# Polynomials and Linear Operators

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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Factor $p$ into $q\cdot r$. Whether the arguments are linear operators or numbers makes no difference formally. –  anon Aug 5 '11 at 18:27
All you need to know is that $T$ commutes with scalar multiplication and itself, both of which should be clear. In other words, $T$ generates a commutative subalgebra of $\text{End}(V)$. –  Qiaochu Yuan Aug 5 '11 at 18:32
@anon: In fact, commutativity is precisely the issue here; see, e.g., see this previous question. It should not be taken for granted. –  Arturo Magidin Aug 6 '11 at 3:28
@anon: Yes, but the point is that the observation needs to be made; by simply saying "factor, doesn't matter what the arguments are" you are in fact skipping over the key observation that needs to be made. The ambient ring of linear operators is not commutative, so one must justify the assertion that evaluation is a morphism when restricted to a particular subring. You say "whether the arguments are linear operators or numbers makes no difference": what if I'm doing 2-var polynomials and plug in two linear operators into a factorization? Suddenly it makes a big difference "formally". –  Arturo Magidin Aug 6 '11 at 3:55
@anon: Statements like "whether the arguments are linear operators or numbers makes no difference formally" are at worst misleading, and at best sloppy and too easy to misinterpret. Especially when addressed at someone who is trying to prove something which far too many people incorrectly believe holds by simple evaluation. –  Arturo Magidin Aug 6 '11 at 4:12

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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The key is of course the claim that $\Phi$ is a morphism. The universal property of the polynomials in one variable guarantees a morphism when mapping into an a ring where the image of the coefficients commutes with the image of the $x$; so you need to make the observation that $\Phi(t)$ centralizes $\Phi(F)$; without that observation you have no warrant for asserting that you have defined a morphism between the polynomials and the linear operators. –  Arturo Magidin Aug 6 '11 at 3:11
You're right, I added a line about the commutativity. I assumed it was the case in the initial question. –  galath Aug 6 '11 at 8:18
You need more than to observe that the coefficients are a field; you need to observe that the image of the coefficients centralizes the image of $t$. –  Arturo Magidin Aug 6 '11 at 17:23
OK, I edited it. –  galath Aug 8 '11 at 10:22