# Tensor identity: $(E_1\otimes F) \cap (E_2 \otimes F) = (E_1\cap E_2)\otimes F.$

I'm trying to understand a proof of the following statement:

Let $E_1, E_2$ be two subspaces of a vector space $E$. Then if $F$ is any other vector space, $(E_1\otimes F) \cap (E_2 \otimes F) = (E_1\cap E_2)\otimes F.$

The book that i'm using starts the proof by saying:

"Clearly, $(E_1\cap E_2)\otimes F \subset (E_1\otimes F) \cap (E_2 \otimes F)$ "...

and well, for me it's not that much clear. I've tried to justify it by myself, but i think that i'm going in the wrong direction, since it is not that much simple. What i've tried so far was to consider a basis to $E_1\cap E_2$ and to extend it to a basis for $E_1$ and $E_2$, respectively. And then writing $E_1\cap E_2 \otimes F$ as a direct sum of one-dimensional vector spaces i've tried to use some of the tensor properties to manage to show that any element of this vector space can be written by elements of the other one, but i was unsucessful.

Can someone give a good explanation? Thanks in advance. The other inclusion i could understand pretty well!

• This is even true for an infinite collection of subspaces of $E$. – D_S Nov 12 '17 at 14:13

You were on the right track, let $B$ be a basis for $E_1 \cap E_2$ and $B_1$, $B_2$ its extensions to bases for $E_1$ and $E_2$, respectively.

Obviously, $B \subseteq B_1, B_2$.

Any vector in $(E_1\cap E_2)\otimes F$ can be written as $$\sum_{i=1}^k \lambda_i (b_i \otimes f_i)$$

for some scalars $\lambda_i$, and vectors $b_i \in B$ and $f_i \in F$. Since $B \subseteq B_1$, $b_i \otimes f_i$ is certainly an element of $E_1 \otimes F$. Then also $\sum_{i=1}^k \lambda_i (b_i \otimes f_i) \in E_1 \otimes F$.

Similarly, since $B \subseteq B_2$, we have $\sum_{i=1}^k \lambda_i (b_i \otimes f_i) \in E_2 \otimes F$.

Hence $$\sum_{i=1}^k \lambda_i (b_i \otimes f_i) \in (E_1 \otimes F) \cap (E_2 \otimes F)$$

which proves the inclusion.