1
$\begingroup$

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 $A$?

$\endgroup$
2
$\begingroup$

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 $n>3$.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Inquest Mar 9 '12 at 18:42
  • $\begingroup$ 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). $\endgroup$ – Davide Giraudo Mar 9 '12 at 18:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.