = B^+A^+$ but A isn't orthonormal.
Thank you so much to anyone that tries to help me.