Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
@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
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 $q\in V$.

share|cite|improve this answer

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.

share|cite|improve this answer

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 $V$.

share|cite|improve this answer
ah okay thanks. – student101 Sep 24 '12 at 5:14

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.