most recent 30 from http://mathoverflow.net 2013-05-22T12:23:13Z http://mathoverflow.net/feeds/question/63027 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/63027/matrices-that-are-hadamard-products-of-x-and-x-t Federico Poloni 2011-04-26T11:59:11Z 2011-04-28T13:41:37Z

<p>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 <em>complex transpose</em> (not conjugate) and $\odot$ is the Hadamard (component-by-component) product?</p> <p>In the $2\times 2$ case, you get the group of matrices in the form $$\begin{bmatrix}a &amp; b\\ b &amp; a \end{bmatrix},$$ 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?</p> <p>Origin: motivated from <a href="http://mathoverflow.net/questions/62390/on-a-tentative-generalization-of-the-schmidt-decomposition" rel="nofollow">this MO question</a>.</p>

http://mathoverflow.net/questions/63027/matrices-that-are-hadamard-products-of-x-and-x-t/63028#63028 greg coxson 2011-04-26T12:30:34Z 2011-04-28T13:41:37Z

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