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This question is related to 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.

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Nothing depends on $k$ in the matrix –  Homology May 9 '10 at 10:11
    
@ Homology: You are right, I modified it. –  Sunni May 9 '10 at 12:52

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up vote 3 down vote accepted

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)

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No, I haven't found counterexample myself. Is it possible rounding error causing one eigenvalue negative? –  Sunni Oct 4 '10 at 3:49
    
The negative eigenvalue seen above seems to be too large to be a byproduct of roundoff error. Nevertheless, I double checked the computation using rational arithmetic (using mathematica), and then eventually solving the eigenvalues to high-precision: the negative eigenvalue is still there. So, I think this is indeed a counterexample. –  Suvrit Oct 4 '10 at 7:32
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Here is an easier to verify ($3 \times 3$) answer (matlab notation): A1=[.21 .72; .72 .30] A2=[.76 .24; .24 .73] A3=[.48 .65; .65 .12] The matrix $M$ has eigenvalues: 68.4403, 11.9033, -0.6254 I generated matrices by doing A=10*rand(2); A=A+A'; A=0.98*A/norm(A). Then, i truncated these to 2 digits (for easier typesetting), while ensuring that the final matrix M, still remains a counterexample. –  Suvrit Oct 4 '10 at 7:54

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