This question is related to [On a positivity of a matrix with trace entries.](https://mathoverflow.net/questions/16983/on-a-positivity-of-a-matrix-with-trace-entries)


Let $A\_1, \cdots, A\_m$ be strictly contractive $n\times n$ complex matrices .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.



A matrix with trace entries.
I guess, in the meanwhile you might have already proved that this matrix is **not** positive-semidefinite. I ran a *brute force* experiment, using $2 \times 2$ symmetric, real matrices, which shows that the above conjecture is not true.


I tried different values of $m$, and indeed, the smaller the $m$, the lower the (empirical) probability for a set of random (e.g., uniform), symmetric, real matrices to yield a counterexample. Here is an explicit example with $m=5$, where each $\|A\_i\|<1$:


$$A\_1= \begin{pmatrix}
0.68 &0.21\\\\ 0.21 &0.84
\end{pmatrix}$$


$$A\_2= \begin{pmatrix}
0.58 &0.31\\\\
0.31 &0.74
\end{pmatrix}
$$


$$A\_3=\begin{pmatrix}
0.20 &0.56\\\\
0.56 &0.58
\end{pmatrix}$$


$$A\_4=\begin{pmatrix}
0.31 &0.39\\\\
0.39 &0.75
\end{pmatrix}$$


$$A\_5=\begin{pmatrix}
0.42 &0.34\\\\
0.34 &0.77
\end{pmatrix}$$


The corresponding matrix $M$ with entries $m\_{ij}=\text{trace}((I-A\_iA\_j)^{-1})$, has the following eigenvalues: 
(127.8507, 7.4835, **-0.3282**, 0.3286, 0.9082)


