If we are given a minimal polynomial for a matrix $B^2$ can we deduce the minimal polynomial for $B$ $?$


if the minimal polynomial for $B^2$ is $m(\lambda) = \lambda^4$ then can we deduce the minimal polynomial for $B$ is $m(\lambda) = \lambda^8$


From my example would I be able to deduce $B$ has a Jordan normal form consisting of an $8\times8$ block with eigenvalue $0$

  • $\begingroup$ Not following. If $m(B^2)=0$ then $B^8=0$ so $B=0$. Thus the minimal polynomial for both $B,B^2$ is just $x$. $\endgroup$ – lulu Dec 31 '18 at 17:36
  • $\begingroup$ As another example, $\left( \sqrt 2 +1\right)^2$ has minimal polynomial $x^2-6x+1$ while $\sqrt 2 +1$ has minimal polynomial $x^2-2x-1$. Hard to see a terribly suggestive pattern there. $\endgroup$ – lulu Dec 31 '18 at 17:40
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    $\begingroup$ Certainly if $m(B^2) =0$ then $B$ is a root of $m(x^2)$, hence it is a multiple of the minimal polynomial of $B$. $\endgroup$ – Berci Dec 31 '18 at 17:41
  • $\begingroup$ @lulu I'm trying to use it to find the Jordan form of the Matrix $B$, so I don't follow how you can deduce B = 0 in your first example $\endgroup$ – Scosh_lr Dec 31 '18 at 17:47
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    $\begingroup$ @Scosh_lr Sorry, I was thinking in the context of algebraic numbers. Yes, for matrices we can't deduce that $B=0$ from $B^8=0$. $\endgroup$ – lulu Dec 31 '18 at 17:49

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