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A finite simple graph is connected if and only if its algebraic connectivity is positive

Statement

Let $G$ be a finite simple graph with at least two vertices. Then $G$ is connected if and only if its algebraic connectivity is positive.

Facts & Assumptions

Given: A finite simple graph $G$ with at least two vertices.

[F1]

The algebraic connectivity of $G$ is the second-smallest Laplacian eigenvalue $μ_{2} (G)$ (The algebraic connectivity of a finite simple graph).

[L1]

The multiplicity of the Laplacian eigenvalue $0$ equals the number of connected components (The multiplicity of the Laplacian eigenvalue $0$ equals the number of connected components).

Proof

technique · direct

1.1

If $G$ is connected, then [L1] says that the eigenvalue $0$ has multiplicity $1$ , so the next Laplacian eigenvalue $μ_{2} (G)$ is strictly positive. By [F1], the algebraic connectivity is positive.

L1F1

1.2

If the algebraic connectivity is positive, then [F1] gives $μ_{2} (G) > 0$ , so $0$ occurs only once in the Laplacian spectrum. By [L1], the number of connected components is therefore $1$ , which means that $G$ is connected.

F1L1

2.1

Steps 1.1 and 1.2 prove the two directions of the equivalence.

step 1.1step 1.2∎

Depends on

Used by

FALSE: positive second Laplacian eigenvalue characterises 2-connectivity False statement

Dependency tree · two levels

7 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

O. Pikhurko, Algebraic Methods in Combinatorics, Section 14.2 (standard reference, not scraped)