= Solution
A <local unitary transformation> preserves a product factorization:
$$
(U\otimes V)(|a\rangle\otimes|b\rangle)=(U|a\rangle)\otimes(V|b\rangle).
$$
The output is therefore another <product state>. If an <entangled state> could be locally mapped to a <product state>, applying the inverse $U^\dagger\otimes V^\dagger$ would make the original <entangled state> a product as well, a contradiction. Hence \b[a <product state> and an <entangled state> are never <local-unitary equivalent>].
In terms of a <tensor-product coefficient matrix>, $C$ changes to $UCV^T$, 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.
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