How can I prove the following statement:

If the characteristic polynomial of matrix $A$ has $n$ zero roots, then $A$ is nilpotent.

Thank you!

  • 2
    $\begingroup$ Is $A$ an $n \times n$ matrix? $\endgroup$ Jul 30 '13 at 13:24
  • 1
    $\begingroup$ See the answers to this question. $\endgroup$
    – jkn
    Jul 30 '13 at 13:25
  • 2
    $\begingroup$ Have you heard of the Cayley-Hamilton theorem? $\endgroup$ Jul 30 '13 at 13:26

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.

  • $\begingroup$ Actually I made that, thanks first of all, now, nothing can be said about the index of nilpotenty, right? $\endgroup$
    – 17SI.34SA
    Jul 30 '13 at 13:31
  • $\begingroup$ You are right, the index of nilpotency is found in the minimal polynomial, not in the characteristic polynomial $\endgroup$
    – vv33d
    Jul 30 '13 at 13:51
  • 1
    $\begingroup$ @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. $\endgroup$
    – quid
    Jul 30 '13 at 14:01
  • $\begingroup$ @quid: many thanks. $\endgroup$
    – 17SI.34SA
    Jul 30 '13 at 14:23

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