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For an arbitrary square matrix $L$ whose symmetric part $\frac{L+L^\mathrm{T}}{2}\succcurlyeq0$ is positive semidefinite, can one always find a decomposition $L=U^\mathrm{T}QU$ where $U=U^\mathrm{T}\succcurlyeq0$ and asymmetric $Q$ has $\frac{Q+Q^\mathrm{T}}{2}\succ0$?

The answer seems to be positive, but I still would like to ask about it.

Note, here, that '$L$ is asymmetric' means $L\neq L^\mathrm{T}$.

Wilbur
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  • by asymmetric do you mean not symmetric or antisymmetric? Also you use $\succcurlyeq$ but you say positive definite, which do you mean? – Set Jun 13 '15 at 06:02
  • Sorry for the misleading. 'Asymmetric' implies 'not symmetric'. I've revised the question. The question assumes that $\frac{L+L^\mathrm{T}}{2}$ is positive semidefinite. – Wilbur Jun 13 '15 at 09:03

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