Smallest eigenvalue of a variant of the line graph Laplacian

Question

I would like to calculate a lower bound on the minimum eigenvalue of the matrix $$M = e_1 e_1^\top + L$$

where $L$ is the Laplacian of the line graph on $T$ vertices: $$1 -2 -3 -4-\ldots -T $$ and $e_1$ is the unit vector $(1,0,0,\ldots, 0)$.

Equivalently, $$M = \begin{pmatrix}2 & -1 & 0 & 0 & \cdots & 0 \\ -1 & 2 & -1 & 0 &\cdots & 0 \\ 0 & \ddots & \ddots & \ddots & & \vdots \\ \vdots & & & & & \vdots \\ 0 & \ldots & \ldots &-1&2& -1 \\ 0 & \ldots & \ldots &0 &-1& 1 \end{pmatrix}$$

Any suggestions?

Well, if its just the Laplacian the min eigenvalue is 0. This is a rank 1-update but most of the literature is about upper-bounding the change in eigenvalue for a rank-1 update — Chinmay Nirkhe, Jun 10 '23 at 18:03
This matrix is actually the matrix of the discretized second derivative operator with Dirichlet-Neumann boundary conditions. The link gives an explicit formula for the eigenvalues. — Ben Grossmann, Jun 10 '23 at 19:02

score 0 · Answer 1 · answered Jun 11 '23 at 19:36

Thanks to the many commentators. An exact solution is given in this result by da Fonseca (eigenvalue, eigenvector) pairs $(\lambda_k, v_k)$ with

$$\lambda_k = 4 \sin^2 \left( \frac{2k+1}{4k+2} \pi \right)$$

and

$$v_k(j) = \sin \left(\frac{2k+1}{2n+1} \cdot j \pi \right).$$

Note that the vector $v_k(j)$ is not normalized but a normalization can be calculated using the Sum of Cosines of Multiples of an Angle formula and the identity $1 - \cos(\theta) = 2 \sin^2(\theta/2)$. The normalization of $v_0$ is $\frac{\sqrt{2n+1}}{2}$.

Smallest eigenvalue of a variant of the line graph Laplacian

1 Answers1