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Assume that $F$ be a field ( in special case let $F=\mathbb C$). Assume that $A$ be a $n \times n $ Jordan block in $M_n(F)$. When $m$ is a natural number, what is the Jordan form of $A^m$?

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    Try calculating $A^m$ for some small value of $m,n$, and you should see a pattern. By that I mean, set $n$ to, say, $4$, and calculate $A^m$ for $m=1,2,3,4$. Notice anything? – 2017-01-24
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    is it true that the jordan form of $A^m$ is a jordan block $n \times n$? – 2017-01-24
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    Did you calculate $A^2$ for at least one $4\times 4$ matrix? – 2017-01-24
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    In https://en.wikipedia.org/wiki/Jordan_normal_form, you can find $A^m$ for a jordan block. – 2017-01-24
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    Yes, but it is not a jordan block. – 2017-01-24
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    My quastion is "what is the Jordan form of $A^m $"? – 2017-01-24
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    Assume that $A$ be $2 \times 2$ jordan block such that $a_{11}=1, a_{12}=1, a_{21}=0, a_{22}=1$, then for any $m$, assume that $A^m=[b_{ij}]$. It is clear that $b_{11}=1, b_{12}=m, b_{21}=0, b_{22}=1$, what is the jordan form of $A^m$? – 2017-01-24
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    I said the jordan form of $A^m$ is $0$? Really? Where did I say that? – 2017-01-24
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    the last comment is not for you, previous comment is deleted. – 2017-01-24

1 Answers 1

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Denote by $J_n(\lambda)$ the Jordan block of size $n$ for an eigenvalue $\lambda$. Hence $A=J_n(\lambda)$ for some $\lambda$.

(I) When $A$ is not nilpotent and the field has characteristic zero, the Jordan form of $A^m$ is just $J_n(\lambda^m)$. That is, the Jordan form itself is a single Jordan block of size $n$.

(II) When $A$ is not nilpotent and the field has finite characteristic, things get messier. For instance, over $GF(2)$, consider the case where $n=3,m=2$ and $$ A=\pmatrix{1&1\\ &1&1\\ &&1},\quad A^m=A^2=\pmatrix{1&0&1\\ 0&1&0\\ 0&0&1} \sim \pmatrix{1&1\\ 0&1\\ &&1}. $$ So, the $A^m$ in this case has Jordan blocks of sizes smaller than $n$. The general case depends on $n,m$, the actual value of the eigenvalue as well as the characteristic of the field. It looks complicated and I'm not sure if there is an easy answer.

(III) When $A$ is nilpotent, there are two possibilities:

  1. $1\le m
  2. $m\ge n$. Then $T=A^m=0$ and the Jordan form of $T$ is just itself, the zero matrix.
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    is there any reference for first case.? "When $A$ is not nilpotent," is it true for any arbitrary field? – 2017-01-24
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    @RezaFallah-Moghaddam It's so trivial that I don't think anyone would state it as a lemma in a reference book. Clearly all eigenvalues of $T=A^m$ are equal to $\lambda^m$. And the index of nilpotency of $N=T-\lambda^m I$ is $n$ (that is, $N^n=0$ but $N^{n-1}\ne0$). So the largest Jordan block of $T$ has size $n$, i.e. the Jordan form of $T$ consists only of a single Jordan block. – 2017-01-24
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    @RezaFallah-Moghaddam Now come to think of it, the answer really isn't trivial if the field has finite characteristic. I'll see think about it when I have time. – 2017-01-24