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Let $p$, $q$, and $r$ be polynomials such that $p(x) = q(x)r(x)$, and let $T$ be a linear operator on a vector space $V$. Is there a simple way to show that $p(T) = q(T)r(T)$ ?

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    @AlexB. Well if he thinks he is right and since he is obviously able to change his opinion when faced with convincing arguments, I don't see the problem with long comments. SE engine shortens it for viewing pleasure anyhow.2011-09-27

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Consider the morphism $\Phi$ between the algebra of polynomials and the algebra of linear operators such that $\Phi(t)=T$.

We have $\Phi(1)=\text{id}$.

Then by linearity, for any polynomial $p$, we have $\Phi(p)=p(T)$.

if $p=q\cdot r$, then $p(T)=\Phi(p)=\Phi(q\cdot r)$

and $\Phi(q\cdot r)=\Phi(q)\Phi(r)$ because $\Phi$ is a morphism.

Hence $p(T)=q(T)r(T)$.

Note: It is assumed that the coefficients of the polynomials belong to a commutative field, and the vector space is on a commutative field. Otherwise there is no guarantee that such a morphism exists.

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    OK, I edited it.2011-08-08