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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(·).

## Matrices

Let λ1, ..., λn be the (real or complex) eigenvalues of a matrix ACn × n. Then its spectral radius ρ(A) is defined as:

$\rho (A){\overset {\underset {\mathrm {def} }{}}{=}}\max _{i}(|\lambda _{i}|)$ The following lemma shows a simple yet useful upper bound for the spectral radius of a matrix:

Lemma: Let $A\in \mathbb {C} ^{n\times n}$ be a complex-valued matrix, ρ(A) its spectral radius and ||·|| a consistent matrix norm; then, for each kN:

Proof: Let (v, λ) be an eigenvector-eigenvalue pair for a matrix A. By the sub-multiplicative property of the matrix norm, we get:

$|\lambda |^{k}\|\mathbf {v} \|=\|\lambda ^{k}\mathbf {v} \|=\|A^{k}\mathbf {v} \|\leq \|A^{k}\|\cdot \|\mathbf {v} \|$ and since v ≠ 0 we have

$|\lambda |^{k}\leq \|A^{k}\|$ and therefore

$\rho (A)\leq \|A^{k}\|^{1/k}\,\,\square$ 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 ACn × n be a complex-valued matrix and ρ(A) its spectral radius; then

$\lim _{k\to \infty }A^{k}=0$ if and only if $\rho (A)<1.$ Moreover, if ρ(A)>1, $\|A^{k}\|$ is not bounded for increasing k values.

Proof:

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

$A^{k}\mathbf {v} =\lambda ^{k}\mathbf {v} ,$ we have:

{\begin{aligned}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{aligned}} and, since by hypothesis v ≠ 0, we must have

$\lim _{k\to \infty }\lambda ^{k}=0$ which implies |λ| < 1. Since this must be true for any eigenvalue λ, we can conclude ρ(A) < 1.

$A=VJV^{-1}$ 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

$A^{k}=VJ^{k}V^{-1}$ $J^{k}={\begin{bmatrix}J_{m_{1}}^{k}(\lambda _{1})&0&0&\cdots &0\\0&J_{m_{2}}^{k}(\lambda _{2})&0&\cdots &0\\\vdots &\cdots &\ddots &\cdots &\vdots \\0&\cdots &0&J_{m_{s-1}}^{k}(\lambda _{s-1})&0\\0&\cdots &\cdots &0&J_{m_{s}}^{k}(\lambda _{s})\end{bmatrix}}$ $J_{m_{i}}^{k}(\lambda _{i})={\begin{bmatrix}\lambda _{i}^{k}&{k \choose 1}\lambda _{i}^{k-1}&{k \choose 2}\lambda _{i}^{k-2}&\cdots &{k \choose m_{i}-1}\lambda _{i}^{k-m_{i}+1}\\0&\lambda _{i}^{k}&{k \choose 1}\lambda _{i}^{k-1}&\cdots &{k \choose m_{i}-2}\lambda _{i}^{k-m_{i}+2}\\\vdots &\vdots &\ddots &\ddots &\vdots \\0&0&\cdots &\lambda _{i}^{k}&{k \choose 1}\lambda _{i}^{k-1}\\0&0&\cdots &0&\lambda _{i}^{k}\end{bmatrix}}$ $\lim _{k\to \infty }J_{m_{i}}^{k}=0\ \forall i$ which implies

$\lim _{k\to \infty }J^{k}=0.$ Therefore,

$\lim _{k\to \infty }A^{k}=\lim _{k\to \infty }VJ^{k}V^{-1}=V(\lim _{k\to \infty }J^{k})V^{-1}=0$ On the other side, if $\rho (A)>1$ , there is at least one element in $J$ which doesn't remain bounded as k increases, so proving the second part of the statement. $\square$ ## Theorem (Gelfand's formula, 1941)

For any matrix norm ||·||, we have

$\rho (A)=\lim _{k\to \infty }\|A^{k}\|^{1/k}.$ In other words, Gelfand's formula shows how the spectral radius of A gives the asymptotic growth rate of the norm of Ak:

$\|A^{k}\|\sim \rho (A)^{k}$ for $k\rightarrow \infty .\,$ Proof: For any ε > 0, consider the matrix

${\tilde {A}}=(\rho (A)+\epsilon )^{-1}A.$ Then, obviously,

$\rho ({\tilde {A}})={\frac {\rho (A)}{\rho (A)+\epsilon }}<1$ and, by the previous theorem,

$\lim _{k\to \infty }{\tilde {A}}^{k}=0.$ That means, by the sequence limit definition, a natural number N1N exists such that

$\forall k\geq N_{1}\Rightarrow \|{\tilde {A}}^{k}\|<1$ which in turn means:

$\forall k\geq N_{1}\Rightarrow \|A^{k}\|<(\rho (A)+\epsilon )^{k}$ or

$\forall k\geq N_{1}\Rightarrow \|A^{k}\|^{1/k}<(\rho (A)+\epsilon ).$ Let's now consider the matrix

${\check {A}}=(\rho (A)-\epsilon )^{-1}A.$ Then, obviously,

$\rho ({\check {A}})={\frac {\rho (A)}{\rho (A)-\epsilon }}>1$ and so, by the previous theorem,$\|{\check {A}}^{k}\|$ is not bounded.

This means a natural number N2N exists such that

$\forall k\geq N_{2}\Rightarrow \|{\check {A}}^{k}\|>1$ which in turn means:

$\forall k\geq N_{2}\Rightarrow \|A^{k}\|>(\rho (A)-\epsilon )^{k}$ or

$\forall k\geq N_{2}\Rightarrow \|A^{k}\|^{1/k}>(\rho (A)-\epsilon ).$ Taking

$N:=\max(N_{1},N_{2})$ and putting it all together, we obtain:

$\forall \epsilon >0,\exists N\in \mathbb {N} :\forall k\geq N\Rightarrow \rho (A)-\epsilon <\|A^{k}\|^{1/k}<\rho (A)+\epsilon$ which, by definition, is

$\lim _{k\to \infty }\|A^{k}\|^{1/k}=\rho (A).\,\,\square$ 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 $\rho (A_{1}A_{2}\ldots A_{n})\leq \rho (A_{1})\rho (A_{2})\ldots \rho (A_{n}).$ 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:

$\forall \epsilon >0,\exists N\in \mathbb {N} :\forall k\geq N\Rightarrow \rho (A)\leq \|A^{k}\|^{1/k}<\rho (A)+\epsilon$ which, by definition, is
$\lim _{k\to \infty }\|A^{k}\|^{1/k}=\rho (A)^{+}.$ Example: Let's consider the matrix

$A={\begin{bmatrix}9&-1&2\\-2&8&4\\1&1&8\end{bmatrix}}$ whose eigenvalues are 5, 10, 10; by definition, its spectral radius is ρ(A)=10. In the following table, the values of $\|A^{k}\|^{1/k}$ for the four most used norms are listed versus several increasing values of k (note that, due to the particular form of this matrix,$\|.\|_{1}=\|.\|_{\infty }$ ):

## Bounded linear operators

For a bounded linear operator A and the operator norm ||·||, again we have

$\rho (A)=\lim _{k\to \infty }\|A^{k}\|^{1/k}.$ 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.

## 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 (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 $G$ let $l^{2}(G)$ denote the space of functions $f\colon V(G)\to {\mathbb {R} }$ with $\sum _{v\in V(G)}\|f(v)^{2}\|<\infty$ . Let $\gamma \colon l^{2}(G)\to l^{2}(G)$ be the adjacency operator of $G$ , i.e., $(\gamma f)(v)=\sum _{(u,v)\in E(G)}f(u)$ . The spectral radius of G is defined to be the spectral radius of the bounded linear operator $\gamma$ .