Jordan canonical form of strictly upper or lower triangular matrices of even order n and rank n-1.

Question

I am trying to find the jordan canonical form of $ A= \left[ {\begin{array}{cccc} 0 & 0 & 0 & 0 \\ 3 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ \end{array} } \right] $. In this process I found $0$ is the only eigen value with algebraic multiplicity $4$ and geometric multiplicity $1$ so jordan canonical form should have only one 0-jordan block of order $4$. But as $A^3 = 0$, we have $x^3$ as minimal polynomial and that suggest the largest $0$-jordan block can be of order $3$.

Please help, where am I wrong?

Answer 1

octave:4> A
A =

   0   0   0   0
   3   0   0   0
   0   2   0   0
   0   0   1   0

octave:5> A^3
ans =

   0   0   0   0
   0   0   0   0
   0   0   0   0
   6   0   0   0

Notice at the lower left corner, the entry is non-zero. The minimal polynomial is indeed $x^4$.