Why is left-inverse $A_L^{-1}$ times any column vector $b$ in row space of $A$?

Question

Let $A$ be a matrix with a left inverse $A_L^{-1}$.
That is, $A_L^{-1} = (A^TA)^{-1}A^T$.

Then for any column vector (with proper dimension) $b$,
$A_L^{-1} b$ is in the row space of $A$.
i.e., $A_L^{-1} b = (A^TA)^{-1}A^Tb \in \mathcal R (A^T)$.

Why is it?
If this is true, then $(A^TA)^{-1}A^Tb = A^T c$ for some column vector $c$, but it seems I can't find the $c$ in terms of $A$ and $b$.

Answer 1

If $A$ is $n\times m$ and $A^TA$ is invertible, then the rows of $A$ span all of $\mathbb{R}^m$ (otherwise there is a $v\neq 0$ orthogonal to all rows of $A$, and $A^TAv = 0$).

Therefore any vector in $\mathbb{R}^m$ (and $A_L^{-1}b$ in particular) must be in the row space of $A$.