Search Results

The smallest eigenvalue can go up or down when an edge is removed. For "down": $G=K_n$ for $n\ge 3$. For "up": Take $K_n$ for $n\ge 1$ and append a new vertex attached to a single vertex of the orig …

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}( …

When I had to do this with a couple of billion matrices, I computed the traces of some powers before going for the full test. A good method is to compute $\mathrm{tr} \,((A+xI)^{2^i})$ for $i=1,2,\ldo …

You are asking for relationships between the maximum independent set and the eigenvalues. If you search with those terms you will find several. … Haemers proved that the maximum size of an independent set is bounded above by $$ n\frac{-\lambda_1\lambda_n}{\delta^2-\lambda_1\lambda_n},$$ where $\lambda_1,\lambda_n$ are the largest and smallest eigenvalues …

Let each $A_i$ be a matrix with all entries 0 except for the $(i,i+1)$ entry which is 1, where $i+1=1$ if $i=n$. The characteristic polynomial of $\sum_j a_j A_j$ is $x^n - \prod_j a_j$. I believe tha …