Let $\lambda$ be the largest eigenvalue of $\boldsymbol{A}\in\mathbb{C}^{n\times n}$ ($\boldsymbol{A}$ is hermitian). Is
$$\lambda = \sum_{\boldsymbol{k}\in\mathbb{Z}^{|\textrm{vec}(\boldsymbol{A})|}}w_{\boldsymbol{k}}\prod_{i=1}^{|\boldsymbol{k}|}\textrm{vec}(\boldsymbol{A})_i^{\boldsymbol{k}_i}$$
where $w_{\boldsymbol{k}}\in\mathbb{C}$ and $\textrm{vec}(\boldsymbol{A})$ vectorizes an array? So if the dimensions of two matrices are the same then the weights of the above sum is also the same?
Is there a way to calculate these weights (I know there will be an infinite amount of them, but lets say I am interested in a specific one). Furthermore is $w_{\boldsymbol{k}} \neq 0 ~\forall ~\boldsymbol{k}\in \mathbb{Z}^{|\textrm{vec}(\boldsymbol{A})|}$?
There cannot be such a linear dependence of eigenvalues on the coefficients of the matrix. Assume there is such a dependence, $$ \lambda =\sum_{i,j=1}^n w_{ij}a_{ij}. $$ First, put all triangular matrices into this formula. As the eigenvalues of a triangular matrix only depend on the diagonal elements, one can prove that $w_{ij}=0$ for all $i,j\ne 1$.
Now, take non-zero matrices with zero diagonal. The formula will predict that the largest eigenvalue is zero, which can be easily drawn into a contradiction.