Generalized inverse of a matrix.

Question

Let $A$ be an $m \times n$ matrix of real numbers with $m\leq n$. I am interested in a generalized inverse $A^+$ of size $n \times m$ such that $A^+A=(I_m \ \bf 0)$. If Rank$(A)=m$, does such an $A^+$ always exists?

Answer 1

The short answer is no, consider $A = (1 \quad 1)$.

To get $(I_m \quad 0)$ you need also a regular matrix $B\in\mathbb R^n$ on the right, that is $A^+ A B = (I_m \quad 0)$. This does follow from the Guassian elinimation.

Answer 2

It depends on the rank of $A$. If $\mathrm{rank}(A)=m$, then it exists, proof is the Gaussian algorithm. If the rank is lower, it does not exist.