I am reading lecture notes on Laplacian matrices of graphs
I don't understand why it should be true the following sentence, below Theorem 3.1.3, which is the famous theorem that links the dimension of the null space with the number of connected components of a graph.
As usual, let $G$ be an undirected graph and $L(G)$ its associated Laplacian matrix, which is a symmetric matrix with the degree of the verices on the diagonal, $0$ in position $i,j$ if two nodes are not connect, and $1$ if two nodes are connected.
Since the matrix $L(G)$ is symmetric and therefore diagonalizable, the multiplicity of zero as a root of its characteristic polynomial is the same as the dimension of the right nullspace of $L(G)$ which is the geometric multiplicity of zero as an eigenvalue of $L(G)$. So in order to prove Theorem 3.1.3 it is enough to consider the right nullspace of $L(G)$.
In particular, why the multiplicity of zero in this case tells me the dimension of the null space? I guess that the algebraic multiplicity of this eigenvalue tell you the geometric multiplicity (i.e. the dimension of the space) only in the case of the null space – that is of the eigenvector $0$.
Best Answer
Every symmetric matrix is diagonalizable, hence the algebraic and geometric multiplicities of eigenvalues coincide.