I have some complex matrix, $M$, which has dimensions ,$m \times 1$.
I can perform a SVD on this matrix and get,
$$ M = ASV^{\dagger} $$
where $S$ and $V$ are a $1 \times 1$ matrices and $V$ is unitary (so it has to look like $e^{i\theta}$) and $A$ has dimensions $m \times 1$ and satisfies $A^{\dagger}A = (1)$.
Is there some way to modify the SVD so that $AA^{\dagger} = I$ and $A$ has dimensions $m \times 1$ instead (where $I$ is the $m \times m$ identity matrix)? If not, is there a decomposition that I can perform on $M$, that has this property?
What you want is not possible: an $m\times1$ matrix has at most rank one, so $AA^\dagger=I$ is impossible.
In the usual Singular Value Decomposition, applied to your $M$, the matrix $A$ is an $m\times m$ unitary (so $A^\dagger A=AA^\dagger=I$), $S$ is $m\times 1$, and $V$ is $1\times1$.