Prove this is a subspace Let $ W_1, W_2$ be subspace of a Vector Space $V$.
Denote $W_1+W_2$ to be the following set
$$W_1+W_2=\left\{u+v, u\in W_1, v\in W_2\right\}$$
Prove that this is a subspace.
I can prove that the set is non emprty (i.e that it houses the zero vector).
pf:  Since $W_1 , W_2$ are subspaces, then the zero vector is in both of them.
$$\mathbb{O}_V+\mathbb{O}_V=\mathbb{O}_V$$
but I can't wrap my head around the closure of addition and scalar multiplication.
 A: Hint: Use closure of $W_1$ and $W_2$.  For instance, if we have two vectors $\vec{v}_1+\vec{v}_2,\vec{u}_1+\vec{u}_2\in W_1+W_2$ (where $\vec{v}_1,\vec{u}_1\in W_1$ and $\vec{v}_2,\vec{u}_2\in W_2$), then we know that $\vec{v}_1+\vec{u}_1\in W_1$ and $\vec{v}_2+\vec{u}_2\in W_2$, by closure of $W_1$ and $W_2$ under addition.  Therefore...
A: If $w_1,w_2 \in W_1+W_2$, then $w_k=u_k+v_k$ for some $u_k \in W_1$ and $v_k \in W_2$. Since $u_1+u_2 \in W_1$ and $v_1+v_2 \in W_2$, we have $w_1+w_2=(u_1+u_2) + (v_1+v_2) \in W_1+W_2$.
Similarly, if $w \in W_1+W_2$, then $w=u+v$ for some $u \in W_1$ and $v \in W_2$, since $\lambda u \in W_1$ and $\lambda v \in W_2$,  we see that $\lambda w = (\lambda u) + (\lambda v) \in W_1+W_2$.
Alternatively, note that the range space of a linear operator is a linear space
and $W_1 \times W_2$ is a vector space with componentwise addition and multiplication. If $L:W_1 \times W_2 \to V$ is the linear operator given by $L((w_1,w_2)) = w_1+w_2$, we see that $W_1+W_2 = L(W_1 \times W_2 )$, hence it is a linear space.
A: Denote $ W = W_1 + W_2 $ and let $v \in W_1$, $w \in W_2$, $W \ni x = v + w $. We need to check that $cx \in W$. But observe $cx = c v + c w$. Since $W_i$ are vector spaces $cv$ and $cw$ are in them. Do something similar for addition (regroup terms, use closure under addition in the subspaces.
