Is there is any relationship between determinant and spectral radius of the matrix?

Question

We all know that determinant is the product of the eigen values of a matrix. I have found some general term for the determinant of the adjacency matrices from some series of graphs. Can i establish some relation between the determinant and spectral radius of those graphs?

Answer 1

Gutman defined the energy of a graph $G$ as the sum of the absolute values of its eigenvalues $$ E(G) = \sum_{\lambda \in \sigma(A)} |\lambda|. $$ If we order the eigenvalues of the adjacency matrix in non-decreasing order $\lambda_1 \leq \lambda_2 \leq \cdots \leq \lambda_n$, the square of the energy satisfies $$ E(G)^2 = \sum_{i=1}^n |\lambda_i|^2 + \sum_{i \neq j} |\lambda_i||\lambda_j|. $$ It's well-known that $\sum_{i=1}^n |\lambda_i|^2$ is twice the number of graph edges, hereafter denoted $2m$. We can use the arithmetic-geometric mean inequality to change the second summation to a product of $n(n-1)$ terms, $$ E(G)^2 \geq 2m + n(n-1)\left(\prod_{i \neq j} |\lambda_i| |\lambda_j|\right)^{1/(n(n-1))}. $$ This product is bounded below by $|\det(A)|^{2/n}$ and the energy satisfies $$ E(G) \geq \sqrt{2m + n(n-1) |\det(A)|^{2/n}}. $$

The largest eigenvalue of the adjacency matrix is always positive, so we can rewrite the above as $$ \lambda_n \geq \sqrt{2m + n(n-1) |\det(A)|^{2/n}} - \sum_{i=1}^{n-1} |\lambda_i|. $$ This is the strongest relationship that I know of between the determinant and spectral radius of an adjacency matrix.