The trace norm of a matrix $A$ is defined as $$ ||A||_1 \doteq \text{Tr} \sqrt{AA^{\dagger}}, $$ thus, if $A$ is hermitian, we yield $$ ||A||_1=\text{Tr} \sqrt{AA^{\dagger}} = \text{Tr} \sqrt{AA} =\text{Tr} \sqrt{D \Lambda D^{-1} D \Lambda D^{-1}} = \text{Tr} \sqrt{D \Lambda \Lambda D^{-1}} = \text{Tr} \sqrt{\Lambda^2}, $$ where I’ve used the trace cyclic property and $\Lambda=\text{diag}(a_1,\cdots,a_N)$. Shouldn’t $$ \text{Tr} \sqrt{\Lambda^2} = \sum_i \pm|a_i| \quad \text{instead of} \quad \text{Tr}\sqrt{\Lambda^2}=\sum_i|a_i| \quad \text{?} $$
Or am I missing something?
Thanks!