If not specifically indicated, the linear spaces discussed below are all finite-dimensional.

Definition of Tensor Product

Universal Property

Let $V, W, Z$ be linear spaces over the field $k$. $\mathcal{L}(V,W;Z)$ denotes the linear space consisting of all bilinear maps from $V \times W$ to $Z$. If there exists a linear space $V \otimes W$ and a bilinear map $\varphi \in \mathcal{L}(V,W;V \otimes W)$ that satisfies the following universal property: for any $g \in \mathcal{L}(V,W;Z)$, there exists a unique linear map $h: V \otimes W \to Z$ such that $g = h \circ \varphi$, i.e., the following diagram commutes

Then this linear space $V \otimes W$ is called the tensor product of $V$ and $W$.

Construction

Let the freely generated linear space of $V \times W$ be

$$\mathrm{Free}_{\mathsf{Vect}_{k}}(V\times W)=\left\{\sum_{i,j}a_{ij}(v_i,w_j) \mid a_{ij}\in k, v_i\in V, w_j\in W \right\}.$$

Define two subspaces of $\mathrm{Free}_{\mathsf{Vect}_{k}}(V\times W)$:

$$S_1=\left\{(\lambda v_1+v_2,w)-\lambda(v_1,w)-(v_2,w) \mid \lambda \in k, v_1, v_2 \in V, w \in W \right\},$$ $$S_2=\left\{(v, \lambda w_1+w_2)-\lambda(v, w_1)-(v, w_2) \mid \lambda \in k, v \in V, w_1, w_2 \in W \right\}.$$

The commutative diagram is shown below:

We define

$$V\otimes W:=\mathrm{Free}_{\mathsf{Vect}_{k}}(V\times W)/\left(S_1\oplus S_2\right)$$

and $f=\pi\circ i$. Since

$$\begin{aligned} f(\lambda v_1+v_2,w)&=\pi(\lambda v_1+v_2,w)\\ &=(\lambda v_1+v_2,w)-\left((\lambda v_1+v_2,w)-\lambda(v_1,w)-(v_2,w)\right)\\ &=\lambda(v_1,w)+(v_2,w)\\ &=\lambda f(v_1,w)+f(v_2,w), \end{aligned}$$ $$\begin{aligned} f( v,\lambda w_1+w_2)&=\pi( v,\lambda w_1+w_2)\\ &=(v,\lambda w_1+w_2)-\left((v,\lambda w_1+w_2)-\lambda(v,w_1)-(v,w_2)\right)\\ &=\lambda(v,w_1)+(v,w_2)\\ &=\lambda f(v,w_1)+f(v,w_2), \end{aligned}$$

we know that $f$ is indeed a bilinear map. According to the universal properties of freely generated linear spaces and quotient spaces, we have $g=\varphi \circ i$ and $\varphi = h \circ \pi$, which implies $g = h \circ f$. Finally, we only need to show the uniqueness of $h$. Suppose there is another linear map $h’: V \otimes W \to Z$ such that $g = h’ \circ f$. Noting that $g = (h’ \circ \pi) \circ i = \varphi \circ i$, the universal property of freely generated linear spaces tells us that $h’ \circ \pi = \varphi$. Using the universal property of quotient spaces again, we know that $h’ = h$, thus proving the uniqueness of $h$. In summary, we have shown that the $V \otimes W$ constructed in this way satisfies the universal property of the tensor product.

The tensor product map $f$ constructed here is customarily denoted by $\otimes$, i.e., $f(v, w) = v \otimes w$.

Properties of Tensor Product

Representable Functor

Define the functor

$$L(V, W; -): \mathsf{Vect}_{k} \to \mathsf{Vect}_{k}$$

whose object mapping is

$$\begin{aligned} L(V, W; -): \mathrm{Ob}(\mathsf{Vect}_{k}) &\longrightarrow \mathrm{Ob}(\mathsf{Vect}_{k})\\ Z &\longmapsto L(V, W; Z), \end{aligned}$$

and whose morphism mapping is

$$\begin{aligned} L(V, W; -): \mathrm{Hom}_{\mathsf{Vect}_{k}}(Z_1, Z_2) &\longrightarrow \mathrm{Hom}_{\mathsf{Vect}_{k}}(L(V, W; Z_1), L(V, W; Z_2))\\ f &\longmapsto \left(g \longmapsto f \circ g\right). \end{aligned}$$

$L(V, W; -)$ is a representable functor. Specifically, $\theta: L(V, W; -) \to \mathrm{Hom}_{\mathsf{Vect}_{k}}\left(V \otimes W, -\right)$ is a natural isomorphism, defined as

$$\begin{aligned} \theta_Z: L(V, W; Z) &\stackrel{\approx}\longrightarrow \mathrm{Hom}_{\mathsf{Vect}_{k}}\left(V \otimes W, Z\right)\\ g &\longmapsto \left(v \otimes w \stackrel{h}\longmapsto g(v, w)\right). \end{aligned}$$

Adjoint Functor

Let $V$ be a linear space. Here we will abbreviate the functor $\mathrm{Hom}_{\mathsf{Vect}_{k}}\left(V, -\right)$ as $\mathrm{Hom}\left(V, -\right)$. This is shown in the following diagram

$(-)\otimes V$ is the left adjoint of $\mathrm{Hom}_{\mathsf{Vect}_{k}}\left(V,-\right)$. Specifically,

$$\eta: \mathrm{Hom}\left((-)\otimes V,-\right)\to \mathrm{Hom}\left(-,\mathrm{Hom}\left(V,-\right)\right)$$

is a natural isomorphism, defined as

$$\begin{aligned} \eta_{U,W}:\mathrm{Hom}\left(U\otimes V,W\right)&\stackrel{\approx}\longrightarrow\mathrm{Hom}\left(U,\mathrm{Hom}\left(V,W\right)\right)\\ h& \longmapsto\left(u\longmapsto h\left(u\otimes -\right)\right) . \end{aligned}$$

Dimension of Tensor Product

Lemma. Let $V$ and $W$ be vector spaces, and let $v_1, \cdots, v_n \in V$ be linearly independent, $w_1, \cdots, w_m \in W$. If

$$\sum_i v_i \otimes w_i = 0,$$

then $w_1 = \cdots = w_m = 0$.

Proof.
Define the standard pairing or evaluation function as follows:

$$\begin{aligned} \langle -,-\rangle: V \times V^* &\longrightarrow k \\ (v_1, v_2^*) &\longmapsto v_2^*(v_1). \end{aligned}$$

Given any $\alpha^* \in V^*, \beta^* \in W^*$, we can verify that the following defined map

$$\begin{aligned} \langle \cdot, \alpha^* \rangle \langle \cdot, \beta^* \rangle: V \times W &\longrightarrow k \\ (v, w) &\longmapsto \langle v, \alpha^* \rangle \langle w, \beta^* \rangle = \alpha^*(v) \beta^*(w) \end{aligned}$$

is a bilinear map. According to the universal property of the tensor product, there exists a unique bilinear map $h: V \otimes W \to k$ such that $h(v \otimes w) = \alpha^*(v) \beta^*(w)$. Note that

$$\begin{aligned} 0 &= h \left( \sum_i v_i \otimes w_i \right) \\ &= \sum_i h(v_i \otimes w_i) \\ &= \sum_i \alpha^*(v_i) \beta^*(w_i) \end{aligned}$$

holds for any $\alpha^* \in V^*, \beta^* \in W^*$. For $k = 1, \cdots, n$, let $\alpha^*(vi) = \delta{i, k}$ respectively, then we can obtain

$$\begin{aligned} \beta^*(w_k) = 0, \quad \forall \beta^* \in W^* \implies w_k = 0. \end{aligned}$$

This proves the lemma. $\square$

Using the lemma, we can find the dimension of the tensor product. In fact, let $B_V = \{\varepsilon_1, \dots, \varepsilon_n\}$ and $B_W = \{\eta_1, \dots, \eta_m\}$ be bases of the vector spaces $V$ and $W$ respectively. We can prove that $B_{V \otimes W} = {\varepsilon_i \otimes \eta_j \mid i = 1, \cdots, n, j = 1, \cdots, m}$ is a basis of $V \otimes W$. On one hand, if

$$0 = \sum_{i=1}^n \sum_{j=1}^m a_{ij} \varepsilon_i \otimes \eta_j = \sum_{i=1}^n \varepsilon_i \otimes \left( \sum_{j=1}^m a_{ij} \eta_j \right),$$

the lemma shows that

$$\sum_{j=1}^m a_{1j} \eta_j = \sum_{j=1}^m a_{2j} \eta_j = \cdots = \sum_{j=1}^m a_{nj} \eta_j = 0.$$

Since $\{\eta_1, \dots, \eta_m\}$ are linearly independent, $a_{ij} = 0$. Therefore, $B_{V \otimes W}$ is linearly independent. On the other hand, for any $\sum_k \lambda_k v_k \otimes w_k \in V \otimes W$, let $v_k = \sum_i a_{ki} \varepsilon_i$, $w_j = \sum_j b_{kj} \eta_j$, then

$$\begin{aligned} \sum_k \lambda_k v_k \otimes w_k &= \sum_k \lambda_k \left( \sum_i a_{ki} \varepsilon_i \right) \otimes \left( \sum_j b_{kj} \eta_j \right) \\ &= \sum_i \sum_j \sum_k \lambda_k a_{ki} b_{kj} \varepsilon_i \otimes \eta_j. \end{aligned}$$

This shows that $B_{V \otimes W}$ spans $V \otimes W$. Therefore, $B_{V \otimes W}$ is a basis of $V \otimes W$. Thus, we obtain the dimension of the tensor product

$$\dim(V \otimes W) = \dim(V) \dim(W).$$