Show if $W = \{ (x_{1},x_{2},\cdots,x_{n})\in \mathbb{R}^{n} \mid x_{1}+x_{2}+...+x_{n-1}-x_{n}=0\} $ is a subspace vector.
I have a doubt in the last term ($-x_{n}$) because I tried to take it as if it were a sum like this: $$ W = \{ (x_{1},x_{2},\cdots,x_{n})\in \mathbb{R}^{n} \mid \sum_{i=1}^{n}x_{i}=0$$ After that just prove if the subset:
- Has vector $0$
$$\displaystyle (0,0,\dots,0) \in W \implies\sum^{n}_{i=1} 0 =0 $$ so $ \quad W \neq 0$.
- Is closed under addition.
Take any $(x_{1},\dots,x_{n})$ and $(y_{1},\dots,y_{n})$, so $$ \sum_{i=1}^{n}x_{i}= 0 \quad \text{and}\quad \sum_{i=1}^{n}y_{i}= 0 $$ so the linear combination is $0$
- Is closed under scalar multiplication.
Take any $(x_{1},\dots,x_{n})\in W$ and $\lambda \in \mathbb{R}$, so
$$ \sum_{i=1}^{n}x_{i}= 0 $$
then
$$ \lambda (x_{1},\dots,x_{n})=(\lambda x_{1},\dots,\lambda x_{n}) \in W$$
therefore
$$ \sum_{i=1}^{n}\lambda x_{i}= \lambda \sum_{i=1}^{n} x_{i} = \lambda \cdot 0=0$$
but in all three cases I don't know how to include the last term in my exercise