Hi I'm having trouble with this homework question:
"Let $F^2 = {0, 1}$ denote the field with 2 elements 0 and 1. Let $V$ be a vector space over $F^2$. Show that every non-empty subset $W$ of $V$ which is closed under addition is a subspace of $V$."
So I know I need to show that it's closed under vector addition (which in the question it is), that it contains the zero vector and that it's closed under scalar multiplication. I was going to argue along the lines of since $V$ is a vector space and $W$ is a subset of $V$ both scalar multiplication and existence of the zero vector are satisfied because if they hold in the bigger thing $V$ then they must hold in the smaller thing $W$. I think this is flawed though because then I don't see why they'd give me that is closed under vector addition as I could argue the same way for that.
I'll show you here why having the $\vec{0}$ element is true because we're using $F^2$:
If we were trying to use $\mathbb{N}$ as a subset of the vector space $\mathbb{R^1}$ - it is closed under addition but surely not a vector space!
Here it holds true because for every vector $\vec{w} \in W$ :
$$ \vec{w} + \vec{w} = (1+1)\cdot \vec{w} = 0 \cdot \vec{w} = \vec{0} $$ and because $W$ is closed under addition, we get that $\vec{0} \in W$ and for each $w\in W$ , $-w = w \in W$ so you have the $\vec{0} $ element and for each vector you have the negative of it in $W$. Also,
$$ 1 \cdot \vec{w} = \vec{w} \in W$$ $$ 0 \cdot \vec{w} = \vec{0} \in W$$
Now you have all of the properties to call it a vector space.
QED