Let $\varphi: \mathbb{F}_{2^m} \longrightarrow \mathbb{F}_2^m,\;\; \sum_{i = 1}^mb_i\xi_i \mapsto (b_i)_i$ with $(\xi_i)_i$ as a basis of $\mathbb{F}_{2^m}$ over $\mathbb{F}_2$. Now let $x, y \in \mathbb{F}_{2^m}^n$. My question is: Does $x^2 = y^2 = 0$ and $xy = \sum_{i = 1}^n x_iy_i = 0$ imply that $\sum_{i = 1}^n \varphi(x_i)\varphi(y_i) = 0$?
I know that $\sum_{i = 1}^n x_i = \sum_{i = 1}^n y_i = 0$ resp. $\sum_{i = 1}^n \varphi(x_i) = \sum_{i = 1}^n \varphi(y_i) = 0$ and $\sum_{i = 1}^n\varphi(x_iy_i) = 0$, because of characteristic 2 and linearity and injectivity of $\varphi$. My problem is, I don't understand the connection between the product in the field and the euclidian product, i.e. between $\varphi(x_iy_i)$ and $\varphi(x_i)\varphi(y_i)$. I tested a couple of cases in $\mathbb{F}_4$ but I didn't find a counter example, so my guess is that the answer to my question is yes.