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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 $\Delta$

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  • $\begingroup$ 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.
    Commented May 1, 2018 at 19:20
  • $\begingroup$ You're right, let's assume $\Delta$ does not have inverse. $\endgroup$
    – dwfa
    Commented 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.
    Commented May 1, 2018 at 20:35

1 Answer 1

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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).

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