Solution
we have the minimal polynomial isÂ
( m(lambda)=igg(lambda-1igg)^3igg(lambda-3igg)^2 )
we take ( m(lambda)=0implies igg(lambda-1igg)^3igg(lambda-3igg)^2 =0 )
since ( lambda_1=1 hspace{2mm}andhspace{2mm}lambda_2=3 )
( implies Index(1)=3 ) (means the bigest Jordan block is ( 3 imes3 ))( implies Index(3)=2 ) Â (means the bigest Jordan block is ( 2 imes2 ))
Therefore. all possible Jordan Canonical form is :
( J_{1:}=egin{pmatrix}1&1&0&0&0&0\ 0&1&1&0&0&0\ 0&0&1&0&0&0\ 0&0&0&3&1&0\ 0&0&0&0&3&0\ 0&0&0&0&0&3end{pmatrix},J_2=egin{pmatrix}1&1&0&0&0&0\ 0&1&1&0&0&0\ 0&0&1&0&0&0\ 0&0&0&1&0&0\ 0&0&0&0&3&1\ 0&0&0&0&0&3end{pmatrix} )
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Answer
Therefore. all possible Jordan Canonical form is :
( J_{1:}=egin{pmatrix}1&1&0&0&0&0\ 0&1&1&0&0&0\ 0&0&1&0&0&0\ 0&0&0&3&1&0\ 0&0&0&0&3&0\ 0&0&0&0&0&3end{pmatrix},J_2=egin{pmatrix}1&1&0&0&0&0\ 0&1&1&0&0&0\ 0&0&1&0&0&0\ 0&0&0&1&0&0\ 0&0&0&0&3&1\ 0&0&0&0&0&3end{pmatrix} )
