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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 $M\ne\left|M\right|$?

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what is square root for matrices? – Evgeny Savinov Feb 5 '12 at 20:32
    
If $U\Lambda U^{-1}$ is a digonalization of X, then $\sqrt{X}=U\sqrt{\Lambda} U^{-1}$. You can refer to en.wikipedia.org/wiki/Square_root_of_a_matrix – af98 Feb 5 '12 at 20:46

I wrote the following in matlab:

function [y] = modm(A)
y = sqrtm(A*A);

Followed by

for i=1:100000
B = triu(randn(3));
A=B+B'-diag(diag(B));
C = triu(randn(3));
D=C+C'-diag(diag(C));
if modm(A) ~= A
    if trace(modm(A-D)) < trace(modm(modm(A)-modm(D)))
       break;
    end
end
end

This lead to several counter-examples to the proposed conjecture.
One specific example is

$M = \left( \begin{array}{ccc} -1.1360 & 1.8503 & 0.3948\\ 1.8503 & -1.9308 & -0.8720\\ 0.3948 & -0.8720 & 0.7281\end{array} \right)$

and

$N = \left( \begin{array}{ccc} 0.8105 & 1.2968 & 0.1548\\ 1.2968 & -0.1578 & -1.5520\\ 0.1548 & -1.5520 & 1.9920\end{array} \right)$

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