Spectral Radius - Matrices

Matrices

Let λ₁, ..., λ_n 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 A ∈ Cn × n 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 for each λ we have

and therefore

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.

Proof:

Let (v, λ) be an eigenvector-eigenvalue pair for matrix A. Since

we have:

$\begin{align} 0 &= \left(\lim_{k \to \infty}A^k\right)\mathbf{v} \\ &= \lim_{k \to \infty}A^k\mathbf{v} \\ &= \lim_{k \to \infty}\lambda^k\mathbf{v} \\ &= \mathbf{v}\lim_{k \to \infty}\lambda^k \end{align}$

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:

with

$J=\begin{bmatrix} J_{m_1}(\lambda_1) & 0 & 0 & \cdots & 0 \\ 0 & J_{m_2}(\lambda_2) & 0 & \cdots & 0 \\ \vdots & \cdots & \ddots & \cdots & \vdots \\ 0 & \cdots & 0 & J_{m_{s-1}}(\lambda_{s-1}) & 0 \\ 0 & \cdots & \cdots & 0 & J_{m_s}(\lambda_s) \end{bmatrix}$

where

$J_{m_i}(\lambda_i)=\begin{bmatrix} \lambda_i & 1 & 0 & \cdots & 0 \\ 0 & \lambda_i & 1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \ddots & \vdots \\ 0 & 0 & \cdots & \lambda_i & 1 \\ 0 & 0 & \cdots & 0 & \lambda_i \end{bmatrix}\in \mathbb{C}^{m_i,m_i}, 1\leq i\leq s.$

It is easy to see that

and, since is block-diagonal,

$J^k=\begin{bmatrix} J_{m_1}^k(\lambda_1) & 0 & 0 & \cdots & 0 \\ 0 & J_{m_2}^k(\lambda_2) & 0 & \cdots & 0 \\ \vdots & \cdots & \ddots & \cdots & \vdots \\ 0 & \cdots & 0 & J_{m_{s-1}}^k(\lambda_{s-1}) & 0 \\ 0 & \cdots & \cdots & 0 & J_{m_s}^k(\lambda_s) \end{bmatrix}$

Now, a standard result on the -power of an Jordan block states that, for :

$J_{m_i}^k(\lambda_i)=\begin{bmatrix} \lambda_i^k & {k \choose 1}\lambda_i^{k-1} & {k \choose 2}\lambda_i^{k-2} & \cdots & {k \choose m_i-1}\lambda_i^{k-m_i+1} \\ 0 & \lambda_i^k & {k \choose 1}\lambda_i^{k-1} & \cdots & {k \choose m_i-2}\lambda_i^{k-m_i+2} \\ \vdots & \vdots & \ddots & \ddots & \vdots \\ 0 & 0 & \cdots & \lambda_i^k & {k \choose 1}\lambda_i^{k-1} \\ 0 & 0 & \cdots & 0 & \lambda_i^k \end{bmatrix}$

Thus, if then, so that

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.

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