Universal Properties of Tensor Products 📂Linear Algebra

Universal Properties of Tensor Products

Buildup¹

Given a finite-dimensional vector space $V_{1}, \dots, V_{r}$ . If $n_{i} = \dim V_{i}$ , and we select a basis for each vector space, we obtain the following coordinate vector as a bijective function $f_{i}$ .

$\begin{align*} f _{i}: & V_{i} \to \mathbb{C}^{n_{i}} \\ & v_{i} \mapsto (a_{i1}, \dots, a_{i n_{i}}) \end{align*}$

From this, the following multilinear transformation $f$ is naturally defined.

$\begin{align*} f : V_{1} \times \cdots \times V_{r} &\to V_{1} \otimes \cdots \otimes V_{r} \\ (v_{1}, \dots, v_{r}) &\mapsto v_{1} \otimes \cdots \otimes v_{r} = \sum_{(j_{1}, \dots, j_{r})}\left( \prod_{i=1}^{r} a_{ij_{i}} \right) e_{j_{1}} \otimes \cdots \otimes e_{j_{r}} \end{align*}$

Here $V_{1} \otimes V_{2}$ is the tensor product of vector spaces, $v_{1} \otimes v_{2}$ is a product vector.

Theorem

Let’s say a multilinear transformation $\phi$ is given for vector space $V_{1}, \dots, V_{r}, W$ .

$\phi : V_{1} \times \cdots \times V_{r} \to W$

Then there exists a unique linear transformation $\psi : V_{1} \otimes \cdots \otimes V_{r} \to W$ that satisfies the following.

$\psi (v_{1} \otimes \cdots \otimes v_{r}) = \phi (v_{1}, \dots, v_{r}),\quad \forall v_{i} \in V_{i},\quad \forall i$

Proof

Let’s call the basis of $V_{i}$ as $\left\{ e_{j_{i}} \right\}_{1 \le j_{i} \le n_{i}}$ . Let’s define $\psi : V_{1} \otimes \cdots \otimes V_{r} \to W$ as the following mapping.

$\psi \left( \sum\limits_{1 \le j_{i} \le n_{i}} a_{j_{1},\dots,j_{r}} e_{j_{1}} \otimes \cdots \otimes e_{j_{r}} \right) = \sum\limits_{1 \le j_{i} \le n_{i}} a_{j_{1},\dots,j_{r}} \phi ( e_{j_{1}}, \dots, e_{j_{r}})$

Since $\phi$ is multilinear, the following holds for $v_{i} = \sum_{1 \le j_{i} \le n_{i}} v_{i}(j_{i})e_{j_{i}}$ .

$\begin{align*} \phi (v_{1}, \dots, v_{r}) &= \sum\limits_{1 \le j_{i} \le n_{i}} \left( \prod_{i=1}^{r} v_{i}(j_{i}) \right) \phi (e_{j_{1}}, \dots, e_{j_{r}}) \\ &= \psi\left( \sum\limits_{1 \le j_{i} \le n_{i}} \left( \prod_{i=1}^{r} v_{i}(j_{i}) \right) e_{j_{1}} \otimes \cdots \otimes e_{j_{r}} \right) \\ &= \psi\left( \left( \sum\limits_{1 \le j_{1} \le n_{1}} v_{1}(j_{1})e_{j_{1}} \right) \otimes \cdots \otimes \left( \sum\limits_{1 \le j_{r} \le n_{1}} v_{1}(j_{r}) e_{j_{r}} \right)\right) \\ &= \psi\left( v_{1} \otimes \cdots \otimes v_{r} \right) \\ \end{align*}$

The third equality is due to the definition of product vector. Assuming there exists another $\psi^{\prime}$ that satisfies this, it is trivially unique since $\psi - \psi^{\prime} = 0$ .

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