1

Inspired by my solution to 2x2 Matrix with no zero entries where $A^k=0$ - Nilpotence? I came up with this problem.

Let $A$ be a $nxn$ matrix which is not the zero matrix $0$ (in which all elements are zero).

Problems:

(i) Find an example for which $A.A\ne0$ but $A.A.A=0$.

(ii) Generalization: defining k-nilpotence for $k\ge4$ as $A^i\ne 0$ for $i=1..(k-1)$ but $A^k=0$ find examples. What can be said about the dimension of the smallest k-nilpotent matrix?

EDIT

Motivation for the second part of (ii): It can be easily shown explicitly that for a 2x2-matrix $A^3=0$ implies $A^2=0$. Hence a 3-nilpotent matrix must have at least dimension 3.

My conjecture is that this holds in general: a $k$-nilpotent matrix must have at least dimension $k$.

Hence the task is to prove of disprove the conjecture.

Dr. Wolfgang Hintze
  • 12,465

1 Answers1

0

You figured out most of it yourself already. Let me anyway provide a little more detailed answer.

First note that a nilpotent square matrix needs to have at least a one-dimensional kernel. Otherwise it would be injective, thus surjective and no power of it could be zero. So you have finitely many possibilities for the rank of such a matrix, namely $\{n-1,\dots,0\}$. Let us, as an example, work out the case $n=3$. Consider $$A=\begin{pmatrix}0& * &*\\0&0&*\\ 0&0&0\end{pmatrix}.$$ It has rank $n-1=2$. $$A^2=\begin{pmatrix}0& 0 &*\\0&0&0\\ 0&0&0\end{pmatrix},\; A^3=\begin{pmatrix}0& 0 &0\\0&0&0\\ 0&0&0\end{pmatrix}.$$ Note that $A$ is $3$-nilpotent, $A^2$ is $2$- nilpotent and $A^3$ is $1$-nilpotent. By $k$-nilpotent I mean that $A^k=0$.

You may repeat the above construction in any dimension in the end showing that for every rank $r\in \{n-1,\dots,0\}$ you find a nilpotent matrix of that rank. Moreover, the matrix $A_r$ of rank $r$ constructed this way is $(r+1)$-nilpotent.

  • @ James: thanks for elaborating your idea. This is an easy contructive way to provide examples of k-nilpotent matrices. But there are other types. – Dr. Wolfgang Hintze Feb 20 '19 at 08:58
  • @Dr.WolfgangHintze You can show that, after basechange, every nilpotent matrix looks like the ones I described. We may therefore reduce to this particular form. –  Feb 20 '19 at 09:01
  • @ James Could you please show the base change for $A=\left( \begin{array}{ccc} 1 & -\frac{1}{2 x} & 0 \ x & 0 & y \ 0 & -\frac{1}{2 y} & -1 \ \end{array} \right)$? – Dr. Wolfgang Hintze Feb 20 '19 at 09:16
  • The question on how to do this is discussed here https://math.stackexchange.com/q/1003525/526015 –  Feb 20 '19 at 09:26
  • Yes, A is not nilpotent, but it is 3-nilpotent, and that's what we are talking about here. And thanks for the link which, however, deals with nilpotent matrices. I have think now that Jordan decomposition seems to be the method of choice. At least in the special case the result is $A' = \left( \begin{array}{ccc} 0 & 1 & 0 \ 0 & 0 & 1 \ 0 & 0 & 0 \ \end{array} \right)$. – Dr. Wolfgang Hintze Feb 20 '19 at 09:57
  • The link deals with nilpotent matrices, yes. which means matrices $A$ such that $A^k=0$ for some $k$. This in particular covers $3$-nilpotent matrices. You are right that Jordan decomposition is the right way. That is, since nilpotent matrices have only zero as eigenvalue. Note that however Jordan decomposition may yield a matrix of the form $$\begin{pmatrix}0&0&0\ 0& 0&1\ 0& 0& 0\end{pmatrix}.$$ In this case base change with a permutation matrix gives the form I wrote in my answer. –  Feb 20 '19 at 10:04
  • @ James: I see that my "k-nilpotence" is already included in the standard definition of nilpotence (of rank k): https://de.wikipedia.org/wiki/Nilpotente_Matrix . Hence it shouldn't be too difficut to prove my conjecture. – Dr. Wolfgang Hintze Feb 20 '19 at 12:45
  • @Dr.WolfgangHintze I don't see what part of your question is still open. From what I wrote it follows that matrix of dimensions $n\times n$ can only be $k$-nilpotent for $k\in{1,\dots,n}$. Hence, $k$-nilpotent matrices only exist from dimension $k$ on... –  Feb 20 '19 at 12:52
  • @ James. Ok, all problems solved. Thanks for the discussion. – Dr. Wolfgang Hintze Feb 20 '19 at 12:58
  • You're welcome. If you are satisfied, please consider accepting the answer ;) –  Feb 20 '19 at 12:59
  • Ok, but please consider upvoting the question first. It once had 1 reputation, now it is down to 0. – Dr. Wolfgang Hintze Feb 25 '19 at 06:29
  • I upvoted your question before answering it. If anyone was giving you a downvote, this is not my fault. 2) The reputation system is not a "quid pro quo". After checking your profile I see that you have never accepted any of various good answers you got for your question. I strongly recommend to start (and to not end) reading the help pages on how the site works, e.g. here https://math.stackexchange.com/help/why-vote
    • –  Feb 25 '19 at 08:08
  • ... you asked 35 questions, 30 of them got at least one answer that has been upvoted by the community and you never accepted a single one. I have no understanding why you don't accept obviously good answers. –  Feb 25 '19 at 08:12
  • I have taken the opportunity to check your remarks. Here is the result. I have asked 30 question (not 35), 7 of them have no answer, this leaves 23 questions with one more answers. It is not true that I have "never accepted" an answer. But indeed I had given only 4 acceptances which, I agree, was not ok. To correct hat I have given just now 32 new acceptances. This includes, however, 12 self answers, leaving 19 answers of other users. – Dr. Wolfgang Hintze Feb 25 '19 at 10:38
  • @Dr.WolfgangHintze I thought it to be 35, sorry. Anyway, I was just doing a quick check this morning and not going through everything in detail and the first impression might very well have been misleading. I am happy to see that you know accepted so many answers (including my own) and hope that you didn't take my comment as personal offence. –  Feb 25 '19 at 10:46