# Does a partial transpose always have real eigenvalues?

I am working with a tripartite system, but when I partially transpose the $$8\times 8$$ density matrix I get two complex eigenvalues. I know the criteria for the positive and negative eigenvalues, but are they always real? If they don't, what do my complex eigenvalues physically mean? (I have already checked my matrix for any mistake.)

Yes, the partial transpose takes Hermitian matrices to Hermitian matrices and consequently the eigenvalues of the partial transpose of a Hermitian matrix are all real.

Proof For concreteness, suppose that $$\rho$$ is a three-qubit state. Expanding $$\rho$$ in the Pauli basis we write

$$\rho = \sum_{P,Q,R\in\{I,X,Y,Z\}} a_{P,Q,R}P\otimes Q \otimes R.$$

Note that $$\rho$$ is Hermitian if and only if the coefficients $$a_{P,Q,R}$$ are real. Now, the partial transpose of $$\rho$$ with respect to say the first subsystem is

\begin{align} \rho^{\Gamma_1} &= \sum_{P,Q,R\in\{I,X,Y,Z\}} a_{P,Q,R}P^T\otimes Q \otimes R \\ &= \sum_{P,Q,R\in\{I,X,Y,Z\}} b_{P,Q,R}P\otimes Q \otimes R. \end{align}

Note that $$I^T=I$$, $$X^T=X$$, $$Y^T=-Y$$ and $$Z^T=Z$$. Therefore, $$b_{P,Q,R} = -a_{P,Q,R}$$ when $$P=Y$$ and $$b_{P,Q,R} = a_{P,Q,R}$$ otherwise. In any case, the coefficients $$b_{P,Q,R}$$ are real and consequently $$\rho^{\Gamma_1}$$ is Hermitian.

We can also convince ourselves of the above in a less formal way by writing down a Hermitian matrix and checking where each matrix element is moved by the partial transpose. It is easy to see that the movements of the elements above and below the main diagonal are mirror images of each other. Therefore, the partial transpose preserves the Hermitian symmetry of the matrix.

Any matrix $$M$$ in the tensor product of two spaces can be written as $$M = \sum_{ijkl} a_{ij,kl} E_{ij} \otimes E_{kl},$$ where $$E_{ij}$$ are matrix units and $$a_{ij,kl}$$ are entries of the matrix $$M$$.

Note that $$E_{ij}^\dagger = \overline{E}_{ij}^T = E_{ji}$$, so the condition $$M = M^\dagger = \sum_{ijkl} \overline{a_{ij,kl}} E_{ji} \otimes E_{lk},$$ means $$\overline{a_{ij,kl}} = a_{ji,lk}.$$ Partial transpose of $$M$$ equals $$(I \otimes T) (M) = M^{T_2} = \sum_{ijkl} a_{ij,kl} E_{ij} \otimes E_{kl}^T =$$ $$=\sum_{ijkl} a_{ij,kl} E_{ij} \otimes E_{lk}.$$ Now, we have $$(M^{T_2})^\dagger = \sum_{ijkl} \overline{a_{ij,kl}} E_{ji} \otimes E_{kl}.$$

Since $$\overline{a_{ij,kl}} = a_{ji,lk}$$ you can see that $$(M^{T_2})^\dagger = M^{T_2}$$, so $$M^{T_2}$$ is Hermitian, thus it has real eigenvalues.