Tricky complex square matrix problem

[weetabixharry]

I have a complex square matrix, \textbf{C}, which satisfies:

\textbf{C}\textbf{C} = (\textbf{I} \odot \textbf{C})

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

Furthermore, \textbf{C} is Hermitian:

\textbf{C}^{H}=\textbf{C}

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

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

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

 

[aesir]

you can take the square root of your equation

since C is positive definite (C=(A^\dagger A)^{-1}) on the left you have C
and you obtain (in components):
C_{ij}=\sqrt{C_{ij}}\delta_{ij}

from which you can conclude that C is the identity matrix

 

[weetabixharry]

you can take the square root of your equation

Brilliant! Thank you very much indeed! I had really be scratching my head over that one. Many thanks again for your help!

 

[AlephZero]

You can show this from "first principles". Let the matrix be
\left(\begin{array}{cc} a & b \\ b* & c \end{array}\right)
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.

 

[weetabixharry]

Thanks AlephZero. That is the approach I took in order to obtain a proof for (2 \times 2) and (3 \times 3) matrices. (If I understand correctly, your a, b and c are scalars.) However, aesir's solution is valid for the general (n \times n) case, which is especially important for me.

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

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

(Therefore aesir's square root still exists in this case).

 

[aesir]

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

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

(Therefore aesir's square root still exists in this case).

I don't think so.
It is true that if \textbf{z} is in the null space of \textbf{A} then \textbf{z}^H\textbf{A}^H\textbf{A}\textbf{z} = \left\Vert{\textbf{Az}}\right\Vert^2 = 0, but this means that (\textbf{A}^{H}\textbf{A}) is semi-positive definite, not its inverse (which does not exists if the null space is non-trivial). BTW if \textbf{A} has linearly independent columns its null space is \{0\}

 

[weetabixharry]

I don't think so.
It is true that if \textbf{z} is in the null space of \textbf{A} then \textbf{z}^H\textbf{A}^H\textbf{A}\textbf{z} = \left\Vert{\textbf{Az}}\right\Vert^2 = 0, but this means that (\textbf{A}^{H}\textbf{A}) is semi-positive definite, not its inverse (which does not exists if the null space is non-trivial). BTW if \textbf{A} has linearly independent columns its null space is \{0\}

Ah yes, of course. Thanks for clearing that up!

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

(2) When the rank of \textbf{A} is less than its number of columns (which includes all fat matrices), (\textbf{A}^H\textbf{A}) is positive semidefinite. In this case, (\textbf{A}^H\textbf{A})^{-1} does not exist.

 

[aesir]

Ah yes, of course. Thanks for clearing that up!

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

(2) When the rank of \textbf{A} is less than its number of columns (which includes all fat matrices), (\textbf{A}^H\textbf{A}) is positive semidefinite. In this case, (\textbf{A}^H\textbf{A})^{-1} does not exist.

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 V_1 you have A^H A = \left(\begin{array}{cc} B^HB & 0 \\ 0 & 0 \end{array} \right)
that has a positive definite inverse if restricted from V_1 to V_1