Hello, all!

I consider Hadamard product $A \circ B$ of matrices $A$, $B$ over finite field. I know $\det{A}$ and $\det{B}$ and want to know about $\det{(A \circ B)}$. Wikipedia and Google let me know properties about determinant for Hadamard product of positive-semidefinite matrices: $det{(A \circ B)} \ge \det{A} \cdot \det{B}$. What happens if matrices are over finite field?

Thank you.

  • $\begingroup$ Could you expand on what "what happens" is supposed to mean? That inequality doesn't make sense over a finite field, and neither does the notion of "positive semi-definite." $\endgroup$ – Ben Webster Mar 20 '11 at 8:48
  • $\begingroup$ I put to "what happens" the same meaning: if we are in finite field then we have no notions for "positive semi-definite" and ordering. I have to discover how it could happen matrix $(A \circ B)$ has zero determinant by determinant values of $A$ and $B$ $\endgroup$ – spk Mar 20 '11 at 8:55
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    $\begingroup$ You seem to be trying to make two generalizations at once: firstly, from (real or complex) matrices that are positive-semidefinite, to (real or complex) matrices that are not necessarily PSD; and secondly, to matrices over finite fields. With all due respect, I think you need to understand the first step before you jump to the second. $\endgroup$ – Yemon Choi Mar 20 '11 at 9:50
  • $\begingroup$ The problem is next: if I work in finite fields I do not know what I should expect from Hadamard product. I hope that there are any relations between $A$, $B$ and $A \circ B$. But I do not know how to open them. $\endgroup$ – spk Mar 20 '11 at 11:15
  • $\begingroup$ I think you need to formulate a more precise question in your own mind before you ask others "what might be true" $\endgroup$ – Yemon Choi Mar 20 '11 at 21:25

Given any $a$ and $b$, let $A=\pmatrix{a&0\cr0&1}$, $B=\pmatrix{0&-b\cr1&0}$, then $\det A=a$, $\det B=b$, and the determinant of the Hadamard product is zero.

  • $\begingroup$ Thank you! But this is not a common case. I could narrow my problem to next one: if it is possible to prove that id $A$ and $B$ have non-zero determinants then $A \circ B$ also has non-zero determinant. $\endgroup$ – spk Mar 20 '11 at 9:21
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    $\begingroup$ ??? Of course it is not possible to prove that if $A$ and $B$ have non-zero determinants then the Hadamard product has a non-zero determinant - my answer shows that $A$ and $B$ can have any determinant you like, and still the Hadamard product can have determinant zero! $\endgroup$ – Gerry Myerson Mar 20 '11 at 9:39
  • $\begingroup$ I'm sorry. But what it could be if matrix $A$ is a Van der Monde one and $B$ is arbitrary matrix? $\endgroup$ – spk Mar 20 '11 at 9:53
  • $\begingroup$ I don't know, but it doesn't seem hard to make up $2\times2$ examples to see for yourself what kind of thing can happen. $\endgroup$ – Gerry Myerson Mar 20 '11 at 9:59
  • $\begingroup$ Yes, I went by that way. I observed wonderful fact for Hadamard product of Van der Monde matrix and arbitrary one using Mathematica for this purpose. For example, let $\begin{array}{ccc} a_{1,1} & a_{1,2} x_1 & a_{1,3} x_1^2 \\ a_{2,1} & a_{2,2} x_2 & a_{2,3} x_2^2 \\ a_{3,1} & a_{3,2} x_3 & a_{3,3} x_3^2 \end{array}$ be a Hadamard product of Van der Monde matrix $X$ and arbitrary one $A$. Then its determinant is equal to just $X$ determinant "cleaned" from elements of $A$. As I saw it is clear for all sizes of matrices $X$ and $A$. I tried to write this correspondence explicitly but failed. $\endgroup$ – spk Mar 20 '11 at 11:18

A nice elementary survey of the properties of Hadamard product (many not requiring that coefficient be real or complex) can be found in


(which looks like an undergraduate project).

  • $\begingroup$ Thank you! But no property from this stuff I could use for my purposes. I will proceed with Yemon Choi's advise and reformulate my problem to more concrete question to Mathoverflow $\endgroup$ – spk Mar 21 '11 at 12:18

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