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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  
\[A= \left( \begin{array}{ccc} 3 & 1 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{array} \right) \]

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

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  
\[m_2\]
.