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    \makebox[\textwidth][l]{\textbf{MATH 6404: Applied [Combinatorics and] Graph Theory} \hfill CU Denver} \\
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    \makebox[\textwidth][l]{Spring 2026 \hfill Instructor: Carlos Mart\'inez}
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\newcommand{\lecturenumber}{$24$}
\newcommand{\lecturetitle}{The Spectrum of the Graph Laplacian}
\newcommand{\scribename}{Simon Ruland}
\newcommand{\lecturedate}{April 22, 2026}

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\section{Graph Laplacian of $K_n$}

Before moving on, consider $K_n$ and $L=L(K_n)$. It consists of 

\begin{enumerate}
    \item $n-1$'s along the diagonal
    \item $-1$'s everywhere else
\end{enumerate}

$$L(K_5) = 
\begin{bmatrix} 4 & -1 & -1 & -1 & -1 \\  
 -1 & 4 & -1 & -1 & -1 \\
  -1 & -1 & 4 & -1 & -1 \\
   -1 & -1 & -1 & 4 & -1 \\
    -1 & -1 & -1 & -1 & 4 \\
    \end{bmatrix}$$

So, $L_{\hat{n},\hat{n}}$ (removing the last column and the last row) is the same matrix except of shape $(n-1) \times(n-1)$.

\begin{example}
    $$L_{\hat{5},\hat{5}} = 
    \begin{bmatrix} 4 & -1 & -1 & -1 \\  
     -1 & 4 & -1 & -1 \\
      -1 & -1 & 4 & -1 \\
       -1 & -1 & -1 & 4\\
        \end{bmatrix}$$
    
    Using elementary row operations, $L_{\hat{5},\hat{5}}$ becomes
    
    $$\begin{bmatrix} 1&1&1&1 \\  
     -1 & 4 & -1 & -1 \\
      -1 & -1 & 4 & -1 \\
       -1 & -1 & -1 & 4\\
        \end{bmatrix}\textasciitilde 
    \begin{bmatrix} 1&1&1&1 \\  
     0 & 5 & 0 & 0 \\
      0 & 0 & 5 & 0 \\
       0 & 0 & 0 & 5\\
       \end{bmatrix}$$

which is triangular and has determinant $\det(L_{\hat{5},\hat{5}}) =5^3 = n^{(n-2)}$
\end{example}

More generally, this argument works for any $n$ and we recover Cayley's formula.

$$|SP(K_n)| = \det(L_{\hat{n}\hat{n}})$$

\section{Properties of the Graph Laplacian}

Moving on, let $G = (V,E)$ be an undirected graph. Consider any orientation of $E$.

For $(i,j) \in E$, $(e_i - e_j)(e_i-e_j)^T$ is a matrix of zeros with $(i,i) = 1, (j,j)=1, (i,j)=-1, (j,i)=-1$.

Much like building the graph one edge at a time, the graph Laplacian can be constructed in the following way:
$$L(G)=\sum_{(i,j)\in E}(e_i - e_j)(e_i-e_j)^T$$

\begin{definition}[Positive semi-definite]
    $A \in \mathbb{R}^{n \times n}$ is positive semi-definite if $$x^TAx \geq 0 \ \forall x \in \mathbb{R}^N$$
    
    In this case, we write $A \succeq 0$.
\end{definition}

For a symmetric $A \in \mathbb{R}^{(n\times n)}$, the following are equivalent 
\begin{enumerate}
    \item $A \succeq 0$
    \item $A = VV^T$ for some $V$
    \item $A$ has non-negative eigenvalues
\end{enumerate}

\begin{remark}
    $L \succeq 0$
\end{remark}

First note that $L$ is symmetric.

\begin{proof}[First proof]
    If $A,B\succeq0$, then

    $A+B \succeq0$ since $x^T(A+B)x=x^TAx+x^TBx\geq 0$ for all $x \in \mathbb{R}^n$

    In particular, $L$ is the sum of positive semi-definite matrices, so $L$ itself is positive semi-definite.
\end{proof}

\begin{proof}[Second proof]
    Our first definition was $L = BB^T$, the directed incidence matrix
\end{proof}

\begin{proof}[Third proof]
    For any $x \in \mathbb{R}^n$

    $$x^TLx = x^T\left[ \sum_{(i,j)\in E} (e_i-e_j)(e_i-e_j)^T \right]x$$

    $$ = \sum_{(i,j)\in E} x^T (e_i-e_j)(e_i-e_j)^T x$$

    $ = \sum_{(i,j)\in E} (x_i - x_j)^2$, which is always non-negative.
\end{proof}

Since $L$ is positive semi-definite, we can write its eigenvalues as

$$0 \leq \lambda_1 \leq \lambda_2 \leq ... \leq \lambda_n$$

Note that $e$ (the all 1's vector) is an eigenvector with the eigenvalue of 0.

$$Le = \sum_{(i,j) \in E} (e_i - e_j)(e_i - e_j)^Te = \sum_{(i,j) \in E} (e_i - e_j) \cdot 0 = 0$$

Thus, $\lambda_1=0$. This checks out since $L$ is linearly dependent $\det(L) = 0$.

$\lambda_2$ is more interesting.

\begin{theorem}

$\lambda_2 = 0$ iff $G$ is disconnected.

\end{theorem}

\begin{proof}
    If $G$ is disconnected, it can be partitioned as $G = G_1 \sqcup G_2$ with no shared edges.

    Thus, after re-indexing, we can write $L= \begin{bmatrix} L_{G_1} & 0 \\ 0 & L_{G_2} \end{bmatrix}$

    Let $e_{G_1}$ and $e_{G_2}$ be the indicator vectors for the edges in $G_1$ and $G_2$ respectively.

    $L_Ge_{G_1} = L_Ge_{G_2} = 0$

    So $e_{G_1}$ and $e_{G_2}$ are eigenvectors of $G$ with the eigenvalue of 0.

    Also, $e_{G_1}e_{G_2}^T = 0$, so $e_{G_1}$ and $e_{G_2}$ are orthogonal. The eigen-space corresponding to $\lambda=0$ has dimension at least 2. 

    $$\implies \lambda_2 = 0$$

    Conversely, let $x_2$ be an eigenvector of $L_G$ with $\lambda_2=0$. 

    We may assume $x_2^Te=0$, possibly after finding an orthogonal basis fo the eigenspace of $\lambda=0$, then

    $$x_2^TLx_2 = x_2^T\lambda_2x_2 = 0$$

    Also, $x_2^TLx_2 = \sum_{(i,j) \in E}[x_2(i)-x_2(j)]^2 = 0$

    $$\implies x_2(i) = x_2(j) \ \forall \ (i,j) \in E$$

    Let $V_1 = \{i \in V x_2(i) \geq 0\}$ and
    $V_2 = \{i \in V x_2(i) < 0\}$

    So, $V_1$ and $V_2$ share no edges.

    But $x_2^Te = 0$ and $x_2 \neq 0$ (since it is an eigenvector), so $x_2$ must have both positive and negative entries.
\end{proof}

$\lambda_2$ is the ``algebraic connectivity" of $G$.

More generally, one can compute this argument to show that $\lambda_k=0$ iff $G$ has at least $k$ components.

We now present several lemmas without proof to demonstrate the utility of the spectrum of $L$.

\begin{lemma}
    $\lambda_2(G) \leq \kappa(G)$ where $\kappa$ is the vertex connectivity of $G$.
\end{lemma}

\begin{lemma}
    If $|V|$ is even, let $b(G)$ be the smallest bisection of $G$ i.e.

    $$b(G) = \min_{S \subset V: |S| = |V-S|}|\delta(s)|$$

    Then,
    $$\frac{n}{4}\lambda_2(G) \leq b(G)$$
\end{lemma}

\begin{lemma}
    Let the maximum cut of G be $\max_{S \subseteq V}|\delta(s)|$, then

    $$\max_{S \subseteq V}|\delta(s)| \leq \lambda_n(L_G)$$
\end{lemma}

Lastly, we return to the matrix-tree theorem.

\begin{theorem}
    Let $L_G$ have the eigenvalues

    $0 = \lambda_1 \leq \lambda_2 \leqq ... \leq \lambda_n$, then

    $$|ST(G)|=\frac{1}{n}\prod_{i=2}^n\lambda_i$$
\end{theorem}

Before proving this, we need a fact from linear algebra:

$$\det(A+B)_ = \sum_{S \subseteq [n]}\det(A_S)$$
where $A_S$ is the matrix obtained from $A$ after replacing each row $a_i$ for $i \in S$ with row $b_i$ of $B$.
This follows from a multi-linear property of determinants.

\begin{proof}
    If $G$ is not connected, then $\lambda_2 = 0$ so

    $$0 = |ST(G)| = \frac{1}{n}\lambda_2\prod_{i=3}^n\lambda_i = 0$$

    Suppose $G$ is connected. Consider the characteristic polynomial of $L_G$

    $$(\lambda-\lambda_1)(\lambda-\lambda_2)...(\lambda-\lambda_n) = \lambda(\lambda-\lambda_2)...(\lambda-\lambda_n)$$

    Thus, the term linear in $\lambda$ is $(-1)^{n-1}\prod_{i=2}^n\lambda_i$

    But also, the characteristic polynomial is defined as $\det(\lambda I-L)$.

    Using the linear algebraic fact, the linear term in $\det(\lambda I-L)$ is

    $$\sum_{S\subseteq[n]:|S|=n-1}\det(L_s) = (-1)^{n-1}\sum_{i=1}^n\det((L_G)_{\hat{i},\hat{i}}) = (-1)^{n-1}n|ST(G)|$$
\end{proof}

\end{document}