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Let $\mathcal{P_n}$ be vector space of all polynomials $\mathbb{R}\to\mathbb{R}$ of degree $\leq n$. A proposition in front of me claims that, if $\pi:\mathbb{R}\to L(\mathcal{P_n})$ is defined by $$[\pi(t)f](s)=f(s-t),\quad t,s\in\mathbb{R},\quad f\in\mathcal{P_n},$$ then $$A=\frac{d}{dt}\pi(t)\Bigg|_{t=0}$$ is given by $$Af=f'.$$

But when I try to get this, I get a minus sign that shouldn't be there. This is my reasoning (I'll write for n=3 so I don't have to meddle with dots): in canonical basis $\{1,s,s^2\}$ the matrix of $\pi(t)$ is $$\pmatrix{1 & -t & t^2\\0 & 1 & -2t\\0&0&1},$$ so the matrix of $\pi'(0)$ is $$\pmatrix{0&-1&0\\0&0&-2\\0&0&0}.$$ But this is the matrix of operator $Bf=-f'$! Where am I wrong?

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I think the “proposition in front of you” is wrong. They’re not infallible you know – Ewan Delanoy Sep 22 '12 at 15:58

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