Jordan Form of a Nilpotent Matrix Example

Suppose we have a nilpotent matrix  
\[M\]
  of degree 4 on  
\[\mathbb{C}^6\]
  of a linear transformation  
\[T\]
. We want to find the Jordan canonical form of  
\[M\]
.
Let  
\[M\]
  be nilpotent of degree  
\[k\]
  on  
\[\mathbb{C}^n\]
,  
\[1 . Let the rank of  
\[M\]
  be  
\[m\]
. Let  
\[k=n-2\]
  then the  
\[m's\]
  associated with  
\[M\]
  (and, therefore, the Jordan Canonical {form of  
\[M\]
) are either uniquely determined or indeterminate according to the following formula:
Let  
\[k=n-2\]
.
1. If  
\[n=4\]
 , then  
\[m_1 =1, \: m_2 =2\]
  or  
\[m_1=2\]
.
2. If  
\[n \geq 5\]
  then  
\[m_1 =m_{n-3}=1\]
  or  
\[m_1=1, \: m_{n-2}=2\]
.
\[M\]
  is nilpotent of degree 4 on  
\[\mathbb{C}^6\]
  hence  
\[k=4\]
  and  
\[n=6\]
  or 
\[k=6-2=n-2\]
. Therefore, we have  
\[m_1=m_{n-3}=1\]
  or  
\[m_1=m_3=1\]
. K = 6 Also,  
\[m_1=1, \: m_{n-2}=2\]
  or  
\[m_1=1, \: m_4=2\]
. We know, too, that the nullity of  
\[M\]
  equals  
\[\sum_{l=1}^k m_l\]
.
Therefore,  
\[N(M)=m_1+m_4=3\]
. We know the first  
\[m_1\]
  elementary Jordan matrices in  
\[M\]
  are each of order  
\[k\]
  (the degree of nilpotent of  
\[M\]
) ; the next  
\[m_2\]
  are each of order  
\[k-1...m_l\]
, each of order  
\[k-l+1, \: (l=1,...,k-1)\]
, and a final zero matrix of order  
\[m_k\]
. Thus  
\[m_1=1\]
  and is of order 4 (since  
\[k=4\]
).
\[m_2=m_4 =0, \: m_3=1\]
  and is of order 2. Hence, the Jordan canonical form of  
\[M\]
  is
\[ \left((\begin{array}{cccccc} 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \end{array} \right) \]
.
When  
\[m_1=1\]
  (order of  
\[k=4\]
),  
\[m_2=m_3=0, \: m_4=2\]
  the Jordan canonical form is
\[ \left(\begin{array}{cccccc} 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{array} \right) \]
.