I try to prove the following result
Let $E,F$ be vector spaces. If $a \otimes b \neq 0 $ for $a \in E$ and $b \in F$, then $a \otimes b = a' \otimes b'$ iff $a' = \lambda a$ and $ b' = \lambda^{-1}b$.
The notion of tensor product is very new for me then I don't understand very well thess objects.
Thanks for all hints.
One implication is clear: obviously we have $ \lambda a \otimes \lambda^{-1} b = a \otimes b $ for $ \lambda \neq 0 $. To obtain the other implication, recall that if $ \{ e_1, e_2, \ldots, e_m \} $ is a basis of $ E $ and $ \{ f_1, f_2, \ldots, f_n \} $ is a basis of $ F $, the tensor products form a basis of $ E \otimes F $. Let
$$ a = \sum_{k=1}^m a_k e_k, \, b = \sum_{k=1}^n b_k f_k $$
and likewise for $ a', b' $. Then, expanding $ a \otimes b $ and $ a' \otimes b' $ in the canonical basis and equating coefficients of $ e_i \otimes f_j $ gives
$$ a_i b_j = a'_i b'_j $$
for all $ i, j $. Since $ a \otimes b \neq 0 $ implies that $ b \neq 0 $, there is some $ b_k \neq 0 $. Using $ j = k $ then gives that $ a_i = \lambda a'_i $ for a constant $ \lambda $ for all $ i $, which implies $ a = \lambda a' $. By symmetry, we also have $ b = \lambda_2 b' $ for some $ \lambda_2 $. But then, $ a' \otimes b' = \lambda \lambda_2 ( a \otimes b) = a \otimes b $ implies $ \lambda_2 = \lambda^{-1} $, since $ a \otimes b \neq 0 $.