Let $A$ be a symmetric matric and q be positive integer. Suppose ${\{a_i\}}_{i\geq0}$ are the eigenvalues of $A$ and $0\leq a_i \leq 1$. I want to bound $$ \sum a_{i}^{2q} $$
Are there some bounds for general sum like above or especially for eigenvalue of symmetric matric?
Let $\|x\|_p$ denote the $p$-norm of the vector $x$, and let $a$ denote the vector of eigenvalues of $A$. Note that your quantity is given by $r = \|a\|_{2q}^{2q}$.
For the case of $q=1$, it is notable that $r = \sum_{i,j = 1}^n a_{ij}^2$, which is to say that $r = \|A\|_F^2$, where $\|\cdot\|_F$ denotes the Frobenius norm, which is easy to work with. It is probably desirable to bound $r$ relative to $\|A\|_F$ more generally.
For $q>1$, let $p = \frac 12 - \frac 1{2q} = \frac{q-1}{2}$. If $x$ denotes the vector of $1$s, then the generalized Hölder's inequality yields $$ r^{1/(2q)} = \|a\|_{2q} = \frac{\|a\|_{2q}\|x\|_{p}}{\|x\|_p} \geq \frac{\|a\|_{2}}{\|x\|_p} = \frac{\|A\|_F}{n^{1/p}} \implies\\ r \geq \frac{\|A\|_F^{2q}}{n^{2q/p}} = \frac{\|A\|_F^{2q}}{n^{4q/(q-1)}} = \frac{\|A\|_F^{2q}}{n^{4(1+\frac 1{q-1})}}. $$ In the other direction, we have $\|a\|_{2q} \leq \|a\|_2$, so that $r \leq \|A\|_F^{2q}$.