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How to Prove a Matrix is Diagonal?

  1. Nov 9, 2011 #1
    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]

    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!
     
    Last edited: Nov 9, 2011
  2. jcsd
  3. Nov 9, 2011 #2

    micromass

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    I might misunderstand your problem but

    [tex]\left(\begin{array}{cc} 1 & 2\\ 3 & 4\end{array}\right)^H\left(\begin{array}{cc} 1 & 2\\ 3 & 4\end{array}\right) = \left(\begin{array}{cc} 10 & 14\\ 14 & 20\end{array}\right)[/tex]

    This is not diagonal. It IS hermition though (as can easily be proven).
     
  4. Nov 9, 2011 #3
    I'm not sure what the consequences of that are in this context. The matrix you suggested cannot satisfy either of the equations:

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

    [itex]\underline{diag}(\textbf{A}\textbf{A}^H) = \underline{1} [/itex]
     
  5. Nov 9, 2011 #4

    I like Serena

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    Does this mean that [itex](\textbf{A}\textbf{A}^H) [/itex] is the identity matrix? :confused:
     
  6. Nov 9, 2011 #5
    It could be any matrix with ones on the diagonal. For example:

    [itex]
    (\textbf{A}\textbf{A}^H) =
    %
    \left[
    \begin{array}{lll}
    1 & 3 & 2 \\
    3 & 1 & 7 \\
    2 & 7 & 1%
    \end{array}%
    \right]


    [/itex]

    would be suitable in this sense.
     
  7. Nov 9, 2011 #6

    I like Serena

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    Oh, okay, so are the Euclidean norms of the rows of A not 1?
     
  8. Nov 9, 2011 #7
    Yes, the Euclidean norms of the rows are 1.

    Consider, for example:

    [itex]
    \textbf{A}=\left[
    \begin{array}{ll}
    1 & 0 \\
    1 & 0 \\
    0 & 1%
    \end{array}%
    \right] ,\quad \underline{\lambda }=\left[
    \begin{array}{l}
    0.25 \\
    0.25 \\
    1%
    \end{array}%
    \right]

    [/itex]

    The norms of all rows of [itex]\textbf{A}[/itex] are equal to one, but [itex]\textbf{A}\textbf{A}^H[/itex] is not the identity matrix:

    [itex]
    \textbf{A}\textbf{A}^H=\left[
    \begin{array}{lll}
    1 & 1 & 0 \\
    1 & 1 & 0 \\
    0 & 0 & 1%
    \end{array}%
    \right]

    [/itex]

    (but [itex]\textbf{A}^H\textbf{A}[/itex] is diagonal, and I want to show that this must always be true).
     
    Last edited: Nov 9, 2011
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