> 
> If $X$ and $Y$ are each a subspace of $\mathbb{R^n}$, is $X+Y$ also a
>  subspace of $\mathbb{R^n}$, where $X+Y$ is the set of all vectors $x+y$ such that $x\in X$ and $y\in Y$?
> 
> 
> 

I've already showed that for some $z\in X+Y$, if $z$ is the sum of two $x$'s $\in X$ or two $y$'s $\in Y$, or $z=x+y$ then $X+Y$ is closed under linear combinations (i.e. the subsets of $X+Y$ that contain $X$ and $Y$). Now I'm trying to show that if $z \in X+Y$ but $z \not \in X$ and $z \not \in Y$, then $X+Y$ is still closed under linear combinations (i.e. the rest of $X+Y$).

I thought of writing $z \in X+Y$ such that $z \not \in X$ and $z \not \in Y$ as $z = a\_1b\_1+a\_2b\_2$ such that $a\_1,b\_1 \in X$ and $a\_2,b\_2 \in Y$, but I'm not entirely sure how I can use this to show that linear combinations of $z$ are also in $X+Y$, thus showing that all of $X+Y$ is closed under linear combinations and proving the positive of the question asked.