\[M\]
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 \[\mathbb{C}^n\]
, \[1 < k \leq n\]
. 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\[ \left(\begin{array}{cccccc} 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{array} \right) \]
.