# Why does the entanglement negativity equal (in magnitude) the sum of the negative eigenvalues?

The entanglement negativity, introduced in (Vidal and Werner 2002), is defined as $$\mathcal N(\rho) \equiv \frac{\|\rho^{T_B}\|_1-1}{2}.$$ It is mentioned there that this equals the sum of the absolute values of the negative values of the eigenvalues of the partial transpose $$\rho^{T_B}$$: $$\mathcal N(\rho)=\sum_{\lambda<0} \lvert\lambda\rvert$$.

How does one show the equivalence of these two quantities?

This is also discussed in the paper linked above. The trace norm of $$X$$ is defined as the sum of the absolute values of the eigenvalues of $$X$$: $$\|X\|_1=\sum_i \lvert\lambda_i\rvert$$. $$\newcommand{\tr}{\operatorname{tr}}$$Given a state $$\rho$$, the normalisation condition amounts to $$\tr(\rho) = \sum_i \lambda_i = \sum_{\lambda\in\sigma_+}\lambda + \sum_{\lambda\in\sigma_-}\lambda = 1,$$ where $$\sigma_\pm$$ is the set of positive (negative) eigenvalues. Given $$\rho=\rho_{ij,k\ell}|ij\rangle\!\langle k\ell|$$, the partial transpose $$\rho^{T_B}$$ reads $$\rho^{T_B}=\sum_{ijk\ell}\rho_{ij,k\ell}|i\ell\rangle\!\langle kj|$$, which means that $$\tr(\rho^{T_B})=\sum_{ij} \rho_{ij,ij} = \tr(\rho) = 1.$$
It follows that $$\|\rho^{T_B}\|_1 = \sum_i \lvert\lambda_i\rvert = \sum_{\lambda\in\sigma_+}\lambda - \sum_{\lambda\in\sigma_-}\lambda = 1 - 2 \sum_{\lambda\in\sigma_-}\lambda.$$ And therefore $$\mathcal N(\rho) = \frac{\|\rho^{T_B}\|_1-1}{2} = -\sum_{\lambda\in\sigma_-}\lambda = \sum_{\lambda\in\sigma_-}\lvert\lambda\rvert.$$