Expression relating second invariant to eigenvalues matrix

Question

Consider a matrix $A\in\mathbb{R}^{3\times 3}$. We know that it has the three invariants $$ I_1 = tr(A),\\ I_2 = 1/2(tr(A)^2-tr(A^2)),\\ I_3 = \det(A).$$ Also, the first and third can be expressed as function of the eigenvalues of $A$ as $$ tr(A)= \Sigma\lambda_i, \\ \det(A) = \Pi\lambda_i.$$ Now, my question is, is there is a general expression relating $I_2$ and $\lambda_i$?

Answer 1

Why not just expand the definition?

$$I_2=\frac{1}{2}(\operatorname{tr}(A)^2-\operatorname{tr}(A^2)) = \frac{1}{2}\left(\left(\sum\lambda_i\right)^2 - \sum\lambda_i^2\right)=\\ =\frac{1}{2}\left(\sum\limits_{i\neq j}\lambda_i \lambda_j\right)=\sum\limits_{i< j}\lambda_i \lambda_j = e_2(\lambda_1,\dots,\lambda_n)$$

Here, $e_2$ is an elementary symmetric polynomial.

In general, for an $n\times n$ matrix, its $k$-th invariant $I_k$ is related to the eigenvalues via the $k$-th elementary symmetric polynomial:

$$I_k=e_k(\lambda_1, \dots, \lambda_n)$$

Answer 2

So I already found the answer for whoever is still interested. We have that $$I_2(A) = \frac{I_1(A^{-1})}{I_3(A^-1)}=\det(A)tr(A^{-1})=(\Pi\lambda_i)(\Sigma\lambda_i^{-1}) = \Sigma_{i<j}\lambda_i\lambda_j.$$