$A\in \mathcal{M}_n (\mathbb R)$ diagonalizable Then $A=P(A^3) \quad /P \in \mathbb{R}[X] $.

Question

Hello I have the following question :

Let $A \in \mathcal{M_n}(\mathbb{R})$ similar to $\operatorname{diag}(\lambda_1 , \ldots ,\lambda_n)$. Show that there exists $P \in \mathbb{R}[X] / A=P(A^3)$. Thanks in advance.

Answer 1

Remark that if $\lambda_i\neq \lambda_j$, $\lambda_i^3\neq \lambda_j^3$, use Lagrange interpolation to find a polynomial such that $Q(\lambda_i^3)=\lambda_i$, write $A=Pdiag(\lambda_1,...,\lambda_n)P^{-1}$, $Q(A^3)=Q(Pdiag(\lambda^3_1,...,\lambda^3_n)P^{-1})=A$.