0
$\begingroup$

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 $\Delta$

$\endgroup$
3
  • $\begingroup$ Is $M$ any matrix? Then set $M=U\Delta^{-1}$. I guess you need some properties of $M$ that are not mentioned above. $\endgroup$
    – A.Γ.
    May 1, 2018 at 19:20
  • $\begingroup$ You're right, let's assume $\Delta$ does not have inverse. $\endgroup$
    – dwfa
    May 1, 2018 at 20:29
  • $\begingroup$ If $\Delta$ is square then it is not possible ($|\det U|=1$, but $\det M\Delta=0$). In general, $\Delta$ must have full column rank necessarily. But for diagonal matrices it means that neglecting zeros below we get an invertible diagonal submatrix. $\endgroup$
    – A.Γ.
    May 1, 2018 at 20:35

1 Answer 1

Reset to default
0
$\begingroup$

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).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.