I've been stuck on this problem for so long it's getting ridiculous. Please help!(adsbygoogle = window.adsbygoogle || []).push({});

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]

mustbe 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!

**Physics Forums - The Fusion of Science and Community**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# How to Prove a Matrix is Diagonal?

Loading...

Similar Threads for Prove Matrix Diagonal | Date |
---|---|

I Proving only 1 normalized unitary vector for normal matrix | Mar 28, 2018 |

I Proving Matrix exponential property | Oct 8, 2016 |

Proving a linear algebra equation | Jul 1, 2015 |

Proving the special property of diagonal matrix? | Jan 8, 2014 |

Proving the scalar matrices are the center of the matrix ring | Sep 12, 2013 |

**Physics Forums - The Fusion of Science and Community**