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Every rank-one projective measurement of one qubit of the three-qubit W state has at least one nonzero-probability outcome whose other two qubits remain entangled. To see this, contract the measured factor against $⟨u|$, where $|u⟩=a|0⟩+b|1⟩$. The unnormalized tensor-product coefficient matrix of the other pair isAt least one vector in a complete measurement basis has $a\ne 0$, giving nonzero determinant and an entangled conditional branch. Its probability $(2|a{|}^2+|b{|}^2)/3$ is positive. This property survives local unitary changes of basis. The GHZ state lacks it because computational-basis measurement of one qubit leaves a product state on each branch, so the property also separates the states under local-unitary equivalence.

Use $S=\mathrm{diag}(1,i)=T^2$ and the Pauli X gate. Starting with the Bell state $|{\Phi }^{+}⟩$, the exact constructions areFor the first, $I\otimes X$ changes the two terms to $|01⟩$ and $|10⟩$, and $S\otimes I$ multiplies just the second term by $i$. For the second,All factors are single-party unitary operators, proving both claimed local-unitary equivalences. The equations hold as vectors, without even an additional global phase.

Fix the two subsystem dimensions. For reflexivity, $|\Psi ⟩=(I_d\otimes I_{d^{\prime }})|\Psi ⟩$, and both identity operators are unitary.

For symmetry, if $|\Psi ⟩=(U\otimes V)|\Phi ⟩$, apply the inverse local unitary transformation:The conjugate transposes of unitary operators are unitary, so the reverse relation holds.

For transitivity, if additionally $|\Phi ⟩=(R\otimes S)|\Theta ⟩$, the tensor product multiplication rule givesBoth products are unitary, for example $(UR{)}^{†}(UR)=R^{†}{U}^{†}UR=I$. Thus local-unitary equivalence is reflexive, symmetric and transitive, and hence is an equivalence relation.