The characteristic polynomial of a 2×2 matrix can be expressed in terms of the trace(T) and determinant(D):
$$\lambda^2 - T \lambda + D = 0$$
The one for 3x3 matrix can be expressed in terms of T and D:
$$\lambda^3 - T \lambda^2 + \frac{T^2 - Tr(A^2)}{2}\lambda - D = 0$$
I am trying to find the one for 4x4 matrix. It looks like this,
$$\lambda^4 - T \lambda^3 + \frac{T^2 - Tr(A^2)}{2}\lambda^2 +(something)\lambda + D = 0$$
But I'm not sure what the coefficient of $\lambda$ is... I think it has $Tr(A), Tr(A^3)$, and $Tr(A)^3$ but not sure..
Please help. What is the coefficient of $\lambda$ in terms of T and D (and/or $T^n$, $Tr(A^n)$?
For those who are interested, from the links in the comments, the coefficients are
Let $T_n$ be $trace(A^n)$ and $D_n$ is a coefficient of characteristic polynomial.
$$\lambda^n + D_1 \lambda^{n-1} + D_2 \lambda^{n-2} + D_3 \lambda^{n-3} ... + D_n = 0$$
$$D_1 = -T_1$$
$$D_2 = -\frac{1}{2!} (T_2 - T_1^2)$$
$$D_3 = -\frac{1}{3!} (2 T_3 - 3 T_2 T_1 + T_1^3)$$
$$D_4 = -\frac{1}{4!} (6 T_4 - 3 T_2^2 - 8 T_1 T_3 + 6 T_1^2 T_2 - T_1^4)$$
$$D_5 = -\frac{1}{5!} (24 T_5 - 20 T_2 T_3 - 30 T_1 T_4 + 15 T_1 T_2^2 + 20 T_1^2 T_3 - 10 T_1^3 T_2 + T_1^5)$$
and so on.