# Prove that $(u_1,u_2,w_1,w_2,w_3)$ is a basis for $U + W.$

Assume that $U$ and $W$ are subspaces of a vector space $V$ and that $U \cap W = \{0\}.$ Assume that $(u_1,u_2)$ is a basis for $U$ and $(w_1,w_2,w_3)$ is a basis for $W.$ Prove that $(u_1,u_2,w_1,w_2,w_3)$ is a basis for $U + W.$

I'm pretty stuck and I would like some hints (no full answers so I can work through myself). Thank you

Suppose $c_1u_1+c_2u_2+c_3w_1+c_4w_2+c_5w_3=0$ then $$c_1u_1+c_2u_2=-c_3w_1-c_4w_2-c_5w_3$$ left hand is $\in U$ and right hand $\in W$ since $U \cap W = \{0\}.$ so $c_1u_1+c_2u_2=-c_3w_1-c_4w_2-c_5w_3=0$ using $(u_1,u_2)$ is a basis for $U$ and $(w_1,w_2,w_3)$ is a basis for $W$, implies $c_i=o, i=1,2,...,5$
It is abvious that $(u_1,u_2,w_1,w_2,w_3)$ generates $U+W$. Because if $x=a+b\in U+W$ where $a\in U$ and $b\in W$ then by definition of basis, there are $e_1, e_2,e_3,e_4,e_5$ such that $a=e_1u_1+e_2u_2$ and $b=e_3w_1+e_4w_2+e_5w_3$ so $$x=e_1u_1+e_2u_2+e_3w_1+e_4w_2+e_5w_3$$
The decomposition $u+w, \enspace u\in U,\enspace w\in W$, of an element of $U+W$ is unique if $U\cap W=0$.
Indeed, if $u+w=u'+w'$, then $u'-u =w-w'\in U\cap W=0$. Thus $u'-u=w-w'=0$, which implies $u'=u$, $w'=w$.
There results that any two vectors $u\in U, w\in W$ are linearly independent, hence, if $\lambda_1 u_1+\lambda_2 u_2+\mu_1 w_1+\mu_2 w_2+\mu_3 w_3=0$ is any relation between $u_1, u_2, w_1, w_2, w_3$,then $$\lambda_1 u_1+\lambda_2 u_2=0,\quad \mu_1 w_1+\mu_2 w_2+\mu_3 w_3=0,$$ whence $\lambda_1=0,\lambda_2=0$ on one hand, $\;\mu_1=0, \mu_2=0, \mu_3=0$ on the other hand. This proves these five vectors are linear independent. As they're obviously a system of generators of $U\oplus W$, they constitute a basis for this direct sum.