Tricky complex square matrix problem

In summary: When the columns of \textbf{A} are independent (which requires that \textbf{A} is tall or square), (\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.
  • #1
weetabixharry
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0
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:

[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!
 
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  • #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):
[tex]C_{ij}=\sqrt{C_{ij}}\delta_{ij}[/tex]

from which you can conclude that C is the identity matrix
 
  • #3
aesir said:
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!
 
  • #4
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
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).
 
  • #6
weetabixharry said:
...
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).

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]
 
  • #7
aesir said:
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]

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.
 
  • #8
weetabixharry said:
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.

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]
 

1. What is a tricky complex square matrix problem?

A tricky complex square matrix problem refers to a mathematical problem involving a square matrix that contains complex numbers. These types of problems often require advanced techniques and strategies to solve.

2. How do I approach solving a tricky complex square matrix problem?

The best approach to solving a tricky complex square matrix problem is to first identify the type of problem it is (e.g. finding determinants, solving systems of equations, etc.) and then use appropriate methods and formulas to solve it. It also helps to break down the problem into smaller, more manageable steps.

3. What are some common techniques used to solve tricky complex square matrix problems?

Some common techniques used to solve tricky complex square matrix problems include Gaussian elimination, Cramer's rule, and eigenvalue decomposition. It is also helpful to have a strong understanding of complex numbers, matrices, and linear algebra concepts.

4. Are there any tips for avoiding common mistakes when solving tricky complex square matrix problems?

It is important to pay close attention to details and double check your work when solving tricky complex square matrix problems. It is also helpful to break down the problem into smaller steps and to use a calculator or computer program to assist with complex calculations.

5. Can I use any software to solve tricky complex square matrix problems?

Yes, there are many software programs and online calculators available that can help with solving tricky complex square matrix problems. However, it is important to understand the concepts and techniques behind the problem-solving process in order to effectively use these tools.

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