It is well-known that, $\left|m-n\right|\ge\left|\left|m\right|-\left|n\right|\right|$ for real numbers. But if one defines $\left|M\right|=\sqrt{M^2}$ for a symmetric matrix $M$, does one have $$\operatorname{trace}\left(\left|M-N\right|\right)\ge\operatorname{trace}\left(\left|\left|M\right|-\left|N\right|\right|\right)$$ if $M\ne\left|M\right|$?
Does the triangle inequality for the absolute value hold for matrix trace?
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linear-algebra
matrices
inequality
absolute-value
trace
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0what is square root for matrices? – 2012-02-05
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0If $U\Lambda U^{-1}$ is a digonalization of X, then $\sqrt{X}=U\sqrt{\Lambda} U^{-1}$. You can refer to http://en.wikipedia.org/wiki/Square_root_of_a_matrix – 2012-02-05