How to find and prove the basis of a subspace? Let $V = \mathbb{R}^4$. 
Consider the subspace
$$U = \{(a_1,a_2,a_3,a_4) \in \mathbb{R}^4 | a_1 +a_2 +a_3 = 0\} \;of\; V$$
Consider the elements $u_1 =(0,0,0,1)$ and $u_2 =(5,−2,−3,0)$ of $U$. Find another element $u_3 \in U$ such that $\{u1,u2,u3\}$ is a basis of $U$, and prove that it is indeed a basis. 
I know the proof of a basis is that the elements must be linearly independent and spans the entire vector space, but how do you do this?
 A: $u_3 = (1,0,-1,0)$ would work. $u_3$ is in $U$, and it is independent from $u_1$ and $u_2$.  
And to prove that the vectors span U in a hand-wavy way.
The dimension U is less than 4, since U is a subspace in $R^4$, and clearly there are vectors in $R^4$ that are not in U, and we have 3 independent vectors in U, the dimension of U must be 3 and these vectors span U.
A: Finding $u_3$ shouldn't be a problem ( cf. JMoravitz's comment/ Doug M's example). Trying to dispel the "hand-wavy way" for proof of span: 
Suppose $u_3 = (a_1, a_2, a_3, a_4)$. An arbitrary vector spanned by the above basis is a linear combination $\alpha_1 u_1 + \alpha_2 u_2 + \alpha_3 u_3$ which results in coordinates: $(5\alpha_2 + \alpha_3 a_1, -2\alpha_2 + \alpha_3 a_2, -3\alpha_2 + \alpha_3 a_3, \alpha_1a_1 + \alpha_3a_4)$.
 Applying the defining condition: $5\alpha_2 + \alpha_3 a_1 -2\alpha_2 + \alpha_3 a_2 -3\alpha_2 + \alpha_3 a_3 = \alpha_3(a_1 + a_2 + a_3) = 0$. For $\alpha_3 \neq 0$ this implies: $a_1 + a_2 + a_3 = 0$. But that condition is already satisfied as $u_1 + u_2 + u_3 $ belongs to the span $\langle u_1, u_2, u_3\rangle$.   
