I have a $16\times 4$ matrix $A$ of rank $4$. Besides its Moore-Penrose pseudoinverse $A^+$, I'm also interested in other generalized inverses $A^g$ that satisfy $A^gA=I_4$.

Is there a way to get all of them (presumably in some analytical form with free variables)?

Are there any special inverses? By "special," I mean $A^+$ gives the solution to $Ax=y$ with minimum $\ell_2$ norm, so is there one such $A^g$ that gives the solution with minimum $\ell_0$ norm, for example?

If the answers to the two questions above are both "No," how do I find any generalized inverse other than the pseudoinverse?

  • $\begingroup$ Here's a hint: Have you thought about putting $A$ in reduced echelon form? What does that tell you? Can you write that in terms of matrix multiplication? $\endgroup$ – Ted Shifrin Apr 22 at 1:35
  • $\begingroup$ Thanks @TedShifrin for the hint! Do you mean expressing the process of transforming $A$ into its reduced row echelon form as matrix multiplication? I know I can use reduced echelon form to find a square matrix's inverse. But can't see other connections between the echelon form and generalized inverse. Thanks! $\endgroup$ – Sibbs Gambling Apr 22 at 1:53
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    $\begingroup$ You can get one left inverse by thinking about the equation $EA=I$, where $E$ is a product of elementary matrices. Notice that once you have one left inverse $B$, you get others $B'$ by making sure that every row of $B-B'$ is orthogonal to all the columns of $A$. $\endgroup$ – Ted Shifrin Apr 22 at 4:04

Denote by $b_i$ the i-th row of $A^g$ and by $e_i$ the i-th row of $I_4$. Then $b_i A=e_i$. This system of four linear equations is solveable by the rank condition for $A$. Following this procedure gives all possible $A^g$.


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