Let $A^\dagger$ be a Moore-Penrose inverse of a matrix $A$.
If $A \in \mathbb{C}^{m\times n}$ and $A=FG^*$, for some $F,G$ and $r(A)=r(F)=r(G)$, prove that
$$A^\dagger = G(F^*AG)^{-1}F^*$$ and $$A^\dagger = (G^\dagger)^*F^\dagger.$$
I need to show this using SVD decomposition and maybe some other properties of a Moore-Penrose inverse.
I tried to show the statement by writing SVD decomposition of all the matrices included, but it just gets messy and I didn't succeed.
Any hints would be really helpful! Thanks in advance!
Best Answer
Both equalities are technically false. The first one is false because in general, $F^\ast AG$ is not necessarily non-singular. For the second one, it is easy to generate a random counterexample when $F,G$ are "fat" matrices with more columns than rows.
However, both equalities are true when $F$ and $G$ are "tall" matrices of full column ranks and of the same sizes. In this case, we have $(F^\ast AG)^{-1}=(F^\ast FG^\ast G)^{-1}=(G^\ast G)^{-1}(F^\ast F)^{-1}$ and $F^\dagger F=G^\dagger G=I$. Now you can easily prove that $A^\dagger=G(G^\ast G)^{-1}(F^\ast F)^{-1}F^\ast=(G^\dagger)^\ast F^\dagger$ by verifying the four defining properties of Moore-Penrose pseudoinverse directly.