TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-28

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The Laplacian is positive semidefinite and sends the all-ones vector to zero

Statement

Let $G$ be a finite simple graph on $n$ vertices, let $L (G)$ be its Laplacian matrix, and let $1 \in R^{n}$ be the all-ones column vector. Then:

$x^{T} L (G) x \geq 0$ for every $x \in R^{n}$ ;
$L (G) 1 = 0$ .

In particular, $L (G)$ is positive semidefinite.

Facts & Assumptions

Given: A finite simple graph $G$ with Laplacian $L$ and an oriented incidence matrix $B$ .

[L1]

The Laplacian satisfies $L = B B^{T}$ (The Laplacian equals $B B^{T}$ for every oriented incidence matrix $B$ ).

[F1]

Every column of an oriented incidence matrix has one $1$ and one $- 1$ (An oriented incidence matrix of a finite simple graph).

Proof

technique · direct

1.1

For every $x \in R^{n}$ , [L1] gives $x^{T} L x = x^{T} B B^{T} x = (B^{T} x)^{T} (B^{T} x)$ . The right-hand side is a sum of squares of real numbers, so it is nonnegative.

L1algebra

1.2

Let $1$ be the all-ones vector. Because each column of $B$ has one $1$ and one $- 1$ , [F1] implies that every column sum of $B$ is $0$ , so $B^{T} 1 = 0$ . Using [L1] again gives $L 1 = B B^{T} 1 = B 0 = 0$ .

F1L1algebra

2.1

Step 1.1 proves positive semidefiniteness, and step 1.2 proves that the all-ones vector lies in the kernel.

step 1.1step 1.2∎

Depends on

Used by

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

Richard P. Stanley, MIT 18.314 handout, Lemma 1.6(a) (standard reference, not scraped)