Suppose $U=Span\{u_{1}, u_{2} \}$ for $u_{1}, u_{2} \in U$ and $V=Span\{ v1, v2\}$ for $v_{1},v_{2} \in V$. Prove that $U+V=Span\{u1,u2,v1,v2\}$.

Question

This is what I have so far, I don't know if this is where I stop or if there is more to prove?

$$U+V = (c_{1}u_{1} + c_{2}u_{2}) + (c_{1}v_{1} + c_{2}v_{2}) = c_{1} (u_{1}+v_{1}) + c_{2}(u_{2}+v_{2})$$

Does this cover the proof on the idea that if they both span U and V separately, then a linear combination of their sum would do the same?

Answer 1

You're sort of on the right track. $\DeclareMathOperator{Span}{Span}$

Note that $w\in U+V$ if and only if there exist $u\in U$ and $v\in V$ such that $w=u+v$. Since \begin{align*} U&=\Span\{u_1,u_2\} & V &= \Span\{v_1,v_2\} \end{align*} it follows that $u\in U$ and $v\in V$ if and only if there are scalars $\alpha_1,\alpha_2,\beta_1,\beta_2\in\Bbb R$ such that \begin{align*} u&=\alpha_1u_1+\alpha_2 u_2 & v&=\beta_1v_1+\beta_2v_2 \end{align*} Hence $w\in U+V$ if and only if there are scalars $\alpha_1,\alpha_2,\beta_1,\beta_2\in\Bbb R$ such that $$ w=\alpha_1u_1+\alpha_2 u_2 +\beta_1v_1+\beta_2v_2 $$ This proves that $$ U+V=\Span\{u_1,u_2,v_1,v_2\} $$