Find a subspace $W$ of $P_{2}(\mathbb{R})$ such that $P_{2}(\mathbb{R})$ = $ U \oplus W$

Question

Let $P_{2}(\mathbb{R})$ be the real vector space of all real polynomials of degree at most 2, equipped with the usual addition and scalar multiplication operations, and ܷ$U$ = { $p(x)\in P_{2}(\mathbb{R}):p'(1)=0$}.
My first answer for this question was $W$ = { $p(x)\in P_{2}(\mathbb{R}):p'(1)\neq0$}. However, I found that zero vector isn't in it. Then I change it to $W$ = { $0$ , $p(x)\in P_{2}(\mathbb{R}):p'(1)\neq0$}. But I still find that $W$ isn't closed under addition. I don't know how to modify it and don't have any direction.

Answer 1

Since $\dim P_2(\Bbb R)=3$ and you can show that $\dim U=2$, then $\dim W=1$, because $ P_2(\Bbb R)=U\oplus W$. Consider $W:=\{kx\vert k\in\Bbb R\}$.

Since any $p(x)$ in $P_2(\Bbb R)$ can be written as $$p(x)=a+bx+cx^2,$$ it follows that $$a+bx+cx^2=\underbrace{a+c(x^2-2x)}_{\displaystyle\in U}+\underbrace{(b+2c)x}_{\displaystyle\in W}$$ Therefore, $$P_2(\Bbb R)=U+W.$$ Now, let $q\in U\cap W$. Thus, $q^\prime(1)=0$ and $q(x)=kx$, which imply $k=0$. Therefore, $q(x)=0$, which means that $$U\cap W=\{0\}.$$ Finally, if $P_2(\Bbb R)=U+W$ and $U\cap W={0}$, then $P_2(\Bbb R)=U\oplus W$.