Let $B = (1, X, X^2)$ be an ordered basis for $\mathbb{R}_2[X]$ and $p ∈ \mathcal{L}(\mathbb{R}_2[X])$ 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 $A = M_B(p)$
Compute $A^2$. Prove that $p ◦ p = p$.
\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}
I computed the multiplication and it appears that $A^2$ = $A$. How should I prove it?
If $A$ is the matrix corresponding to the linear operator $p$, then $p(v)=A\cdot v\ \forall v\in\Bbb R_2[X]$.
$\implies(p\circ p)(v)=p(p(v))=A\cdot(p(v))=A\cdot(Av)=(A\cdot A)v=A^2\cdot v$
This means the matrix corresponding to the linear operator $p\circ p$ is $A^2$. Since $A^2=A$, the matrices corresponding to $p$ and $p\circ p$ are identical, which means $p=p\circ p$.