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# Tensor product

Suppose we are given two vector spaces $V$ and $W$ over field $R$ (or modules over ring $R$).

First, the product $V \times W$ is the set consisting of pairs $(v, w)$ with $v\in V$ and $w\in W$.

Second, we define the free vector space $< V \times W >$ of this space as the vector space of all formal linear combinations of the elements of $V \times W$. In other words, $V \times W$ is its basis. Observe that the relations between the elements of $V,W,V \times W$ are lost in the new vector space.

Third, the tensor product $V \otimes W$ of $V$ and $W$ is defined as a certain quotient of $< V \times W >$.

We consider the subspace $Z$ of $< V \times W >$ generated by the following elements: \begin{align} (v_1, w) + (v_2, w) - (v_1 + v_2, w),\\ (v, w_1) + (v, w_2) - (v, w_1 + w_2),\\ c \cdot (v, w) - (cv, w), \\ c \cdot (v, w) - (v, cw), \end{align} where $$v, v_1,v_2\in V, w, w_1,w_2\in W, c\in R.$$ Note: To simplify notation, we drop the coefficient if it is equal to $1$, i.e., $(v, w)$ stands for $1 \cdot (v, w) \in < V \times W >$.

The tensor product is defined as the quotient vector space $$V \otimes W := < V \times W > / Z.$$

So, the tensor product of two vectors $v$ and $w$ is the equivalence class $(v,w) + Z$ of $(v,w) \in V⊗W$. It is denoted $v⊗w$. These are called elementary tensors.

Note: not all of the elements of the tensor product have the form $v \otimes w$ but rather linear combinations of these. So, if $B_V,B_W$ are bases of $V,W$, then $\{v \otimes w: v\in B_V,w\in B_W\}$ is a basis of $V \otimes W$.

It follows that these equations hold in $V⊗W$: \begin{align} (v_1 + v_2) \otimes w &= v_1 \otimes w + v_2 \otimes w;\\ v \otimes (w_1 + w_2) &= v \otimes w_1 + v \otimes w_2;\\ cv \otimes w &= v \otimes cw = c(v \otimes w). \end{align}

Given two linear operators $$S : V → X,T : W → Y,$$ the tensor product of the two linear operators $S$ and $T$ is a linear operator $$S\otimes T:V\otimes W\rightarrow X\otimes Y$$ defined by $$(S\otimes T)(v\otimes w)=S(v)\otimes T(w).$$