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    \makebox[\textwidth][l]{\textbf{MATH 6404: Applied [Combinatorics and] Graph Theory} \hfill CU Denver} \\
    \rule{\textwidth}{0.5pt} \\
    \makebox[\textwidth][l]{Spring 2026 \hfill Instructor: Carlos Mart\'inez}
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\begin{tcolorbox}[colback=cugold!40,colframe=black,left=6pt,right=6pt,top=10pt,bottom=10pt]
    \centering
    \textbf{Lecture #1:} #2 \\
    \textbf{Date:} #3 \hfill     \textbf{Scribe:} #4
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\newcommand{\lecturenumber}{22}
\newcommand{\lecturetitle}{The Graph Laplacian}
\newcommand{\scribename}{Hanbyul (Han) Lee}
\newcommand{\lecturedate}{April 15, 2026}

\begin{document}

\thispagestyle{firstpage}
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\section{Spanning Trees}

\begin{definition}
A subgraph $H \subseteq G$ is \emph{spanning} if $V(H)=V(G)$.
\end{definition}

\begin{definition}
A \emph{spanning tree} of $G$ is a subgraph $T$ that is both spanning and a tree. Let
\[
\mathcal{ST}(G)
\]
be the set of spanning trees of $G$.
\end{definition}

\begin{example}
Consider the graph $G$ shown below, with vertex set $\{v,w,x,y\}$ and edge set
\[
E(G)=\{e,f,s,h,i\},
\]
where $e=\{v,w\}$, $f=\{v,y\}$, $s=\{v,x\}$, $h=\{w,x\}$, and $i=\{y,x\}$.
\[
\begin{tikzpicture}[scale=1.05]
\Vertex[x=0,y=1,label={v},position={above left},Math,size=.25,color=white]{v}
\Vertex[x=2.4,y=1,label={w},position={above right},Math,size=.25,color=white]{w}
\Vertex[x=2.4,y=0,label={x},position={below right},Math,size=.25,color=white]{x}
\Vertex[x=0,y=0,label={y},position={below left},Math,size=.25,color=white]{y}
\Edge[label={e},position={above},Math](v)(w)
\Edge[label={f},position={left},Math](v)(y)
\Edge[label={h},position={right},Math](w)(x)
\Edge[label={i},position={below},Math](y)(x)
\Edge[label={s},position={below},Math](v)(x)
\end{tikzpicture}
\]
Then $\mathcal{ST}(G)$ consists of the following eight spanning trees:
\begin{center}
\begin{tabular}{@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}}
\begin{tikzpicture}[scale=.95,baseline=-.5ex]
\Vertex[x=0,y=1,NoLabel,size=.24,color=white]{v}
\Vertex[x=1.8,y=1,NoLabel,size=.24,color=white]{w}
\Vertex[x=1.8,y=0,NoLabel,size=.24,color=white]{x}
\Vertex[x=0,y=0,NoLabel,size=.24,color=white]{y}
\Edge(v)(w)\Edge(v)(y)\Edge(v)(x)
\end{tikzpicture}
&
\begin{tikzpicture}[scale=.95,baseline=-.5ex]
\Vertex[x=0,y=1,NoLabel,size=.24,color=white]{v}
\Vertex[x=1.8,y=1,NoLabel,size=.24,color=white]{w}
\Vertex[x=1.8,y=0,NoLabel,size=.24,color=white]{x}
\Vertex[x=0,y=0,NoLabel,size=.24,color=white]{y}
\Edge(v)(w)\Edge(v)(y)\Edge(y)(x)
\end{tikzpicture}
&
\begin{tikzpicture}[scale=.95,baseline=-.5ex]
\Vertex[x=0,y=1,NoLabel,size=.24,color=white]{v}
\Vertex[x=1.8,y=1,NoLabel,size=.24,color=white]{w}
\Vertex[x=1.8,y=0,NoLabel,size=.24,color=white]{x}
\Vertex[x=0,y=0,NoLabel,size=.24,color=white]{y}
\Edge(v)(w)\Edge(w)(x)\Edge(y)(x)
\end{tikzpicture}
&
\begin{tikzpicture}[scale=.95,baseline=-.5ex]
\Vertex[x=0,y=1,NoLabel,size=.24,color=white]{v}
\Vertex[x=1.8,y=1,NoLabel,size=.24,color=white]{w}
\Vertex[x=1.8,y=0,NoLabel,size=.24,color=white]{x}
\Vertex[x=0,y=0,NoLabel,size=.24,color=white]{y}
\Edge(v)(w)\Edge(v)(x)\Edge(y)(x)
\end{tikzpicture}
\\[.1em]
$\{e,f,s\}$ & $\{e,f,i\}$ & $\{e,h,i\}$ & $\{e,s,i\}$
\\[1.2em]
\begin{tikzpicture}[scale=.95,baseline=-.5ex]
\Vertex[x=0,y=1,NoLabel,size=.24,color=white]{v}
\Vertex[x=1.8,y=1,NoLabel,size=.24,color=white]{w}
\Vertex[x=1.8,y=0,NoLabel,size=.24,color=white]{x}
\Vertex[x=0,y=0,NoLabel,size=.24,color=white]{y}
\Edge(v)(y)\Edge(w)(x)\Edge(y)(x)
\end{tikzpicture}
&
\begin{tikzpicture}[scale=.95,baseline=-.5ex]
\Vertex[x=0,y=1,NoLabel,size=.24,color=white]{v}
\Vertex[x=1.8,y=1,NoLabel,size=.24,color=white]{w}
\Vertex[x=1.8,y=0,NoLabel,size=.24,color=white]{x}
\Vertex[x=0,y=0,NoLabel,size=.24,color=white]{y}
\Edge(v)(y)\Edge(v)(x)\Edge(w)(x)
\end{tikzpicture}
&
\begin{tikzpicture}[scale=.95,baseline=-.5ex]
\Vertex[x=0,y=1,NoLabel,size=.24,color=white]{v}
\Vertex[x=1.8,y=1,NoLabel,size=.24,color=white]{w}
\Vertex[x=1.8,y=0,NoLabel,size=.24,color=white]{x}
\Vertex[x=0,y=0,NoLabel,size=.24,color=white]{y}
\Edge(w)(x)\Edge(y)(x)\Edge(v)(x)
\end{tikzpicture}
&
\begin{tikzpicture}[scale=.95,baseline=-.5ex]
\Vertex[x=0,y=1,NoLabel,size=.24,color=white]{v}
\Vertex[x=1.8,y=1,NoLabel,size=.24,color=white]{w}
\Vertex[x=1.8,y=0,NoLabel,size=.24,color=white]{x}
\Vertex[x=0,y=0,NoLabel,size=.24,color=white]{y}
\Edge(v)(w)\Edge(v)(y)\Edge(w)(x)
\end{tikzpicture}
\\[.1em]
$\{f,h,i\}$ & $\{f,s,h\}$ & $\{h,i,s\}$ & $\{e,f,h\}$
\end{tabular}
\end{center}
Thus
\[
|\mathcal{ST}(G)|=8.
\]
\end{example}

The general goal for the next few lectures is to derive a formula for $|\mathcal{ST}(G)|$. If $G=K_n$, then we know from quite a few lectures ago that
\[
|\mathcal{ST}(K_n)|=n^{n-2}.
\]
But we want a more general formula.

\section{Matrices Associated to an Undirected Graph}

\begin{definition}
Let $G=(V,E)$ be an undirected graph. Its \emph{adjacency matrix} is
\[
A(G)=A\in \mathbb{R}^{V\times V},
\]
where
\[
 a_{uv}=\begin{cases}
1, & \text{if } \{u,v\}\in E,\\
0, & \text{otherwise.}
\end{cases}
\]
We must keep a consistent labeling of the rows and columns; that is, we fix an ordering of $V$ and stick to it. The matrix $A$ is symmetric.
\end{definition}

For the graph $G$ above, using the vertex order $(v,w,x,y)$, the adjacency matrix is
\[
A(G)=
\begin{array}{c|cccc}
 & v & w & x & y\\ \hline
v & 0 & 1 & 1 & 1\\
w & 1 & 0 & 1 & 0\\
x & 1 & 1 & 0 & 1\\
y & 1 & 0 & 1 & 0
\end{array}.
\]

\begin{definition}
The \emph{undirected incidence matrix} of $G$ is
\[
B(G)=B\in \mathbb{R}^{V\times E},
\]
where
\[
 b_{v,e}=\begin{cases}
1, & \text{if } v \text{ is an endpoint of } e,\\
0, & \text{otherwise.}
\end{cases}
\]
This differs slightly from the incidence matrix for directed graphs.
\end{definition}

Using the vertex order $(v,w,x,y)$ and the edge order $(e,f,s,h,i)$, the undirected incidence matrix for the example is
\[
B(G)=
\begin{array}{c|ccccc}
 & e & f & s & h & i\\ \hline
v & 1 & 1 & 1 & 0 & 0\\
w & 1 & 0 & 0 & 1 & 0\\
x & 0 & 0 & 1 & 1 & 1\\
y & 0 & 1 & 0 & 0 & 1
\end{array}.
\]

\begin{definition}
The \emph{degree matrix} of $G$ is the diagonal matrix
\[
C(G)=C\in \mathbb{R}^{V\times V}
\]
whose diagonal entries are
\[
 c_{uu}=\deg(u).
\]
\end{definition}

For the example,
\[
C(G)=
\begin{array}{c|cccc}
 & v & w & x & y\\ \hline
v & 3 & 0 & 0 & 0\\
w & 0 & 2 & 0 & 0\\
x & 0 & 0 & 3 & 0\\
y & 0 & 0 & 0 & 2
\end{array}.
\]

The matrices $A$, $B$, and $C$ are nicely related.

\begin{proposition}
For an undirected graph $G$,
\[
B(G)B(G)^T=A(G)+C(G).
\]
\end{proposition}

\begin{proof}
Recall that $B\in\mathbb{R}^{V\times E}$, so $BB^T\in\mathbb{R}^{V\times V}$. Thus $BB^T$ is indexed by pairs of vertices.

For $u,v\in V$, the entry $(BB^T)_{uv}$ is obtained by taking the dot product of the $u$-row of $B$ with the $v$-row of $B$:
\[
\begin{tikzpicture}[scale=.55,baseline=-.5ex]
\node at (-.45,1.15) {$u$};
\draw[fill=black!8] (0,1) rectangle (3,.72);
\draw (0,0) rectangle (3,2);
\node at (1.5,-.35) {$e$};
\node at (4.05,2.35) {$v$};
\draw[fill=black!8] (4,.15) rectangle (4.28,2.15);
\draw (3.4,.15) rectangle (4.9,2.15);
\node at (5.25,1.15) {$e$};
\draw[fill=black!25] (4,.72) rectangle (4.28,1);
\node at (2.1,2.35) {$B$};
\node at (4.15,-.25) {$B^T$};
\end{tikzpicture}
\qquad
(BB^T)_{uv}=\sum_{e\in E} b_{u,e}b_{v,e}.
\]
If $u=v$, then
\[
(BB^T)_{uu}=\sum_{e\in E} b_{u,e}^2
=\sum_{e\in E} \mathbf{1}\{u\text{ is an endpoint of }e\}
=\deg(u)=C_{uu}.
\]
If $u\neq v$, then
\begin{align*}
(BB^T)_{uv}
&=\sum_{e\in E} b_{u,e}b_{v,e}\\
&=\sum_{e\in E}\mathbf{1}\{u\text{ is an endpoint of }e\}\mathbf{1}\{v\text{ is an endpoint of }e\}\\
&=\mathbf{1}\{\{u,v\}\in E\}=A_{uv}.
\end{align*}
Therefore the diagonal entries of $BB^T$ agree with $C$, and the off-diagonal entries agree with $A$. Hence $BB^T=A+C$.
\end{proof}

\section{Orientations and Directed Incidence Matrices}

We will need a directed version of $G$ going forward.

\begin{definition}
An \emph{orientation} of an undirected graph $G$ is a directed graph $D$ such that
\[
V(D)=V(G),
\]
and for each edge $\{u,v\}\in E(G)$, exactly one of $(u,v)$ or $(v,u)$ is in $E(D)$. We say that $G$ is the underlying undirected graph of $D$.
\end{definition}

For the example, one orientation $D$ is
\[
\begin{tikzpicture}[scale=1.05]
\Vertex[x=0,y=1,label={v},position={above left},Math,size=.25,color=white]{v}
\Vertex[x=2.4,y=1,label={w},position={above right},Math,size=.25,color=white]{w}
\Vertex[x=2.4,y=0,label={x},position={below right},Math,size=.25,color=white]{x}
\Vertex[x=0,y=0,label={y},position={below left},Math,size=.25,color=white]{y}
\Edge[Direct,label={e},position={above},Math](v)(w)
\Edge[Direct,label={f},position={left},Math](v)(y)
\Edge[Direct,label={s},position={below},Math](x)(v)
\Edge[Direct,label={h},position={right},Math](x)(w)
\Edge[Direct,label={i},position={below},Math](y)(x)
\end{tikzpicture}
\]
so the directed edges are $e=(v,w)$, $f=(v,y)$, $s=(x,v)$, $h=(x,w)$, and $i=(y,x)$.

\begin{definition}
Recall the incidence matrix for a directed graph $D$:
\[
B(D)=B\in\mathbb{R}^{V\times E},
\]
where
\[
 b_{v,e}=\begin{cases}
1, & \text{if } e=(u,v) \text{ for some } u,\\
-1, & \text{if } e=(v,u) \text{ for some } u,\\
0, & \text{otherwise.}
\end{cases}
\]
\end{definition}

For the oriented graph $D$ above,
\[
B(D)=
\begin{array}{c|ccccc}
 & e & f & s & h & i\\ \hline
v & -1 & -1 & 1 & 0 & 0\\
w & 1 & 0 & 0 & 1 & 0\\
x & 0 & 0 & -1 & -1 & 1\\
y & 0 & 1 & 0 & 0 & -1
\end{array}.
\]

Here are some properties of $D$.

\begin{proposition}
Let $G$ be an undirected graph and let $D$ be an orientation of $G$. Then:
\begin{enumerate}
\item[(a)] If the rows of $B(D)$ are $b_1,b_2,\ldots,b_n$, then
\[
b_1+b_2+\cdots+b_n=\mathbf{0}.
\]
\item[(b)] We have
\[
B(D)B(D)^T=C(G)-A(G).
\]
\end{enumerate}
\end{proposition}

\begin{proof}
For part (a), each column of $B(D)$ has exactly one $1$, exactly one $-1$, and zeros everywhere else. Hence the sum of the rows is the zero vector.

For part (b), use the same case analysis as in the previous proof. If $u=v$, then
\[
(B(D)B(D)^T)_{uu}=\sum_{e\in E} b_{u,e}^2=\deg(u)=C_{uu}.
\]
If $u\neq v$, then a nonzero contribution can occur only when $u$ and $v$ are endpoints of the same edge. In that case one endpoint contributes $1$ and the other contributes $-1$, so
\[
(B(D)B(D)^T)_{uv}=-\mathbf{1}\{\{u,v\}\in E(G)\}=-A_{uv}.
\]
Therefore $B(D)B(D)^T=C(G)-A(G)$.
\end{proof}

In part (b), the left-hand side is computed from an arbitrary orientation $D$, but the right-hand side only depends on the underlying undirected graph. Therefore $B(D)B(D)^T$ is independent of the choice of orientation. This matrix is extremely useful.

\section{The Graph Laplacian}

\begin{definition}
The \emph{Laplacian matrix} of an undirected graph $G$ is
\[
L(G)=C(G)-A(G).
\]
\end{definition}

For the example,
\[
L(G)=
\begin{array}{c|cccc}
 & v & w & x & y\\ \hline
v & 3 & 0 & 0 & 0\\
w & 0 & 2 & 0 & 0\\
x & 0 & 0 & 3 & 0\\
y & 0 & 0 & 0 & 2
\end{array}
-
\begin{array}{c|cccc}
 & v & w & x & y\\ \hline
v & 0 & 1 & 1 & 1\\
w & 1 & 0 & 1 & 0\\
x & 1 & 1 & 0 & 1\\
y & 1 & 0 & 1 & 0
\end{array}
=
\begin{bmatrix}
3 & -1 & -1 & -1\\
-1 & 2 & -1 & 0\\
-1 & -1 & 3 & -1\\
-1 & 0 & -1 & 2
\end{bmatrix}.
\]
The sum of any row of $L(G)$ is zero: there is one $\deg(u)$ entry, $\deg(u)$ many $-1$'s, and zeros everywhere else.

For example,
\[
\begin{bmatrix}
3 & -1 & -1 & -1\\
-1 & 2 & -1 & 0\\
-1 & -1 & 3 & -1\\
-1 & 0 & -1 & 2
\end{bmatrix}
\begin{bmatrix}
4\\2\\5\\3
\end{bmatrix}
=
\begin{bmatrix}
2\\-5\\6\\-3
\end{bmatrix}.
\]
The corresponding labeled graph is
\[
\begin{tikzpicture}[scale=1.05]
\Vertex[x=0,y=1,label={v},position={left},Math,size=.25,color=white]{v}
\Vertex[x=2.4,y=1,label={w},position={right},Math,size=.25,color=white]{w}
\Vertex[x=2.4,y=0,label={x},position={right},Math,size=.25,color=white]{x}
\Vertex[x=0,y=0,label={y},position={left},Math,size=.25,color=white]{y}
\node at (0,1.35) {$4$};
\node at (2.4,1.35) {$2$};
\node at (2.4,-.35) {$5$};
\node at (0,-.35) {$3$};
\Edge(v)(w)
\Edge(v)(y)
\Edge(w)(x)
\Edge(y)(x)
\Edge(v)(x)
\end{tikzpicture}
\]

In particular, since every row sum of $L(G)$ is zero,
\[
L(G)\mathbf{1}=\mathbf{0}.
\]
Thus the matrix has linearly dependent columns, so
\[
\det(L(G))=0.
\]

Now let $L_{i,j}$ be the submatrix of $L$ obtained by removing the $i$-th row and the $j$-th column. For the example above,
\[
L_{1,1}=\begin{bmatrix}
2 & -1 & 0\\
-1 & 3 & -1\\
0 & -1 & 2
\end{bmatrix}
\qquad\text{and}\qquad
L_{1,3}=\begin{bmatrix}
-1 & 2 & 0\\
-1 & -1 & -1\\
-1 & 0 & 2
\end{bmatrix}.
\]
Their determinants are
\[
\det(L_{1,1})
=2\bigl(3\cdot 2-(-1)(-1)\bigr)-(-1)\bigl((-1)(2)-0\bigr)=8,
\]
and
\[
\det(L_{1,3})
=-1\bigl((-1)(2)\bigr)-2\bigl((-1)(2)-(-1)(-1)\bigr)=8.
\]
Therefore
\[
\det(L_{1,1})=\det(L_{1,3})=8=|\mathcal{ST}(G)|.
\]
Next time, we will show that this is not a coincidence.

\end{document}