# Characteristic polynomial depends only on the trace and determinant of the matrix?

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.
• Maybe I'm wrong but shouldn't you be saying for $n \ge 4$ since, for a $3 \times 3$ case, the characteristics polynomial still stays as a function of trace and determinant. Wiki stands as reference. – Inquest Mar 9 '12 at 18:42
• It's the trace of $A^2$ not the trace of $A$ (I guess the OP wanted an answer with only the trace of $A$ and the determinant of $A$, but maybe I misunderstood the problem). – Davide Giraudo Mar 9 '12 at 18:44