# minimal polynomial of a matrix B given minimal polynomial of $B^2$

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

Example:

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

Edit

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

• 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 $x$. – lulu Dec 31 '18 at 17:36
• 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. – lulu Dec 31 '18 at 17:40
• 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 $B$. – Berci Dec 31 '18 at 17:41
• @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 – Scosh_lr Dec 31 '18 at 17:47
• @Scosh_lr Sorry, I was thinking in the context of algebraic numbers. Yes, for matrices we can't deduce that $B=0$ from $B^8=0$. – lulu Dec 31 '18 at 17:49