Determine/prove what is the dimension of the subspace $S=\{p\in P_2\ |\ p(0)=p(1) \}$ of $P_2$

I believe that I need to find a basis for the subspace to determine the dimension but I need help.

  • $\begingroup$ What is $\bP_2$? $\endgroup$ Commented Mar 5, 2019 at 20:44
  • $\begingroup$ polynomials of degree 2 $\endgroup$
    – zt-thuggy
    Commented Mar 5, 2019 at 20:45
  • $\begingroup$ Polynomials of degree $2$ are not a subspace. $\endgroup$
    – Bernard
    Commented Mar 5, 2019 at 20:46
  • $\begingroup$ OK. So yes, you could find a basis for $S$. Do you know the rank-nullity theorem? You could show that $S$ is the kernel of some linear transformation. $\endgroup$ Commented Mar 5, 2019 at 20:46

2 Answers 2


So you have $$p(x)=ax^2+bx+c$$ and $$p(0)=p(1)\implies c=a+b+c\implies b=-a$$

so $$S = \{ax^2-ax+c;a,c\in\mathbb{C}\}$$ So we have exactly two independent parameters so $\dim S =2$


As mentioned in the comments, polynomials of degree $2$ don't form a vector space (consider $f\in P_2$, then $-f\in P_2$ but their sum is $0$ which is not a degree $2$ polynomial). Rather, it should be "polynomials of degree at most $2$" which makes it a vector space.

Now, consider $f\in P_2$ as $f(x)=ax^2+bx+c$ where $a,b,c\in\Bbb R$. For $f\in S$, we have $f(0)=f(1)$, ie, $c=a+b+c$, ie, $a+b=0$

So, if $a=k$, then $b=-k$ and $c$ is unrestricted, so for $f\in S$, it must be of the form $kx^2-kx+c=k(x^2-x)+c\cdot 1$

Can you show that $\{x^2-x,1\}$ forms a basis for $S$ and hence conclude that $\dim S=\ldots$


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