What is it?
Most activity matrices represent the connections between n entities, e.g., users and m entities, such as films, \(n<>m\).
Sometimes the connections is between the same entities, such as endorsement, teams defeating other teams, friends or followers on social networks etc.
In such cases, the matrix is square.
Hence, standard Geometry holds and we can extracts the Eigenpairs.
Eigenpairs
Matrix \(A\) has a real \(\lambda\) and a vector \(\mathbf{e}\) s.t.
\[A\mathbf{e} = \lambda \mathbf{e}\]
\(\lambda\) is an eigenvalue and \(\mathbf{e}\) an eigenvector of A.
If A has rank n, then there could be up to n eigenpairs. In practice,
they might not be real, nor \(\neq 0\), and
are always costly (at least quadratic time in the size of the m., \(\Omega(n^2)\)) to find.
Interesting square matrices
A is called symmetric when \(A=A^T\)
Also called positive semidefinite when for any x we have
\[\mathbf{x}^T A \mathbf{x} \ge 0\]
In such case its eigenvalues are non-negative: \(\lambda_i\ge 0\).
Applications of Spectral analysis
Spectral properties
Adjacency matrices represent connections between entities in a network (graph), e,g., the Web.
The eigenvalues of adjacency matrices provide bounds for several network features.
The Google PageRank algorithm is spectral network analysis.
![]()
Early applications in Psychology, Social science, Bibliometrics, Economy, and Choice theory (seriously).
Spectral ranking
Given a matrix representing preference or likeability between people, can we rank the participants (from best to worst) on the basis of their general, intrinsic likeability?
[Seely, 1949] created an index of likeability based on the ideas of diffusion: it is important to be liked by people who in turn are well-liked and so on.
Let \(M\) be a square matrix where \(m_{ij}\) represents approval or endorsement (negative values represent disapproval)
my likeability index should be equal to the weighted sum of of the indices of the people who like me.
my likeability index should be equal to the weighted sum of of the indices of the people who like me.
But their likeability is turn will depend on mine…
Let’s use row vectors \(\mathbf{r} = [ r_1, r_2, \dots r_n]\):
\[\mathbf{r} = \mathbf{r} M\]
i.e., \(\mathbf{r}\) is a left eigenvector of M.
This formula might have no solution, but matrix preprocessing can assure that one exists.
