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Spectrum of non-invertible operator

Can spectrum “specify” an operator?Does an unbounded operator $T$ with non-empty spectrum have an unbounded spectrum?decomposition of spectrum in case of a Banach algebraExample of a self-adjoint bounded operator on a Hilbert space with empty point spectrumSpectrum of an Operator on a Banachspacefinite spectrum eigenvalueA spectrum of an invertible operator does not contain $0$Spectrum for a bounded linear operator and its adjoint on a Banach space are same.Finding spectrum of given operatorCounterexample of non invertible operator

1

$begingroup$

By an operator on Banach space $X$ we mean a bounded linear map $T:Xto X$. The spectrum of an operator $T$ on a complex Banach space $X$ is the set

beginequation
sigma(T)= lambdainmathbbC: T-lambda I textis not invertible
endequation

We denote by $mathbbD$ the open unit disc in the complex plane $mathbbC$.

Can I say that:

1) If $sigma(T)subset mathbbD$, then there is $0<t<1$ and $Cgeq 0$ such that $||T^n||leq Ct^n$ for all $ninmathbbN$?

2) If $T:Xto X$ is non-invertible, then
$sigma(T)cap (mathbbC-overlinemathbbD)=emptyset$ ?

Please help me to know them.

functional-analysis operator-theory

share|cite|improve this question

asked Mar 11 at 21:15

user479859user479859

987

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What have you tried? Did you play around with any 2x2 matrices?
  $endgroup$
  – Daniel McLaury
  Mar 11 at 21:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @DanielMcLaury, In my research, $X$ is a Banach space and $T:Xto X$ is an operator. My field is dynamical system and I want to study dynamic of operator on $X$ and I do not know properties of spectrum.
  $endgroup$
  – user479859
  Mar 11 at 21:37
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  The Euclidean plane is a perfectly good Banach space, and every 2x2 matrix gives you a bounded linear map on it.
  $endgroup$
  – Daniel McLaury
  Mar 11 at 21:41
  

  
  
  
  

add a comment | 

1

$begingroup$

By an operator on Banach space $X$ we mean a bounded linear map $T:Xto X$. The spectrum of an operator $T$ on a complex Banach space $X$ is the set

beginequation
sigma(T)= lambdainmathbbC: T-lambda I textis not invertible
endequation

We denote by $mathbbD$ the open unit disc in the complex plane $mathbbC$.

Can I say that:

1) If $sigma(T)subset mathbbD$, then there is $0<t<1$ and $Cgeq 0$ such that $||T^n||leq Ct^n$ for all $ninmathbbN$?

2) If $T:Xto X$ is non-invertible, then
$sigma(T)cap (mathbbC-overlinemathbbD)=emptyset$ ?

Please help me to know them.

functional-analysis operator-theory

share|cite|improve this question

asked Mar 11 at 21:15

user479859user479859

987

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What have you tried? Did you play around with any 2x2 matrices?
  $endgroup$
  – Daniel McLaury
  Mar 11 at 21:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @DanielMcLaury, In my research, $X$ is a Banach space and $T:Xto X$ is an operator. My field is dynamical system and I want to study dynamic of operator on $X$ and I do not know properties of spectrum.
  $endgroup$
  – user479859
  Mar 11 at 21:37
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  The Euclidean plane is a perfectly good Banach space, and every 2x2 matrix gives you a bounded linear map on it.
  $endgroup$
  – Daniel McLaury
  Mar 11 at 21:41
  

  
  
  
  

add a comment | 

$begingroup$

By an operator on Banach space $X$ we mean a bounded linear map $T:Xto X$. The spectrum of an operator $T$ on a complex Banach space $X$ is the set

beginequation
sigma(T)= lambdainmathbbC: T-lambda I textis not invertible
endequation

We denote by $mathbbD$ the open unit disc in the complex plane $mathbbC$.

Can I say that:

1) If $sigma(T)subset mathbbD$, then there is $0<t<1$ and $Cgeq 0$ such that $||T^n||leq Ct^n$ for all $ninmathbbN$?

2) If $T:Xto X$ is non-invertible, then
$sigma(T)cap (mathbbC-overlinemathbbD)=emptyset$ ?

Please help me to know them.

functional-analysis operator-theory

share|cite|improve this question

asked Mar 11 at 21:15

user479859user479859

987

$endgroup$

share|cite|improve this question

asked Mar 11 at 21:15

user479859user479859

987

share|cite|improve this question

asked Mar 11 at 21:15

user479859user479859

987

asked Mar 11 at 21:15

user479859user479859

987

asked Mar 11 at 21:15

user479859user479859

987

2 Answers
2

active

oldest

votes

3

$begingroup$

Yes to 1. If $sigma(T)subseteqlambda$, then, because the spectrum is closed, there exists $0 < r < 1$ such that $sigma(T)subseteq lambda : $, and there exists $N$ large enough that

$$
|T^n|^1/n le frac1+r2 < 1,;;; n ge N,\
|T^n| le left(frac1+r2right)^n,;; n ge N.
$$

There is a constant $C > 1$ such that $|T^n| le Cleft(frac1+r2right)^n$ for all $1 le n < N$. So $|T^n| le Cleft(frac1+r2right)^n$ for all $n ge 1$.

No to 2. If $T$ is non-invertible, then the only thing you can say is that $0insigma(T)$. The spectrum can be any compact subset of $mathbbC$ that includes $0$, which would alow $sigma(T)cap(mathbbCsetminusmathbbD)$ to be non-empty.

share|cite|improve this answer

answered Mar 11 at 22:55

DisintegratingByPartsDisintegratingByParts

59.8k42681

$endgroup$

add a comment | 

1

$begingroup$

Hint: For 1) if the subset is not strict consider the identity operator on $mathbb R^2$. If the subset is strict then remember we know that $rho(T)=lim_ntoinfty|T^n|^1/n$, where $rho(T)$ is the spectral radius, and we also know $sigma(T)$ is compact.

For 2) consider the operator $T_A:mathbb R^2to mathbb R^2$ given by the matrix:
$$A=beginpmatrix2&0\0&0endpmatrix.$$

share|cite|improve this answer

edited Mar 11 at 23:15

answered Mar 11 at 22:59

K.PowerK.Power

3,345926

$endgroup$

add a comment | 

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3

$begingroup$

Yes to 1. If $sigma(T)subseteqlambda$, then, because the spectrum is closed, there exists $0 < r < 1$ such that $sigma(T)subseteq lambda : $, and there exists $N$ large enough that

$$
|T^n|^1/n le frac1+r2 < 1,;;; n ge N,\
|T^n| le left(frac1+r2right)^n,;; n ge N.
$$

There is a constant $C > 1$ such that $|T^n| le Cleft(frac1+r2right)^n$ for all $1 le n < N$. So $|T^n| le Cleft(frac1+r2right)^n$ for all $n ge 1$.

No to 2. If $T$ is non-invertible, then the only thing you can say is that $0insigma(T)$. The spectrum can be any compact subset of $mathbbC$ that includes $0$, which would alow $sigma(T)cap(mathbbCsetminusmathbbD)$ to be non-empty.

share|cite|improve this answer

answered Mar 11 at 22:55

DisintegratingByPartsDisintegratingByParts

59.8k42681

$endgroup$

add a comment | 

3

$begingroup$

Yes to 1. If $sigma(T)subseteqlambda$, then, because the spectrum is closed, there exists $0 < r < 1$ such that $sigma(T)subseteq lambda : $, and there exists $N$ large enough that

$$
|T^n|^1/n le frac1+r2 < 1,;;; n ge N,\
|T^n| le left(frac1+r2right)^n,;; n ge N.
$$

There is a constant $C > 1$ such that $|T^n| le Cleft(frac1+r2right)^n$ for all $1 le n < N$. So $|T^n| le Cleft(frac1+r2right)^n$ for all $n ge 1$.

No to 2. If $T$ is non-invertible, then the only thing you can say is that $0insigma(T)$. The spectrum can be any compact subset of $mathbbC$ that includes $0$, which would alow $sigma(T)cap(mathbbCsetminusmathbbD)$ to be non-empty.

share|cite|improve this answer

answered Mar 11 at 22:55

DisintegratingByPartsDisintegratingByParts

59.8k42681

$endgroup$

add a comment | 

$begingroup$

Yes to 1. If $sigma(T)subseteqlambda$, then, because the spectrum is closed, there exists $0 < r < 1$ such that $sigma(T)subseteq lambda : $, and there exists $N$ large enough that

$$
|T^n|^1/n le frac1+r2 < 1,;;; n ge N,\
|T^n| le left(frac1+r2right)^n,;; n ge N.
$$

There is a constant $C > 1$ such that $|T^n| le Cleft(frac1+r2right)^n$ for all $1 le n < N$. So $|T^n| le Cleft(frac1+r2right)^n$ for all $n ge 1$.

No to 2. If $T$ is non-invertible, then the only thing you can say is that $0insigma(T)$. The spectrum can be any compact subset of $mathbbC$ that includes $0$, which would alow $sigma(T)cap(mathbbCsetminusmathbbD)$ to be non-empty.

share|cite|improve this answer

answered Mar 11 at 22:55

DisintegratingByPartsDisintegratingByParts

59.8k42681

$endgroup$

share|cite|improve this answer

answered Mar 11 at 22:55

DisintegratingByPartsDisintegratingByParts

59.8k42681

answered Mar 11 at 22:55

DisintegratingByPartsDisintegratingByParts

59.8k42681

answered Mar 11 at 22:55

DisintegratingByPartsDisintegratingByParts

59.8k42681

1

$begingroup$

Hint: For 1) if the subset is not strict consider the identity operator on $mathbb R^2$. If the subset is strict then remember we know that $rho(T)=lim_ntoinfty|T^n|^1/n$, where $rho(T)$ is the spectral radius, and we also know $sigma(T)$ is compact.

For 2) consider the operator $T_A:mathbb R^2to mathbb R^2$ given by the matrix:
$$A=beginpmatrix2&0\0&0endpmatrix.$$

share|cite|improve this answer

edited Mar 11 at 23:15

answered Mar 11 at 22:59

K.PowerK.Power

3,345926

$endgroup$

add a comment | 

1

$begingroup$

Hint: For 1) if the subset is not strict consider the identity operator on $mathbb R^2$. If the subset is strict then remember we know that $rho(T)=lim_ntoinfty|T^n|^1/n$, where $rho(T)$ is the spectral radius, and we also know $sigma(T)$ is compact.

For 2) consider the operator $T_A:mathbb R^2to mathbb R^2$ given by the matrix:
$$A=beginpmatrix2&0\0&0endpmatrix.$$

share|cite|improve this answer

edited Mar 11 at 23:15

answered Mar 11 at 22:59

K.PowerK.Power

3,345926

$endgroup$

add a comment | 

$begingroup$

Hint: For 1) if the subset is not strict consider the identity operator on $mathbb R^2$. If the subset is strict then remember we know that $rho(T)=lim_ntoinfty|T^n|^1/n$, where $rho(T)$ is the spectral radius, and we also know $sigma(T)$ is compact.

For 2) consider the operator $T_A:mathbb R^2to mathbb R^2$ given by the matrix:
$$A=beginpmatrix2&0\0&0endpmatrix.$$

share|cite|improve this answer

edited Mar 11 at 23:15

answered Mar 11 at 22:59

K.PowerK.Power

3,345926

$endgroup$

share|cite|improve this answer

edited Mar 11 at 23:15

answered Mar 11 at 22:59

K.PowerK.Power

3,345926

answered Mar 11 at 22:59

K.PowerK.Power

3,345926

answered Mar 11 at 22:59

K.PowerK.Power

3,345926