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Tricky complex square matrix problem

  1. Oct 26, 2011 #1
    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:


    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!
  2. jcsd
  3. Oct 27, 2011 #2
    you can take the square root of your equation

    since C is positive definite ([itex]C=(A^\dagger A)^{-1}[/itex]) on the left you have C
    and you obtain (in components):

    from which you can conclude that C is the identity matrix
  4. Oct 27, 2011 #3
    Brilliant! Thank you very much indeed! I had really be scratching my head over that one. Many thanks again for your help!
  5. Oct 27, 2011 #4


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    You can show this from "first principles". Let the matrix be
    [tex]\left(\begin{array}{cc} a & b \\ b* & c \end{array}\right)[/tex]
    where a and c are real.

    Mlutiplying the matrices out gives 3 equations

    a^2 + bb* = a
    c^2 + bb" = c
    ab + bc = 0

    Subtracting the first two equations, either a = c, or a+c = 1
    From the third equation, either b = 0, or a+c = 0

    So either b = 0, or a = c = 0

    But from the first two equations, if a = c = 0 then b = 0 also.

    So, the first two equations reduce to a^2 = a and c^2 = c, and the only solution which gives a matrix of full rank is C = I.
  6. Oct 28, 2011 #5
    Thanks AlephZero. That is the approach I took in order to obtain a proof for [itex](2 \times 2)[/itex] and [itex](3 \times 3)[/itex] matrices. (If I understand correctly, your [itex]a[/itex], [itex]b[/itex] and [itex]c[/itex] are scalars.) However, aesir's solution is valid for the general [itex](n \times n)[/itex] case, which is especially important for me.

    A final question on positive definiteness:
    If [itex]\textbf{A}[/itex] is not square, but instead is tall (with linearly independent columns) then is it correct to say that [itex](\textbf{A}^{H}\textbf{A})^{-1}[/itex] is now positive semi-definite?

    My reasoning is that [itex]\textbf{z}^H\textbf{A}^H\textbf{A}\textbf{z} = \left\Vert{\textbf{Az}}\right\Vert^2 \geq 0[/itex] for any [itex]\textbf{z}[/itex] (with equality when [itex]\textbf{z}[/itex] lies in the null space of [itex]\textbf{A}[/itex]).

    (Therefore aesir's square root still exists in this case).
  7. Oct 28, 2011 #6
    I don't think so.
    It is true that if [itex]\textbf{z}[/itex] is in the null space of [itex]\textbf{A}[/itex] then [itex]\textbf{z}^H\textbf{A}^H\textbf{A}\textbf{z} = \left\Vert{\textbf{Az}}\right\Vert^2 = 0[/itex], but this means that [itex](\textbf{A}^{H}\textbf{A})[/itex] is semi-positive definite, not its inverse (which does not exists if the null space is non-trivial). BTW if [itex]\textbf{A}[/itex] has linearly independent columns its null space is [itex]\{0\}[/itex]
  8. Oct 28, 2011 #7
    Ah yes, of course. Thanks for clearing that up!

    So are the following two statements correct?
    (1) [itex] (\textbf{A}^H\textbf{A}) [/itex] is positive definite when the columns of [itex]\textbf{A}[/itex] are independent (which requires that [itex]\textbf{A}[/itex] is tall or square). Therefore [itex] (\textbf{A}^H\textbf{A})^{-1} [/itex] is also positive definite.

    (2) When the rank of [itex]\textbf{A}[/itex] is less than its number of columns (which includes all fat matrices), [itex](\textbf{A}^H\textbf{A})[/itex] is positive semidefinite. In this case, [itex](\textbf{A}^H\textbf{A})^{-1}[/itex] does not exist.
  9. Oct 28, 2011 #8
    Yes, that's true.
    In case (2) you can say a little more. If you split the vector space in null{A} and its orthogonal complement [itex]V_1[/itex] you have [tex]A^H A = \left(\begin{array}{cc} B^HB & 0 \\ 0 & 0 \end{array} \right)[/tex]
    that has a positive definite inverse if restricted from [itex]V_1[/itex] to [itex]V_1[/itex]
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