3 years ago

A decomposition of the Schur block product on matrices of operators.

Erik Christensen

Given two $m \times n $ matrices $A = (a_{ij})$ and $B=(b_{ij}) $ with entries in $B(H)$ for some Hilbert space $H,$ the Schur block product is the $m \times n$ matrix $ A\square B := (a_{ij}b_{ij}).$ There exists an $m \times n$ matrix $S = (s_{ij})$ with entries from $B(H)$ such that $S$ is a contraction operator and $ A \square B = \big(\mathrm{diag}(AA^*)\big)^{\frac{1}{2}}S\big(\mathrm{diag}(B^*B)\big)^{\frac{1}{2}}.$ This decomposition of the Schur product seems to be unknown, even for scalar matrices.

Publisher URL: http://arxiv.org/abs/1811.03668

DOI: arXiv:1811.03668v1

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