sum of the eigenvalues = trace($A$)? Is it true that for a square matrix $A$, all of whose eigenvalues exist in the base field, sum of the eigenvalues = trace($A$)?
The result holds in all the matrices I've studied.
 A: Yes. Just look at the characteristic polynomial (say of degree n). Trace=-the coefficient of the term of $x^{(n-1)}$ which is also the sum of the roots of the characteristic polynomial (the coefficient of the term $x^{(n-1)}$ of any monic polynomial of degree $n$ is the sum of its roots with a minus sign.).
A: Here is the way I would show it. $A = VDV^{-1}$ which means $tr(A) = tr(VDV^{-1}).$ But trace has a cyclic property. so $tr(VDV^{-1}) = tr(DV^{-1}V) = tr(D)$ so I think your assertion is correct!
A: Let $A \in M_n(k)$ and let $\overline{k}$ be an algebraic closure of $k$. Because the charateristic polynomial $p_A$ of $A$ splits into linear factors in $\overline{k}[X]$, $A$ is trigonalizable in $M_n(\overline{k})$: there exists $P \in GL_n(\overline{k})$ such that $A=PTP^{-1}$ for some triangular matrix $T \in M_n(\overline{k})$. Now $\text{tr}(A)=\text{tr}(T)$ is the sum of the eigenvalues of $T$, and a fortiori of $A$.
Notice that $\text{tr}(A)$ is a symmetric polynomial with respect to the eigenvalues of $A$ so you can express $\text{tr}(A)$ with respect to the elementary symmetric polynomials and finally with respect to the coefficients of $p_A$. Because $p_A \in k[X]$, $\text{tr}(A) \in k$.
A: Since all eigenvalues exist in the base field, $A$ will be similar to a matrix $J$ in Jordan Canonical Form, with eigenvalues of $A$ along the diagonal, and assertion easily follows. (Hint: Use $tr(AB) = tr(BA)$ for $A, B$ square matrices)
