# Partitions in the tensor product

Let $$V$$ and $$W$$ be two real vector spaces, $$v \in V$$ and $$w \in W$$. I'm getting some trouble in the following problem:

Let $$u_1$$ and $$u_2$$ be two elements in the tensor $$V \otimes W$$ such that $$u_1 + u_2 = v \otimes w$$. I'm asking myself if this implies that

• $$u_1 = v_1 \otimes w$$ and $$u_2 = v_2 \otimes w$$ with $$v_1 + v_2 = v$$, or
• $$u_1 = v \otimes w_1$$ and $$u_2 = v \otimes w_2$$ with $$w_1 + w_2 = w$$.

This seems reasonable to happen, but I'm not so used to work with tensor products. I'd like some reference to study this kind of situation if possible.

• Notice your question is equivalent to "is every tensor a simple tensor?" (i.e. of the form $v \otimes w$) to which the answer is no. Assuming that every tensor is simple is the number 1 mistake made when first working with them. – RghtHndSd Jul 1 at 15:00

No you could have $$v\otimes w=(v\otimes w -v_1\otimes w_1-v_2\otimes w_2)+(v_1\otimes w_1+v_2\otimes w_2)$$. This does not in general have either of the forms you suggested.
• If you note the two tensors you took and sum $v - v_1 - v_2 + v_1 + v_2 = v$ and $w - w_1 - w_2 + w_1 + w_2$. This is a fenomenum that always happen? – user 242964 Jul 1 at 15:33
• No you could have $v\otimes w=(v\otimes w -v_1\otimes 2w_1-v_2\otimes w_2)+(2v_1\otimes w_1+v_2\otimes w_2)$. – tkf Jul 1 at 15:46