Let $X$ be a nonzero $n × n$ matrix such that $X^m = 0$ for some $m ∈ N$. Let $b ∈ N$ be the smallest number for which $X^b = 0$. Show that $b≤ n$.

Question

I figure it has something to do with minimal polynomials and the Jordan canonical form, I just can't piece it together. I would really appreciate a hint on how to start it.

score 2 · Answer 1 · edited May 09 '21 at 14:55

2

If the range of $X^k$ is equal to the range of $X^{k+1}$ then the range of $X^{k+i}$ is equal to the range of $X^k$ for all positive $k$.

It follows the ranks of $X,X^2,\dots,X^b$ are strictly decreasing, and since the rank of $X$ is at most $n$ we have $b$ is at most $n$.

edited May 09 '21 at 14:55

Arthur

199,419

answered May 09 '21 at 14:54

Asinomás

105,651

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Excellent! I thought this was much deeper, using Cayley-Hamilton to bound the degree of the minimal polynomial... – David C. Ullrich May 09 '21 at 15:01
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I think Cayley Hamilton uses something similar in one of the parts of the proof with $(X-\lambda I)^k$ on some of the subspaces but I don't remember very well. – Asinomás May 09 '21 at 15:03

score 0 · Answer 2 · answered May 09 '21 at 15:00

0

Your hunch is correct: The minimal polynomial must be divide $t^m$ and have degree at most $n$.

answered May 09 '21 at 15:00

lhf

216,483

Let $X$ be a nonzero $n × n$ matrix such that $X^m = 0$ for some $m ∈ N$. Let $b ∈ N$ be the smallest number for which $X^b = 0$. Show that $b≤ n$.

2 Answers2