## Jordan Canonical Form of a Matrix Example

To find the Jordan canonical form for a matrix, first find the minimum polynomial. then the Jordan canonical form can be written down.
For example let
$A= \left( \begin{array}{ccc} 1 & 5 & 7 \\ 0 & 4 & 3 \\ 0 & 0 & 1 \end{array} \right)$
.
The eigenvalues are the solutions to
$det(A- \lambda I)=0 \rightarrow \left| \begin{array}{ccc} 1- \lambda & 5 & 7 \\ 0 & 4- \lambda & 3 \\ 0 & 0 & 1- \lambda \end{array} \right| =0$
.
We have
$(1- \lambda )(4- \lambda )(1- \lambda )=0 \rightarrow \lambda =1, \:4$

The minimum polynomial is either
$m_1 (\lambda)=(1- \lambda )(4- \lambda), \: m_2 (\lambda )=(1- \lambda)^2 (4 - \lambda)$

If the minimum polynomial is
$m_1 (\lambda )$
then
$m_1(A)$
is the zero matrix.
\begin{aligned} m_1 (A) &=( \left( \begin{array}{ccc} 1 & 5 & 7 \\ 0 & 4 & 3 \\ 0 & 0 & 1 \end{array} \right)-\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right) )( \left( \begin{array}{ccc} 1 & 5 & 7 \\ 0 & 4 & 3 \\ 0 & 0 & 1 \end{array} \right)-\left( \begin{array}{ccc} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{array} \right) ) \\ &= \left( \begin{array}{ccc} 0 & 5 & 7 \\ 0 & 3 & 3 \\ 0 & 0 & 0 \end{array} \right) \left( \begin{array}{ccc} -3 & 5 & 7 \\ 0 & 0 & 3 \\ 0 & 0 & -3 \end{array} \right) \\ &= \left( \begin{array}{ccc} 0 & 0 & -6 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right) \end{aligned}

Hence the minimum polynomial is
$m_2 (\lambda)$
.
Hence the submatrix of the Jordan canonical form of
$A$
corresponding to the eigenvalue 1 is the two by two matrix
$\left( \begin{array}{cc} 1 & 0 \\ 1 & 1 \end{array} \right)$
.
The Jordan canonical form of
$A$
is
$\left( \begin{array}{ccc} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 4 \end{array} \right)$
.