Let N be an n*n matrix on a field F with nilpotency index n i.e. $\displaystyle N^n=0$ but $\displaystyle N^{n-1}\neq 0$ . Now show that N doesn't have square root i.e. there is no matrix like A where $\displaystyle A^2=N$.
Let N be an n*n matrix on a field F with nilpotency index n i.e. $\displaystyle N^n=0$ but $\displaystyle N^{n-1}\neq 0$ . Now show that N doesn't have square root i.e. there is no matrix like A where $\displaystyle A^2=N$.
Since the conclusion is negative, that is, "there does not exist", an indirect proof is indicated. Suppose the conclusion where false. That is, suppose there exist A such that $\displaystyle A^2= N$. Then $\displaystyle A^{2n}= N^n= 0$. Now, what does that say about $\displaystyle A^n$?