I’m currently considering the question in this popular post: Traces of all positive powers of a matrix are zero implies it is nilpotent, namely:
Let $F$ be a field with characteristic zero and $A$ an $n \times n$ matrix with entries in $F$. If $\text{tr}(A^n)=0$ for all positive integers $n$, then $A$ is nilpotent.
Almost all comments and answers related to this question considered eigenvalues and applied conclusions that proved by theories related to Jordan canonical forms (as far as the proofs that I know).
My question is: Why such methods can be used when we haven’t required that the base field $F$ to be algebraically closed?
Since on non-algebraically closed fields, the eigenvalues may not exists, or the number of them may be less then the size of the matrix $A$, even if we count multiplicity. Moreover, it seems that the theories on JNF cannot be applied in non-algebraically closed fields.
Thank you for your answers and comments! 🙂
Best Answer
There is a finite extension $K$ of $F$ in which the characteristic polynomial of $A$ splits. Therefore, $A$ is (in $K$ similar to a triangular matrix $T$). Thus for each $n \geq 1$, $T^n$ has trace zero.
Now, let $\lambda_1,\ldots,\lambda_r$ be the distinct eigenvalues of $T$ with multiplicities (on the diagonal) $a_1,\ldots,a_r$.
We know that for any polynomial $P$ with $P(0)=0$, $\sum_{i=1}^r{a_iP(\lambda_i)}=0$ (it’s the trace of $P(T)$).
Assume that $\lambda_i \neq 0$. Then there is a polynomial $P$ such that $P(\lambda_j)=0$ if $j \neq i$, $P(\lambda_i) \neq 0$, $P(0)=0$. So $0=\sum_{j=1}^r{a_jP(\lambda_j)}=a_iP(\lambda_i)$ so $a_i=0$, a contradiction since we are in characteristic zero. So every $\lambda_i$ is zero.
Therefore, $T^n=0$. So $A^n=0$.