Let f be a linear operator on a ﬁnite dimensional vector space V over a ﬁeld K Let $f$ be a linear operator on a ﬁnite dimensional vector space $V$ over a ﬁeld $K$.
I need to show that the characteristic polynomial, $C_f$, of $f$ is divisible by the minimal polynomial, $M_f$, of $f$.
I know that if $x$ is a root of the characteristic polynomial then it is a root of the minimal polynomial too but I'm not sure this will lead to a sufficient proof for this 5 mark question.
 A: Well, according to the Cayley-Hamilton theorem (see link below), letting $C_f(x) \in K[x]$ be the characteristic polynomial of $f$, we have $C_f(f) = 0$.  Furthermore, $M_f(f) = 0$ by definition, $M_f(x) \in K[x]$ being the minimal polynomial of $f(x)$, and $M_f(x)$ is of minimal degree amongst all polynomials $p(x) \in K[x]$ with $p(f) = 0$; that is, we have $\deg M_f \le \deg p$ for any such $p$.  We now invoke the division algorithm for polynomials to write
$C_f(x) = M_f(x)q(x) + r(x) \tag{1}$
for unique polynomials $q(x), r(x) \in K[x]$ and if $r(x) \ne 0$, $\deg r < \deg M_f$.    We evaluate (1) on $f$, yielding
$0 = C_f(f) = M_f(f)q(f) + r(f) = r(f) \tag{2}$
since $M_f(f) = 0$; thus $f$ satisfies $r(x)$, and thus we must have $r(x) = 0$; the alternative, $r(x) \ne 0$ requires $\deg r < \deg M_f$, in contradiction to the minimality of $\deg M_f$.  Thus
$C_f(x) = M_f(x)q(x), \tag{3}$
or
$M_f(x) \mid C_f(x), \tag{4}$
as was to be shown.  QED.
Link to Cayley-Hamilton: http://en.m.wikipedia.org/wiki/Cayley%E2%80%93Hamilton_theorem
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
A: Using the division algorithm we have
$C_f=qM_f+r$   for some polynomials $q$ and $r$
Now as $C_f(f)=0$ and $M_f(f)=0$, we get that
$0=q.0+r \implies r=0$
and hence $M_f$ divides $C_f$
