This note, I copied from Yoshi on MathStackExchange:
Let be an complex nilpotent matrix. Then we know that because all eigenvalues of must be 0, it follows that for all positive integers .
What I would like to show is the converse, that is, if for all positive integers , then is nilpotent. I tried to show that 0 must be an eigenvalue of , then try to show that all other eigenvalues must be equal to 0. However, I am stuck at the point where I need to show that .
May I know of the approach to show that is nilpotent?
The answer of Alvarez:
If the eigenvalues of are , , , then the eigenvalues of are , , . It follows that if all powers of have zero trace, then .
Using [Newton’s identities] to express the elementary symmetric functions of the ‘s in terms of their power sums, we see that all the coefficients of the characteristic polynomial of (except that of greatest degree, of course) are zero. This means that is nilpotent.
The answer of JBC:
Assume that for all , where is a matrix. We consider the eigenvalues in .
Suppose is not nilpotent, so has some non-zero eigenvalues .
Let the multiplicity of then
So we have
So the system has a unique solution which is . Contradiction.
The answer of Qiaochu Yuan: Here is an argument that does not involve Newton’s identities, although it is still closely related to symmetric functions. Write
where are the eigenvalues of . As a meromorphic function, has poles at the reciprocals of all of the nonzero eigenvalues of . Hence if identically, then there are no such nonzero eigenvalues.
The argument using Newton’s identities, however, proves the stronger statement that we only need to require for . Newton’s identities are in fact equivalent to the identity
where . To prove this identity it suffices to observe that
and differentiating both sides gives
(The argument using Newton’s identities is also valid over any field of characteristic zero.)