When do the eigenvectors of a Laplacian matrix form a basis?

Question

Eigenvectors do not always form a basis. When do the eigenvectors of a Laplacian matrix form a basis? When the associated adjacency matrix is symmetric? Why?

Answer 1

I assume we are talking about a graph Laplacian matrix.

This is defined as $$L_{i,j} := \begin{cases} \deg(v_i) & \mbox{if}\ i = j \\ -1 & \mbox{if}\ i \neq j\ \mbox{and}\ v_i \mbox{ is adjacent to } v_j \\ 0 & \mbox{otherwise} \end{cases}$$ where the $v_i$'s are the vertices in the graph.

This Laplacian matrix is a symmetric matrix. (This assumes that if $v_i$ is adjacent to $v_j$, then $v_j$ is adjacent to $v_i$, which you have stated.)

By the spectral theorem for symmetric matrices, this Laplacian matrix is diagonalisable, i.e. it has an eigenvector basis.