let $V=\mathbb{R^2}$, and let $W$ be the subspace generated by $(2,1)$. Let $U$ be the subspace generated by $(0,1)$. Show that $V$ is a direct sum of $U$ and $W$.
My proof. Let $(x,y)\in\mathbb{R^2}$. If $a=x/2$ and $b=y-x/2$ then
$a(2,1)+b(0,1)=(x,y)$.
So, $V=U+W$. Now, we need to show that for every element $v$ in $V$ there exist unique elements $u\in U$ and $w\in W$ such that $v=u+w$. Since $U$ and $W$ are subspaces $0\in U$ and $0\in W$. Hence, $0\in U\cap W$. Now, we will show that $U\cap W\in \left\{ 0\right\}$. Let $t\in U\cap W$. so, XXX How can I show this statement, can you help?
This isn't too difficult. You were almost there.
Let $x \in U \cap W$. Then, we know by definition that $x \in U$ and $x \in W$. As $U = span(0,1)$, we can write $x = a(0,1)$. Similarly, because $W = span(2,1)$, we find that $x$ can be written as $b(2,1)$. We find that $x = a(0,1) = b(2,1)$ for some $a,b \in \mathbb{R}$. Hence, we obtain that $(0,a) = (2b,b) \iff 0 = 2b \land a = b \iff a = b = 0$. We deduce that $x = (0,0)$ and this ends the proof.