Question on Subspaces

1. We show that $U$ is a non-zero subspace. If $n \geq 2$, then the vector $(1,-1,0,\dots,0) \in U$, so $U$ is non-zero. Let $\underline{x}, \underline{y} \in U$, and $\lambda, \mu \in \mathbb{R}$. We want to show that $\lambda \underline{x} + \mu \underline{y} \in U$. But \[ \sum_{i=1}^n (\lambda \underline{x} + \mu \underline{y})_i = \sum_{i=1}^n (\lambda x_i + \mu y_i) = \lambda \sum_{i=1}^n x_i + \mu \sum_{i=1}^n y_i = \lambda \cdot 0 + \mu \cdot 0 = 0 \] (because $\underline{x}, \underline{y} \in U$), so $\lambda \underline{x} + \mu \underline{y} \in U$, as desired.

2. We show that $V$ is a non-zero subspace. The vector $(1,1,\dots,1) \in V$, so $V$ is non-zero. Let $\underline{x}, \underline{y} \in W$, and $\lambda, \mu \in \mathbb{R}$. We want to show that $\lambda \underline{x} + \mu \underline{y} \in V$. But for all $i,j$, \[ (\lambda \underline{x} + \mu \underline{y})_i = \lambda x_i + \mu y_i = \lambda x_j + \mu y_j= (\lambda \underline{x} + \mu \underline{y})_j \] (because $\underline{x}, \underline{y} \in W$), so $\lambda \underline{x} + \mu \underline{y} \in V$, as desired.

3. We want to show that $U \cap W = \{0\}$. Let $\underline{x} \in U \cap W$. Since $\underline{x} \in W$, all its coordinates are equal, so $\underline{x} = (x, x, \dots, x)$. Since $\underline{x} \in U$, the sum of its coordinates must be $0$, so \[ 0 = \sum_{i=1}^n x_i = \sum_{i=1}^n x = nx, \] but since $n \neq 0$ then $x=0$, so $\underline{x} = (0,0,\dots,0)$, as desired.

4. We want to show $U \oplus W = V$. Let $\underline{x} \in V$, and let $s = \sum_{i=1}^n x_i$. Now consider the vectors $\underline{u} = (x_1 - s/n, x_2 - s/n, \dots, x_n - s/n)$ and $\underline{w} = (s/n, s/n, \dots, s/n)$. Clearly $\underline{x} = \underline{u} + \underline{w}$, and clearly $w \in W$ because its coordinates are all equal. Now, \[ \sum_{i=1}^n u_i = \sum_{i=1}^n (x_i - s/n) = \sum_{i=1}^n x_i - \sum_{i=1}^n s/n = s - n \cdot s/n = s-s = 0, \] so $\underline{u} \in U$, as desired.

5. Let $p \mid n$ a prime, and let $F = \mathbb{F}_p$ the finite field with $p$ elements. Consider the vector $\underline{x} = (1,1,\dots,1) \in F^n$. Clearly $\underline{x} \neq 0$ and $\underline{x} \in W$ because its coordinates are all equal. Now, \[ \sum_{i=1}^n x_i = \sum_{i=1}^n 1 = n = 0 \] because $p \mid n$, so $\underline{x} \in U$ as well, so $U \cap W \neq \{0\}$ and the two subspaces are not in direct sum.

The problem with the previous proof is that, if $n=0$, then the expression $s/n$ does not make sense, and so that argument does not hold for generic fields.