Let $A,B\in\mathbb{R}[X]$ with $\deg B=n+1$. Let $\phi(P)$ be the remainder of the euclidian division of $AP$ by $B$.
a) Show that $\phi$ is an endomorphism of $\mathbb{R}_n[X]$ (I have done that one)
b) We suppose that $B$ has $(n+1)$ distinct real roots. Show that $\phi$ is diagonalizable.
What I have done :
$AP=BQ+\phi(P)$
Let $\lambda_1,\dots,\lambda_{n+1}$ be the roots of $B$.
$(AP)(\lambda_i)=\phi(P)(\lambda_i)$ and that holds for $n+1$ values (whereas $\deg\phi( P)<n+1$).
Let $(AP)(\lambda_i)=\alpha_i$
Hence $\phi(P)=\displaystyle\sum_{i=1}^{n+1}\alpha_i \prod_{j=1,j\neq i}^{n+1}\dfrac{X-\lambda_j}{\lambda_i-\lambda_j}$
We can then express its matrix in the base $L$ of the Lagrange polynomials :
$$\mathrm{mat}(\phi)_L=\begin{bmatrix}0&\alpha_2&\alpha_3&\dots&\alpha_n \\\alpha_1&0&\alpha_3&\dots&\alpha_n \\\alpha_1&\alpha_2&0&\dots&\alpha_n \\\vdots&&&&\vdots \\\alpha_1&\alpha_2&\alpha_3&\dots&0 \end{bmatrix}$$
Which I am not able to diagonalize ...
As $\text{deg }\phi(P)\le n$ you can write \begin{align}\phi(P) &= \sum_{k=0}^n \color{red}{\phi(P)(\lambda_k)} \color{blue}{\prod_{j=0,j\neq k}^{n}\dfrac{X-\lambda_j}{\lambda_k-\lambda_j}}\\ &= \sum_{k=0}^n \color{red}{A(\lambda_k) P(\lambda_k)} \color{blue}{\Phi_k} \end{align}
In particular, with $P$ being a Lagrange polynomial $\Phi_k$: $$ \phi(\Phi_k) =A(\lambda_k) \Phi_k $$ So this is a basis made of eigenvalues.