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 O where O denotes null matrix of same order as of A
Now A n = O det ( A ) m = 0 det ( A ) = 0
but A n 1 O det ( A ) 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 O ?
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Answers (1)

trabadero2l
Answered 2022-09-12 Author has 15 answers
The statement A n 1 O det ( A ) 0 is wrong. In other words, there can be non-zero matrices with zero determinants. For example, consider the matrix
( 1 0 0 0 )

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