How can I prove the following statement:
If the characteristic polynomial of matrix $A$ has $n$ zero roots, then $A$ is nilpotent.
Thank you!
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2Is $A$ an $n \times n$ matrix? – Mark Bennet Jul 30 '13 at 13:24
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1See the answers to this question. – jkn Jul 30 '13 at 13:25
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2Have you heard of the Cayley-Hamilton theorem? – Mark Bennet Jul 30 '13 at 13:26
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If the characteristic polynomial has only zero roots, it has the form $\lambda^n = 0$ (or $(-\lambda)^n = 0$, depending on convention). By the theorem of Cayley-Hamilton, a square matrix $A$ fulfills its own characteristic equation (even if it's not diagonalizable), therefore $A^n = 0$ (zero matrix), therefore $A$ is nilpotent.
ccorn
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Actually I made that, thanks first of all, now, nothing can be said about the index of nilpotenty, right? – 17SI.34SA Jul 30 '13 at 13:31
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1@17SI.34SA: no, nothing (except of course it is at most n). Note that a matrix is nilpotent if and only if its minimal polynomial is of this form (over a field and a domain, at least, but I assume you are in this situation). Also, consider a matrix, 0 except for 1 on the fist upper diagonal, compute the sqaure of this matrix and observe you get one 0 except for 1 on second upper diagonal and so on. – quid Jul 30 '13 at 14:01
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You are right, the index of nilpotency is found in the minimal polynomial, not in the characteristic polynomial – vv33d Jul 30 '13 at 13:51