We covered matrix inverses the other day and wanting to know about singular and non-square cases I read about the Moore-Penrose pseudo inverse. Using their definition there pseudoinverse is unique.

However lets consider the case $A \in \mathbb{R}^{m \times n}$ with $m < n$. Is there more than one matrix $B \in \mathbb{R}^{n \times m}$ that satifies

$$ A \cdot B = I $$

Where $I$ is the identity matrix of $\mathbb{R^{m \times m}}$.

If so, how do you go about finding them? If not, why?


1 Answer 1


Since $A\in\Bbb R^{m\times n}$, rank of $A$ (and $B$) is at most $n$. So, rank of $AB$ is also at most $n$, and if $m>n$, it cannot ever be the identity matrix.

For the case $m<n$, we can have $AB=I$, and this means that $\mathrm{rank\,} A=m$, and in such a case indeed there are infinitely many solutions for $B$:

Considering $A$ as a linear map $\Bbb R^n\to\Bbb R^m\ $ [namely, $\Bbb R^n\ni v\mapsto A\cdot v\,\in\Bbb R^m$], $\,$ the condition '$\mathrm{rank\,}A=m$' means that $A$ is surjective onto $\Bbb R^m$.
A matrix $B$ is a right inverse of $A$ if and only if $Ab_i=e_i$ for all $i$ where $b_i$ is the $i$th column of $B$ and $e_i$ is the $i$th element of the standard basis of $\Bbb R^m$ (having $1$ at coordinate $i$ and $0$ else).

  • $\begingroup$ Yes sorry, I meant to say $m < n$. Thank you for the throughful answer. Is there a method to find one or multiple of those solutions? (method as in gram-schmidt for orthogonalization) $\endgroup$
    – Kamiwaga
    Sep 23, 2017 at 22:25
  • $\begingroup$ Well, the row elimination also produces the inverse for square matrices, I guess, it can be similarly used for nonsquare matrices as well. $\endgroup$
    – Berci
    Sep 23, 2017 at 22:36

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