# On the spectral radius of bounded linear operators

• MHB
• sarrah1
In summary, the homogeneous linear Fredholm integral equation has a nontrivial solution if and only if $\mu$ is an eigenvalue of the integral operator $K$. By using Picard successive approximations, one can prove convergence of the non-homogeneous equation under the condition $|\lambda| < 1/\rho(K)$, where $\rho(K)$ is the spectral radius of the operator $K$. This condition is both necessary and sufficient for convergence, as shown in a similar way to matrix theory.
sarrah1
Hi EVERYBODY:

General knowledge: The homogeneous linear Fredholm integral equation

$\mu\ \varPsi(x)=\int_{a}^{b} \,k(x,s) \varPsi(s) ds$ (1)

has a nontrivial solution if and only if $\mu$ is an eigenvalue of the integral operator $K$. By multiplying (1) by $k(x,s)$ and integrating while changing the order of integration one obtains

${\mu}^{2}\varPsi(x)=\int_{a}^{b} \,{[k(x,s)]}^{(2)}\varPsi(s) ds$

where ${[k(x,s)]}^{(2)}=\int_{a}^{b} \,{[k(x,t)]}^{(1)}{[k(t,s)]}^{(1)}dt$ , where ${[k(x,s)]}^{(1)}=k(x,s)$

after $n$ iterations ${\mu}^{n} \varPsi(x)=\int_{a}^{b} \,{[k(x,s)]}^{(n)} \varPsi(s) ds$ (2)

where ${[k(x,s)]}^{(n)}=\int_{a}^{b} \,{[k(x,t)]}^{(1)}{[k(t,s)]}^{(n-1)}dt$

Now my Question:

Solving the non-homogeneous equation

${\varPsi}(x)=f(x)+\lambda\int_{a}^{b} \,k(x,s) {\varPsi}(s) ds$ , (where $\lambda$ is a parameter)

by Picard successive approximations, results in the following equation after $n$ iterations

${\varPsi}_{n+1}(x)=f(x)+\lambda\int_{a}^{b} \,k(x,s) {\varPsi}_{n}(s) ds$

to prove convergence (I can use norms to reach the classical condition $|\lambda|<1/||K||$) however I want to reach a condition based on the spectral radius of the operator $K$ i.e. $\rho(K)$ . So one can show by imitating the above that after $n$ iterations

${\varPsi}_{n}(x)-{\varPsi}(x)={\lambda}^{n}\int_{a}^{b} \,{[k(x,s)]}^{(n)} [{\varPsi}_{0}(s)-{\varPsi}(s)] ds$ (3)

if $|\mu|<1$ then Eq. (2) gives $\lim_{{n}\to{\infty}}\int_{a}^{b} \,{[k(x,s)]}^{(n)} \varPsi(s) ds=0$ , $\lor x$
can this help me equally in Eq. (3) by saying that convergence occurs if

$|\lambda| < 1/\rho(K)$

grateful
Sarrah

sarrah said:
${\varPsi}_{n}(x)-{\varPsi}(x)={\lambda}^{n}\int_{a}^{b} \,{[k(x,s)]}^{(n)} [{\varPsi}_{0}(s)-{\varPsi}(s)] ds$ (3)

if $|\mu|<1$ then Eq. (2) gives $\lim_{{n}\to{\infty}}\int_{a}^{b} \,{[k(x,s)]}^{(n)} \varPsi(s) ds=0$ , $\lor x$
can this help me equally in Eq. (3) by saying that convergence occurs if

$|\lambda| < 1/\rho(K)$

grateful
Sarrah
If I understand this correctly, you can write equation (3) as ${\varPsi}_{n}(x)-{\varPsi}(x)={\lambda}^{n}K^n({\varPsi}_{0}(s)-{\varPsi}(s))$. In that case, $$\|{\varPsi}_{n}(x)-{\varPsi}(x)\| \leqslant |{\lambda}|^{n}\|K^n\|\|{\varPsi}_{0}(s)-{\varPsi}(s)\|.\quad(4)$$ But $$\displaystyle \rho(K) = \lim_{n\to\infty}\|K^n\|^{1/n}.$$ If $\rho(K) < 1/|\lambda|$, choose $\nu$ with $\rho(K) < \nu < 1/|\lambda|.$ Then $\|K^n\|^{1/n} < \nu$ for all sufficiently large $n$, so that $\|K^n\| < \nu^n.$ You can then conclude from $(4)$ that convergence occurs in $(3)$, because $\nu|\lambda| < 1$ and so $(\nu|\lambda|)^n \to0$ as $n\to\infty.$

Opalg said:
If I understand this correctly, you can write equation (3) as ${\varPsi}_{n}(x)-{\varPsi}(x)={\lambda}^{n}K^n({\varPsi}_{0}(s)-{\varPsi}(s))$. In that case, $$\|{\varPsi}_{n}(x)-{\varPsi}(x)\| \leqslant |{\lambda}|^{n}\|K^n\|\|{\varPsi}_{0}(s)-{\varPsi}(s)\|.\quad(4)$$ But $$\displaystyle \rho(K) = \lim_{n\to\infty}\|K^n\|^{1/n}.$$ If $\rho(K) < 1/|\lambda|$, choose $\nu$ with $\rho(K) < \nu < 1/|\lambda|.$ Then $\|K^n\|^{1/n} < \nu$ for all sufficiently large $n$, so that $\|K^n\| < \nu^n.$ You can then conclude from $(4)$ that convergence occurs in $(3)$, because $\nu|\lambda| < 1$ and so $(\nu|\lambda|)^n \to0$ as $n\to\infty.$

I am most grateful Opalg.

Eq. (3) can -as you did - be written of course as
${\varPsi}_{n}(x)-{\varPsi}(x)={\lambda}^{n}K^n({\varPsi}_{0}(s)-{\varPsi}(s))$
from which Eq. (4) also follows correctly

The problem is - as it seems - from me. What I understood from your proof is that the condition you wrote as $\nu|\lambda| < 1$ is another way of writing a sufficient condition. Because from (4) if $|\lambda ||K||<1$ it follows directly that ${\varPsi}_{n}\implies\varPsi$ in the limit.

my intention was to obtain a necessary condition based on the spectral radius though like we do in matrix theory. if $K$ is a matrix then
${T}^{-1}{K}^{n}T= diag({{\lambda}_{1}}^{n},...,{{\lambda}_{m}}^{n})$ (5)
so $\lambda\rho(K)<1$ becomes both a necessary and sufficient condition whereas $|\lambda|||K||<1$ becomes only a sufficient condition because $\rho(K)\le||K||$. I hope you got me.

So I thought there can be a relation like (5) for integral operators, something similar instead of $$\displaystyle \rho(K) = \lim_{n\to\infty}\|K^n\|^{1/n}.$$ If or $\rho(K)\le||K||$

still very grateful
Sarrah

sarrah said:
I am most grateful Opalg.

Eq. (3) can -as you did - be written of course as
${\varPsi}_{n}(x)-{\varPsi}(x)={\lambda}^{n}K^n({\varPsi}_{0}(s)-{\varPsi}(s))$
from which Eq. (4) also follows correctly

The problem is - as it seems - from me. What I understood from your proof is that the condition you wrote as $\nu|\lambda| < 1$ is another way of writing a sufficient condition. Because from (4) if $|\lambda ||K||<1$ it follows directly that ${\varPsi}_{n}\implies\varPsi$ in the limit.

my intention was to obtain a necessary condition based on the spectral radius though like we do in matrix theory. if $K$ is a matrix then
${T}^{-1}{K}^{n}T= diag({{\lambda}_{1}}^{n},...,{{\lambda}_{m}}^{n})$ (5)
so $\lambda\rho(K)<1$ becomes both a necessary and sufficient condition whereas $|\lambda|||K||<1$ becomes only a sufficient condition because $\rho(K)\le||K||$. I hope you got me.

So I thought there can be a relation like (5) for integral operators, something similar instead of $$\displaystyle \rho(K) = \lim_{n\to\infty}\|K^n\|^{1/n}.$$ If or $\rho(K)\le||K||$

still very grateful
Sarrah

Hi Oplag
Upon reading your post again, and your easy and sound proof, it seems to me there is no such necessary condition, only a sufficient one. Perhaps I didn't express myself well. I meant convergence is guaranteed if $|\lambda|\rho(K)<1$ ok, which is tight but could it be that $|\lambda| ||K||>1$ and convergence still exists ? based on the former condition spectral radius as $\rho(K)\le||K||$
many thanks
Sarrah

## 1. What is the spectral radius of a bounded linear operator?

The spectral radius of a bounded linear operator is the largest absolute value of its eigenvalues. In other words, it is the maximum distance of any eigenvalue from the origin on the complex plane.

## 2. How is the spectral radius calculated?

The spectral radius can be calculated using the spectral mapping theorem, which states that the spectral radius of a bounded linear operator is equal to the limit of the norm of the operator raised to the power of n as n approaches infinity.

## 3. What is the significance of the spectral radius in linear algebra?

The spectral radius is an important measure in linear algebra as it provides information about the stability and behavior of a bounded linear operator. It is also used in applications such as numerical analysis and control theory.

## 4. Can the spectral radius be greater than the norm of a bounded linear operator?

Yes, the spectral radius can be greater than the norm of a bounded linear operator. This can occur when the operator has complex eigenvalues that are not aligned with the norm of the operator.

## 5. How is the spectral radius related to the growth rate of a linear system?

The spectral radius of a bounded linear operator is directly related to the growth rate of a linear system. A larger spectral radius indicates a faster growth rate, whereas a smaller spectral radius indicates a slower growth rate.

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