**Theorem 1.** Given a graph G, its associated matrices In(G) and L(G) have the following
properties.

- L(G) is a symmetric matrix. This means the eigenvalues of L(G) are real, and its eigenvectors are real and orthogonal.
- Let e=[1,...,1]', where ' means transpose, i.e. the column vector of all ones. Then L(G)*e = 0.
- In(G)*(In(G))' = L(G). This is independent of the signs chosen in each column of In(G).
- Suppose L(G)*v = lambda*v, where v is nonzero. Then
lambda = norm(In(G)'*v)

^{2}/ norm(v)^{2}where norm(z)^{2}= sum_{i}z(i)^{2}= sum_{{all edges e=(i,j)}}(v(i)-v(j))^{2}/ sum_{i}v(i)^{2} - The eigenvalues of L(G) are nonnegative:
0 <= lambda
_{1}<= lambda_{2}<= ... <= lambda_{n}. - The number of of connected components of G is equal to the number of
lambda
_{i}equal to 0. In particular, lambda_{2}!= 0 if and only if G is connected.

*Proof of part 1.* Symmetry follows from the definition of L(G): Since
G is an undirected graph, (i,j) is an edge if and only if (j,i) is an edge.

*Proof of part 2.* The i-th entry of L(G)*e is just the sum of
the entries of the i-th row of L(G). This equals the degree of node
i, L(G)(i,i), minus 1 for each incident edge (L(G)(i,j), or exactly zero.

*Proof of part 3.*

(In(G)*(In(G))')(i,i) = sum_{{all edges e, such that i is an endpoint of e}}(+-1)^{2}= degree of node i and (In(G)*(In(G))')(i,j) = sum_{{all edges e = (i,j)}}(-1)*(+1) = -1 if an edge e=(i,j) exists

*Proof of part 4.* Suppose L(G)*v = lambda*v, where lambda is an
eigenvalues and v is a nonzero eigenvector. Then v'*L(G)*v = lambda*v'*v, where
v'*v is a positive scalar. Thus

lambda = ( v'* L(G) *v ) / ( v'*v ) = ( v'* (In(G)*(In(G))') *v ) / ( v'*v ) = ( v'* In(G)) * (In(G))' *v ) / ( v'*v ) = ( y') * ( y ) / ( v'*v ) where y = In(G)'*v = sumIf edge e = (i,j), it is easy to see by construction that y(e) = v(i)-v(j) or its negative, depending on the arbitrary choice of signs in column e of In(G). Thus y(e)_{e}y(e)^{2}/ sum_{i}v(i)^{2}

*Proof of part 5.* By part 4, each eigenvalue lambda is the quotient
of two nonnegative quantities, and so must be nonnegative.

*Proof of part 6.* For lambda to equal 0, each y(e) in
the expression

lambda = summust be zero. This means v(i)=v(j) for each edge e=(i,j). Starting with any node i and applying the fact v(i)=v(j) repeatedly, one can see that any node k reachable from i also satisfies v(k)=v(i)=c. In other words, the eigenvector v is a constant c on each connected component. Since L(G) is symmetric, the number of independent eigenvectors corresponding to lambda=0 is equal to the number of eigenvalues equal to 0. If there are exactly d connected components, there are exactly d independent eigenvectors, since choosing d constants c(1),...,c(d) (one for each connected component) determines each eigenvector uniquely._{e}y(e)^{2}/ sum_{i}v(i)^{2}