Let $A$ be a $n\times n$ such that $A=PBP^{-1}$ where $B$ is in Jordan normal form with $\lambda_i(k)_j$ Where $i$ is the size, $k$ is the eigenvalue and $j$ the order.

If $A$ was diagonal($i=1$) then $A^n$ in Jordan form has $\lambda_1(k^n)_j$.
If the Jordan form has Jordan blocks bigger then 1, how do we find $A^n$ In Jordan form?


A Jordon block of size $i$ and eigenvalue $k$ has the form $(kI+t)$ where $t^i=0$. Thus $(kI+t)^n$ will just be the truncated binomial expansion:$$\sum_{r=0}^{i-1} {n \choose r} k^{n-r}t^r$$

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  • $\begingroup$ Can you explain why we can treat each Jordan block on his own? Aren’t they all a part of one matrix? $\endgroup$ – razivo Jul 19 at 9:22
  • $\begingroup$ Also, $t$ is a very specific matrix, is there a nice expression or pattern to its powers? $\endgroup$ – razivo Jul 19 at 9:25
  • $\begingroup$ The non-zero entries of $t^i$ will be the diagonal row of $1$'s starting at $(1,1+i)$ and moving down and right. $\endgroup$ – tkf Jul 19 at 9:29
  • $\begingroup$ If I got you, does it mean it will always be an upper diagonal matrix? $\endgroup$ – razivo Jul 19 at 9:33
  • $\begingroup$ Yes. To answer you first question, when you multiply a row by a column, the non-zero entries on the row will never overlap with entries on the column belonging to a different block. $\endgroup$ – tkf Jul 19 at 9:36

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