# Spectral radius

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In mathematics, the spectral radius of a square matrix or a bounded linear operator is the largest absolute value of its eigenvalues (i.e. supremum among the absolute values of the elements in its spectrum). It 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:

${\displaystyle \rho (A)=\max \left\{|\lambda _{1}|,\dotsc ,|\lambda _{n}|\right\}.}$

The spectral radius is a sort of infimum of all norms of a matrix. Indeed, on the one hand, ${\displaystyle \rho (A)\leqslant \|A\|}$ for every natural matrix norm ${\displaystyle \|\cdot \|}$; and on the other hand, Gelfand's formula states that ${\displaystyle \rho (A)=\lim _{k\to \infty }\|A^{k}\|^{1/k}}$. Both these results are shown below.

However, the spectral radius does not necessarily satisfy ${\displaystyle \|A\mathbf {v} \|\leqslant \rho (A)\|\mathbf {v} \|}$ for arbitrary vectors ${\displaystyle \mathbf {v} \in \mathbb {C} ^{n}}$. To see why, let ${\displaystyle r>1}$ be arbitrary and consider the matrix

${\displaystyle C_{r}={\begin{pmatrix}0&r^{-1}\\r&0\end{pmatrix}}}$.

The characteristic polynomial of ${\displaystyle C_{r}}$ is ${\displaystyle \lambda ^{2}-1}$, so its eigenvalues are ${\displaystyle \{-1,1\}}$ and thus ${\displaystyle \rho (C_{r})=1}$. However, ${\displaystyle C_{r}\mathbf {e} _{1}=r\mathbf {e} _{2}}$. As a result, for any ${\displaystyle \ell ^{p}}$ norm,

${\displaystyle \|C_{r}\mathbf {e} _{1}\|=r>1=\rho (C_{r})\|\mathbf {e} _{1}\|.}$

As an illustration of Gelfand's formula, note that ${\displaystyle \|C_{r}^{k}\|^{1/k}\to 1}$ as ${\displaystyle k\to \infty }$, since ${\displaystyle C_{r}^{k}=I}$ if ${\displaystyle k}$ is even and ${\displaystyle C_{r}^{k}=C_{r}}$ if ${\displaystyle k}$ is odd.

A special case in which ${\displaystyle \|A\mathbf {v} \|\leqslant \rho (A)\|\mathbf {v} \|}$ for all ${\displaystyle \mathbf {v} \in \mathbb {C} ^{n}}$ is when ${\displaystyle A}$ is a Hermitian matrix and ${\displaystyle \|\cdot \|}$ is the Euclidean norm. This is because any Hermitian Matrix is diagonalizable by a unitary matrix, and unitary matrices preserve vector length:

${\displaystyle \|A\mathbf {v} \|=\|U^{*}DU\mathbf {v} \|=\|DU\mathbf {v} \|\leqslant \rho (A)\|U\mathbf {v} \|=\rho (A)\|\mathbf {v} \|.}$

## Bounded linear operators

For a bounded linear operator A on a Banach space, the eigenvalues are replaced with the spectrum of the operator, those values ${\displaystyle \lambda }$ for which ${\displaystyle A-\lambda I}$ fails to be injective; we denote the spectrum by

${\displaystyle \sigma (A)=\left\{\lambda \in \mathbb {C}$ :\ker(A-\lambda I)\neq \emptyset \right\}}

The spectral radius is then defined as the supremum of the magnitudes of elements in the spectrum:

${\displaystyle \rho (A)=\sup _{\lambda \in \sigma (A)}|\lambda |}$

If we denote the operator norm by ${\displaystyle \|\cdot \|}$, then we have the spectral radius formula or Gelfand's formula:

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

A bounded operator (on a complex Hilbert space) is 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 define:

${\displaystyle \ell ^{2}(G)=\left\{f:V(G)\to \mathbf {R} \$ :\ \sum \nolimits _{v\in V(G)}\left\|f(v)^{2}\right\|<\infty \right\}.}

Let γ be the adjacency operator of G:

${\displaystyle {\begin{cases}\gamma$ :\ell ^{2}(G)\to \ell ^{2}(G)\\(\gamma f)(v)=\sum _{(u,v)\in E(G)}f(u)\end{cases}}}

The spectral radius of G is defined to be the spectral radius of the bounded linear operator γ.

## Upper bound

### Upper bounds for spectral radius of a matrix

The following proposition shows a simple yet useful upper bound for the spectral radius of a matrix:

Proposition. Let ACn×n with spectral radius ρ(A) and a consistent matrix norm ||⋅||. Then for each integer ${\displaystyle k\geqslant 1}$:

${\displaystyle \rho (A)\leq \|A^{k}\|^{\frac {1}{k}}.}$

Proof

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

${\displaystyle |\lambda |^{k}\|\mathbf {v} \|=\|\lambda ^{k}\mathbf {v} \|=\|A^{k}\mathbf {v} \|\leq \|A^{k}\|\cdot \|\mathbf {v} \|}$

and since v ≠ 0 we have

${\displaystyle |\lambda |^{k}\leq \|A^{k}\|}$

and therefore

${\displaystyle \rho (A)\leq \|A^{k}\|^{\frac {1}{k}}.}$

### 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

${\displaystyle {\frac {(k-2)(k-3)}{2}}\leq m-n\leq {\frac {k(k-3)}{2}}}$

where ${\displaystyle 3\leq k\leq n}$ is an integer, then [1]

${\displaystyle \rho (G)\leq {\sqrt {2m-n-k+{\frac {5}{2}}+{\sqrt {2m-2n+{\frac {9}{4}}}}}}}$

## 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:

Theorem. Let ACn×n with spectral radius ρ(A). Then ρ(A) < 1 if and only if
${\displaystyle \lim _{k\to \infty }A^{k}=0.}$
On the other hand, if ρ(A) > 1, ${\displaystyle \lim _{k\to \infty }\|A^{k}\|=\infty }$. The statement holds for any choice of matrix norm on Cn×n.

### Proof of theorem

Assume the limit in question is zero, we will show that ρ(A) < 1. Let (v, λ) be an eigenvector-eigenvalue pair for A. Since Akv = λkv we have:

{\displaystyle {\begin{aligned}0&=\left(\lim _{k\to \infty }A^{k}\right)\mathbf {v} \\&=\lim _{k\to \infty }\left(A^{k}\mathbf {v} \right)\\&=\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

${\displaystyle \lim _{k\to \infty }\lambda ^{k}=0}$

which implies |λ| < 1. Since this must be true for any eigenvalue λ, we can conclude ρ(A) < 1.

Now assume the radius of A is less than 1. From the Jordan normal form theorem, we know that for all ACn×n, there exist V, JCn×n with V non-singular and J block diagonal such that:

${\displaystyle A=VJV^{-1}}$

with

${\displaystyle J={\begin{bmatrix}J_{m_{1}}(\lambda _{1})&0&0&\cdots &0\\0&J_{m_{2}}(\lambda _{2})&0&\cdots &0\\\vdots &\cdots &\ddots &\cdots &\vdots \\0&\cdots &0&J_{m_{s-1}}(\lambda _{s-1})&0\\0&\cdots &\cdots &0&J_{m_{s}}(\lambda _{s})\end{bmatrix}}}$

where

${\displaystyle J_{m_{i}}(\lambda _{i})={\begin{bmatrix}\lambda _{i}&1&0&\cdots &0\\0&\lambda _{i}&1&\cdots &0\\\vdots &\vdots &\ddots &\ddots &\vdots \\0&0&\cdots &\lambda _{i}&1\\0&0&\cdots &0&\lambda _{i}\end{bmatrix}}\in \mathbf {C} ^{m_{i}\times m_{i}},1\leq i\leq s.}$

It is easy to see that

${\displaystyle A^{k}=VJ^{k}V^{-1}}$

and, since J is block-diagonal,

${\displaystyle J^{k}={\begin{bmatrix}J_{m_{1}}^{k}(\lambda _{1})&0&0&\cdots &0\\0&J_{m_{2}}^{k}(\lambda _{2})&0&\cdots &0\\\vdots &\cdots &\ddots &\cdots &\vdots \\0&\cdots &0&J_{m_{s-1}}^{k}(\lambda _{s-1})&0\\0&\cdots &\cdots &0&J_{m_{s}}^{k}(\lambda _{s})\end{bmatrix}}}$

Now, a standard result on the k-power of an ${\displaystyle m_{i}\times m_{i}}$ Jordan block states that, for ${\displaystyle k\geq m_{i}-1}$:

${\displaystyle J_{m_{i}}^{k}(\lambda _{i})={\begin{bmatrix}\lambda _{i}^{k}&{k \choose 1}\lambda _{i}^{k-1}&{k \choose 2}\lambda _{i}^{k-2}&\cdots &{k \choose m_{i}-1}\lambda _{i}^{k-m_{i}+1}\\0&\lambda _{i}^{k}&{k \choose 1}\lambda _{i}^{k-1}&\cdots &{k \choose m_{i}-2}\lambda _{i}^{k-m_{i}+2}\\\vdots &\vdots &\ddots &\ddots &\vdots \\0&0&\cdots &\lambda _{i}^{k}&{k \choose 1}\lambda _{i}^{k-1}\\0&0&\cdots &0&\lambda _{i}^{k}\end{bmatrix}}}$

Thus, if ${\displaystyle \rho (A)<1}$ then for all i${\displaystyle |\lambda _{i}|<1}$. Hence for all i we have:

${\displaystyle \lim _{k\to \infty }J_{m_{i}}^{k}=0}$

which implies

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

Therefore,

${\displaystyle \lim _{k\to \infty }A^{k}=\lim _{k\to \infty }VJ^{k}V^{-1}=V\left(\lim _{k\to \infty }J^{k}\right)V^{-1}=0}$

On the other side, if ${\displaystyle \rho (A)>1}$, there is at least one element in J which doesn't remain bounded as k increases, so proving the second part of the statement.

## Gelfand's formula

### Theorem

The next theorem gives the spectral radius as a limit of matrix norms.

Theorem (Gelfand's Formula; 1941). For any matrix norm ||⋅||, we have
${\displaystyle \rho (A)=\lim _{k\to \infty }\left\|A^{k}\right\|^{\frac {1}{k}}.}$. [2]

### Proof

For any ε > 0, first we construct the following two matrices:

${\displaystyle A_{\pm }={\frac {1}{\rho (A)\pm \varepsilon }}A.}$

Then:

${\displaystyle \rho \left(A_{\pm }\right)={\frac {\rho (A)}{\rho (A)\pm \varepsilon }},\qquad \rho (A_{+})<1<\rho (A_{-}).}$

First we apply the previous theorem to A+:

${\displaystyle \lim _{k\to \infty }A_{+}^{k}=0.}$

That means, by the sequence limit definition, there exists N+N such that for all k ≥ N+,

{\displaystyle {\begin{aligned}\left\|A_{+}^{k}\right\|<1\end{aligned}}}

so

{\displaystyle {\begin{aligned}\left\|A^{k}\right\|^{\frac {1}{k}}<\rho (A)+\varepsilon .\end{aligned}}}

Applying the previous theorem to A implies ${\displaystyle \|A_{-}^{k}\|}$ is not bounded and there exists NN such that for all k ≥ N,

{\displaystyle {\begin{aligned}\left\|A_{-}^{k}\right\|>1\end{aligned}}}

so

{\displaystyle {\begin{aligned}\left\|A^{k}\right\|^{\frac {1}{k}}>\rho (A)-\varepsilon .\end{aligned}}}

Let N = max{N+, N}, then we have:

${\displaystyle \forall \varepsilon >0\quad \exists N\in \mathbf {N} \quad \forall k\geq N\quad \rho (A)-\varepsilon <\left\|A^{k}\right\|^{\frac {1}{k}}<\rho (A)+\varepsilon }$

which, by definition, is

${\displaystyle \lim _{k\to \infty }\left\|A^{k}\right\|^{\frac {1}{k}}=\rho (A).}$

## 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

${\displaystyle \rho (A_{1}\cdots A_{n})\leq \rho (A_{1})\cdots \rho (A_{n}).}$

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:

${\displaystyle \forall \varepsilon >0,\exists N\in \mathbf {N} ,\forall k\geq N\quad \rho (A)\leq \|A^{k}\|^{\frac {1}{k}}<\rho (A)+\varepsilon }$

which, by definition, is

${\displaystyle \lim _{k\to \infty }\left\|A^{k}\right\|^{\frac {1}{k}}=\rho (A)^{+},}$

where the + means that the limit is approached from above.

## Example

Consider the matrix

${\displaystyle A={\begin{bmatrix}9&-1&2\\-2&8&4\\1&1&8\end{bmatrix}}}$

whose eigenvalues are 5, 10, 10; by definition, ρ(A) = 10. In the following table, the values of ${\displaystyle \|A^{k}\|^{\frac {1}{k}}}$ for the four most used norms are listed versus several increasing values of k (note that, due to the particular form of this matrix,${\displaystyle \|.\|_{1}=\|.\|_{\infty }}$):

k${\displaystyle \|.\|_{1}=\|.\|_{\infty }}$${\displaystyle \|.\|_{F}}$${\displaystyle \|.\|_{2}}$
11415.36229149610.681145748
212.64911064112.32829434810.595665162
311.93483191911.53245066410.500980846
411.50163316911.15100298610.418165779
511.21604315110.92124223510.351918183
${\displaystyle \vdots }$${\displaystyle \vdots }$${\displaystyle \vdots }$${\displaystyle \vdots }$
1010.60494442210.45591043010.183690042
1110.54867768010.41370221310.166990229
1210.50192183510.37862093010.153031596
${\displaystyle \vdots }$${\displaystyle \vdots }$${\displaystyle \vdots }$${\displaystyle \vdots }$
2010.29825439910.22550444710.091577411
3010.19786089210.14977692110.060958900
4010.14803164010.11212368110.045684426
5010.11825103510.08959882010.036530875
${\displaystyle \vdots }$${\displaystyle \vdots }$${\displaystyle \vdots }$${\displaystyle \vdots }$
10010.05895175210.04469950810.018248786
20010.02943256210.02232483410.009120234
30010.01961209510.01487769010.006079232
40010.01470546910.01115619410.004559078
${\displaystyle \vdots }$${\displaystyle \vdots }$${\displaystyle \vdots }$${\displaystyle \vdots }$
100010.00587959410.00446098510.001823382
200010.00293936510.00223024410.000911649
300010.00195948110.00148677410.000607757
${\displaystyle \vdots }$${\displaystyle \vdots }$${\displaystyle \vdots }$${\displaystyle \vdots }$
1000010.00058780410.00044600910.000182323
2000010.00029389810.00022300210.000091161
3000010.00019593110.00014866710.000060774
${\displaystyle \vdots }$${\displaystyle \vdots }$${\displaystyle \vdots }$${\displaystyle \vdots }$
10000010.00005877910.00004460010.000018232

## Notes and references

1. Guo, Ji-Ming; Wang, Zhi-Wen; Li, Xin (2019). "Sharp upper bounds of the spectral radius of a graph". Discrete Mathematics. 342 (9): 2559–2563. doi:10.1016/j.disc.2019.05.017.
2. The formula holds for any Banach algebra; see Lemma IX.1.8 in Dunford & Schwartz 1963 and Lax 2002 , pp. 195–197

## Bibliography

• Dunford, Nelson; Schwartz, Jacob (1963), Linear operators II. Spectral Theory: Self Adjoint Operators in Hilbert Space, Interscience Publishers, Inc.
• Lax, Peter D. (2002), Functional Analysis, Wiley-Interscience, ISBN   0-471-55604-1

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