linear algebra - 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)