core27 min studyLesson 220 in Mathematics

Graph Laplacians and connectivity

Express edge disagreements as a positive quadratic form and read connected components from the Laplacian kernel.

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What you’ll understand

  • Construct the combinatorial Laplacian and derive its energy identity.
  • Prove that the zero-eigenvalue multiplicity equals the component count.
  • Interpret the second eigenvalue and a finite two-state averaging model without conflating conventions.

Graphs can carry numbers as well as connections: temperatures at rooms, opinions at agents, or voltages at junctions. The Laplacian turns the difference between neighboring values into a linear operator. Its quadratic form measures total disagreement across edges, and its zero modes reveal precisely where a signal can remain constant independently. This gives an algebraic way to recover connectedness from the network structure.

Build an operator from edge differences

Fix a finite simple undirected graph with vertices in a chosen order. Let A be its symmetric adjacency matrix and D the diagonal matrix of vertex degrees. The combinatorial Laplacian is L=D−AL=D-A. For a real vector x of vertex values, the coordinate (Lx)v(Lx)_v is the sum of xv−xux_v-x_u over neighbors u. Every row sum is zero, so the constant vector lies in the kernel. This sign convention makes L positive semidefinite; some physical evolution equations use its negative.

Expand the quadratic form: degree terms contribute one square for each endpoint of an edge, while the symmetric adjacency terms subtract the two cross terms. Grouping by unordered edges produces a sum of squared differences. Each term is nonnegative, proving positive semidefiniteness without computing eigenvalues. With positive edge weights, multiply each squared difference by its weight and define D using weighted degrees. Zero-weight connections do not enforce agreement and should be omitted when discussing connectivity through positive-weight edges.

L=D−A,xTLx=∑{u,v}∈E(xu−xv)2≥0L=D-A,\qquad x^{\mathsf T}Lx=\sum_{\{u,v\}\in E}(x_u-x_v)^2\ge0
Each unordered edge appears once. This is the unnormalized combinatorial Laplacian for a finite simple undirected graph.↔ Scroll wide equations horizontally.

Zero energy identifies the components

If Lx=0Lx=0, then its quadratic form is zero, forcing every edge difference to vanish. Values agree along every path, so x is constant on each connected component. Conversely, any vector constant separately on each component has zero differences at every edge and therefore lies in the kernel. The component indicator vectors are independent and span all such signals. Consequently the dimension of the kernel, equivalently the multiplicity of eigenvalue zero for this real symmetric matrix, equals the number of components.

For a graph with at least two vertices, order the eigenvalues increasingly. Connectedness is equivalent to the second eigenvalue being positive, since zero is always present and no eigenvalue is negative. On a connected graph, minimize the energy among nonzero vectors orthogonal to the constant vector and divide by squared norm: the result is the second eigenvalue. Orthogonality removes the trivial constant signal, and normalization prevents shrinking a signal toward zero merely to reduce its energy.

dim⁡ker⁡L=c(G),λ2=min⁡x≠0, x⊥1xTLxxTx(n≥2)\dim\ker L=c(G),\qquad \lambda_2=\min_{x\ne0,\,x\perp\mathbf1}\frac{x^{\mathsf T}Lx}{x^{\mathsf T}x}\quad(n\ge2)
The variational formula follows from the spectral theorem; connectedness holds exactly when λ₂>0 in this unnormalized convention.↔ Scroll wide equations horizontally.

Dynamics reveal what the eigenvalues measure

Continuous-time diffusion obeys x′(t)=−Lx(t)x'(t)=-Lx(t). An eigenvector with eigenvalue λ\lambda evolves by the factor e−λte^{-\lambda t}, while component-wise constants remain unchanged. Thus a connected graph approaches one constant level, and a disconnected graph can retain different constants in different components. The conserved quantity is the ordinary sum of vertex values for this symmetric combinatorial evolution. Rates depend on the chosen edge weights and time units, while moving vertices in a drawing does not change the fixed graph’s Laplacian spectrum.

Discrete averaging uses a different operator, such as P=I−αLP=I-\alpha L with a step size making all entries nonnegative. Its eigenvalues are 1−αλ1-\alpha\lambda, so a mode may oscillate rather than decay monotonically. On the two-vertex one-edge graph, the symmetric two-state chain with transition probability alpha has eigenvalues one and 1−2α1-2\alpha. At alpha equal to one it alternates forever. Connectivity guarantees a unique constant direction but does not by itself guarantee convergence of every discrete-time update rule.

The spectrum of a three-vertex path

For the path a−b−ca-b-c, construct L, verify three eigenvectors, and interpret the smallest positive eigenvalue.

  1. The degrees are one, two, one, so L=(1−10−12−10−11)L=\begin{pmatrix}1&-1&0\\-1&2&-1\\0&-1&1\end{pmatrix}. Multiplying by (1,1,1)T(1,1,1)^{\mathsf T} gives zero, as expected from the row sums.

  2. Multiplication sends (1,0,−1)T(1,0,-1)^{\mathsf T} to itself, giving eigenvalue one, and sends (1,−2,1)T(1,-2,1)^{\mathsf T} to three times itself, giving eigenvalue three. The three nonzero mutually orthogonal vectors form a basis.

  3. The spectrum is therefore 0,1,3. The single zero mode confirms one connected component. Among mean-zero unit signals, the least energy is one, achieved along the first nonconstant eigenvector; in diffusion this mode decays as exp(−t), slower than the eigenvalue-three mode.

The result

The path is connected, its Laplacian eigenvalues are 0, 1, and 3, and its algebraic connectivity is one.

Explore the model

Interactive lab

Interactive lab · markov

Two-vertex averaging and periodic failure

Pause and Reset, then set p = 0.25 and q = 0.25. Predict P(state 1) is 0.25 after one Step and 0.375 after two; check the readout. Reset and set both p = 1 and q = 1; Step repeatedly to compare permanent alternation. The symmetric chain equals I−αL for the two-vertex edge with α=p=q. This is discrete averaging, not a solver for an arbitrary graph Laplacian.

Two-vertex averaging and periodic failureExact finite-step probabilities for a two-state chain starting in state 0, not sampled trajectories. Playback advances two transitions per second. When p = q = 0 every distribution is stationary; when p = q = 1 the chain alternates forever. A single zero transition creates an absorbing state. Current readings: P(state 1) at step 0: 0%; Stationary P(state 1): 60%; Chain behavior: Monotone convergence. All parameters are adjustable with the labeled controls below.Two-state Markov distribution after 0 steps: state zero probability 1, state one probability 0.Initial distribution (1, 0) · step 0p = 0.3q = 0.201100%0%1−p1−qP(0)P(1)Monotone convergence
P(state 0)P(state 1)Stationary distribution
P(state 1) at step 0
0%
Stationary P(state 1)
60%
Chain behavior
Monotone convergence
t = 0 s

Exact finite-step probabilities for a two-state chain starting in state 0, not sampled trajectories. Playback advances two transitions per second. When p = q = 0 every distribution is stationary; when p = q = 1 the chain alternates forever. A single zero transition creates an absorbing state. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThe nonconstant eigenvalue 1−2α controls two-state averaging; a connected graph can still support periodic discrete-time dynamics.

Check your understanding

The combinatorial Laplacian of a finite simple undirected graph has zero as an eigenvalue of multiplicity four. What follows?

Try it yourself

For the triangle K3K_3, show that its Laplacian is 3I−J3I-J, where J is the all-ones matrix. Find its spectrum and compute the energy of x=(1,−1,0)Tx=(1,-1,0)^{\mathsf T} directly from edges.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

J sends a constant vector to three times itself and annihilates every vector whose coordinates sum to zero. Count each of the triangle's edges once in the energy sum.

Compare with the solution

The Laplacian has diagonal two and off-diagonal minus one, exactly 3I−J3I-J. The constant vector has eigenvalue zero, and the two-dimensional mean-zero subspace has eigenvalue three. Thus the spectrum is 0,3,30,3,3. For x the squared differences are (1−(−1))2=4(1-(-1))^2=4, (−1−0)2=1(-1-0)^2=1, and (0−1)2=1(0-1)^2=1, summing to six. Its squared norm is two, so its Rayleigh quotient is three, agreeing with the mean-zero eigenvalue.

Take it with you

  • The combinatorial Laplacian energy sums one squared difference per unordered edge.
  • The zero-eigenvalue multiplicity equals the number of connected components.
  • Continuous diffusion and discrete averaging have related but different convergence conditions.

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