For a bipartite state \rho, information about the spectrum of its partial transpose \rho^{\Gamma_B} can be inferred from measurements on multiple copies of \rho, without full state tomography. This raises a natural question: which eigenvalue lis
For a bipartite state \rho, information about the spectrum of its partial transpose \rho^{\Gamma_B} can be inferred from measurements on multiple copies of \rho, without full state tomography. This raises a natural question: which eigenvalue lists can arise as \operatorname{spec}(\rho^{\Gamma_B}) for a density operator \rho? We completely solve this inverse eigenvalue problem for two qubits. Every nonnegative trace-one spectrum is realized as \operatorname{spec}(\rho^{\Gamma_B}) by some PPT state \rho, whereas an ordered candidate eigenvalue list (x,y,z,-q), with x\ge y\ge z\ge0, q>0, and x+y+z-q=1, is realized by an NPT state iff q\le y and qy\le xz. Sufficiency in the latter case is established by an explicit X state whose quantum steering ellipsoid has center c=(y-q)/(1-z) and normalized volume V/V_{\max}(c)=qy/(xz), providing a geometric interpretation of the inequalities q\le y and qy\le xz as the allowed ellipsoid-center region and the fixed-center volume bound. Beyond this geometric picture, the two-qubit inverse theorem also yields exact negativity bounds from the two lowest nontrivial PT moments. Given fixed values of p_2=Tr[(\rho^{\Gamma_B})^2] and p_3=Tr[(\rho^{\Gamma_B})^3], we determine the exact minimum and maximum negativity over all two-qubit states subject to these moment constraints. When no PPT state is consistent with the pair (p_2,p_3), the minimum is attained either at x=y or qy=xz, while the maximum is attained either at y=z or q=y. Finally, we show how the two-qubit inequalities persist as necessary constraints for the inverse eigenvalue problem in qubit–qudit systems.