Proving that the span$((1,1,0),(1,0,1)) =$ span$(1,1,0) +$ span$(1,0,1)$ If I'd like to prove that span$((1,1,0),(1,0,1)) = $span$(1,1,0) +$ span$(1,0,1)$, is it enough to simply state that:
span$(1,1,0)$ = {$x (1,1,0)| x \in R$}
span$(1,0,1)$ = {$y (1,0,1)| y \in R$}
{$x (1,1,0)| x \in \epsilon$} + {$y (1,0,1)| y \in \epsilon$} = {$x(1,1,0) + y(1,0,1)| x,y \in \epsilon$} = span$((1,1,0),(1,0,1))$ ?
I feel like something is missing, or is this enough to prove the above statement? Thank you.
 A: span$((1,1,0),(1,0,1))=a(1,1,0)+b(1,0,1)=$ span$(1,1,0) + $span$(1,0,1)$ for some $a,b\in\mathbb{R}$.
A: *

*Step 1


Let $z \in \text{span}((1,1,0),(1,0,1))$, then there exists $\lambda_1, \lambda_2 \in \mathbb{R}$ such that $z = \lambda_1 (1,1,0) + \lambda_2 (1,0,1)$
Since $\lambda_1 (1,1,0) \in \text{span}((1,1,0))$ and $\lambda_2 (1,0,1) \in \text{span}((1,0,1))$, we deduce $$z \in \text{span}((1,1,0)) + \text{span}((1,0,1))$$
Thus $\text{span}((1,1,0),(1,0,1)) \subset \text{span}((1,1,0)) + \text{span}((1,0,1))$


*

*Step 2


Let $z \in \text{span}((1,1,0)) + \text{span}((1,0,1)))$, then there exists $x \in \text{span}((1,1,0)), y \in \text{span}((1,0,1))$ such that $z = x+y$.
Since $x \in \text{span}((1,1,0))$, there exists $\mu_1 \in \mathbb{R}$ such that $x = \mu_1 (1,1,0)$. Similarly there exists $\mu_2 \in \mathbb{R}$ such that $y = \mu_2 (1,0,1)$.
Since $z = \mu_1 (1,1,0) + \mu_2 (1,0,1)$, we deduce $z \in \text{span}((1,1,0),(1,0,1))$.
Thus $\text{span}((1,1,0)) + \text{span}((1,0,1)) \subset \text{span}((1,1,0),(1,0,1))$


*

*Conclusion 


$\text{span}((1,1,0)) + \text{span}((1,0,1)) = \text{span}((1,1,0),(1,0,1))$
