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Prove that a square matrix over an algebraic closed field is nilpotent if and only if all their eigenvalues are zero.

A nilpotent matrix is that $A^k = 0 $ for some k.

Some idea?

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    $\begingroup$ Caley Hamilton Theorem. $\endgroup$ May 8 '19 at 8:46
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Hint: If all the eigenvalues are zero, then what is the characteristic polynomial of the matrix? Apply Cayley-Hamilton Theorem now.

Conversely, if $A^k = 0$ for some $k> 0, k \in \mathbb{Z}$, then suppose $ \lambda$ is an eigenvalue with eigenvector $v \neq 0 $. So, $ Av = \lambda v$. Keep applying A both sides of the equation, then you get $ A^k = \lambda^k v$. Conclude.

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  • $\begingroup$ The conversely is clear to me. But, for the first part, the characteristic polynomial of a square matrix is given by $ \det(A-\lambda I_n) $ right? so we have just a number if $\lambda = 0$, the determinant of the matrix, so, how to conclude? $\endgroup$ May 8 '19 at 9:15
  • $\begingroup$ No, it is a polynomial in $'\lambda'$, remember $\lambda$ is unknown here. If you plug in some value of $\lambda$, then ofcourse it is a number. In fact, that's why it is called a polynomial in the first place. Better use $t$ for the variable in the polynomial instead of using $\lambda$ $\endgroup$
    – P-addict
    May 8 '19 at 9:17
  • $\begingroup$ well, I understand as follows: The polynomial is $\det(A-\lambda I_n)$ if $\lambda = 0$ then $\det(A)$ is the polynomial. $\endgroup$ May 8 '19 at 9:20
  • $\begingroup$ Now, what are the roots of the characterstic polynomial? You'll observe that they are precisely the eigenvalues of $A$ by definition. So, if the only eigenvalue of a matrix is $0$, then the characteristic polynomial is of course $t^n$. Now, what does Cayley-Hamilton theorem tell you? $\endgroup$
    – P-addict
    May 8 '19 at 9:21
  • $\begingroup$ @José Marín det(A) is not a polynomial, when you plug in some value in the place of variable in a polynomial, it gives you a number. Here, you are plugging $0$ for the variable $\lambda$ in the characteristic polynomial, so it gives you a number, not a polynomial $\endgroup$
    – P-addict
    May 8 '19 at 9:24

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