In mathematics, the spectral radius of a square matrix or a bounded linear operator is the supremum among the absolute values of the elements in its spectrum, which is sometimes denoted by ρ(·).
Let λ1, ..., λn be the (real or complex) eigenvalues of a matrix A ∈ Cn × n. Then its spectral radius ρ(A) is defined as:
The following lemma shows a simple yet useful upper bound for the spectral radius of a matrix:
Lemma: Let be a complex-valued matrix, ρ(A) its spectral radius and ||·|| a consistent matrix norm; then, for each k ∈ N:
Proof: Let (v, λ) be an eigenvector-eigenvalue pair for a matrix A. By the sub-multiplicative property of the matrix norm, we get:
and since v ≠ 0 we have
The spectral radius is closely related to the behaviour of the convergence of the power sequence of a matrix; namely, the following theorem holds:
Theorem: Let A ∈ Cn × n be a complex-valued matrix and ρ(A) its spectral radius; then
- if and only if
Moreover, if ρ(A)>1, is not bounded for increasing k values.
Let (v, λ) be an eigenvector-eigenvalue pair for matrix A. Since
and, since by hypothesis v ≠ 0, we must have
which implies |λ| < 1. Since this must be true for any eigenvalue λ, we can conclude ρ(A) < 1.
From the Jordan normal form theorem, we know that for any complex valued matrix , a non-singular matrix and a block-diagonal matrix exist such that:
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 , so that
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.
Theorem (Gelfand's formula, 1941)
For any matrix norm ||·||, we have
In other words, Gelfand's formula shows how the spectral radius of A gives the asymptotic growth rate of the norm of Ak:
Proof: For any ε > 0, consider the matrix
and, by the previous theorem,
That means, by the sequence limit definition, a natural number N1 ∈ N exists such that
which in turn means:
Let's now consider the matrix
and so, by the previous theorem, is not bounded.
This means a natural number N2 ∈ N exists such that
which in turn means:
and putting it all together, we obtain:
which, by definition, is
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
Example: Let's consider the matrix
whose eigenvalues are 5, 10, 10; by definition, its spectral radius is ρ(A)=10. In the following table, the values of for the four most used norms are listed versus several increasing values of k (note that, due to the particular form of this matrix,):
Bounded linear operators
For a bounded linear operator A and the operator norm ||·||, again we have
A bounded operator (on a complex Hilbert space) called a spectraloid operator if its spectral radius coincides with its numerical radius. An example of such an operator is a normal operator.
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 (i.e. there exists some real number C such that the degree of every vertex of the graph is smaller than C). In this case, for the graph let denote the space of functions with . Let be the adjacency operator of , i.e., . The spectral radius of G is defined to be the spectral radius of the bounded linear operator .