# Prove that for a general tri-partite state $\rho_{ABE}$, $H(AB) = H(E)$

How do I prove that for a general tri-partite state $$\rho_{ABE}$$, the following holds: $$H(\rho_{AB}) = H(\rho_{E}), H(\rho_{AE}) = H(\rho_{B}),$$ where, $$H$$ is the Von Neumann entropy. Would Schmidt decomposition help? But I can only do it in a bi-partite scenario. Thanks!

This is not true for general tripartite states. Take the trivial example where $$ABE$$ share a maximally mixed state and each parties subsystem is of dimension $$d$$. The reduced states of a two-party subsystem are maximally mixed of dimension $$d^2$$ and of a single party system are maximally mixed of dimension $$d$$. As the dimensions are different and they are maximally mixed they cannot have the same entropy.
However, the result does hold if $$\rho_{ABE} = |\psi\rangle \langle \psi|$$ is a pure state. Moreover, we can use the Schmidt decomposition by identifying a two-party subsystem with just a single party. For example lets call $$H_{A} \otimes H_B$$ just $$H_{D}$$. Then we can view $$|\psi\rangle$$ as a state in $$H_D \otimes H_E$$ and use the Schmidt decomposition. That is we know there exists orthonormal bases $$\{|i\rangle_D\}$$ and $$\{|i\rangle_E\}$$ of $$D$$ and $$E$$ respectively such that $$|\psi \rangle_{DE} = \sum_i \sqrt{\lambda_i} |ii\rangle.$$ As a consequence the reduced states have the same spectrum, $$\rho_D = \sum_i \lambda_i |i \rangle \langle i |$$ and $$\rho_E = \sum_i \lambda_i |i \rangle \langle i |$$ and hence $$H(D) = H(E)$$. If you want to make this proof more formal you can do the identifying $$H_A \otimes H_B$$ with a single Hilbert space $$H_D$$ step using an isometry map $$V : H_{A} \otimes H_{B} \rightarrow H_D$$ and then note that the von Neumann entropy is invariant under isometries.