TITLE: Does the triangle inequality for the absolute value hold for matrix trace?
QUESTION [5 upvotes]: 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 $M\ne\left|M\right|$?

REPLY [2 votes]: 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  
$N = \left( \begin{array}{ccc}
0.8105 &   1.2968 &   0.1548\\
    1.2968 &  -0.1578 &  -1.5520\\
    0.1548  & -1.5520 &   1.9920\end{array} \right)$