tensor product of subspaces of vector spaces

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 $\{e^{\prime}_{j}\}_{j\in P}$ 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 $\{e^{\prime}_{j}\}_{j\in P^{\prime}}$ 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

 $q=\sum_{i\in I,j\in P}\alpha_{i,j}\,e_{i}\otimes e^{\prime}_{j}\ +\ \sum_{i\in I% ^{\prime},j\in P}\beta_{i,j}\,e_{i}\otimes e^{\prime}_{j}\ +$
 $+\ \sum_{i\in I,j\in P^{\prime}}\delta_{i,j}\,e_{i}\otimes e^{\prime}_{j}\ +\ % \sum_{i\in I^{\prime},j\in P^{\prime}}\gamma_{i,j}\,e_{i}\otimes e^{\prime}_{j}.$

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 $i\neq t$. 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

 $q=\sum_{i\in I,j\in P}\alpha_{i,j}\,e_{i}\otimes e^{\prime}_{j}\in A\otimes B,$

which completes the proof. $\square$

Title tensor product of subspaces of vector spaces TensorProductOfSubspacesOfVectorSpaces 2013-03-22 18:49:16 2013-03-22 18:49:16 joking (16130) joking (16130) 4 joking (16130) Theorem msc 15A69