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I've been stuck on this problem for so long it's getting ridiculous. Please help!

I have a complex matrix, [itex]\textbf{A}[/itex], whose columns are linearly independent. In other words, [itex]\textbf{A}[/itex] is either tall or square and [itex] \left( \textbf{A}^H\textbf{A}\right)^{-1}[/itex] exists (where [itex]\left(\right)^H[/itex] denotes conjugate transpose). I am trying to prove that the matrix:

[itex]\textbf{B} \triangleq \left( \textbf{A}^H\textbf{A}\right)[/itex]

[itex]\textbf{A}= diag(\underline{\lambda})\textbf{A}\textbf{A}^H \textbf{A} \textbf{A}^H \textbf{A}[/itex]

for some real diagonal matrix [itex]diag(\underline{\lambda})[/itex]. It may or may not also be useful to note that [itex]\textbf{A}[/itex] is also subject to the constraint:

[itex]\underline{diag}(\textbf{A}\textbf{A}^H) = \underline{1} [/itex]

by which I mean that all the diagonal entries of [itex](\textbf{A}\textbf{A}^H) [/itex] are equal to 1 (i.e. the Euclidean norms of the rows of [itex]\textbf{A}[/itex] are all 1).

I have deduced all sorts of properties of [itex]\textbf{A}[/itex], but strongly believe that it should be possible to show that [itex]\textbf{B}[/itex] is diagonal... but a proof escapes me. Any help is greatly appreciated!

I have a complex matrix, [itex]\textbf{A}[/itex], whose columns are linearly independent. In other words, [itex]\textbf{A}[/itex] is either tall or square and [itex] \left( \textbf{A}^H\textbf{A}\right)^{-1}[/itex] exists (where [itex]\left(\right)^H[/itex] denotes conjugate transpose). I am trying to prove that the matrix:

[itex]\textbf{B} \triangleq \left( \textbf{A}^H\textbf{A}\right)[/itex]

**must**be diagonal, based on the following:[itex]\textbf{A}= diag(\underline{\lambda})\textbf{A}\textbf{A}^H \textbf{A} \textbf{A}^H \textbf{A}[/itex]

for some real diagonal matrix [itex]diag(\underline{\lambda})[/itex]. It may or may not also be useful to note that [itex]\textbf{A}[/itex] is also subject to the constraint:

[itex]\underline{diag}(\textbf{A}\textbf{A}^H) = \underline{1} [/itex]

by which I mean that all the diagonal entries of [itex](\textbf{A}\textbf{A}^H) [/itex] are equal to 1 (i.e. the Euclidean norms of the rows of [itex]\textbf{A}[/itex] are all 1).

I have deduced all sorts of properties of [itex]\textbf{A}[/itex], but strongly believe that it should be possible to show that [itex]\textbf{B}[/itex] is diagonal... but a proof escapes me. Any help is greatly appreciated!

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