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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.

Two bipartite pure states on fixed subsystem spaces are local-unitary equivalent when related by the displayed product of unitary operators. Identity operators give reflexivity, inverse operators give symmetry, and products of local operators give transitivity. Thus this is an equivalence relation, whose classes are the orbits under local unitaries.

Every pair of normalized product states is local-unitary equivalent: extend each factor to an orthonormal basis and use a unitary operator carrying one basis to the other. Local operators preserve a product factorization, and their inverses do too, so quantum entanglement cannot be created or removed this way. More generally, the tensor-product coefficient matrix changes as $C\mapsto UC{V}^T$, which preserves its singular values and therefore its Schmidt coefficients. The transpose on the second factor is ordinary, not a conjugate transpose, because coefficients of a tensor product depend bilinearly on the two input vectors.

Consider one normalized measurement vector $|u⟩=a|0⟩+b|1⟩$ on C. The unnormalized partial inner product on AB isIts tensor-product coefficient matrix and determinant areThe two-qubit product-state determinant criterion therefore says that this branch is entangled whenever $a\ne 0$. Its probability is $(|b{|}^2+2|a{|}^2)/3>0$, so such a branch is physically possible and can be normalized without changing whether its determinant vanishes.

For an orthonormal basis $\{|u_0⟩,|u_1⟩\}$, completeness gives $|⟨0|u_0⟩{|}^2+|⟨0|u_1⟩{|}^2=1$. At least one vector has $a\ne 0$. Hence at least one nonzero-probability outcome leaves AB entangled, proving conditional entanglement of the three-qubit W state. Some bases give two entangled branches; the computational basis gives one entangled and one product branch. The claim is not that every outcome must entangle AB.

Taking partial inner products with $⟨0|$ and $⟨1|$ on the first wire gives $(|0⟩+|1⟩)/2$ and $(-|0⟩+|1⟩)/2$. They are nonzero and orthogonal, hence linearly independent. This contradicts the necessary product-state property proved in part (a), so the second state is entangled.

Equivalently, its tensor-product coefficient matrix is $C=\frac{1}{2}\left(\begin{array}{cc}1& 1\\ -1& 1\end{array}\right)$, with $\det C=1/2\ne 0$. The two-qubit product-state determinant criterion gives the same conclusion. In fact $C{C}^{†}=I/2$, so its Schmidt coefficients are both $1/\sqrt{2}$: it is maximally entangled.

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.

In a two-factor tensor product over a field, a pure tensor is also called a separable or decomposable tensor: it can be expressed as one product $v\otimes w$. Every tensor is a finite sum of pure tensors, but a sum need not itself be a pure tensor.^[ref]

For finite-dimensional vector spaces, the tensor-product coefficient matrix gives a decisive test: a nonzero tensor is a pure tensor exactly when its matrix rank is one. Indeed, every column of $v{w}^T$ is a scalar multiple of $v$. Conversely, if all columns of a nonzero matrix are multiples of one nonzero column $v$, collect their multipliers in $w$ to obtain $C=v{w}^T$. The zero tensor is also a pure tensor, for example $0\otimes w$, but its matrix rank is zero.

In ${\mathbb{R}}^2{\otimes }_{\mathbb{R}}{\mathbb{R}}^2$, the tensoris nonseparable: its matrix rank is two and its determinant is one, whereas $\det (v{w}^T)=0$. Directly, trying $t=(a{e}_1+b{e}_2)\otimes (c{f}_1+d{f}_2)$ requires $ac=bd=1$ and $ad=bc=0$. The first two equations force all four scalars nonzero, contradicting the others. In contrast, products sharing a factor do combine: $e_1\otimes f_1+e_1\otimes f_2=e_1\otimes (f_1+f_2)$. Thus being written as two summands alone does not establish nonseparability.

A tensor product turns a bilinear map into a linear map. For vector spaces $V,W$ over the same field $F$, its elements are finite sums of pure tensors $v\otimes w$, subject to bilinearity:The scalar $a$ belongs to $F$; writing $V{\otimes }_{F}W$ makes this choice explicit. These rules preserve separate linearity while allowing arbitrary linear combinations of products.^[ref]

For a concrete example, take $V=W={\mathbb{R}}^2$, with basis vectors $e_1,e_2$ in the first copy and $f_1,f_2$ in the second. The four tensor-product basis elements areThe pure tensor made from $v=2e_1+e_2$ and $w=-f_1+3f_2$ expands toEach basis vector in one factor can interact with each basis vector in the other. This is why dimensions multiply; the tensor-product coefficient matrix displays these four interactions as a two-by-two grid.

The tensor product is different from the Cartesian product $V\times W$. With componentwise vector addition and scalar multiplication, the latter contains pairs and has dimension $\dim V+\dim W$. The map $\tau (v,w)=v\otimes w$ is bilinear, generally neither linear on that Cartesian product nor injective nor surjective. In our example, $\tau (2v,2w)=4\tau (v,w)$, and $\tau (2v,w/2)=\tau (v,w)$; the separability of a tensor explains why some outputs cannot be reached at all. Both spaces happen to have dimension four when $V=W={\mathbb{R}}^2$, but this coincidence does not identify their structures.

The two-factor tensor rank is the smallest number of pure tensors whose sum equals $t$, with zero allowed as the empty sum. For two finite-dimensional vector spaces over the same field, it equals the matrix rank of the tensor-product coefficient matrix. A sum of $r$ outer products has column space contained in the span of their $r$ first factors, so its matrix rank is at most $r$. Conversely, choose a basis $u_1,\dots ,u_r$ for the column space, express each column in that basis, and collect the coefficients into row vectors $w_k^T$. This yields $C={\sum }_{k=1}^r{u}_k{w}_k^T$, hence a sum of $r$ pure tensors.

The pure tensor $(2,1)\otimes (-1,3)$ therefore has two-factor tensor rank one, whereas $e_1\otimes f_1+e_2\otimes f_2$ has two-factor tensor rank two. Both have two factors: the number of factors and the minimum number of product summands are different counts. Over the real numbers, the singular value decomposition supplies a useful decomposition $C={\sum }_k{\sigma }_k{u}_k{v}_k^T$, with one nonzero term per matrix rank. The two-factor rank equality does not extend to three or more factors by merely choosing one matrix layout.