I have the following question about finding a basis for this polynomial space.
Let $V=\{p \in P_3 |P(1)=P(-1)\}.$
a) If $p=a_3x^3+a_2x^2+a_1x+a_0 \in P_3$, find a condition on the coefficients of $p$ for $p$ to be in $V$.
b) Find a basis for $V$, and write down dim(V).
I already did part a)
a) $p(a)=a_3+a_2+a_1+a_0=-a_3+a_2-a_1+a_0$ which directly implies that $a_3=-a_1$ so we're done.
b) Since $a_3=-a_1$, this implies that any general $p(x)=-a_1x^3+a_2x^2+a_1x+a_0$.
If I rearrange this, I get $p(x)=(x-x^3)a_1+x^2a_2+xa_1+a_0$ so would my basis just be $\{1,x^2,x-x^3\}$? This implies that the dimension is 3 which is not consistent with what should happen in $P_3$. Is this method correct?
In general, I'm a little confused about how to generally find the basis for a vector space subject to a condition of some kind. I realize there are multiple possible bases but it's really hard knowing where to kind of start.
Also for part b), If my answer is correct, don't I have to prove that the vectors are linearly independent and span $P_3$ as well?
EDIT: I removed $x$ from my basis. I accidently put $\{1,x,x^2,x-x^3\}$ at first.
Best Answer
You have $\dim P_3 = 4$ and $\varphi(p) = p(-1)-p(1)$ is a non zero linear form. As $V = (\ker \varphi )\cap P_3$, you have $\dim V=3$.
$x$ is not an element of $V$ as you said. $\{x^2,x-x^3,1\}$ is a basis of $V$.