Computing the sum of $\log^2(\text{eigenvalues})$

Question

Is there a method to find $$\sum_{\{\lambda \text{ is an eigenvalue}\}} \log^2(\lambda)$$ of a symmetric matrix without explicitly calculating the eigenvalues?

Answer 1

We need to deal with positive definite real symmetric matrices, otherwise $\log^2(\lambda)$ for $\lambda\in\text{Spec}(M)$ might be undefined. We may notice that for any $k\in\mathbb{N}$ the identity $$ \sum_{\lambda\in\text{Spec}(M)}\lambda^k = \text{Tr}(M^k) $$ holds as a consequence of the Hamilton-Cayley theorem. We have $$ \log^2(1-x)=\sum_{n\geq 2}\frac{2 H_{n-1}}{n}\,x^n,\qquad H_n=\frac{1}{1}+\frac{1}{2}+\ldots+\frac{1}{n} $$ for any $x\in(-1,1)$, hence by assuming that the spectral radius of $M$ is less than $2$ we have $$ \sum_{\lambda\in\text{Spec}(M)}\log^2(\lambda) = \text{Tr}\sum_{n\geq 2}\frac{2H_{n-1}}{n}(I-M)^n. $$