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Some basic observations lead me to ask the following quesiton

Let $A_1, \cdots, A_m$ be $n\times n$ complex matrices. For positive integer $k\ge 1$, show $$\left(\begin{array}{cccc}Tr\{(A_1^*A_1)^k\}&Tr\{(A_1^*A_2)^k\}&\cdots &Tr\{(A_1^*A_m)^k\}\\Tr\{(A_2^*A_1)^k\}&Tr\{(A_2^*A_2)^k\}&\cdots &Tr\{(A_2^*A_m)^k\}\\\cdots&\cdots&\cdots&\cdots\\Tr\{(A_m^*A_1)^k\}&Tr\{(A_m^*A_2)^k\}&\cdots &Tr\{(A_m^*A_m)^k\} \end{array}\right)$$ is positive semidefinite.

Remark

1). When $m=2$, it suffices to show $|Tr\{(A_1^*A_2)^k\}|^2\le Tr\{(A_1^*A_1)^k\}\cdot Tr\{(A_2^*A_2)^k\}$, which is a consequence of a unitarily invariant norm inequality appeared in p.81 of X.Zhan, Matrix inequalities, Springer, 2002.

2). It is easy to show $$\left(\begin{array}{cccc}(Tr\{A_1^*A_1\})^k&(Tr\{A_1^*A_2\})^k&\cdots &(Tr\{A_1^*A_m\})^k\\(Tr\{A_2^*A_1\})^k&(Tr\{A_2^*A_2\})^k&\cdots &(Tr\{A_2^*A_m\})^k\\\cdots&\cdots&\cdots&\cdots\\(Tr\{A_m^*A_1\})^k&(Tr\{A_m^*A_2\})^k&\cdots &(Tr\{A_m^*A_m\})^k \end{array}\right)$$ is positive semidefinite, since it is $k$ Hadamard product of a Gram matrix.

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I fixed your latex, which was not rendering correctly. –  Mariano Suárez-Alvarez Mar 3 '10 at 17:52
    
Even though it is a closely related problem I don't see why you couldn't ask it in a new question. That way you wouldn't have had to unaccept Pavel Etingof's answer. It is common to ask follow-up questions and link to the previous ones, and I like that better than editing when the original questions have already been answered. –  Gjergji Zaimi May 9 '10 at 1:06
    
Thank you. I shall follow your advice. –  Sunni May 9 '10 at 3:02
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1 Answer

up vote 9 down vote accepted

It seems that this is not true. Here is a counterexample. Consider a regular $d$-gon, where $d\ge 3$ is an odd number. Let $S,T$ be the permutation matrices on the vertices of this $d$-gon, induced by reflections in two adjacent symmetry axes. Let $A_1=1, A_2=S, A_3=T$, which are $d$ by $d$ matrices. We have $S^2=T^2=1$, and $ST$ is a rotation of order $d$ (so $(ST)^2$ has no fixed points).

Let $k=2$. Then the matrix in the question seems to be the following: $$ \left(\begin{matrix} d & d & d\\ d & d & 0\\ d & 0 & d\\ \end{matrix}\right) $$ The determinant of this matrix is $-d^3$.

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I have just misread your solution, ignoring the word "permutation" (so I understood your S and T as the 2x2 matrices corresponding to the reflections of the real plane R^2). Then, the matrix in the question looks the same, except that the d's should be replaced by 2's and the 0's should be replaced by 2 cos (2pi/d) or something like that. Anyway, the matrix still has negative determinant, and thus we get a counterexample for 2x2 matrices. –  darij grinberg Mar 3 '10 at 19:27
    
I'd love to know what suggested the example :P –  Mariano Suárez-Alvarez Mar 3 '10 at 19:28
1  
Well, I decided to look at unitary matrices so that I could apply group theory. In this case, if the matrices commute, the inequality holds, since it reduces to the case $k=1$. So I took the simplest noncommutative situation. –  Pavel Etingof Mar 3 '10 at 19:39
    
Oh, sorry. I should have put 'show' as 'prove or disprove'. I did not want to mislead you. I should be more careful about this. Anyway, thank you, readers. –  Sunni Mar 4 '10 at 0:38
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