Let $A$ be an $n\times n$ matrix with entries in an arbitrary field $k$.
Is the characteristic polynomial $\det(tI_n-A)$ dependent only on the trace and determinant of $A$?
Let $A$ be an $n\times n$ matrix with entries in an arbitrary field $k$.
Is the characteristic polynomial $\det(tI_n-A)$ dependent only on the trace and determinant of $A$?
No, look at $\left[\matrix{1&0&0\cr0&1&0\cr0&0&0 }\right]$ and $\left[\matrix{2&0&0\cr0&0&0\cr0&0&0 }\right]$. Both have determinant 0 and trace 2 but they do not have the same eigenvalues. Similar examples show it's not true for any $n>3$.
It's not true for $n\geq 3$. Take the matrix $A:=\pmatrix{0&1&0\\\ 0&0&1\\\ a_0&a_1&a_2}$ (companion matrix). Its characteristic polynomial is $-X^3+a_2X^2+a_1X+a_0$ and its determinant and trace are respectively $a_0$ and $a_2$. But the characteristic polynomial depends on $a_1$, which is independent of the trace and the determinant.