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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.

Updated The above question was denied by Pavel Etingof. Now I'd like to ask a closely related problem. Let $A_1, \cdots, A_m$ be strictly contractive $n\times n$ complex matrices . For positive integer $k\ge 1$, is it true that $$\left(\begin{array}{cccc}Tr\{(I-A_1^*A_1)^{-1}\}&Tr\{(I-A_1^*A_2)^{-1}\}&\cdots &Tr\{(I-A_1^*A_m)^{-1}\}\\Tr\{(I-A_2^*A_1)^{-1}\}&Tr\{(I-A_2^*A_2)^{-1}\}&\cdots &Tr\{(I-A_2^*A_m)^{-1}\}\\ \cdots&\cdots&\cdots&\cdots\\Tr\{(I-A_m^*A_1)^{-1}\}&Tr\{(I-A_m^*A_2)^{-1}\}&\cdots &Tr\{(I-A_m^*A_m)^{-1}\} \end{array}\right)$$ is positive semidefinite.

show/hide this revision's text 3 update a problem.; edited body

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.

Updated The above question was denied by Pavel Etingof. Now I'd like to ask a closely related problem. Let $A_1, \cdots, A_m$ be strictly contractive $n\times n$ complex matrices . For positive integer $k\ge 1$, is it true that $$\left(\begin{array}{cccc}Tr\{(I-A_1^*A_1)^{-1}\}&Tr\{(I-A_1^*A_2)^{-1}\}&\cdots &Tr\{(I-A_1^*A_m)^{-1}\}\\Tr\{(I-A_2^*A_1)^{-1}\}&Tr\{(I-A_2^*A_2)^{-1}\}&\cdots &Tr\{(I-A_2^*A_m)^{-1}\}\\ \cdots&\cdots&\cdots&\cdots\\Tr\{(I-A_m^*A_1)^{-1}\}&Tr\{(I-A_m^*A_2)^{-1}\}&\cdots &Tr\{(I-A_m^*A_m)^{-1}\} \end{array}\right)$$ is positive semidefinite.

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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 $$\left(\begin{array}{cccc}Tr{(A_1^*A_1)^k}&Tr{(A_1^*A_2)^k}&\cdots &Tr{(A_1^*A_m)^k}\Tr{(A_2^*A_1)^k}&Tr{(A_2^*A_2)^k}&\cdots &Tr{(A_2^*A_m)^k}\\cdots&\cdots&\cdots&\cdots\Tr{(A_m^*A_1)^k}&Tr{(A_m^*A_2)^k}&\cdots &Tr{(A_m^*A_m)^k} $\left(\begin{array}{cccc}Tr\{(A_1^*A_1)^k\}&Tr\{(A_1^*A_2)^k\}&\cdots &Tr\{(A_1^*A_m)^k\}\\Tr\{(A_2^*A_1)^k\}&Tr\{(A_2^*A_2)^k\}&\cdots &Tr\{(A_2^*A_m)^k\}\\\cdots&\cdots&\cdots&\cdots\\Tr\{(A_m^*A_1)^k\}&Tr\{(A_m^*A_2)^k\}&\cdots &Tr\{(A_m^*A_m)^k\} \end{array}\right)$$ is positive semidefinite.

{\bf Remark.}

Remark

1). When $m=2$, it suffices to show $|Tr{(A_1^*A_2)^k}|^2\le Tr{(A_1^*A_1)^k}\cdot Tr{(A_2^*A_2)^k}$|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 $\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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