 # Consider a nilpotent matrix A of index of nilpotency n , then A^n=O and A^(n−1) != O where O denotes null matrix of same order as of A niouzesto 2022-09-11 Answered
Consider a nilpotent matrix A of index of nilpotency n , then ${A}^{n}=O$ and ${A}^{n-1}\ne O$ where O denotes null matrix of same order as of A
Now ${A}^{n}=O\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}det\left(A{\right)}^{m}=0\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}det\left(A\right)=0$
but ${A}^{n-1}\ne O\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}det\left(A\right)\ne 0$
I know I am wrong in the second statement because a matrix not being equal to a null matrix doesn't imply that it has to be necessarily non-singular.
Therefore aren't we allowed to take determinant on both sides for the inequation $A\ne O$ ?
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The statement ${A}^{n-1}\ne O\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}det\left(A\right)\ne 0$ is wrong. In other words, there can be non-zero matrices with zero determinants. For example, consider the matrix
$\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)$

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