Formulas for the (top) coefficients of the characteristic polynomial of a matrix The characteristic polynomial of a matrix $A$ is defined as:
$$\chi(A) = \det(xI-A) = \sum_{i=0}^n (-1)^i\cdot \operatorname{tr}^{(i)}(A) \cdot x^{n-i}$$
The trace is the sum of the eigenvalues of a matrix, tr(A) = tr(1)(A).  It is also the sum of the diagonal entries:
$$\operatorname{tr}(A) = \sum_{i=1}^n A_{ii}$$
The sum of the products of pairs of eigenvalues is like the next trace.
Is this formula valid?
$$\operatorname{tr}^{(2)}(A) = \sum_{1 \leq i < j \leq n } A_{ii} A_{jj} - A_{ij} A_{ji}$$
What about this one?
$$\operatorname{tr}^{(2)}(A) = \tfrac12(\operatorname{tr}(A)^2 - \operatorname{tr}(A^2))$$
Are there corresponding formulas for the next one, tr(3)?

Regarding your last question on $\text{tr}^{(3)}$, there is a kind of recursive definition for all $\text{tr}^{(k)}$ given by:
$$
\text{tr}^{(k)}=-(1/k)\left(\sum_{j=1}^k (-1)^{j} \text{Tr}(A^j)\text{tr}^{(k-j)}\right).
$$
