Let be the eigenvalues of a matrix. Then its spectral radius is defined as: The condition number of can be expressed using the spectral radius as. The spectral radius is a sort of infimum of all norms of a matrix. On the one hand, for every natural matrix norm, and on the other hand, Gelfand's formula states that ; both these results are shown below. However, the spectral radius does not necessarily satisfy for arbitrary vectors. To see why, let be arbitrary and consider the matrix. The characteristic polynomial of is, hence its eigenvalues are, and thus. However, so for being any norm on. What still allows as is that, making as. does hold when is a Hermitian matrix and is the Euclidean norm.
Graphs
The spectral radius of a finite graph is defined to be the spectral radius of its adjacency matrix. This definition extends to the case of infinite graphs with bounded degrees of vertices. In this case, for the graph define: Let be the adjacency operator of : The spectral radius of is defined to be the spectral radius of the bounded linear operator.
The following proposition shows a simple yet useful upper bound for the spectral radius of a matrix: Proposition. Let with spectral radius and a consistent matrix norm. Then for each integer : Proof Let be an eigenvector-eigenvalue pair for a matrix A. By the sub-multiplicative property of the matrix norm, we get: and since we have and therefore
Upper bounds for spectral radius of a graph
There are many upper bounds for the spectral radius of a graph in terms of its number n of vertices and its number m of edges. For instance, if where is an integer, then
Power sequence
Theorem
The spectral radius is closely related to the behaviour of the convergence of the power sequence of a matrix; namely, the following theorem holds:
Proof of theorem
Assume the limit in question is zero, we will show that. Let be an eigenvector-eigenvalue pair for A. Since we have: and, since by hypothesis, we must have which implies |λ| < 1. Since this must be true for any eigenvalue λ, we can conclude ρ < 1. Now assume the radius of is less than. From the Jordan normal form theorem, we know that for all, there exist with non-singular and block diagonal such that: with where It is easy to see that and, since is block-diagonal, Now, a standard result on the -power of an Jordan block states that, for : Thus, if then for all . Hence for all we have: which implies Therefore, On the other side, if, there is at least one element in which doesn't remain bounded as k increases, so proving the second part of the statement.
Gelfand's formula
Theorem
The next theorem gives the spectral radius as a limit of matrix norms.
Proof
For any, first we construct the following two matrices: Then: First we apply the previous theorem to : That means, by the sequence limit definition, there exists such that for all k ≥ N+, so Applying the previous theorem to implies is not bounded and there exists such that for all k ≥ N−, so Let then we have: which, by definition, is
Gelfand corollaries
Gelfand's formula leads directly to a bound on the spectral radius of a product of finitely many matrices, namely assuming that they all commute we obtain Actually, in case the norm is consistent, the proof shows more than the thesis; in fact, using the previous lemma, we can replace in the limit definition the left lower bound with the spectral radius itself and write more precisely: which, by definition, is where the + means that the limit is approached from above.
Example
Consider the matrix whose eigenvalues are ; by definition,. In the following table, the values of for the four most used norms are listed versus several increasing values of k :
