# $AM=I$, where $M$ is a rectangular matrix with full column rank, prove that $A=M^+$?

$AM=I$, where $M$ is a rectangular matrix with full column rank, then $A=M^+$(Moore-Penrose pseudoinverse)?

• What is $M^{+}$? The pseudo-inverse? – EuYu Nov 5 '14 at 7:16
• @EuYu: yes, MP pseudoinverse – Robert Fan Nov 5 '14 at 7:18
• If that's the case, then the statement in your question is incorrect. The Moore-Penrose pseudoinverse is uniquely defined, whereas $A$ is just a left-inverse of $M$. One-sided inverses of matrices are not unique; there are multiple matrices which can play the role of $A$. – EuYu Nov 5 '14 at 7:37

It appears that all but one of the defining properties of the Moore-Penrose pseudo-inverse are satisfied. The missing one is $$(MA)^H = MA.$$ As EuYu pointed out, $A$ is just a left-inverse.
A counter-example would be: $$M = \pmatrix{1 \\ 0}.$$ Then all matrices of type $$A = \pmatrix{ 1 & a}$$ satisfy your assumptions, but only for $a=0$ the product $MA$ is Hermitian, thus $$M^+ = \pmatrix{ 1 & 0}.$$