6
$\begingroup$

What are the matrices that you can write in the form $X \odot X^{-T}$, for a complex square matrix $X$, where $X^{-T}$ is the inverse of the complex transpose (not conjugate) and $\odot$ is the Hadamard (component-by-component) product?

In the $2\times 2$ case, you get the group of matrices in the form $$\begin{bmatrix}a & b\\\\ b & a \end{bmatrix},$$ with $a+b=1$, which are closed under matrix multiplication and would form a group were it not for the matrix $a=b=\frac12$ which admits no inverse [EDIT: corrected this assertion, thanks to Denis Serre]. In larger dimension, one sees that all the obtained matrices have the vector of all ones as both a right and left eigenvector. Is this the only restriction? Is the resulting set of matrices closed under multiplication? Is this problem known and studied?

Origin: motivated from this MO question.

Improve this question
$\endgroup$
4
  • 2
    $\begingroup$ In the $2\times2$ case, there is a restriction on the output: $a+b=0$ (maybe you forgot to divide by the determinant of $X$, which occurs from $X^{-T}$). This makes an affine set that is stable under $\times$, but is not a multiplicative group, because of the matrix with $a=b=\frac12$. $\endgroup$
    – Denis Serre
    Commented Apr 26, 2011 at 12:42
  • $\begingroup$ I think you mean $a+b=1$ --- anyway good point, I omitted it in my computations because it didn't matter for the problem and then forgot about it. $\endgroup$
    – Federico Poloni
    Commented Apr 26, 2011 at 13:16
  • $\begingroup$ yes, $a+b=1$. Sorry for the misprint. $\endgroup$
    – Denis Serre
    Commented Apr 26, 2011 at 13:44
  • $\begingroup$ The question is whether every matrix $A$ such $A\vec e=A^T\vec e=\vec e$ (with $\vec e=(1,\ldots,1)^T$) can be written as $X\odot X^{-T}$. When $A$ has non-negative entries, this is a question about bistochastic matrices. A special case is that of ortho-stochastic matrices, for which such an $X$ exists, a unitary one. But it is known that ortho-stochastic matrices form a small part of bistochastic ones. In addition, we don't want to restrict to non-negative entries. $\endgroup$
    – Denis Serre
    Commented Apr 26, 2011 at 14:01

1 Answer 1

Reset to default
3
$\begingroup$

There are some properties of this product in Horn and Johnson, "Topics in Matrix Analysis", Cambridge Univ Press 1991.

Improve this answer
$\endgroup$
1
  • 3
    $\begingroup$ In fact I checked chapter 5 and there is quite a detailed discussion of the Hadamard product $X \odot X^{-T}$ --- for instance, they cite the fact the row and column sums are 1. Some of that material was new to me, it does not contain only the "usual" textbook properties of the Hadamard product. $\endgroup$
    – Federico Poloni
    Commented Apr 27, 2011 at 7:08

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .