Let $P$ be the vector space of all polynomials, and let $V=\{p\in P:p(2)=p(3)\}$; we want to prove that $V$ is a vector space, and the easiest way to do this is to prove that it’s a subspace of the known vector space $P$. This requires that you prove three things:
- $V\ne\varnothing$. ($V$ is non-empty.)
- If $p,q\in V$, then $p+q\in V$. ($V$ is closed under vector addition.)
- If $p,q\in V$ and $\alpha\in\Bbb R$, then $\alpha p\in V$. ($V$ is closed under scalar multiplication.
Proving (1) is easy: just exhibit a polynomial $p$ such that $p(2)=p(3)$. The simplest one is the constant polynomial $p(x)=0$, which also happens to be the zero vector in $P$ and in $V$.
To prove (2), you must start with arbitrary polynomials $p$ and $q$ in $V$. In other words, you have polynomials $p(x)$ and $q(x)$ such that $p(2)=p(3)$ and $q(2)=q(3)$. (Note that you don’t know what $p(2)$ and $q(2)$ actually are.) To help keep the notation straight, let $t=p+q$; $t$ is a polynomial, and for every $x\in\Bbb R$ it satisfies $t(x)=p(x)+q(x)$. In particular,
$$\begin{align*}
t(2)&=p(2)+q(2)\\
&=p(3)+q(3)\qquad\text{ because }p,q\in V\\
&=t(3)\;,
\end{align*}$$
so $t\in V$. This shows that $V$ is closed under vector addition.
You prove (3) in very much the same way. Let $p$ be any polynomial in $V$, let $\alpha$ be any real number, let $q=\alpha p$ (i.e., $q(x)=\alpha p(x)$ for all $x\in\Bbb R$), and show that $q\in V$.
I believe your addition does not have inverse element, so the following is not true:
$$\forall v \in V, \exists w(\text{also called }-v) : v \oplus w = 0$$
because $max(v,w)$ can be only $v$ or $w$ but not $0$.
Best Answer
Polynomials of degree $n$ does not form a vector space because they don't form a set closed under addition.
For instance:
$$X^n-X^n=0$$
which is not of degree $n$.
So, don't get confused with the set of polynomials of degree less or equal then $n$, which form a vector space of dimension $n+1$. We often work with this space.