For vector spaces $U, V$ there exits a unique (up to isomorphism) vector space, denoted by $U \otimes V$, and a bilinear map $\eta : U \times V \to U \otimes V$ such that for every bilinear map $\xi : U \times V \to W$ there exists a unique linear map $\xi_{\otimes} : U \otimes V \to W$ such that $\xi(u,v) = \xi_{\otimes}(\eta(u,v))$ for all $u \in U, v \in V$.
This vector space $U \otimes V$ is called the tensor product of $U$ and $V$. One way to construct it (just sketched) and the associated bilinear map $\eta : U \times V \to U \otimes V$ is by first choosing bases $\{ u_i \}_{i\in I}$ and $\{ v_j \}_{j \in J}$ of $U$ and $V$ and setting $$ U \otimes V := F( \{ (u_i, v_j) \}_{i \in I, j \in J} ) $$ i.e. the free vector space over $(u_i, v_j)$ for $i \in I, j \in J$. Then we can define $$ \eta(u_i, v_j) := (u_i, v_j) $$ and by extension this gives a bilinear map $\eta : U\times V \to U\otimes V$. For the other property, let $\xi : U \times V \to W$ be some bilinear map and set $w_{ij} := \xi(u_i, v_j)$, then $$ \xi_{\otimes}( (u_i, v_j) ) := u_{ij} $$ defines a linear map with the mentioned property.
Now as I see it, for this universal property to hold, all that is relevant is the image $\eta( U \times V ) \subseteq U \otimes V$, so there are unneccessary elements in $U \otimes V$.
For example consider $U = \mathbb R^2, V = \mathbb R^2$. Then $U \otimes V = \mbox{span}\{ (e_1, e_1), (e_1, e_2), (e_2, e_1), (e_2, e_2)\}$ and with $u = \alpha e_1 + \beta e_2$ and $v = \gamma e_1 + \delta e_2$ we have \begin{align*} \eta( u, v ) & = \alpha \gamma \eta(e_1, e_1) + \alpha \delta \eta(e_1, e_2) + \beta \gamma \eta(e_2, e_1) + \beta \delta \eta(e_2,e_2) \\ & = \alpha \gamma (e_1, e_1) + \alpha \delta (e_1, e_2) + \beta \gamma a(e_2, e_1) + \beta \delta (e_2,e_2) \end{align*} i.e. all $4$-tuples (to speak in coordinates in $U\otimes V$) that could be written as $(\alpha\gamma, \alpha\delta, \beta\gamma, \beta\delta)$ with $\alpha, \beta, \gamma,\delta \in \mathbb R$. For example $(0,1,1,0)$ could not be written that way.
To be more precise we have $$\eta(U\times V) = \{ \lambda_1 (e_1,e_1) + \lambda_2 (e_1, e_2) + \lambda_3 (e_2, e_1) + \lambda_4 (e_2, e_2) : \lambda_1 \lambda_4 - \lambda_2 \lambda_3 = 0 \}.$$ For if $(\alpha\gamma, \alpha\delta, \beta\gamma, \beta\delta) \in \eta(U\times V)$ of course $\alpha\gamma\beta\delta - \alpha\delta\beta\gamma = 0$ (again I wrote coordinate tupels instead of linear combinations of basis elements), for the other direction let $(\lambda_1, \lambda_2,\lambda_3,\lambda_4)$ with $\lambda_1 \lambda_4 = \lambda_2 \lambda_3$ and suppose w.l.o.g. $\lambda_1 \ne 0$, then $\eta( (1, \frac{\lambda_3}{\lambda_1}), (\lambda_1, \lambda_2) )$ gets mapped on $(\lambda_1, \lambda_2, \lambda_3, \lambda_4)$.
Might have something to do with the fact that $\eta(U\times V)$ is not a vector space anymore, but still this seems to be some unsatisfactory artefact of this contruction. So why these unnecessary elements?