From nilpotent matrix to Jordan matrix

Question

Let's say I have a matrix $A\in M_{n\times n}(\mathbb F)$, and it's given that $A$ is a nilpotent matrix. How Can I find a Jordan matrix that is similar to $A$?

because $A$ is nilpotent, I can find a basis $B$ s.t. $A$ represented by the basis $B$ is a Jordan matrix with eigenvalue $0$. How can I proceed from here?

Answer 1

If we're not required to find the basis $B$, then our job is a lot easier.

Note that $J$ is the Jordan form corresponding to $A$, then $\operatorname{rank}(A^k) = \operatorname{rank}(J^k)$. Moreover, for $k = 1,2,\dots,n$, $\operatorname{rank}(J^{k-1}) - \operatorname{rank}(J^k)$ is the number of Jordan blocks in $J$ with size at least $k$. Finally, note that $A$ is nilpotent, which means that $0$ is its only eigenvalue, which means that we only need to worry about blocks for the eigenvalue $0$.

Putting all this together, it suffices to determine $\operatorname{rank}(A^k)$ for all $k \leq m$, where $m$ is the first integer such that $A^m = 0$.

This process is described in Horn and Johnson's Matrix Analysis.

Answer 2

We consider an explicit example in dimension $7$; let (in the Maple format)

$A= Matrix(7, 7, {(1, 1) = -1, (1, 2) = 1, (1, 3) = 3, (1, 4) = 7, (1, 5) = -3, (1, 6) = 5, (1, 7) = -1, (2, 1) = -3, (2, 2) = 2, (2, 3) = 1, (2, 4) = 1, (2, 5) = 2, (2, 6) = -7, (2, 7) = 0, (3, 1) = 0, (3, 2) = -1, (3, 3) = -1, (3, 4) = -2, (3, 5) = 0, (3, 6) = -1, (3, 7) = 0, (4, 1) = 1, (4, 2) = 0, (4, 3) = 1, (4, 4) = 3, (4, 5) = -2, (4, 6) = 5, (4, 7) = 0, (5, 1) = 2, (5, 2) = -1, (5, 3) = -1, (5, 4) = -1, (5, 5) = -1, (5, 6) = 3, (5, 7) = 1, (6, 1) = 0, (6, 2) = 0, (6, 3) = -1, (6, 4) = -2, (6, 5) = 1, (6, 6) = -3, (6, 7) = 1, (7, 1) = 1, (7, 2) = -1, (7, 3) = -2, (7, 4) = -4, (7, 5) = 1, (7, 6) = -2, (7, 7) = 1})$

We obtain $dim(\ker(A))=3,dim(\ker(A^2))=6,dim(\ker(A^3))=7$.

Step 1. We seek $7-6=1$ linearly independent (L.I.) vectors in $\ker(A^3)\setminus\ker(A^2)$: $u=[4,-33,0,16,22,1,11]^T$. $Au\in\ker(A^2)\setminus\ker(A),A^2u\in\ker(A)$. We put (in reverse order) $e_1=A^2u,e_2=Au,e_3=u$.

Step 2. It remains to obtain $6-3-1=2$ vectors L.I. (with $e_2$) in $\ker(A^2)\setminus\ker(A)$: $v=[3,-33,0,16,22,1,11]^T,w=[2,-27,0,13,18,1,9]^T$. $Av\in\ker(A)$; in reverse order, $e_4=Av,e_5=v$. $Aw\in\ker(A)$; in reverse order, $e_6=Aw,e_7=w$.

It adds up! Let $Q=[e_1,\cdots,e_7]$. Then $Q^{-1}AQ=diag(J_3,J_2,J_2)$ where $J_k$ is the nilpotent Jordan block of dimension $k$.

Remark. This method is valid for any eigenvalue $\lambda$ of any matrix $A$: replace $\ker(A^k)$ with $\ker((A-\lambda I)^k)$.