A is a $20×20$ nilpotent matrix.
$$rank(A)=11, rank(A^2)=5, rank(A^3)=2, rank(A^4)=0$$
I know that the minimal polynomial is $m_{\lambda}=\lambda^4$.
There's one eigenvalue which is $0$ (because A is nilpotent).
Because the rank is 11 the nullity is 9 so there are 9 Jordan blocks, the first one of size 4 and the rest are either all 2 or one of size 3, 6 blocks of 2 and one block of size 1.
How can I determine which is the correct form?
Best Answer
On a quick google I find the following source for this result,
The nullities are 9, 15, 18, 20, 20 ,…, 20 so that the consecutive differences are 9, 6, 3, 2, 0,…, 0. This means that there are 9 Jordan blocks of size at least 1 (as you noted), 6 of size at least 2, 3 of size at least 3, and 2 of size at least 4. Hence there are 2 of size 4, 1 of size 3, 3 of size 2 and 3 of size 1.