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

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?

• Caley Hamilton Theorem. May 8 '19 at 8:46

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.
• 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? May 8 '19 at 9:15
• 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 $\lambda$ May 8 '19 at 9:17
• well, I understand as follows: The polynomial is $\det(A-\lambda I_n)$ if $\lambda = 0$ then $\det(A)$ is the polynomial. May 8 '19 at 9:20
• 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? May 8 '19 at 9:21
• @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 May 8 '19 at 9:24