Tensor product
Linearly disjoint

Like the universal property above, the following characterization may also be used to determine whether or not a given vector space and given bilinear map form a tensor product.

Theorem — Let X, Y {\displaystyle X,Y}, and Z {\displaystyle Z} be complex vector spaces and let T: X × Y → Z {\displaystyle T:X\times Y\to Z} be a bilinear map. Then (Z, T) {\displaystyle (Z,T)} is a tensor product of X {\displaystyle X} and Y {\displaystyle Y} if and only if the image of T {\displaystyle T} spans all of Z {\displaystyle Z} (that is, span T (X × Y) = Z {\displaystyle \operatorname {span} \;T(X\times Y)=Z}), and also X {\displaystyle X} and Y {\displaystyle Y} are T {\displaystyle T} -linearly disjoint, which by definition means that for all positive integers n {\displaystyle n} and all elements x 1, …, x n ∈ X {\displaystyle x_{1},\ldots,x_{n}\in X} and y 1, …, y n ∈ Y {\displaystyle y_{1},\ldots,y_{n}\in Y} such that ∑ i = 1 n T (x i, y i) = 0 {\displaystyle \sum _{i=1}^{n}T\left(x_{i},y_{i}\right)=0},

Equivalently, X {\displaystyle X} and Y {\displaystyle Y} are T {\displaystyle T} -linearly disjoint if and only if for all linearly independent sequences x 1, …, x m {\displaystyle x_{1},\ldots,x_{m}} in X {\displaystyle X} and all linearly independent sequences y 1, …, y n {\displaystyle y_{1},\ldots,y_{n}} in Y {\displaystyle Y}, the vectors { T (x i, y j): 1 ≤ i ≤ m, 1 ≤ j ≤ n } {\displaystyle \left\{T\left(x_{i},y_{j}\right):1\leq i\leq m,1\leq j\leq n\right\}} are linearly independent.

For example, it follows immediately that if m {\displaystyle m} and n {\displaystyle n} are positive integers then Z:= C m n {\displaystyle Z:=\mathbb {C} ^{mn}} and the bilinear map T: C m × C n → C m n {\displaystyle T:\mathbb {C} ^{m}\times \mathbb {C} ^{n}\to \mathbb {C} ^{mn}} defined by sending (x, y) = ((x 1, …, x m), (y 1, …, y n)) {\displaystyle (x,y)=\left(\left(x_{1},\ldots,x_{m}\right),\left(y_{1},\ldots,y_{n}\right)\right)} to (x i y j) j = 1, …, n i = 1, …, m {\displaystyle \left(x_{i}y_{j}\right)_{\stackrel {i=1,\ldots,m}{j=1,\ldots,n}}} form a tensor product of X:= C m {\displaystyle X:=\mathbb {C} ^{m}} and Y:= C n {\displaystyle Y:=\mathbb {C} ^{n}}. Often, this map T {\displaystyle T} will be denoted by ⊗ {\displaystyle \,\otimes \,} so that x ⊗ y:= T (x, y) {\displaystyle x\otimes y\;:=\;T(x,y)} denotes this bilinear map's value at (x, y) ∈ X × Y {\displaystyle (x,y)\in X\times Y}.

As another example, suppose that C S {\displaystyle \mathbb {C} ^{S}} is the vector space of all complex-valued functions on a set S {\displaystyle S} with addition and scalar multiplication defined pointwise (meaning that f + g {\displaystyle f+g} is the map s ↦ f (s) + g (s) {\displaystyle s\mapsto f(s)+g(s)} and c f {\displaystyle cf} is the map s ↦ c f (s) {\displaystyle s\mapsto cf(s)}). Let S {\displaystyle S} and T {\displaystyle T} be any sets and for any f ∈ C S {\displaystyle f\in \mathbb {C} ^{S}} and g ∈ C T {\displaystyle g\in \mathbb {C} ^{T}}, let f ⊗ g ∈ C S × T {\displaystyle f\otimes g\in \mathbb {C} ^{S\times T}} denote the function defined by (s, t) ↦ f (s) g (t) {\displaystyle (s,t)\mapsto f(s)g(t)}. If X ⊆ C S {\displaystyle X\subseteq \mathbb {C} ^{S}} and Y ⊆ C T {\displaystyle Y\subseteq \mathbb {C} ^{T}} are vector subspaces then the vector subspace Z:= span { f ⊗ g: f ∈ X, g ∈ Y } {\displaystyle Z:=\operatorname {span} \left\{f\otimes g:f\in X,g\in Y\right\}} of C S × T {\displaystyle \mathbb {C} ^{S\times T}} together with the bilinear map: X × Y → Z (f, g) ↦ f ⊗ g {\displaystyle {\begin{alignedat}{4}\;&&X\times Y&&\;\to \;&Z\\&&(f,g)&&\;\mapsto \;&f\otimes g\\\end{alignedat}}} form a tensor product of X {\displaystyle X} and Y {\displaystyle Y}.
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