Let $A^+$ denote the Moore-Penrose inverse of a matrix $A$.

Problem. Let $A$ and $B$ be two compatible matrices where $B$ has full row rank. Show that $$AB(AB)^+=AA^+.$$

The verification is easy: Both sides are symmetric and $AA^+AB=AB=AB(AB)^+AB$. Of course we cannot cancel $AB$ on both sides, but I can't figure out how to start and what the condition $B$ has full row rank is used for. Hope anyone has good suggestions.

Update. Here are some of my recent findings but they are not sufficient to solve this problem.

Since $B$ has full row rank, we have $B^+=B^T(BB^T)^{-1}$ and thus $BB^+=I$. Thus, we have $$AA^+=AIA^+=ABB^+A^+.$$ If we can show that $(AB)^+=B^+A^+$. Then we are done. Nonetheless, this does not generally hold. We can see here that $B^+A^+$ is merely a $(1,2,3)$-inverse of $AB$:

  1. $ABB^+A^+AB=AA^+AB=AB$.
  2. $B^+A^+ABB^+A^+=B^+A^+AA^+=B^+A^+$.
  3. $ABB^+A^+=AA^+$ is symmetric.

The last condition $B^+A^+AB$ is symmetric cannot be verified (at least by me). Of course, if $A$ has full column rank, then it would be easier because $A^+A=I$ provided $A$ has full column rank.


1 Answer 1


Here is an almost one-line proof obtained just now from my instructor.

Note that $AA^+$ is a projection matrix onto the space $R(A)$.

Since $B$ has full row rank, we indeed have $R(AB)=R(A)$. Thus we must have $(AB)(AB)^+=AA^+$.


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