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## Main Question or Discussion Point

I have a complex square matrix, [itex]\textbf{C}[/itex], which satisfies:

[itex]\textbf{C}\textbf{C} = (\textbf{I} \odot \textbf{C})[/itex]

where [itex]\textbf{I}[/itex] is the identity matrix and [itex]\odot[/itex] denotes the Hadamard (element-by-element) product. In other words, [itex]\textbf{C}\textbf{C}[/itex] is a diagonal matrix whose diagonal entries are the same as the diagonal entries of [itex]\textbf{C}[/itex], which is not necessarily diagonal itself.

Furthermore, [itex]\textbf{C}[/itex] is Hermitian:

[itex]\textbf{C}^{H}=\textbf{C}[/itex]

and [itex]\textbf{C}[/itex] must be full rank (because actually, in my problem, [itex]\textbf{C} \triangleq (\textbf{A}^{H}\textbf{A})^{-1}[/itex], where [itex]\textbf{A}[/itex] is complex square invertible).

I want to determine whether [itex]\textbf{C} = \textbf{I}[/itex] is the only solution (because this would imply that [itex]\textbf{A}[/itex] is unitary). (This is equivalent to proving that [itex]\textbf{C}[/itex] is diagonal). By expanding out terms, I've shown that [itex]\textbf{C} = \textbf{I}[/itex] is the only invertible solution for [itex](3 \times 3)[/itex] matrices, but I can't seem to obtain a general proof.

Any help or insight would be very much appreciated - I'm completely stumped!

[itex]\textbf{C}\textbf{C} = (\textbf{I} \odot \textbf{C})[/itex]

where [itex]\textbf{I}[/itex] is the identity matrix and [itex]\odot[/itex] denotes the Hadamard (element-by-element) product. In other words, [itex]\textbf{C}\textbf{C}[/itex] is a diagonal matrix whose diagonal entries are the same as the diagonal entries of [itex]\textbf{C}[/itex], which is not necessarily diagonal itself.

Furthermore, [itex]\textbf{C}[/itex] is Hermitian:

[itex]\textbf{C}^{H}=\textbf{C}[/itex]

and [itex]\textbf{C}[/itex] must be full rank (because actually, in my problem, [itex]\textbf{C} \triangleq (\textbf{A}^{H}\textbf{A})^{-1}[/itex], where [itex]\textbf{A}[/itex] is complex square invertible).

I want to determine whether [itex]\textbf{C} = \textbf{I}[/itex] is the only solution (because this would imply that [itex]\textbf{A}[/itex] is unitary). (This is equivalent to proving that [itex]\textbf{C}[/itex] is diagonal). By expanding out terms, I've shown that [itex]\textbf{C} = \textbf{I}[/itex] is the only invertible solution for [itex](3 \times 3)[/itex] matrices, but I can't seem to obtain a general proof.

Any help or insight would be very much appreciated - I'm completely stumped!