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Let $V$ be a real vector space. Let $T$ be a linear operator on a vector space $V$. Let $\mathbb R[X]$ be the polynomial ring over $\mathbb R$. Let $q,r \in \mathbb R[X]$. Let $p := qr \in \mathbb R[X]$, i.e., $p$ is the product/convolution of $q, r$.

It is mentioned in this thread that $p(T) = q(T)r(T)$. In my understanding, $q(T)r(T) := q(T) \circ r(T)$ where $\circ$ is the composition operation. Because $\mathbb R$ is commutative, so is $\mathbb R[X]$. Then $p= r q$ and thus $p(T) = r(T) q(T)$. Hence $$ q(T) \circ r(T) = r(T) \circ q(T) \quad \forall q,r \in \mathbb R[X]. $$

Could you confirm if my above understanding is correct? Thank you so much for your elaboration!

Akira
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    Yes, this is correct ! – TheSilverDoe Jun 23 '23 at 15:18
  • @TheSilverDoe Thank you so much for your verification! The conclusion that the linear operators $q(T)$ and $r(T)$ commute for every polynomials $q,r$ are so surprising for me. – Akira Jun 23 '23 at 15:21
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    @Akira well, T commutes with T and scalars, and when you compute $q(T)r(T)$ and expand out the product, you just get a bunch of products involving $T$ and scalars, which will all individually commute. It’s a very nice fact, and once you get used to it, it’ll feel obvious (in a good way). I agree that it’s surprising at first though! – Nicholas Priebe Jun 23 '23 at 15:34

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