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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?

José Marín
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1 Answers1

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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.

P-addict
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  • 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? – José Marín May 08 '19 at 09:15
  • 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 $\lambda$ – P-addict May 08 '19 at 09:17
  • well, I understand as follows: The polynomial is $\det(A-\lambda I_n)$ if $\lambda = 0$ then $\det(A)$ is the polynomial. – José Marín May 08 '19 at 09:20
  • 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? – P-addict May 08 '19 at 09:21
  • @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 – P-addict May 08 '19 at 09:24
  • is it clear now? – P-addict May 08 '19 at 09:24
  • in my opinion, if it is not clear to you, perhaps you should go for one more read of the theory. – P-addict May 08 '19 at 09:28
  • It is not clear for me the part $t^n$ but I will working on it. – José Marín May 08 '19 at 09:29
  • If $0$ is the only root of a n-degree polynomal, tell me what the polynomial can be? – P-addict May 08 '19 at 09:33
  • Let us continue this discussion in chat. – P-addict May 08 '19 at 09:35