Editing Tensor product (section)

=== 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.{{sfn|Trèves|2006|pp=403-404}}

{{math theorem|math_statement=
Let {{tmath|1= X, Y }}, and <math>Z</math> be complex vector spaces and let <math>T : X \times Y \to Z</math> be a bilinear map. Then <math>(Z, T)</math> is a tensor product of <math>X</math> and <math>Y</math> if and only if{{sfn|Trèves|2006|pp=403-404}} the image of <math>T</math> spans all of <math>Z</math> (that is, {{tmath|1= \operatorname{span} \; T(X \times Y) = Z }}), and also <math>X</math> and <math>Y</math> are {{em|<math>T</math>-linearly disjoint}}, which by definition means that for all positive integers <math>n</math> and all elements <math>x_1, \ldots, x_n \in X</math> and <math>y_1, \ldots, y_n \in Y</math> such that {{tmath|1= \sum_{i=1}^n T\left(x_i, y_i\right) = 0 }}, 
# if all <math>x_1, \ldots, x_n</math> are [[linearly independent]] then all <math>y_i</math> are {{tmath|1= 0 }}, and 
# if all <math>y_1, \ldots, y_n</math> are linearly independent then all <math>x_i</math> are {{tmath|1= 0 }}.

Equivalently, <math>X</math> and <math>Y</math> are <math>T</math>-linearly disjoint if and only if for all linearly independent sequences <math>x_1, \ldots, x_m</math> in <math>X</math> and all linearly independent sequences <math>y_1, \ldots, y_n</math> in {{tmath|1= Y }}, the vectors <math>\left\{T\left(x_i, y_j\right) : 1 \leq i \leq m, 1 \leq j \leq n\right\}</math> are linearly independent.
}}

For example, it follows immediately that if {{tmath|1=X=\C^m}} and {{tmath|1=Y=\C^n}}, where <math>m</math> and <math>n</math> are positive integers, then one may set <math>Z = \Complex^{mn}</math> and define the bilinear map as 
<math display=block>\begin{align}
T : \Complex^m \times \Complex^n &\to \Complex^{mn}\\
(x, y) = ((x_1, \ldots, x_m), (y_1, \ldots, y_n)) &\mapsto (x_i y_j)_{\stackrel{i=1,\ldots,m}{j=1,\ldots,n}}\end{align}</math>
to form the tensor product of <math>X </math> and {{tmath|1= Y }}.{{sfn|Trèves|2006|pp=407}} Often, this map <math>T</math> is denoted by <math>\,\otimes\,</math> so that <math>x \otimes y = T(x, y).</math> 

As another example, suppose that <math>\Complex^S</math> is the vector space of all complex-valued functions on a set <math>S</math> with addition and scalar multiplication defined pointwise (meaning that <math>f + g</math> is the map <math>s \mapsto f(s) + g(s)</math> and <math>c f</math> is the map {{tmath|1= s \mapsto c f(s) }}). Let <math>S</math> and <math>T</math> be any sets and for any <math>f \in \Complex^S</math> and {{tmath|1= g \in \Complex^T }}, let <math>f \otimes g \in \Complex^{S \times T}</math> denote the function defined by {{tmath|1= (s, t) \mapsto f(s) g(t) }}. 
If <math>X \subseteq \Complex^S</math> and <math>Y \subseteq \Complex^T</math> are vector subspaces then the vector subspace <math>Z := \operatorname{span} \left\{f \otimes g : f \in X, g \in Y\right\}</math> of <math>\Complex^{S \times T}</math> together with the bilinear map: 
<math display=block>\begin{alignat}{4}
 \;&& X \times Y &&\;\to    \;& Z \\[0.3ex]
     && (f, g) &&\;\mapsto\;& f \otimes g \\
\end{alignat}</math>
form a tensor product of <math>X</math> and {{tmath|1= Y }}.{{sfn|Trèves|2006|pp=407}}