Past exam of the computer science course of the University of Oxford / 2020 / Quantum information / 2 / d / iv / Solution

A local unitary transformation preserves a product factorization:The output is therefore another product state. If an entangled state could be locally mapped to a product state, applying the inverse $U^{†}\otimes V^{†}$ would make the original entangled state a product as well, a contradiction. Hence a product state and an entangled state are never local-unitary equivalent.

In terms of a tensor-product coefficient matrix, $C$ changes to $UC{V}^T$, with an ordinary transpose on the second factor. Both multiplying matrices are invertible, so the matrix rank, equivalently the Schmidt rank, is unchanged. This supplies an algebraic form of the same argument: a nonzero pure state is a product state exactly when that rank is one.