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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!

mathbbandstuff
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1 Answers1

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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.

user1551
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    Where did you use the information about the ranks? Doesn't that maybe change the fact that $F^AG$ is not always non-singular? I'm not sure that it does, but that information should be used. And also, I just wanted to say that $F^AG$ is always well-defined because of $A=FG^*$. – mathbbandstuff Jan 07 '19 at 13:24
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    @mathbbandstuff You are right that $F^\ast AG$ is well defined. The ranks conditions are used, as said in the answer: $F^\ast AG$ is always non-singular when $F$ and $G$ have full column ranks, but not so otherwise. E.g. $F^\ast AG=1$ is nonsingular when $F=G=\pmatrix{1\ 0}$, but $F^\ast AG=\pmatrix{1&0\ 0&0}$ is singular when $F=G=\pmatrix{1&0}$. In both cases $F,G$ and $A$ are rank-1 matrices. – user1551 Jan 07 '19 at 14:17