Question

This question is related to http://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.

Answer 1

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)