Prove that if X is a subspace and Y is a subspace, then X+Y is a subspace.

Question

I'm doing a proof and I'm not sure if the logic is completely sound.
Proposition:
If X is a subspace of V and Y is a subspace of V, then X + Y is a subspace of V.
Proof:
Let $ a,b \in X,$ and $l,k,\phi,\lambda \in \Re $

Since X is a subspace, $ l\vec{a} + k\vec{b} \in X $
Since Y is a subspace, $ \phi\vec{p} + \lambda\vec{q} \in Y $

X + Y implies $ (l\vec{a} + k\vec{b}) + (\phi\vec{p} + \lambda\vec{q}) \in $ X + Y
Thus, X + Y is closed under vector addition and scalar multiplication.

Since X is a subspace, $ \vec{0} \in X $
Since Y is a subspace, $ \vec{0} \in Y $
X + Y implies $ \vec{0} + \vec{0} = \vec{0} $
Thus, X + Y is nonempty.

Hence, X + Y is a subspace of V.

Answer 1

To show $X+Y$ is closed under addition and scalar multiplication you need to let $v_1, v_2 \in X+Y$ and show that $\alpha v_1 + \beta v_2 \in X+Y$. Since $v_1, v_2 \in X+Y$, there exists $x_1,x_2 \in X$ and $y_1,y_2 \in Y$ such that $v_1 = x_1 + y_1$ and $v_2 = x_2 + y_2$ then it follows that $\alpha v_1 + \beta v_2 = (\alpha x_1 + \beta x_2)+(\alpha y_1 + \beta y_2) \in X+Y$ since $(\alpha x_1 + \beta x_2) \in X$ and $(\alpha y_1 + \beta y_2)\in Y$.