Distance matrix of connected graph always invertible?

Asked Mar 26 '14 at 14:46

Active Mar 27 '14 at 06:37

Viewed 622 times

I know there's a question elsewhere about distance matrix for points on Euclidean plane, but I'm not sure if that one was relevant.

Anyway, given a connected (simple) graph G with $n$ vertices $v_1,...,v_n$, the distance matrix is a $n \times n$ matrix whose $(i,j)$ entry is the distance between $v_i$ and $v_j$ (note: this is NOT the adjacency matrix!). The distance of a vertex to itself is taken to be $0$.

So I know that the matrix is real symmetric and hence diagonalizable. However, I was wondering if $0$ is ever an eigenvalue. That is, is the distance matrix always invertible?

asked Mar 26 '14 at 14:46

suncup224

2,859

4

Consider the cycle graph on 4 vertices. – Mar 26 '14 at 19:40
It is invertible when $G$ is a tree though. – Chris Godsil Mar 27 '14 at 01:07
http://www.sciencedirect.com/science/article/pii/0024379574900317 – Christiaan Hattingh Mar 27 '14 at 06:46
http://math.stackexchange.com/questions/79779/properties-of-zero-diagonal-symmetric-matrices – Christiaan Hattingh Mar 27 '14 at 06:49

Distance matrix of connected graph always invertible?

0 Answers0

Linked