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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?

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Non emptiness is easy to check. Clearly $p(x)=(x-2)(x-3)$ is a member of the set. –  RSG Sep 24 '12 at 5:08
2  
@rsg: Even more clearly(?) the zero polynomial is. –  Hagen von Eitzen Sep 24 '12 at 5:16
    
so even though we cant sub in '2' or '3' as in P(2) or P(3); the zero polynomial is still sufficient to prove non emptiness? –  student101 Sep 24 '12 at 5:21
    
@HagenvonEitzen lol. Of course yes. Wanted to give a relatively (more?) nontrivial example. –  RSG Sep 24 '12 at 5:54

3 Answers 3

up vote 7 down vote accepted

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:

  1. $V\ne\varnothing$. ($V$ is non-empty.)
  2. If $p,q\in V$, then $p+q\in V$. ($V$ is closed under vector addition.)
  3. 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 $q\in V$.

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Besides direct proof (as in Makoto's answer), one can simply note that the evaluation maps $\rm\:E_{\,r}\!:\, p(x)\to p(r)\:$ are $\,\Bbb R$-linear, hence so too is the difference $\rm\:D = E_{\,3} - E_{\,2}\!:\ p(x)\to p(3)-p(2).\:$ Your set is the kernel of $\rm\,D,\,$ hence it is a subspace ot the vector space of real polynomials.

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Let $V$ be the space of all polynomials over $\mathbb{R}$. Let $W = \{p \in V\colon p(2) = p(3)\}$. Since $0 \in W$, $W$ is not empty. Let $f, g \in W$. Let $a, b \in \mathbb{R}$. Then $af(2) + bg(2) = af(3) + bg(3)$. Hence $af + bg \in W$. Thus $W$ is a subspace of $V$.

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ah okay thanks. –  student101 Sep 24 '12 at 5:14

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