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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!

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    $\begingroup$ it's not clear exactly what you are asking... your title is "Why is the trace norm of a hermitian matrix equal to the sum of its eigenvalues" which is wrong. The "trace norm" aka Schatten 1 norm is equal to the sum of the modulus of each eigenvalue. $\endgroup$ Dec 9, 2020 at 18:28
  • $\begingroup$ I edited that. I am asking why is it like that, or better, why is $\text{Tr}\sqrt{AA^{\dagger}}$ equal to that. Bonus: is the matrix square root associatice? $\endgroup$ Dec 9, 2020 at 18:50
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    $\begingroup$ if A is Hermtian, its eignevalues are real and $|a| = \sqrt{a^2}$ in the real numbers. $\endgroup$
    – Doug M
    Dec 9, 2020 at 18:51
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    $\begingroup$ @FriendlyLagrangian Yes, but the square root of a matrix (as in $\sqrt{AA^\dagger}$) is defined such that the resulting eigenvalues are positive. In other words, this is effectively a "principal" square root $\endgroup$ Dec 9, 2020 at 19:00
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    $\begingroup$ @FriendlyLagrangian For a Hermitian positive semidefinite matrix $M$, $P = \sqrt{M}$ is defined to be the unique positive semidefinite matrix for which $P^2 = M$ $\endgroup$ Dec 9, 2020 at 19:01

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