Show that the restriction is diagonalizable

Question

Let $\mathbb{K}$ be a field and for each $\displaystyle{f=\sum_{i=0}^ma_it^i\in \mathbb{K}[t]}$ let $\displaystyle{f'=\sum_{i=1}^mia_it^{i-1}\in \mathbb{K}[t]}$ be the formal derivative of $f$.

Let \begin{equation*}\phi : \mathbb{K}[t]\rightarrow \mathbb{K}[t], \ f\mapsto tf' \ \text{ und } \ \psi : \mathbb{K}[t]\rightarrow \mathbb{K}[t], \ f\mapsto (tf)'\end{equation*}

I have shown that $\phi$ and $\psi$ are linear operators of $\mathbb{K}$-vector space $\mathbb{K}[t]$.

Let $n \in \mathbb{N}$. We consider the set $V_n:=\left \{f\in \mathbb{K}[t]\mid \deg (f)\leq n\right \}$.

I have shown that $V_n$ is a $(n+1)$-dimensional subspace of $\mathbb{K}[t]$, with $V_n\leq_{\phi}\mathbb{K}[t]$ and $V_n\leq_{\psi}\mathbb{K}[t]$.

How can we show that for all $n\in \mathbb{N}$ the restriction $\phi_n$ of $\phi$ on $V_n$ is diagonalizable (and respective for $\psi_n$) ?

Answer 1

Hint.
What's the matrix of $\Phi_n$ in the basis $(1,t,t^2,\ldots,t^n)$ ?