Show that two complex nilpotent matrices are similar

Question

We consider he following proposition.

Two $n\times n$ complex nilpotent matrices are similar if and only if they have same nilpotence index.

How do I show that the statement is correct for $n=3$ ? And why isn't it correct for $n=4$?

Does this answer your question? Prove that all nxn nilpotent matrices of order n are similar.. The counterexample for $n=4$ is also given there. — Dietrich Burde, Jun 25 '20 at 14:23
To the OP. i) The Jordan norml forms of the nilpotent matrices are the same for any underlying field. ii) When Dietrich writes something, the probability that it is true is 1. iii) Below, you obtain a perfect answer. You can give to the author the green ticket. — , Jun 25 '20 at 21:45

score 1 · Answer 1 · answered Jun 25 '20 at 14:21

This question can be solved using the Jordan normal form. For complex matrices, this normal form always exists, and two complex matrices are similar if and only if their Jordan normal forms are the same, up to order of the Jordan blocks.

Since nilpotent matrices only have the single eigenvalue $0$, the Jordan normal form of such a matrix will always have only zeroes on the diagonal, and then ones and zeroes below.

Now for nilpotent $3\times3$ matrices, we can find all possible Jordan normal forms (again, up to order of the Jordan blocks):

$\begin{pmatrix}0&0&0\\0&0&0\\0&0&0\end{pmatrix},\begin{pmatrix}0&0&0\\1&0&0\\0&0&0\end{pmatrix},\begin{pmatrix}0&0&0\\1&0&0\\0&1&0\end{pmatrix}$

These all have different index, so if two nilpotent matrices have the same index, they have the same Jordan normal form, and are thus similar. And if they are similar, they have the same Jordan normal form, and thus the same index.

This argument breaks down for $4\times4$ matrices. Here, we have the following normal forms:

$\begin{pmatrix}0&0&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&0\end{pmatrix},\begin{pmatrix}0&0&0&0\\1&0&0&0\\0&0&0&0\\0&0&0&0\end{pmatrix},\begin{pmatrix}0&0&0&0\\1&0&0&0\\0&1&0&0\\0&0&0&0\end{pmatrix},\begin{pmatrix}0&0&0&0\\1&0&0&0\\0&0&0&0\\0&0&1&0\end{pmatrix},\begin{pmatrix}0&0&0&0\\1&0&0&0\\0&1&0&0\\0&0&1&0\end{pmatrix}$

Numbers 2 and 4 have the same index (it's 2), but are not similar. So two nilpotent matrices with index 2 could have different Jordan normal forms, and would thus not be similar.

score 1 · Answer 2 · answered Jun 25 '20 at 22:53

Note that for any nilpotent matrix in Jordan form, the nilpotence index is the size of the largest block.

It is only possible to have $2$ distinct normal forms for $n \geq 4$ because $4$ is the smallest number that has two partitions with the same maximal number. In particular, we can write both $4 = 2 + 2$ and $4 = 2 + 1 + 1$.

Show that two complex nilpotent matrices are similar

2 Answers2