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

  • $\begingroup$ What is $M^{+}$? The pseudo-inverse? $\endgroup$ – EuYu Nov 5 '14 at 7:16
  • $\begingroup$ @EuYu: yes, MP pseudoinverse $\endgroup$ – Robert Fan Nov 5 '14 at 7:18
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    $\begingroup$ 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 $A$. $\endgroup$ – 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 $$ M^+ = \pmatrix{ 1 & 0}. $$

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