## 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
$\left(\begin{array}{cccccc} 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{array} \right)$
.