# Expressing the trace of the adjugate matrix in terms of eigenvalues

Can the trace of $$A^*$$ (the adjugate of $$A$$) be expressed in terms of the eigenvalues of $$A$$?

For example, the characteristic polynomial of a matrix $$A \in M_3$$ is:

$$\det(XI_3 - A) = X^3 - \mathrm{tr}(A)X^2 + \mathrm{tr}(A^*)X - \det(A)$$ and using Viete we get:
$$\mathrm{tr}(A^*) = \lambda_1 \lambda_2 + \lambda_2 \lambda_3 + \lambda_3 \lambda_1$$

Also, for $$A \in M_2$$ we have $$\mathrm{tr}(A^*)=\mathrm{tr}(A)=\lambda_1 + \lambda_2$$

But is there any way to express $$\mathrm{tr}(A^*)$$ in terms of the eigenvalues of A for a $$4 \times 4$$ matrix?

Or even for a $$n \times n$$ matrix?

If $$A$$ is invertible, $$A^* = \det(A) A^{-1}$$, so $$\text{trace}(A^*) = \det(A) \text{trace}(A^{-1}) = \left(\prod_j \lambda_j \right) \sum_j 1/\lambda_j = \sum_j \prod_{i \ne j} \lambda_i$$ (where the eigenvalues of $$A$$ are $$\lambda_j$$, counted by algebraic multiplicity). By continuity, $$\text{trace}(A^*) = \sum_j \prod_{i\ne j} \lambda_i$$ for all square matrices.
Thus for $$n=4$$ it's $$\lambda_2 \lambda_3 \lambda_4 + \lambda_1 \lambda_3 \lambda_4 + \lambda_1 \lambda_2 \lambda_4 + \lambda_1 \lambda_2 \lambda_3$$.
\begin{align} \det(A)^n\det(tI-A)&=\det(t\det(A)I-\det(A)A)\\ &=\det(tA^*-\det(A))\det(A)\\ &=(-1)^n\det(A)\det(\det(A)-tA^*) \end{align}
Therefore, if $$p(t)$$ is the characteristic polynomial of $$A$$, then $$q(t)=(-1)^{n}\det(A)^{n-1}t^np(\det(A)/t)$$ is the characteristic polynomial of $$A^*$$.
If $$\lambda_i$$ are the roots of $$p(t)$$, then $$\det(A)/\lambda_i=\frac{\prod_{k=1}^n\lambda_k}{\lambda_i}$$ are the roots of $$q(t)$$.