Is there a meaningful pseudo-inverse of a singular projection matrix?

Question

Hello linear algebra experts. In my research I'd like to solve (or approximate) for $B$, in the form

$$ A = GBG $$

where A and B are symmetric, square matrices and G is a symmetric, square, singular projection matrix ($G^2 = G$). A is also singular. I've done a bit of research on pseudo and generalized inverses. However, since G is a projection matrix, the Moore-Penrose and Drazin inverses trivially imply $G = G^+$. In my case, $G^+AG^+ = B$ is empirically false for these generalized inverses.

Here's a reproducible example coded up in R.

require(Matrix)
B <- matrix(seq(.01, .25, length.out=25), 5)
B <- B + t(B)
diag(B) <- 1
G <- diag(5) - matrix(rep(1,25), 5)/5
A <- G %*% B %*% G
Gpi <- ginv(G)
Gpi %*% A %*% Gpi  # this == A, not B
Gpi %*% Gpi   # == Gpi, not the Identity 

From what I understand, the identity matrix is the only projection matrix that is invertible. So the question: is there any pseudo-inverse that can approximate the inverse in this setting? In a least-squares setting perhaps? Or perhaps there's no way to recover the lost 'dimension' by performing the singular projection...

Thanks!

Answer 1

Due to the idempotence property of projections that you quote, a projection is its own unique pseudoinverse, as may be shown by substituting in conditions (I) to (IV) of the definition of a Moore-Penrose pseudoinverse. Other generalised inverses satisfying properties (I) and (II) will in general exist, but the MP pseudoinverse is optimal in a minimum norm sense, that is to say there is nothing better. You already have the least squares solution you are looking for I'm afraid.

Multiplying a matrix by a generalised inverse will not in general give the identity matrix on either side, viz. it may be lacking in rank on both sides, and this is in general the case for projections. The identity matrix is the exception as you observe.

You are correct in suggesting that the lost dimensions cannot be recovered.

Answer 2

I assume, that you want to solve for $B$ at given $A$ and $G$. First we note, that this problem is only solvable, if

$GAG = A$

(evident when we multiply both sides of $A = GBG$ with $G$ and use $G^2=G$).

In that case, $B=A$ is a solution to your problem.

I dont know if that is a solution you had in mind. I guess the approach you are using to solve whatever problem has some fundamental issue.

So, concluding, either your problem is not solvable for given $A$, or, if it is, has the solution $B=A$.