What special significance does the eigendecomposition of a mixed density operator hold over other pure state decompositions? It is known that in general, a mixed state can have multiple pure state decompositions. However, it has a unique eigendecomposition in the absence of degenerate eigenvalues. What is the special significance of this eigendecomposition over other pure state decompositions for the same density matrix?
 A: It turns out it has a special significance in terms of compressibility of information. For a state ${\rho}$ with an eigendecomposition, 
${\rho=\sum_{i}\lambda_{i}|\psi_{i}\rangle\langle \psi_{i}|}$
the Von Neumann entropy ${S(\rho)}$is defined as, 
${S(\rho)=-\mathrm{tr} \rho \log \rho=\sum_{i} \lambda_{i} \log \lambda_{i}}$
For any general pure state decomposition $\rho=\sum_{k} p_{k}\rho_{k}$, the Shannon entropy ${H(\{p_{k}\})}$ for the particular ensemble is defined as, 
${H(\{p_{k}\})=-\sum_{k} p_{k} \log p_{k}}$
and the following inequality always holds true, 
${H(\{p_{k}\}) \geq S(\rho)}$
with the equality being satisfied only when ${\{p_{k}\}}$ decomposition is the eigendecomposition itself.
Thus, the Shannon entropy of any pure state decomposition of a mixed state can never be less than the Von Neumann entropy of the state. This is due to the fact that in the eigendecomposition, the constituent states, being orthogonal, are perfectly distinguishable and hence correspond to minimal Shannon entropy. 
