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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?

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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 $\lambda_2 \lambda_3 \lambda_4 + \lambda_1 \lambda_3 \lambda_4 + \lambda_1 \lambda_2 \lambda_4 + \lambda_1 \lambda_2 \lambda_3$.

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$$\begin{align} \det(A)^n\det(tI-A)&=\det(t\det(A)I-\det(A)A)\\ &=\det(tA^*-\det(A))\det(A)\\ &=(-1)^n\det(A)\det(\det(A)-tA^*) \end{align}$$

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 $q(t)$.

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