MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
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)

share|cite|improve this answer
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
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.