Jordan Canonical Form of a Matrix Example
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.\[\begin{equation} \begin{aligned} m_1 (A) &=( \left( \begin{array}{ccc} 1 & 5 & 7 \\ 0 & 4 & 3 \\ 0 & 0 & 1 \end{array} \right)-\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right) )( \left( \begin{array}{ccc} 1 & 5 & 7 \\ 0 & 4 & 3 \\ 0 & 0 & 1 \end{array} \right)-\left( \begin{array}{ccc} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{array} \right) ) \\ &= \left( \begin{array}{ccc} 0 & 5 & 7 \\ 0 & 3 & 3 \\ 0 & 0 & 0 \end{array} \right) \left( \begin{array}{ccc} -3 & 5 & 7 \\ 0 & 0 & 3 \\ 0 & 0 & -3 \end{array} \right) \\ &= \left( \begin{array}{ccc} 0 & 0 & -6 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right) \end{aligned} \end{equation}\]
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 \[ \left( \begin{array}{ccc} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 4 \end{array} \right) \]
.
