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?
Best Answer
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 $\lambda_2 \lambda_3 \lambda_4 + \lambda_1 \lambda_3 \lambda_4 + \lambda_1 \lambda_2 \lambda_4 + \lambda_1 \lambda_2 \lambda_3$.