Let $V$ be a finite-dimensional vector space over a field $\mathbb{F}$, and $T:\,V \longmapsto V$ a linear map.
The trace of $T$, $tr(T)$, is the trace of any matrix $A \in M_n(\mathbb{F})$ related to $T$ with respect any basis of $V$.
I have to show that
Lemma If $T$ is a nilpotent linear map, then $tr(T)=0$.
proof. We know that if $T:\,V \longmapsto V$ a linear nilpotent map of a a finite-dimensional vector space, then then there exists a basis of $V$ such that the matrix representation of $T$ is upper triangular with zero diagonal elements. Then $tr(T)=0$.
QUESTION: My teacher uses a different proof of this lemma, but I think there is something missing...
proof.2 First of all it is easy to show that the trace of a linear transformation is also equal to the sum of the roots of his characteristic polynomial $p_T$ in a splitting field $K$ of $p_t$ over $\mathbb{F}$.
Also if $T:\,V \longmapsto V$ a linear nilpotent map then its only eigenvalue is 0.
Then $tr(T)=0$.
I think this proof is true only if the field $\mathbb{F}$ is algebraically closed: if so, $\mathbb{F}$ contains all the roots of $p_T$. Then every root of $p_T$ is an eigenvalue of $T$ and so it must be zero.
But if $\mathbb{F}$ is not algebraically closed, it is possible that there are some roots of $p_T$ that are not in $\mathbb{F}$. So we are not sure that every root of $p_T$ is an eigenvalue of $T$, and then I cannot use the fact that every eigenvalue of $T$ is 0 to show that $tr(T)=0$. Am I wrong??
Thanks a lot!!
The argument you describe is perfectly correct. Your matrix can be viewed as having elements in $\mathbf{F}$ or as in $\overline{\mathbf{F}},$ since $\mathbf{F} \subseteq \overline{\mathbf{F}}.$ The argument shows that if you think of the elements as lying in the closure, the trace is zero. But however you think of them, $0\to 0$ under the inclusion $\mathbf{F} \to \overline{\mathbf{F}}.$