If left inverse exists, what does it imply about right inverse? $A, B$ are $n \times m$ complex matrices ($n > m$) with the following property:
\begin{equation}
B^\dagger A = I
\end{equation}
where $I$ is a $m \times m$ identity matrix.
What do we know about $A B^\dagger$?
I know it cannot be the $n \times n$ identity matrix. My intuition is that $A B^\dagger$ is a $n \times n$ matrix with zeros on the off-diagonals, zeros on the $n - m$ main diagonal entries, and $1$ on the $m$ main diagonal entries. If this is true how would one go about proving this? If this is not true is there something we can say in general about this product?
 A: I'm going to replace $B^{\dagger}$ by $B$.
If $BA = I$ then we have $(AB)^2 = A(BA)B = AB$ so $AB$ is idempotent; in particular it has eigenvalues $0$ and $1$ and is diagonalizable.  We also have the general result that $AB$ and $BA$ always have the same characteristic polynomial ignoring factors of $t$, from which it follows that the multiplicity of $1$ as an eigenvalue is $m$. (Alternatively, we can just observe that $\text{tr}(AB) = \text{tr}(BA) = m$, but the above argument continues to work in positive characteristic.)
So $AB$ it is similar to a matrix of the form you describe but need not be a matrix of the form you describe on the nose. In fact $AB$ can be any idempotent matrix of rank $m$.
To see this, geometrically the condition $BA = I$ says that $A$ is an injective linear map $V \to W$ and $B$ is a surjective linear map $W \to V$ which projects back down to $V$; this implies that if we take $U = \text{ker}(B)$ then $W$ decomposes as a direct sum $V \oplus U$, and in terms of this direct sum $A$ is the inclusion of the first summand and $B$ is the projection to the first summand. So $AB$ is the map $(v, u) \to (v, 0)$ which projects from $W$ down to $V$ but regarded as a subspace of $W$. Every idempotent linear map $P$ arises in this way for a unique choice of $V$, $U$, namely $P = \text{im}(P), U = \text{ker}(P)$.
A: We know a few things, but not much. In particular, your intuition is false as stated. For example,
$$
\begin{pmatrix}
1 & 2 & 2 \\
0 & 1 & 1
\end{pmatrix}
\begin{pmatrix}
1 & -2 \\
0 & 2 \\
0 & -1
\end{pmatrix} = 
\begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix},
$$
but
$$
\begin{pmatrix}
1 & -2 \\
0 & 2 \\
0 & -1
\end{pmatrix}
\begin{pmatrix}
1 & 2 & 2 \\
0 & 1 & 1
\end{pmatrix} = 
\begin{pmatrix}
1 & 0 & 0 \\
0 & 2 & 2 \\
0 & -1 & -1
\end{pmatrix},
$$
Even this block diagonal form came from me being a little lazy with the counterexample.
There's always some slack when choosing left inverses of rectangular matrices. If we consider them as linear maps, having $BA = I$ implies $A$ is an injective linear map (in matrix terms, the columns of $A$ are linearly independent). In fact, this property is equivalent to admitting a left-inverse.
If our matrix is not square, say $n \times m$ for $m < n$, then the linear map is injective, but not surjective. On the image of this map (in matrix terms, the columnspace), which is an $m$-dimensional subspace of $\Bbb{C}^n$, there is a unique inverse map back to $\Bbb{C}^m$. To get a left inverse, a map from $\Bbb{C}^n$ to $\Bbb{C}^m$, we get to freely extend this unique inverse on the image of the map to the entirety of $\Bbb{C}^n$. This gives us freedom of choice, which we can use to mess up your intuition.
