Problem about mp inverse

Question

Show that $ABB^+(ABB^+)^+=AB(AB)^+$.

This is an exercise from Matrix Differential Calculus with Applications in Statistics and Econometrics. There are no other assumptions. I know if $|A|\neq 0$ , then $(AB)^+=B^+(ABB^+)^+$, and the above equation establishes.

Answer 1

What precisely is the question?

You provide that for $\mathbf{A}\ne\mathbf{0}$, then $$ % \left( \mathbf{AB} \right)^{+} = % \mathbf{B}^{+} \left( \mathbf{ABB}^{+} \right)^{+} % \tag{1} $$

You note an immediate consequence of $(1)$ is $$ % \left( \mathbf{AB} \right) \left( \mathbf{AB} \right)^{+} = % \left( \mathbf{AB} \right) \mathbf{B}^{+} \left( \mathbf{ABB}^{+} \right)^{+} = % \left( \mathbf{ABB}^{+} \right) \left( \mathbf{ABB}^{+} \right)^{+} % \tag{2} $$ which seems to be the result you are looking for.

Answer 2

For any real matrix $X$, $XX^{+}$ is the orthogonal projection matrix onto the column space of $X$.

It follows consequently that the LHS is the orthogonal projection matrix on the column space of $ABB^{+}$ and the RHS is the orthogonal projection matrix on the column space of $AB$.

So it is enough to show that $ABB^{+}$ and $AB$ have the same column space.

Clearly column space of $ABB^{+}$ is contained in column space of $AB$. Now $AB = ABB^{+}B$, so the column space of $AB$ is also contained in the column space of $ABB^{+}$. Hence $AB$ and $ABB^{+}$ have the same column space and the result follows.