We know that given a $n$-by-$n$ matrix $A$, and
- its eigen values $\{\lambda_i\}_1^n$
- its trace $\text{tr}(A)$.
then the following holds:
$$ \sum\limits_{i=1}^{n} \lambda_i = \text{tr}(A) $$
then I read from this paper (page 2, equations on the upper left, staring with $\log \det B$ ), which implies the following
given any function $T_k(x)$ defined by Chebyshev polynomial, the following also holds:
$$ \sum\limits_{i=1}^{n} T_k(\lambda_i) = \text{tr}(T_k(A)) \quad (1)$$
where for matrix input, $T_k(A)=\{T_k(A_{ij})\}_{ij}$ (equivalent to applying $T_j$ on $A$ element-wisely.)
My question is:
If we replace $T_j$ with any function $f$ in $(1)$, does $(1)$ still hold?