Prove that this set is a vector space (by proving that it is a subspace of a known
vector space).
The set of all polynomials p with p(2) = p(3).
I understand I need to satisfy, vector addition, scalar multiplication and show that it is non empty.
I'm new to this concept so not even sure how to start. Do i maybe use P(2)-P(3)=0 instead?
My concern is I'm not sure what two polynomials I need to add to prove vector addition; proving scalar multiplication seems okay though.
Thankyou
also a follow up question; if I prove something is a subspace of a known vector space does this imply the subspace is a vector space. Or does that subspace has to span the entire vector space first? how would i prove this in this case?
Best Answer
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:
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 $q\in V$.