Motivate with $\text{Hom}_{K}(U^{*},V)$ (TBC).
# Definitions
Throughout $V,W$ are $K$-vector spaces.
###### Universal property
A universal property prescribes a vector space up to isomorphism.
A tensor product $V \otimes W$ is a $K$-vector space together with a bilinear map $\beta: V \times W \to V\otimes W;(v,w) \mapsto v\otimes w$ with the following universal property: for any $K$-vector space $S$ and a bilinear map $\tau :V\times W \to S$, there exists a unique map $\bar{\tau}: V \otimes W \to S$ such that $\tau = \bar{\tau} \circ \beta$, i.e. the following diagram commutes:
```tikz
\usepackage{tikz-cd}
\begin{document}
\begin{tikzcd}
V \times W
\arrow[r, "\beta"]
\arrow[dr, "\tau"']
& V \otimes W
\arrow[d, "!\exists \bar{\tau}"] \\
& S
\end{tikzcd}
\end{document}
```
We will we show that the tensor product is well-defined.
###### Construction as quotient space
Let $F(V,W)$ denote the $K$-vector space whose basis is $V \times W$ (the set of $K$-linear combinations of elements of $V \times W$ regarded as indeterminates).
TBC
# Properties
###### Finite-dimensional case (basis)
Claim: if $v_{1},\dots,v_{n}$ is basis for $V$ and $w_{1},\dots,w_{m}$ are basis for $W$ then $\{ (v_{i} \otimes w_{j}): 1 \leq i\leq n, 1\leq j \leq m \}$ is a basis for $V \otimes W$. So $\dim (U \otimes V)=\dim U \dim V$.
Proof:
- Clearly a $K$-generating set, use bilinearity of $\otimes$ to expand $\left( \sum_{i=1}^{n} \lambda_{i} u_{i} \right) \otimes \left( \sum_{j=1}^{m} \mu_{j} v_{j} \right)$.
- To show linear independence use linear map $V \otimes W \to V$ defined by $v\otimes w \mapsto v^*(v)w$ for $v^*\in V^*$.
###### Isomorphism to $\text{Hom}_{K}(V^{*},W)$
Consider the bilinear mapping $\tau: V \times W \to \mathrm{Hom}_{K}(V^{*},W)$ defined by $(\tau(v,w))(v^{*}) = v^{*}(v)w$. The universal property yields a unique $\bar{\tau}: V \otimes W \to \mathrm{Hom}_{K}(V^{*},W)$ so that the following diagram commutes.
```tikz
\usepackage{tikz-cd}
\begin{document}
\begin{tikzcd}
V \times W
\arrow[r, "\beta"]
\arrow[dr, "\tau"']
& V \otimes W
\arrow[d, "!\exists \bar{\tau}"] \\
& \mathrm{Hom}_{K}(V^{*},W)
\end{tikzcd}
\end{document}
```
Let $v_{1},\dots,v_{n}$ be a basis to for $V$ and $w_{1},\dots,w_{m}$ a basis for $U$. To show that $\bar{\tau}$ is an isomorphism, we show that $\{ \bar{\tau}(v_{i} \otimes w_{j}): 1 \leq i\leq n, 1\leq j \leq m \}$ is a linearly independent set. Indeed $\sum_{i,j} \alpha_{ij} \tau(v_{i} \otimes w_{j})$ , $\alpha_{ij}\in K$ yields $0 = \sum_{i,j} \alpha_{ij} v^{*}_{k}(v_{i})w_{j} = \sum_{j} \alpha_{kj} v_{j} $hence all $\alpha_{ij}$ vanish. Since $\dim U \otimes V= nm = \dim \text{Hom}_{K}(U^{*}, V)$, this completes the proof.
TBC:
- $V\otimes W$ is isomorphic to $W \otimes V$
- $V^{*} \otimes W^{*}\cong (V \otimes W)^{*}$
- Distributivity over direct sum. Motivate [[Tensor algebra of vector space]].