Let $v \in \operatorname{Im}(p)$. Compute $p(v)$.

Question

Let $B = (1, X, X^2)$ be an ordered basis for $\Bbb R_2[X]$ and $p ∈ \mathcal{L}\big(\Bbb R_2[X]\big)$ be the linear map defined by $p(1) = \frac{1}{3}(2 − X − X^2)$, $p(X) = \frac{1}{3}(−1 + 2X − X^2)$ and $p(X^2) = \frac{1}{3} (−1 − X + 2X^2)$.

By definition of the task, I found that $A=M_B(p)$, which is the matrix of $p$ with respect to the basis $B$.

\begin{bmatrix} \frac{2}{3} & -\frac{1}{3} & -\frac{1}{3} \\ -\frac{1}{3} & \frac{2}{3} & -\frac{1}{3} \\ -\frac{1}{3} & -\frac{1}{3} &\frac{2}{3} \\ \end{bmatrix}

Let $v\in \operatorname{Im}(p)$. Compute $p(v)$.

I know that if $v$ is in $\operatorname{Im}(p)$, there is some $w$ such that $v$ = $p(w)$. I'm quite stuck on this problem.

Answer 1

$A^2=A\implies p\circ p=p$

So $p(v)=p(p(w))=(p\circ p)(w)=p(w)=v$.