# Find the Jordan Canonical Form of a nilpotent matrix

A is a $20x20$ 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?

On a quick google I find the following source for this result,

The$\DeclareMathOperator{null}{null}$ differences $\null[(A − λI)^j] − \null[(A − λI)^{j−1}]$ is the number of Jordan blocks associated to λ that are of size at least $j × j$.

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.

• @PanthersFan92 there's 3 of size at least 3, but there are also 2 of size at least 4 (which implies 2 of size 4 since the minimal polynomial tells us the maximal size of a block). 3-2=1. – Calvin Khor May 12 '16 at 15:22