# Spectral theorem¶

Course: Math 833 - Modern Discrete Probability (MDP)
Author: Sebastien Roch, Department of Mathematics, University of Wisconsin-Madison
Updated: Nov 20, 2020

## 1 Statement and proof of the spectral theorem¶

Theorem (Spectral Theorem): Let $A \in \mathbb{R}^{d \times d}$ be a symmetric matrix, that is, $A^T = A$. Then $A$ has $d$ orthonormal eigenvectors $\mathbf{q}_1, \ldots, \mathbf{q}_d$ with corresponding (not necessarily distinct) real eigenvalues $\lambda_1 \geq \lambda_2 \geq \cdots \geq \lambda_d$. In matrix form, this is written as the matrix factorization

$$A = Q \Lambda Q^T = \sum_{i=1}^d \lambda_i \mathbf{q}_i \mathbf{q}_i^T$$

where $Q$ has columns $\mathbf{q}_1, \ldots, \mathbf{q}_d$ and $\Lambda = \mathrm{diag}(\lambda_1, \ldots, \lambda_d)$. We refer to this factorization as a spectral decomposition of $A$.

Proof idea (Spectral Theorem): Use a greedy sequence maximizing the quadratic form $\langle \mathbf{v}, A \mathbf{v}\rangle$. How is this quadratic form is related to eigenvalues? Note that, for a unit eigenvector $\mathbf{v}$ with eigenvalue $\lambda$, we have $\langle \mathbf{v}, A \mathbf{v}\rangle = \langle \mathbf{v}, \lambda \mathbf{v}\rangle = \lambda$.

Exercise: Consider the block matrices

$$\begin{pmatrix} \mathbf{y}\\ \mathbf{z} \end{pmatrix} \quad \text{and} \quad \begin{pmatrix} A & B\\ C & D \end{pmatrix}$$

where $\mathbf{y}\in\mathbb{R}^{d_1}$, $\mathbf{z}\in\mathbb{R}^{d_2}$, $A \in \mathbb{R}^{d_1 \times d_1}$, $B \in \mathbb{R}^{d_1 \times d_2}$, $C \in \mathbb{R}^{d_2 \times d_1}$, and $D \in \mathbb{R}^{d_2 \times d_2}$. Show that

$$\begin{pmatrix} \mathbf{y}\\ \mathbf{z} \end{pmatrix}^T \begin{pmatrix} A & B\\ C & D \end{pmatrix} \begin{pmatrix} \mathbf{y}\\ \mathbf{z} \end{pmatrix} = \mathbf{y}^T A \mathbf{y} + \mathbf{y}^T B \mathbf{z} + \mathbf{z}^T C \mathbf{y} + \mathbf{z}^T D \mathbf{z}.$$

$\lhd$

Proof (Spectral Theorem): We proceed by induction.

A first eigenvector: Let $A_1 = A$ and

$$\mathbf{v}_1 \in \arg\max\{\langle \mathbf{v}, A_1 \mathbf{v}\rangle:\|\mathbf{v}\| = 1\}$$

and

$$\lambda_1 = \max\{\langle \mathbf{v}, A_1 \mathbf{v}\rangle:\|\mathbf{v}\| = 1\}.$$

Complete $\mathbf{v}_1$ into an orthonormal basis of $\mathbb{R}^d$, $\mathbf{v}_1$, $\hat{\mathbf{v}}_2, \ldots, \hat{\mathbf{v}}_d$, and form the block matrix

$$\hat{W}_1 = \begin{pmatrix} \mathbf{v}_1 & \hat{V}_1 \end{pmatrix}$$

where the columns of $\hat{V}_1$ are $\hat{\mathbf{v}}_2, \ldots, \hat{\mathbf{v}}_d$. Note that $\hat{W}_1$ is orthogonal by construction.

Getting one step closer to diagonalization: We show next that $W_1$ gets us one step closer to a diagonal matrix by similarity transformation. Note first that

$$\hat{W}_1^T A_1 \hat{W}_1 = \begin{pmatrix} \lambda_1 & \mathbf{w}_1^T \\ \mathbf{w}_1 & A_2 \end{pmatrix}$$

where $\mathbf{w}_1 = \hat{V}_1^T A_1 \mathbf{v}_1$ and $A_2 = \hat{V}_1^T A_1 \hat{V}_1$. The key claim is that $\mathbf{w}_1 = \mathbf{0}$. This follows from a contradiction argument.

Indeed, suppose $\mathbf{w}_1 \neq \mathbf{0}$ and consider the unit vector

$$\mathbf{z} = \hat{W}_1 \times \frac{1}{\sqrt{1 + \delta^2 \|\mathbf{w}_1\|^2}} \begin{pmatrix} 1 \\ \delta \mathbf{w}_1 \end{pmatrix}$$

which, by the exercise above, achieves objective value

\begin{align*} \mathbf{z}^T A_1 \mathbf{z} &= \frac{1}{1 + \delta^2 \|\mathbf{w}_1\|^2} \begin{pmatrix} 1 \\ \delta \mathbf{w}_1 \end{pmatrix}^T \begin{pmatrix} \lambda_1 & \mathbf{w}_1^T \\ \mathbf{w}_1 & A_2 \end{pmatrix} \begin{pmatrix} 1 \\ \delta \mathbf{w}_1 \end{pmatrix}\\ &= \frac{1}{1 + \delta^2 \|\mathbf{w}_1\|^2} \left( \lambda_1 + 2 \delta \|\mathbf{w}_1\|^2 + \delta^2 \mathbf{w}_1^T A_2 \mathbf{w}_1 \right). \end{align*}

By a Taylor expansion, for $\epsilon \in (0,1)$ small,

$$\frac{1}{\sqrt{1 + \epsilon^2}} \approx 1 - \epsilon^2/2.$$

So, for $\delta$ small,

\begin{align} \mathbf{z}^T A_1 \mathbf{z} &\approx (\lambda_1 + 2 \delta \|\mathbf{w}_1\|^2 + \delta^2 \mathbf{w}_1^T A_2 \mathbf{w}_1) (1 - \delta^2 \|\mathbf{w}_1\|^2/2)\\ &\approx \lambda_1 + 2 \delta \|\mathbf{w}_1\|^2 + C \delta^2\\ &> \lambda_1 \end{align}

where $C \in \mathbb{R}$ depends on $\mathbf{w}_1$ and $A_2$. That gives the desired contradiction.

So, letting $W_1 = \hat{W}_1$,

$$W_1^T A_1 W_1 = \begin{pmatrix} \lambda_1 & \mathbf{0} \\ \mathbf{0} & A_2 \end{pmatrix}.$$

Finally note that $A_2 = \hat{V}_1^T A_1 \hat{V}_1$ is symmetric

$$A_2^T = (\hat{V}_1^T A_1 \hat{V}_1)^T = \hat{V}_1^T A_1^T \hat{V}_1 = \hat{V}_1^T A_1 \hat{V}_1 = A_2$$

by the symmetry of $A_1$ itself.

Next step of the induction: Apply the same argument to the symmetric matrix $A_2 \in \mathbb{R}^{(d-1)\times (d-1)}$, let $\hat{W}_2 \in \mathbb{R}^{(d-1)\times (d-1)}$ be the corresponding orthogonal matrix, and define $\lambda_2$ and $A_3$ through the equation

$$\hat{W}_2^T A_2 \hat{W}_2 = \begin{pmatrix} \lambda_2 & \mathbf{0} \\ \mathbf{0} & A_3 \end{pmatrix}.$$

Define the block matrix

$$W_2 = \begin{pmatrix} 1 & \mathbf{0}\\ \mathbf{0} & \hat{W}_2 \end{pmatrix}$$

and observe that

$$W_2^T W_1^T A_1 W_1 W_2 = W_2^T \begin{pmatrix} \lambda_1 & \mathbf{0} \\ \mathbf{0} & A_2 \end{pmatrix} W_2 = \begin{pmatrix} \lambda_1 & \mathbf{0}\\ \mathbf{0} & \hat{W}_2^T A_2 \hat{W}_2 \end{pmatrix} =\begin{pmatrix} \lambda_1 & 0 & \mathbf{0} \\ 0 & \lambda_2 & \mathbf{0} \\ \mathbf{0} & \mathbf{0} & A_3 \end{pmatrix}.$$

Proceeding similarly by induction gives the claim. $\square$

## 2 Extremal characterization of eigenvalues¶

Definition (Rayleigh Quotient): Let $A \in \mathbb{R}^{d \times d}$ be a symmetric matrix. The Rayleigh quotient is defined as

$$\mathcal{R}_A(\mathbf{u}) = \frac{\langle \mathbf{u}, A \mathbf{u} \rangle}{\langle \mathbf{u}, \mathbf{u} \rangle}$$

which is defined for any $\mathbf{u} \neq \mathbf{0}$ in $\mathbb{R}^{d}$. $\lhd$

Exercise: Let $A \in \mathbb{R}^{d \times d}$ be a symmetric matrix. Let $\mathbf{v}$ be a (not necessarily unit) eigenvector of $A$ with eigenvalue $\lambda$. Show that $\mathcal{R}_A(\mathbf{v}) = \lambda$. $\lhd$

Exercise: Let $\mathcal{U}, \mathcal{V} \subseteq \mathbb{R}^d$ be subspaces such that $\mathrm{dim}(\mathcal{U}) + \mathrm{dim}(\mathcal{V}) > d$. Show there exists a non-zero vector in the intersection $\mathcal{U} \cap \mathcal{V}$. [Hint: Take the union of a basis for $\mathcal{U}$ and a basis for $\mathcal{V}$, then use linear dependence.] $\lhd$

Theorem (Courant-Fischer): Let $A \in \mathbb{R}^{d \times d}$ be a symmetric matrix with spectral decomposition $A = \sum_{i=1}^d \lambda_i \mathbf{v}_i \mathbf{v}_i^T$ where $\lambda_1 \geq \cdots \geq \lambda_d$. For each $k = 1,\ldots,d$, define the subspace

$$\mathcal{V}_k = \mathrm{span}(\mathbf{v}_1, \ldots, \mathbf{v}_k) \quad\text{and}\quad \mathcal{W}_{d-k+1} = \mathrm{span}(\mathbf{v}_k, \ldots, \mathbf{v}_d).$$

Then, for all $k = 1,\ldots,d$,

$$\lambda_k = \min_{\mathbf{u} \in \mathcal{V}_k} \mathcal{R}_A(\mathbf{u}) = \max_{\mathbf{u} \in \mathcal{W}_{d-k+1}} \mathcal{R}_A(\mathbf{u}).$$

Furthermore we have the following min-max formulas, which do not depend on the choice of spectral decomposition, for all $k = 1,\ldots,d$

$$\lambda_k = \max_{\mathrm{dim}(\mathcal{V}) = k} \min_{\mathbf{u} \in \mathcal{V}} \mathcal{R}_A(\mathbf{u}) = \min_{\mathrm{dim}(\mathcal{W}) = d-k+1} \max_{\mathbf{u} \in \mathcal{W}} \mathcal{R}_A(\mathbf{u}).$$

Proof idea: For the local formula, we expand a vector in $\mathcal{V}_k$ into the basis $\mathbf{v}_1,\ldots,\mathbf{v}_k$ and use the fact that $\mathcal{R}_A(\mathbf{v}_i) = \lambda_i$ and that eigenvalues are in non-increasing order. The global formulas then follow from a dimension argument.

Proof: We first prove the local formula, that is, the one involving a specific decomposition.

Local formulas: Since $\mathbf{v}_1, \ldots, \mathbf{v}_k$ form an orthonormal basis of $\mathcal{V}_k$, any nonzero vector $\mathbf{u} \in \mathcal{V}_k$ can be written as $\mathbf{u} = \sum_{i=1}^k \langle \mathbf{u}, \mathbf{v}_i \rangle \mathbf{v}_i$ and it follows that

$$\langle \mathbf{u}, \mathbf{u} \rangle = \sum_{i=1}^k \langle \mathbf{u}, \mathbf{v}_i \rangle^2$$$$\langle \mathbf{u}, A \mathbf{u} \rangle = \left\langle \mathbf{u}, \sum_{i=1}^k \langle \mathbf{u}, \mathbf{v}_i \rangle \lambda_i \mathbf{v}_i \right\rangle = \sum_{i=1}^k \lambda_i \langle \mathbf{u}, \mathbf{v}_i \rangle^2.$$

Thus,

$$\mathcal{R}_A(\mathbf{u}) = \frac{\langle \mathbf{u}, A \mathbf{u} \rangle}{\langle \mathbf{u}, \mathbf{u} \rangle} = \frac{\sum_{i=1}^k \lambda_i \langle \mathbf{u}, \mathbf{v}_i \rangle^2}{\sum_{i=1}^k \langle \mathbf{u}, \mathbf{v}_i \rangle^2} \geq \lambda_k \frac{\sum_{i=1}^k \langle \mathbf{u}, \mathbf{v}_i \rangle^2}{\sum_{i=1}^k \langle \mathbf{u}, \mathbf{v}_i \rangle^2} = \lambda_k$$

where we used $\lambda_1 \geq \cdots \geq \lambda_k$ and the fact that $\langle \mathbf{u}, \mathbf{v}_i \rangle^2 \geq 0$. Moreover $\mathcal{R}_A(\mathbf{v}_k) = \lambda_k$. So we have established

$$\lambda_k = \min_{\mathbf{u} \in \mathcal{V}_k} \mathcal{R}_A(\mathbf{u}).$$

The expression in terms of $\mathcal{W}_{d-k+1}$ is proved similarly.

Global formulas: Since $\mathcal{V}_k$ has dimension $k$, it follows from the local formula that

$$\lambda_k = \min_{\mathbf{u} \in \mathcal{V}_k} \mathcal{R}_A(\mathbf{u}) \leq \max_{\mathrm{dim}(\mathcal{V}) = k} \min_{\mathbf{u} \in \mathcal{V}} \mathcal{R}_A(\mathbf{u}).$$

Let $\mathcal{V}$ be any subspace with dimension $k$. Because $\mathcal{W}_{d-k+1}$ has dimension $d - k + 1$, we have that $\dim(\mathcal{V}) + \mathrm{dim}(\mathcal{W}_{d-k+1}) > d$ and there must be non-zero vector $\mathbf{u}_0$ in the intersection $\mathcal{V} \cap \mathcal{W}_{d-k+1}$ by the exercise above.

We then have by the other local formula that

$$\lambda_k = \max_{\mathbf{u} \in \mathcal{W}_{d-k+1}} \mathcal{R}_A(\mathbf{u}) \geq \mathcal{R}_A(\mathbf{u}_0) \geq \min_{\mathbf{u} \in \mathcal{V}} \mathcal{R}_A(\mathbf{u}).$$

Since this inequality holds for any subspace of dimension $k$, we have

$$\lambda_k \geq \max_{\mathrm{dim}(\mathcal{V}) = k} \min_{\mathbf{u} \in \mathcal{V}} \mathcal{R}_A(\mathbf{u}).$$

Combining with inequality in the other direction above gives the claim. The other global formula is proved similarly. $\square$

## 3 Application to spectral graph theory¶

### 3.1 Matrices associated to graphs¶

Definition (Adjacency Matrix): Assume the undirected graph $G = (V,E)$ has $n = |V|$ vertices. The adjacency matrix $A$ of $G$ is the $n\times n$ symmetric matrix defined as

\begin{align} A_{xy} = \begin{cases} 1 & \text{if \{x,y\} \in E}\\ 0 & \text{o.w.} \end{cases} \end{align}

$\lhd$

(Source)

$$A= \begin{pmatrix} 0 & 1 & 1\\ 1 & 0 & 1\\ 1 & 1 & 0 \end{pmatrix}.$$

$\lhd$

Definition (Incidence Matrix): The incidence matrix of an undirected graph $G = (V, E)$ is the $n \times m$ matrix $B$, where $n = |V|$ and $m =|E|$ are the numbers of vertices and edges respectively, such that $B_{ij} = 1$ if the vertex $i$ and edge $e_j$ are incident and 0 otherwise. $\lhd$

(Source)

$$B = \begin{pmatrix} 1 & 1 & 1 & 0\\ 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 1\\ 0 & 0 & 1 & 1 \end{pmatrix}.$$

$\lhd$

In the digraph case, we have the definitions are adapted as follows. The adjacency matrix $A$ of a digraph $G = (V, E)$ is the matrix defined as

\begin{align} A_{xy} = \begin{cases} 1 & \text{if (x,y) \in E}\\ 0 & \text{o.w.} \end{cases} \end{align}

The oriented incidence matrix of a digraph $G$ with vertices $1,\ldots,n$ and edges $e_1, \ldots, e_m$ is the matrix $B$ such that $B_{ij} = -1$ if egde $e_j$ leaves vertex $i$, $B_{ij} = 1$ if egde $e_j$ enters vertex $i$, and 0 otherwise.

An orientation of an (undirected) graph $G = (V, E)$ is the choice of a direction for each of its edges, turning it into a digraph.

Definition (Oriented Incidence Matrix): An oriented incidence matrix of an (undirected) graph $G = (V, E)$ is the incidence matrix of an orientation of $G$. $\lhd$

(Source)

$\lhd$

In [2]:
g = complete_graph(3)

Out[2]:
{3, 3} undirected simple Int64 graph
In [3]:
A = adjacency_matrix(g)

Out[3]:
3Ã—3 SparseArrays.SparseMatrixCSC{Int64,Int64} with 6 stored entries:
[2, 1]  =  1
[3, 1]  =  1
[1, 2]  =  1
[3, 2]  =  1
[1, 3]  =  1
[2, 3]  =  1
In [4]:
Array(A)

Out[4]:
3Ã—3 Array{Int64,2}:
0  1  1
1  0  1
1  1  0
In [5]:
g = smallgraph(:petersen)

Out[5]:
10Ã—10 Array{Int64,2}:
0  1  0  0  1  1  0  0  0  0
1  0  1  0  0  0  1  0  0  0
0  1  0  1  0  0  0  1  0  0
0  0  1  0  1  0  0  0  1  0
1  0  0  1  0  0  0  0  0  1
1  0  0  0  0  0  0  1  1  0
0  1  0  0  0  0  0  0  1  1
0  0  1  0  0  1  0  0  0  1
0  0  0  1  0  1  1  0  0  0
0  0  0  0  1  0  1  1  0  0
In [6]:
Array(incidence_matrix(g))

Out[6]:
10Ã—15 Array{Int64,2}:
1  1  1  0  0  0  0  0  0  0  0  0  0  0  0
1  0  0  1  1  0  0  0  0  0  0  0  0  0  0
0  0  0  1  0  1  1  0  0  0  0  0  0  0  0
0  0  0  0  0  1  0  1  1  0  0  0  0  0  0
0  1  0  0  0  0  0  1  0  1  0  0  0  0  0
0  0  1  0  0  0  0  0  0  0  1  1  0  0  0
0  0  0  0  1  0  0  0  0  0  0  0  1  1  0
0  0  0  0  0  0  1  0  0  0  1  0  0  0  1
0  0  0  0  0  0  0  0  1  0  0  1  1  0  0
0  0  0  0  0  0  0  0  0  1  0  0  0  1  1

### 3.2 Laplacian matrix¶

Definition (Laplacian Matrix): Let $G = (V,E)$ be a graph with vertices $V = \{1, \ldots, n\}$ and adjacency matrix $A \in \mathbb{R}^{n \times n}$. Let $D = \mathrm{diag}(\delta(1), \ldots, \delta(n))$ be the degree matrix. The Laplacian matrix associated to $G$ is defined as $L = D - A$. Its entries are

$$l_{ij} = \begin{cases} \delta(i) & \text{if i = j}\\ -1 & \text{if \{i,j\} \in E}\\ 0 & \text{o.w.} \end{cases}$$

$\lhd$

Observe that the Laplacian matrix $L$ of a graph $G$ is symmetric:

$$L^T = (D- A)^T = D^T - A^T = D - A$$

where we used that both $D$ and $A$ are themselves symmetric. The associated quadratic form is particularly simple and will play an important role.

Lemma (Laplacian Quadratic Form): Let $G = (V, E)$ be a graph with $n = |V|$ vertices. Its Laplacian matrix $L$ is positive semi-definite and furthermore we have the following formula for the Laplacian quadratic form

$$\mathbf{x}^T L \mathbf{x} = \sum_{e = \{i,j\} \in E} (x_i - x_j)^2$$

for any $\mathbf{x} = (x_1, \ldots, x_n)^T \in \mathbb{R}^n$.

Proof: Let $B$ be an oriented incidence matrix of $G$. We claim that $L = B B^T$. Indeed, for $i \neq j$, entry $(i,j)$ of $B B^T$ is a sum over all edges containing $i$ and $j$ as endvertices, of which there is at most one. When $\{i,j\} \in E$, that entry is $-1$, since one of $i$ or $j$ has a $1$ in the column of $B$ corresponding to $e$ and the other one has a $-1$. For $i = j$, letting $b_{xy}$ be entry $(x,y)$ of $B$,

$$(B B^T)_{ii} = \sum_{e = \{x, y\} \in E: i \in e} b_{xy}^2 = \delta(i).$$

Thus, for any $\mathbf{x}$, we have $(B^T \mathbf{x})_k = x_v - x_u$ if the edge $e_k = \{u, v\}$ is oriented as $(u,v)$ under $B$. That implies

$$\mathbf{x}^T L \mathbf{x} = \mathbf{x}^T B B^T \mathbf{x} = \|B^T \mathbf{x}\|^2 = \sum_{e = \{i,j\} \in E} (x_i - x_j)^2.$$

Since the latter is always nonnegative, it also implies that $L$ is positive semidefinite. $\square$

As a convention, we denote the eigenvalues of a Laplacian matrix $L$ by

$$0 \leq \mu_1 \leq \mu_2 \leq \cdots \leq \mu_n.$$

Another important observation:

Lemma ($\mu_1 = 0$): Let $G = (V, E)$ be a graph with $n = |V|$ vertices and Laplacian matrix $L$. The constant unit vector

$$\mathbf{y}_1 = \frac{1}{\sqrt{n}} (1, \ldots, 1)^T$$

is an eigenvector of $L$ with eigenvalue $0$.

Proof: Let $B$ be an oriented incidence matrix of $G$ recall that $L = B B^T$. By construction $B^T \mathbf{y}_1 = \mathbf{0}$ since each column of $B$ has exactly one $1$ and one $-1$. So $L \mathbf{y}_1 = B B^T \mathbf{y}_1 = \mathbf{0}$ as claimed. $\square$

By our proof of the Spectral Theorem, the largest eigenvalue $\mu_n$ of the matrix Laplacian $L$ is the solution to the optimization problem

$$\mu_n = \max\{\langle \mathbf{x}, L \mathbf{x}\rangle:\|\mathbf{x}\| = 1\}.$$

Such extremal characterization is useful in order to bound the eigenvalue $\mu_n$, since any choice of $\mathbf{x}$ with $\|\mathbf{x}\| =1$ gives a lower bound through the quantity $\langle \mathbf{x}, L \mathbf{x}\rangle$. That perspective will be key to our application to graph partitioning.

Lemma (Laplacian and Degree): Let $G = (V, E)$ be a graph with maximum degree $\bar{\delta}$. Let $\mu_n$ be the largest eigenvalue of its matrix Laplacian $L$. Then

$$\mu_n \geq \bar{\delta}+1.$$

Proof idea: As explained before the statement of the lemma, it suffices to find a good test unit vector $\mathbf{x}$ to plug into $\langle \mathbf{x}, L \mathbf{x}\rangle$. A clever choice does the trick.

Proof: Let $u \in V$ be a vertex with degree $\bar{\delta}$. Let $\mathbf{z}$ be the vector with entries

$$z_i = \begin{cases} \bar{\delta} & \text{if i = u}\\ -1 & \text{if \{i,u\} \in E}\\ 0 & \text{o.w.} \end{cases}$$

and let $\mathbf{x}$ be the unit vector $\mathbf{z}/\|\mathbf{z}\|$. By definition of the degree of $u$, $\|\mathbf{z}\|^2 = \bar{\delta}^2 + \bar{\delta}(-1)^2 = \bar{\delta}(\bar{\delta}+1)$.

Using the Laplacian Quadratic Form Lemma,

$$\langle \mathbf{z}, L \mathbf{z}\rangle = \sum_{e = \{i, j\} \in E} (z_i - z_j)^2 \geq \sum_{i: \{i, u\} \in E} (z_i - z_u)^2 = \sum_{i: \{i, u\} \in E} (-1 - \bar{\delta})^2 = \bar{\delta} (\bar{\delta}+1)^2$$

where we restricted the sum to those edges incident with $u$ and used the fact that all terms in the sum are nonnegative. Finally

$$\langle \mathbf{x}, L \mathbf{x}\rangle = \left\langle \frac{\mathbf{z}}{\|\mathbf{z}\|}, L \frac{\mathbf{z}}{\|\mathbf{z}\|}\right\rangle = \frac{1}{\|\mathbf{z}\|^2} \langle \mathbf{z}, L \mathbf{z}\rangle = \frac{\bar{\delta} (\bar{\delta}+1)^2}{\bar{\delta}(\bar{\delta}+1)} = \bar{\delta}+1$$

so that

$$\mu_n = \max\{\langle \mathbf{x}', L \mathbf{x}'\rangle:\|\mathbf{x}'\| = 1\} \geq \langle \mathbf{x}, L \mathbf{x}\rangle = \bar{\delta}+1$$

as claimed. $\square$

Corollary (Extremal Characterization of $\mu_2$): Let $G = (V, E)$ be a graph with $n = |V|$ vertices. Assume the Laplacian $L$ of $G$ has spectral decomposition $L = \sum_{i=1}^n \mu_i \mathbf{y}_i \mathbf{y}_i^T$ with $0 = \mu_1 \leq \mu_2 \leq \cdots \leq \mu_n$ and $\mathbf{y}_1 = \frac{1}{\sqrt{n}}(1,\ldots,1)^T$. Then

$$\mu_2 = \min\left\{ \frac{\sum_{\{u, v\} \in E} (x_u - x_v)^2}{\sum_{u = 1}^n x_u^2} \,:\, \mathbf{x} = (x_1, \ldots, x_n)^T \neq \mathbf{0}, \sum_{u=1}^n x_u = 0 \right\}.$$

One application of this extremal characterization is a graph drawing heuristic. Consider the entries of the second Laplacian eigenvector $\mathbf{y}_2$ normalized to have unit norm. The entries are centered around $0$ by the condition $\sum_{u=1}^n x_{u} = 0$. Because it minimizes the quantity

$$\frac{\sum_{\{u, v\} \in E} (x_u - x_v)^2}{\sum_{u = 1}^n x_u^2}$$

over all centered unit vectors, $\mathbf{y}_2$ tends to assign similar coordinates to adjacent vertices. A similar reasoning applies to the third Laplacian eigenvector, which in addition is orthogonal to the second one.

(Source)

In [7]:
g = grid([4,7])

Out[7]:
{28, 45} undirected simple Int64 graph
In [8]:
function rand_locs(g)
return rand(nv(g)), rand(nv(g)) # random x and y coord for each vertex
end

Out[8]:
rand_locs (generic function with 1 method)
In [9]:
(x, y) = rand_locs(g)
hcat(x,y)

Out[9]:
28Ã—2 Array{Float64,2}:
0.372401   0.115604
0.963378   0.707802
0.914642   0.970695
0.575798   0.822461
0.287746   0.434905
0.635274   0.0614735
0.0327425  0.939314
0.187151   0.752754
0.974652   0.574964
0.930674   0.986834
0.120294   0.365029
0.41426    0.0593342
0.583661   0.706189
â‹®
0.208685   0.815978
0.0302106  0.130846
0.21552    0.82209
0.542607   0.910461
0.733773   0.695888
0.391655   0.38637
0.326295   0.154331
0.212848   0.896353
0.699021   0.200717
0.767283   0.490476
0.509114   0.460842
0.509541   0.718471
In [10]:
gplot(g, layout=rand_locs)

Out[10]:
In [11]:
function spec_locs(g)
L = Array(laplacian_matrix(g)) # laplacian L of g
F = eigen(L) # spectral decomposition
v = F.vectors # eigenvectors
return v[:,2], v[:,3] # returns 2nd and 3rd eigenvectors
end

Out[11]:
spec_locs (generic function with 1 method)
In [12]:
gplot(g, layout=spec_locs)

Out[12]: