tensor product of subspaces of vector spaces

Proposition. Let $V,W$ be vector spaces over a field $k$. Moreover let $A\subseteq V$, $B\subseteq W$ be subspaces. Then $V\otimes B\cap A\otimes W=A\otimes B$.

Proof. The inclusion ,,$\supseteq$” is obvious. We will show the inclusion ,,$\subseteq$”.

Let ${\{e_i\}}_{i\in I}$ and ${\{e_j^{\prime }\}}_{j\in P}$ be bases of $A$ and $B$ respectively. Moreover let ${\{e_i\}}_{i\in I^{\prime }}$ be a completion of given basis of $A$ to the basis of $V$, i.e. ${\{e_i\}}_{i\in I\cup I^{\prime }}$ is a basis of $V$. Analogously let ${\{e_j^{\prime }\}}_{j\in P^{\prime }}$ be a completion of a basis of $B$ to the basis of $W$. Then each element $q\in V\otimes W$ can be uniquely written in a form

$$q=\sum _{i\in I,j\in P}{\alpha }_{i,j}{e}_i\otimes e_j^{\prime }+\sum _{i\in I^{\prime },j\in P}{\beta }_{i,j}{e}_i\otimes e_j^{\prime }+$$

$$+\sum _{i\in I,j\in P^{\prime }}{\delta }_{i,j}{e}_i\otimes e_j^{\prime }+\sum _{i\in I^{\prime },j\in P^{\prime }}{\gamma }_{i,j}{e}_i\otimes e_j^{\prime }.$$

Assume that $q\in V\otimes B\cap A\otimes W.$ Let $i\in I^{\prime }$ and $j\in P^{\prime }$. Consider the following linear map: $f_i:V\to k$ such that $f_i(e_t)=1$ if $i=t$ and $f_i(e_t)=0$ if $i\ne t$. Analogously we define $g_j:W\to k$. Then we combine these two mappings into one, i.e.

$$f_i\otimes g_j:V\otimes W\to k;$$

$$(f_i\otimes g_j)(v\otimes w)=f_i(v)g_j(w).$$

Furthermore we have

$$(f_i\otimes g_j)(q)={\gamma }_{i,j}.$$

Note that since $q\in V\otimes B$, then $(f_i\otimes g_j)(q)=0$ and thus

$${\gamma }_{i,j}=0.$$

Similarly we obtain that all ${\beta }_{i,j}$ and ${\delta }_{i,j}$ are equal to $0$. Thus

$$q=\sum _{i\in I,j\in P}{\alpha }_{i,j}{e}_i\otimes e_j^{\prime }\in A\otimes B,$$

which completes the proof. $\mathrm{\square }$