Let $\Delta$ be a diagonal, non-invertible matrix with complex entries. Is it possible to come up with a matrix $M$ such that
$U\equiv M.\Delta$
is unitary?
Though I don't know about the proof, I heard that there is a theorem stating that any invertible matrix $A$ with complex entries can be written as
$A=U.T$
where $T$ is upper triangular. Since diagonal matrices are a subset of upper triangular, I basically want to know if the converse of this theorem above is true, and if there's an algorithmic way to find $A$ such that $A.T^{-1}$ defines a unitary.
PS.: $M$ has the same dimension as $\Delta$
Consider a general case of $n\times m$ matrix $M$ and $m\times n$ matrix $\Delta$ (maybe even non-diagonal). Then we have two cases:
Case 1: $\text{rank}\,\Delta<n$.
Then $\text{rank}\,U=\text{rank}\,M\Delta\le\text{rank}\,\Delta<n$ $\Rightarrow$ impossible for a unitary $U$.
Case 2: $\text{rank}\,\Delta=n$.
Then $\Delta$ has a left inverse, and choosing $M$ to be any left inverse (for example, $M=(\Delta^*\Delta)^{-1}\Delta^*$) gives $U=I$ (unitary).