# Tricky complex square matrix problem

 P: 108 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!
 P: 27 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
P: 108
 Quote by aesir 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!

 Engineering Sci Advisor HW Helper Thanks P: 7,172 Tricky complex square matrix problem 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.
 P: 108 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).
P: 27
 Quote by weetabixharry ... 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 $\{0\}$
P: 108
 Quote by aesir 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.
P: 27
 Quote by weetabixharry 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 $V_1$

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