2
$\begingroup$

Reading some papers about spectral graph analysis and graph neural networks I have found the following notation which I'm not too sure how to expand:

Given matrices $F, L \in \mathbb{R}^{n \times n}$, let's consider $tr(F^TLF)$.

Assuming that $ L:=D-A$ is the laplacian associated to some adjacency matrix $A$ and $F \in \mathbb{R}^{n \times n}$,

  1. How to expand the above formula?
  2. What it actually represents?
$\endgroup$
2
  • $\begingroup$ Your notation is inconsistent, because $\mathbb{R}^n$ is a set of vectors, not matrices, and unless $n=1$ the expression "$tr(F^\prime L F)$" is not defined. $\endgroup$
    – whuber
    Commented Oct 2, 2021 at 19:21
  • $\begingroup$ Sorry it’s just a typo. I meant that $F, L \in \mathbb{R}^{n \ times n} for both the cases. $\endgroup$ Commented Oct 2, 2021 at 19:33

1 Answer 1

3
$\begingroup$

Let $F_i$ denote the $i$th column of $F$. Then the $i$th element of the diagonal of $F^TLF$ is $F_i^TLF_i$. It’s a standard result that a quadratic form with the Laplacian can be written as $$ x^TLx = \sum_{u\sim v} A_{uv}(x_u - x_v)^2 $$ where "$u\sim v$" means the sum is over all unordered pairs of edges, and this quantity can be interpreted as measuring how often $x$ assigns very different values to nearby nodes. This will be small when $x$ only changes over weak edges, and this is the idea behind spectral clustering: the bottom eigenvectors of $L$ give a way to assign values to the nodes that don’t have huge changes over strongly connected nodes.

This means $$ \text{tr}(F^TLF) = \sum_i F_i^TLF_i $$ can be interpreted as the the total alignment of the columns of $F$ with the graph structure as encoded by $A$. If all of the columns of $F$ only change over loosely connected nodes, i.e. lie in the eigenspace of $L$ for small eigenvalues, then this will be small.

If we can diagonalize $L$ as $L = Q\Lambda Q^T$ (when $A$ is the adjacency matrix of an undirected graph this will always be possible via the spectral theorem for square and symmetric real-valued matrices) then we can write each column of $F$ with respect to the eigenbasis $Q$ as $F_i = Qc_i$ and then $F = QC$ for $C = (c_1 \mid \dots \mid c_n)$. This means $$ \text{tr}(F^TLF) = \text{tr}(C^TQ^TQ\Lambda Q^TQC) = \text{tr}(C^T\Lambda C) = \|\Lambda^{1/2} C\|_F^2 $$ where $\|\cdot\|_F$ denotes the Frobenius norm, and this similarly shows that $\text{tr}(F^TLF)$ will be small when the coordinates of the columns of $F$ w.r.t. the eigenbasis are concentrated on directions with small eigenvalues.

For more, I'd recommend Fan Chung's Spectral Graph Theory (chapter one here) as a great place to get more comfortable working with and interpreting graph Laplacians.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.