Tricky complex square matrix problemby weetabixharry Tags: complex matrix, hermitian matrix, involutory matrix, matrix inverse, unitary matrix 

#1
Oct2611, 06:46 AM

P: 96

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 (elementbyelement) 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: [itex]\textbf{C}^{H}=\textbf{C}[/itex] 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
Oct2711, 08:42 AM

P: 27

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): [tex]C_{ij}=\sqrt{C_{ij}}\delta_{ij}[/tex] from which you can conclude that C is the identity matrix 



#3
Oct2711, 01:22 PM

P: 96





#4
Oct2711, 07:42 PM

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P: 6,344

Tricky complex square matrix problem
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. 



#5
Oct2811, 02:40 AM

P: 96

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 semidefinite? 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). 



#6
Oct2811, 03:20 AM

P: 27

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 semipositive definite, not its inverse (which does not exists if the null space is nontrivial). BTW if [itex]\textbf{A}[/itex] has linearly independent columns its null space is [itex]\{0\}[/itex] 



#7
Oct2811, 05:08 AM

P: 96

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. 



#8
Oct2811, 07:10 AM

P: 27

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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