# Is this set a subspace?

Let $\mathbb{F}[X]$ be the vector space of all polynomials over the field $\mathbb{F}.$ I'm trying to determine whether the following set is a subspace: $$W=\{p:p(0)=p(1)\}$$ where $p$ is a polynomial.

My working:

Let us write $p(x)=a_0+a_1x+a_2x^2\cdots+a_nx^n$. Then $$\begin{cases}p(0)=a_0 \\ p(1)=a_0+a_1+a_2+\cdots +a_n \end{cases}$$ Now, $P(0)=p(1) \iff a_0=a_0+a_1+\cdots+a_n \iff a_1+a_2+\cdots +a_n=0$.

Therefore, we can write $W$ as $$W=\left\{a_0+a_1x+\cdots+a_nx^n:a_1+a_2+\cdots+a_n=0\right\}.$$ Note that this is independent of $a_0.$ Now, the question is: is $W$ closed under addition and scalar multiplication?

Let $$b=b_0+b_1x+\cdots+b_nx^n \in W$$ and let $$c=c_0+c_1x+\cdots+c_nx^n \in W \quad .$$

Then (as we've just shown): $$\begin{cases} b_1+b_2+\cdots+b_n=0 \\ c_1+c_2+\cdots+c_n=0\end{cases} \quad .$$

Now, $b+c=(b_0+b_1x+\cdots+b_nx^n)+(c_0+c_1x+\cdots+c_nx^n)$.

Is this right so far? What should I do now?

• You are in the right way. Now show the same for the scalar multiplication.
– mfl
Commented Oct 13, 2014 at 19:16
• I think you're over-complicating it (though what you write does give a proof). Given $p,q\in W$, so that $p(0)=p(1)$ and $q(0)=q(1)$, then $(p+q)(0)=p(0)+q(0)=p(1)+q(1)=(p+q)(1)$, and thus $p+q\in W$. Scalar multiplication follows similarly. Commented Oct 13, 2014 at 19:18
• @Hayden Is this right? Given $p, q \in W, \lambda \in \mathbb{F},$ so that $p(0)=p(1).$ Then $\lambda p(0)=\lambda p(1)$. Commented Oct 13, 2014 at 19:23
• @SpongebobSquarepants That looks right, though you don't need to mention anything about $q\in W$. The nice thing about this more simplistic approach is that if the values $0$ and $1$ were changed to any arbitrary elements of $\mathbb{F}$, the methodology you were trying to use would become much more cumbersome, but these arguments would continue to hold. Commented Oct 13, 2014 at 19:25
• @Hayden Also, I note that your proof does not assume $p$ and $q$ to be polynomials. Is your statement true for all (not necessarily polynomial) functions, $p$ and $q$ then? Commented Oct 13, 2014 at 19:25

Although the methodology of proof that you are using works in spirit, it is overcomplicating the central details that the sum of two functions $f$ and $g$ that take values in $\mathbb{F}$ be defined by $$(f+g)(x)=f(x)+g(x),$$ and similarly that the scalar multiple of $f$ by $\lambda\in \mathbb{F}$ be defined by $$(\lambda f)(x)=\lambda f(x).$$
From this it follows that if $p,q\in W$, and thus $p(0)=p(1)$ and $q(0)=q(1)$, we have $$(p+q)(0)=p(0)+q(0)=p(1)+q(1)=(p+q)(1) \quad \text{and} \quad (\lambda p)(0)=\lambda p(0)=\lambda p(1)=(\lambda p)(1).$$
More generally, given any linear subspace $Y$ of the function space $$\mathcal{F}(X,\mathbb{F})= \{f \mid \text{f is a function f:X\rightarrow \mathbb{F}}\}$$ and two (not necessarily distinct) values $a,b\in X$, the set $$W=\{f \in \mathcal{F}(X,\mathbb{F}) \mid f(a)=f(b)\}$$ defines a linear subspace of $Y$ by the same argument as above.