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Show that the restriction is diagonalizable

Mathematics Asked on December 6, 2021

Let $mathbb{K}$ be a field and for each $displaystyle{f=sum_{i=0}^ma_it^iin mathbb{K}[t]}$ 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], fmapsto tf’ text{ und } psi : mathbb{K}[t]rightarrow mathbb{K}[t], fmapsto (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 $V_n:=left {fin mathbb{K}[t]mid deg (f)leq nright }$.

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

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

One Answer

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

Answered by perroquet on December 6, 2021

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