Show that $(AB)^+ = B^+(ABB^+)^+$ when $|A| \neq 0$. I came across the following exercise: If $|A|\neq 0$, then $(AB)^+ = B^+(ABB^+)^+$. Here, $X^+$ denotes the MP inverse of the matrix $X$ (assumed to be real). Just walking through the conditions: * $(AB)(AB)^+(AB)=ABB^+(ABB^+)^+AB=ABB^+(ABB^+)^+ABB^+B=ABB^+B=AB$ * $(AB)^+(AB)(AB)^+=B^+(ABB^+)^+ABB^+(ABB^+)^+=B^+(ABB^+)^+=(AB)^+$ * $((AB)(AB)^+)^T=(ABB^+(ABB^+)^+)^T=ABB^+(ABB^+)^+=(AB)(AB)^+$ So, since if MP inverses exist, they are unique, I think this shows the result, but nowhere did I use that $|A|\neq 0$. Did I do something wrong? Am I missing something?

Your three bullet points look fine, but you forgot to verify the fourth property, namely, $(AB)^+(AB)$ is symmetric. Note that when $A$ is singular, it is possible that the alleged pseudoinverse $X=B^+(ABB^+)^+$ of $AB$ satisfies all assertions in your three bullet points but fails to satisfy the fourth property. For instance, consider $$ A=\pmatrix{1&0\\\ 1&0},\ B=\pmatrix{1&1\\\ 0&1},\ AB=\pmatrix{1&1\\\ 1&1},\ \color{red}{(AB)^+=\frac14\pmatrix{1&1\\\ 1&1}}. $$ Then $$ B^+=\pmatrix{1&-1\\\ 0&1},\quad ABB^+=A,\quad (ABB^+)^+=\frac12\pmatrix{1&1\\\ 0&0}. $$ Hence the alleged pseudoinverse of $AB$ is given by $$ \color{red}{X=B^+(ABB^+)^+=\frac12\pmatrix{1&1\\\ 0&0}} \
eq(AB)^+. $$ It is easy to verify that $(AB)X(AB)=AB,\ X(AB)X=X$ and $(AB)X$ is symmetric. Yet $X(AB)=A^T$ is not symmetric.