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 $V_1$