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Schrodinger operators, operators on manifolds, general differential operators, numerical studies, integral operators, discrete models, resonances, non-self-adjoint operators, random operators/matrices

19 votes

Reasons for difficulty of Graph Isomorphism and why Johnson graphs are important?

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 …
Brendan McKay's user avatar
13 votes

How are eigenvalues and eigenvectors affected by adding the all-ones matrix?

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}( …
Brendan McKay's user avatar
7 votes

An orbit of symmetric polynomials

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 …
Brendan McKay's user avatar
6 votes
Accepted

Computation to differentiate a determinant

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 …
Brendan McKay's user avatar
5 votes

Can we count isospectral graphs?

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 …
Brendan McKay's user avatar
4 votes

Matching polynomials and Ramanujan graphs

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 …
Brendan McKay's user avatar
3 votes

Invertibility of a certain matrix indexed by the Hamming cube

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 …
Brendan McKay's user avatar
3 votes

What are the eigenvectors of the graph Laplacian of a Johnson graph J(n,k)?

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 …
Brendan McKay's user avatar