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The eigenvalues of $A+\lambda$ are $\{\mu_j+\lambda\}$ which are positive by assumption. So $$\frac{d}{d\lambda} \log\det(A+\lambda) = \frac{d}{d\lambda} \sum_j \log (\lambda+\mu_j) = \sum_j (\lambd …

These questions are all answered in the Wikipedia article Johnson graph. As Chris noted, it doesn't matter if you consider the adjacency matrix or the Laplacian matrix. The eigenvectors stay the same …

Noting Pietro's finite check, I can report that all symmetric polynomials $p(x_1,x_2,x_3)$ up to degree 18 inclusive such that $\mathcal{L}p$ is symmetric are linear combinations of $1,\mathcal{L}1,\m …

A cute fact that is trivial to prove is this: define the characteristic polynomial of a matrix $M$ by $\phi_M(x) = |xI-M|$. Then for any $A$ and any $s$, $$\phi_{A+sJ}(x) = (1-s)\phi_A(x)+s\phi_{A+J}( …

Johnson graphs do not cause difficulty to existing programs. Actually they are rather easy; nauty can handle them up to tens of millions of vertices, and so can other programs such as Traces and Bliss …

The moments of the adjacency matrix eigenvalues count closed walks in the graph, while the moments of the matching polynomial roots count tree-like closed walks. When the graph has few short cycles, a …

The answer to the first question is unknown. It is even unknown if the fraction of graphs on $n$ vertices determined by their spectra converges to 0, or 1, or something between, or even converges at …

For any $i,j,k$, the automorphism group of $A$ is transitive on the set of pairs $(I,J)$ such that $|I|=i, |J|=j, |I\cap J|=k$. Therefore the same is true of the inverse (if it exists). That is, the …