Finding a condition(s) on a left inverse of a tall matrix that makes it unique. Let $A$ be a tall matrix, $n\times m$ with $n>m$. Suppose it has full rank. It is a fact that $A$ will have infinitely many left inverses. I would like to know if there are interesting conditions that we can impose on $B$ to make it unique.
For instance the condition that $AB$ be symmetric seems fairly strong, and is at least satisfied by the pseudo-inverse. Does this in fact completely determine $B$? I haven't been able to find a counterexample.
I'm also interested in other interesting ideas along these lines, though this is more open ended and less well-defined a question.
 A: Using the SVD decomposition of the matrices you can see that $N_s=0$ must be true for symmetry to be true:
$$A=U\begin{pmatrix}\Sigma\\0\end{pmatrix}V^T$$
$$B=V\begin{pmatrix}\Sigma^{-1} & N_s\end{pmatrix}U^T$$
$$BA = I$$
$$AB=U\begin{pmatrix}I& \Sigma N_s \\ 0 & 0 \end{pmatrix}U^T$$
$$(AB)^\top=U\begin{pmatrix}I& 0 \\ N_s^\top \Sigma & 0 \end{pmatrix}U^T$$
The following is another path to see it.
Let $N\ne 0$ and $NA = 0$ so that $N$ is in the null space of $A$. Then $(B + N)A = BA + 0= I$ and
$$A(B+N) = AB + AN$$
Since $AB$ is symmetric, only the symmetry of $AN$ needs to be inspected. Compare it to its transpose $N^\top A^\top$. They must be equal for symmetry to be true. 
$$N^\top A^\top \overset{?}{=} AN$$
If they both are left multiplied with $N$:
$$NN^\top A^\top \overset{?}{=} 0$$
Since $NN^\top \ne 0$ (unless we are dealing with complex numbers) then $NN^\top$ would have to be the null column space of $A$ for this to be true (the null row space of $A^\top$). This contradicts that $A$ has full span.
A: Let $C=B-A^+$ (the symbol $A^+$ means the Moore-Penrose pseudoinverse of $A$; hence $CA=0$). To make the left inverse $B$ unique means to fix the choice of $C$ for each $A$.
One condition that fixes this choice is to simply pick $C=0$. This is equivalent to the condition that $AB$ is real symmetric (or Hermitian in the complex case). Indeed, if $A=U\begin{pmatrix}\Sigma\\0\end{pmatrix}V^T$ is a singular value decomposition of $A$, then all left inverses of $A$ can be written in the form of $B=V(\Sigma^{-1},R)U^T$ (here $C=V(0,R)U^T$). If $AB$ is symmetric, $R$ must be zero and therefore $B=A^+$.
Since every function $f:M_{n\times m}(\mathbb{R})\to M_{m\times n}(\mathbb{R})$ that satisfies $f(A)A=0$ gives rise to a left inverse function $B=A^++f(A)$, the previous condition is not the sole condition that one can impose. However, apart from the zero function, I don't know any simple way to explicitly specify $f$ (especially when $A$ has repeated singular values).
