The question is an extension of the Prove that $A$ is similar to $A^n$ based on A's Jordan form.

Let $J$ be Jordan block of any form.
In what circumstances Jordan form of power $J^n$ has the same superdiagonal as $J$?

  • 1
    $\begingroup$ One should maybe mention that the result implies the following observation: On a finite dimensional vector space, any linear operator with $1$ as sole eigenvalue is similiar to all its powers. $\endgroup$ – MooS Jul 12 '17 at 13:58

Let $m$ be the size of the block. Then $J = \lambda + N$ with $N^{m-1}\neq 0$ and $N^m=0$.

For $\lambda = 0$ the answer is clearly negative.

For $\lambda \neq 0$, you can just extend the given proof:

$$J^n = \lambda^n + n\lambda^{n-1}N + \dotsb + \binom{n}{m-1}N^{m-1},$$ hence $(J^n-\lambda^n)^{m-1} = (n\lambda^{n-1}N)^{m-1} \neq 0$, i.e. the Jordan form of $J^n$ is given by one Jordan block of size $m$, i.e. the superdiagonal is the same as for $J$, namely $N$.

  • $\begingroup$ Thank you, Moos. only identity matrix is lost beside $\lambda^n$, namely $\lambda^nI$, $\endgroup$ – Widawensen Jul 12 '17 at 14:00
  • $\begingroup$ It is pretty common to write $A+\lambda$ instead of $A+\lambda I$, because you simply take the polynomial $X+\lambda$ and plug in $A$. In dimension one, would you write $a+\lambda \cdot 1$ instead of $a+\lambda$? $\endgroup$ – MooS Jul 12 '17 at 14:02
  • $\begingroup$ ok. I didn' know that.. however it looks a little strange.. $\endgroup$ – Widawensen Jul 12 '17 at 14:03

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