## The Minimum Polynomial

The minimal polynomial of a square matrix
$A$
is the monic polynomial
$p$
in
$A$
of smallest degree
$n$
such that
$p(A)=0$
.
The characteristic polynomial of a matrix
$A$
is
$det(A- \lambda I)$
. By the Cayley - Hamilton Theory, every matrix is a zero of its characteristic polynomial. Because the minimum polynomial is a polynomial of smallest degree, it must divide the characteristic polynomial.
Suppose
$A= \left( \begin{array}{ccc} 3 & 1 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{array} \right)$

\begin{aligned} det( \left( \begin{array}{ccc} 3 & 1 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{array} \right)- \lambda \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right) ) &= det (\left( \begin{array}{ccc} 3- \lambda & 1 & 0 \\ 0 & 3- \lambda & 0 \\ 0 & 0 & 3- \lambda \end{array} \right) ) \\ &= (3- \lambda)^3 \end{aligned}

The minimum polynomial must be one of
$m_1=(A- \lambda I ), \: m_2=(A- \lambda I)^2 , \: m_3=(A- \lambda I)^3$
.
Put
$\lambda =3$
.
$m_1 (A)= \left( \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right) ) \neq 0$

$m_2 (A)= \left( \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right) \left( \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & \end{array} \right) = \left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right)$

Hence the minimum polynomial is
$m_2$
.