Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


Let $V$ be a finite-dimensional vector space over a field $K$ and let $T$ be a linear transformation of $V$ to itself. We define the minimum polynomial $m(x)$ of $T$. Suppose that $q(x)$ is any polynomial in $K[x]$ and that $r(x)$ is the highest common factor of $q(x)$ and $m(x)$. Show that Ker($q(T)$) = Ker($r(T)$).


$r(x)$ hcf of $q(x)$ amd $m(x)$ $\Rightarrow$ $q=f_1r$ and $m=f_2r$ for some $f_1, f_2 \in K[x]$.

Take $v \in$ Ker ($r(T)$), then $q(T)v=(f_1r)(T)v=f_1(T)(r(T)v)=f_1(T)(0)=0$.

That is, Ker$r(T) \subseteq$ Ker $q(T)$. Not sure how to show the opposite relation; any help would be greatly appreciated. Regards.

share|cite|improve this question
up vote 3 down vote accepted

Use Bézout's identity: $r(x) = a(x) q(x) + b(x) m(x)$, for some $a,b \in K[x]$.

Now take $v\in \ker q(T)$. Then $$r(T)v = a(T)(q(T)v)+b(T)(m(T)v) = a(T)(0)+b(T)(0) = 0$$ and $v\in \ker r(T)$.

share|cite|improve this answer
Can we say then that, for $v \in$ Ker$r(T)$, $r(T)v=0=aq(T)v+bm(T)v=0$, and so, as $r(x)|q(x)$ and $r(x)|m(x)$, that this forces $q(T)v=0$ and concludes the argument? – Mathmo Dec 22 '11 at 13:05
@TJO, the relation is for the reverse inclusion: if $v\in \ker q(T)$ then $v\in \ker r(T)$. – lhf Dec 22 '11 at 13:53

Note that $r = f_1 q + f_2 m$ (why? Think Euclid).

share|cite|improve this answer
$f_1$ and $f_2$ are not the same as in the question. – lhf Dec 22 '11 at 12:56
Cheers! Hadn't thought of dividing through by $r(x)$ and applying Bezout's result. – Mathmo Dec 22 '11 at 13:07

It is a matter of taste, but one can do without explicit mention of polynomial division (Euclid, Bezout). Since $q(t)$ is a multiple of $r(t)$ one certainly has $\ker(q(T)) \supseteq \ker(r(T))$ (as you remarked). For the opposite inclusion write $q(t)=p(t)r(t)$ and suppose $v\in\ker(q(T))$ so $p(T)(r(T)(v))=\vec0$; we wish to show that already $w=r(T)(v)=\vec0$. By definition of $r(t)$, there is no non-unit common factor of $p(t)$ and $(m/r)(t)$ but $w$ is annihilated by both $p(T)$ and by $(m/r)(T)$, so the minimal polynomial in $T$ that annihilates $w$ can only be $1$, which means that $w=\vec0$, as we wished to show.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.