For, $$U=\{(x_1,...x_n)\in\mathbb{F}^n\vert x_1+...+x_n=0\} \text{ and }V=\{(x_1,...x_n)\in\mathbb{F}^n\vert x_1=...=x_n\},$$ I am trying to prove that $\mathbb{F}^n=U\oplus V$.
I know I need to show that $(1)$ $\mathbb{F}^n=U+ V$ and $(2)$ $V\cap U=\{0\}$. I can easily show $(2)$, but I have no idea how to show $(1)$. Thanks in advance for the help.
Consider $T: \mathbb{F}^n \to U \oplus V$ given by $(x_1, \ldots, x_n) \mapsto (x_1 - \bar x, \ldots, x_n - \bar x, \bar x, \ldots, \bar x)$ where $\bar x = n^{-1}(x_1 + \ldots + x_n)$, and show that this is an isomorphism. This works if $\mathbb{F}$ has characteristic $0$ or the characteristic does not divide $n$. The given proposition is false if the characteristic does divide $n$: take for example $\mathbb{F} = \mathbb{Z}/2\mathbb{Z}$ and $n=2$. Then $U = V = \operatorname{span}\{(1, 1)\} \neq \mathbb{F}^2$.
Edit: The basic idea is that every $x \in \mathbb{F}^n$ can be written as a sum $u + v$ where $v$ is the vector having as its components the average of the components of $x$, and $u$ is the component-wise deviation from the average. Then it is easy to show that $u \in U$ and $v \in V$.