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:
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.
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
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 .
- Spectral gap
- The Joint spectral radius is a generalization of the spectral radius to sets of matrices.