This is one of my linear algebra problems:

Prove that polynomials of degree $n$ does not (The professor made these words bold intentionally) form a vector space.

From what I read, the set of polynomials of degree $n$ should be a vector space, because:

  1. There is an "One" and a "Zero" in this set;
  2. We can find inverse for addition and multiplication from this set;
  3. It follows all the axioms of addition.
  4. It follows all the axioms of scalar multiplication.

Then can someone give me some hints to prove it does not form a vector space?

  • 2
    $\begingroup$ What would be the zero? Also, vector spaces do not have a "one". $\endgroup$ Commented Sep 18, 2016 at 20:03
  • 1
    $\begingroup$ Not closed under difference of polynomials. $\endgroup$ Commented Sep 18, 2016 at 20:04
  • 1
    $\begingroup$ p=0 haven't degree n for all n $\endgroup$ Commented Sep 18, 2016 at 20:05
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    $\begingroup$ Your professor is talking about polynomials exactly of degree $n$, but you are thinking of polynomials of degree at most $n$ (which indeed form a vector space). $\endgroup$ Commented Sep 18, 2016 at 20:06
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    $\begingroup$ @B.Goddard $-1$ is a rather poor choice of degree for the zero polynomial because then the degree function isn't additive. The degree I usually see is $-\infty$. $\endgroup$ Commented Sep 18, 2016 at 20:26

2 Answers 2


Polynomials of degree $n$ does not form a vector space because they don't form a set closed under addition.

For instance:


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.

  • 1
    $\begingroup$ That's based on convention. It's actually very useful to say zero has every degree. For example, let $[k[x]]_d$ denote the set of all homogeneous polynomials of degree $d$ in one variable $x$ with coefficients in $k$. Then we usually consider $[k[x]]_d$ a $k$-vector space. Hence, we are assuming zero has every degree. But the idea of your example is absolutely correct, just take $(x^n+1)-x^n=1$ instead. Now we can all agree $1$ is a degree zero polynomial, regardless of conventions. $\endgroup$ Commented Jul 18, 2020 at 23:02

Polynomials of degree $n$ is a set which is not closed under addition. For example, if $n=3$, then $x^3+x^2$ and $-x^3$ are both $3$rd degree polynomials but their sum is not: $$ x^3+x^2-x^3=x^2 $$ (which is not a $3$rd degree polynomial).


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