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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 $p(T) = q(T)r(T)$ ?

Jodles
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    Factor $p$ into $q\cdot r$. Whether the arguments are linear operators or numbers makes no difference formally. – anon Aug 05 '11 at 18:27
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    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 05 '11 at 18:32
  • @anon: That's not true; the evaluation morphism does not work well when the arguments are not commutative. For example, over the quaternions we have that $x^2+1 = (x-i)(x+i)$, but if you evaluate at $j$, $j^2+1$ does not equal $(j-i)(j+i)$. That's why the proof that a polynomial of degree $n$ has at most $n$ roots breaks down over noncommutative division rings. – Arturo Magidin Aug 06 '11 at 03:04
  • @Arturo: Well, yes, I am aware, but I was taking commutativity for granted because I assumed it was trivial in this context. – anon Aug 06 '11 at 03:17
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    @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 06 '11 at 03:28
  • @Arturo: I'm sorry, but I still agree with Qiaochu here: the fact linear operators commute with scalar multiplication (over the base field, and fields are always commutative rings) and themselves should be obvious, and that's all that's relevant here, no? – anon Aug 06 '11 at 03:38
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    @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 06 '11 at 03:55
  • @Arturo: Sure, you have to see that $T\cdot T=T\cdot T$ and what that means for all algebra involving just $T$ and scalars. Whether you want to say "seeing" it is enough or actually go through the motions of formal justification I guess depends on desired level of rigor, which I admit is perhaps needed for a class or book exercise but not for me personally. – anon Aug 06 '11 at 04:08
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    @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 06 '11 at 04:12
  • @Arturo: You've convinced me on that point, but I also believe you seemed more poised to misinterpret my statement than others would be, perhaps related to your greater experience with the relevant background math. I apologize if that comes off insincerely. – anon Aug 06 '11 at 04:23
  • @anon For what it's worth, here is another data point: I completely agree with Arturo and I think that the comment you made "doesn't matter..." was misleading and unhelpful, obfuscating the main issue. Defending it in this long comment thread doesn't make it more correct or helpful. – Alex B. Aug 28 '11 at 23:35
  • @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. –  Sep 27 '11 at 23:06

1 Answers1

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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.

galath
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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 06 '11 at 03:11
  • You're right, I added a line about the commutativity. I assumed it was the case in the initial question. – galath Aug 06 '11 at 08:18
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    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 06 '11 at 17:23
  • OK, I edited it. – galath Aug 08 '11 at 10:22