The Perron-Frobenius theorem: some of its applications

Abstract

For matrices with all nonnegative entries, the Perron-Frobenius theorem guarantees the existence of an eigenvector with all nonnegative components. We show that the existence of such an eigenvector is also guaranteed for a very different class of matrices, namely real symmetric matrices with exactly two eigenvalues. We also prove a partial converse, that among real symmetric matrices with any more than two eigenvalues there exist some having no nonnegative eigenvector.A nonnegative vector is one whose components are all nonnegative. This concept has no place in pure linear algebra, as it is highly basis dependent. However, nonnegative vectors (and their cousins, positive vectors) sometimes crop up and prove useful in applications. For example, one of the consequences of the Perron-Frobenius theorem is that a matrix with nonnegative entries has a nonnegative (or even positive, under appropriate hypotheses) eigenvector, which fact is of great consequence for, e.g., ranking pages in search engine results [2].In this note, we prove that the existence of a nonnegative eigenvector is also guaranteed for a very different class of matrices, namely real symmetric matrices having only two distinct eigenvalues. Recall that a symmetric matrix has a set of orthogonal eigenvectors that span the ambient space. This is the only fact about symmetric matrices that we will need.Let M ∈ R n×n be our matrix of interest. Since we suppose M has only two eigenvalues, it has two eigenspaces V and W which are orthogonal and satisfy V + W = R n . Hence W = V ⊥ (with respect to the standard inner product on R n ) and vice versa. Thus the existence of a nonnegative eigenvector of M is an immediate corollary of the following proposition.Proposition. For any subspace V ⊆ R n , either V contains a nonzero, nonnegative vector or V ⊥ does.Some commentary before commencing with the proof: Although this is ostensibly a result about linear algebra, we have noted already that the notion of nonnegativity is inherently not a purely linear algebraic property. Hence it should not be surprising that the proof should require other ideas. It turns out that convexity is the key here. Proof of Proposition. Define setsR n ≥0 is the set of all nonnegative vectors, and proving the proposition amounts to showing that V or V ⊥ intersects R n ≥0 in a nonzero vector. Because V and R n ≥0 are both January 2014] NOTES 1

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“…, p Kr ) ∈ A that gives this point. This corresponds to the conditions SINR k = g −1 k (r k ) ∀k, which can be formulated as K r linear equations and solved using the approach in [205]. Finally, the existence of a (p 1 , .…”