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Can a square non-zero and non-diagonal $n\times n$ complex unitary matrix be decomposed into non-zero and non-diagonal square $m\times m$ complex unitary matrices (such that $n > m$)?

For example if U is a square non-zero and non-diagonal complex unitary matrix, can U be decomposed into square non-zero and non-diagonal square matrices A, B, C, and D such that

$$ \newcommand{\bigzero}{\mbox{\normalfont\Large\bfseries 0}} U = \left(\begin{array}{@{}cc@{}} A & B \\ C & D \end{array}\right) $$

I know various decomposition methods exist where if we assume that the matrices in question are diagonal then we can write down a decomposition of U, but in this case I'm considering the restrictions that we place on the matrices to include that U, A, B, C, and D are non-zero, non-diagonal, and unitary (i.e. to be unitary but to take the most general approach as to the form of the matrices). It makes sense to me that the above statement would be true, but I can't find this explicitly stated or explained anywhere...

Any reference of the proof or statement in a published source (arXiv paper, textbook, journal article, etc...) would also be greatly appreciated.

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If $A,B,C,D$ are unitary, then it can never be the case that $U$ is unitary. One way to see this is to note that for an $n \times n$ unitary matrix $M$, we must have $\|M\|_F^2 = n$ (where $\|\cdot\|_F$ denotes the Frobenius norm, so that $\|M\|_F^2 = \sum_{i,j}|m_{ij}|^2$). It follows that if $A,B,C,D$ are unitary with size $m = n/2$, then $$ \|U\|_F^2 = \|A\|_F^2 + \|B\|_F^2 + \|C\|_F^2 + \|D\|_F^2 = 4 \cdot m = 2n. $$ But if $U$ is an $n \times n$ unitary matrix, then $\|U\|_F^2 = n$.

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  • $\begingroup$ Very elegant! Thank you for the explanation and for correcting my incorrect intuition. $\endgroup$ Commented Mar 9, 2022 at 2:53

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