Does an infinite dimensional operator $T$ having an eigenvalue of $0$ imply that $T^-1$ doesn't exist?

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I can't seem to find anywhere a discussion on whether or not the idea of a zero eigenvalue (value in the point spectrum) implies the inverse doesn't exists for an infinite dimensional operator.

We know for a finite operator, this is the case (a 0 eigenvalue means a matrix is not invertible). I am wondering if this same condition (or some other condition on the eigenvalues or spectrum) gives us that the inverse of an infinite dimensional operator doesn't exist.

functional-analysis spectral-theory

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asked Jul 30 at 4:10

MathIsHard

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up vote
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down vote

favorite

I can't seem to find anywhere a discussion on whether or not the idea of a zero eigenvalue (value in the point spectrum) implies the inverse doesn't exists for an infinite dimensional operator.

We know for a finite operator, this is the case (a 0 eigenvalue means a matrix is not invertible). I am wondering if this same condition (or some other condition on the eigenvalues or spectrum) gives us that the inverse of an infinite dimensional operator doesn't exist.

functional-analysis spectral-theory

share|cite|improve this question

asked Jul 30 at 4:10

MathIsHard

1,122415

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

I can't seem to find anywhere a discussion on whether or not the idea of a zero eigenvalue (value in the point spectrum) implies the inverse doesn't exists for an infinite dimensional operator.

We know for a finite operator, this is the case (a 0 eigenvalue means a matrix is not invertible). I am wondering if this same condition (or some other condition on the eigenvalues or spectrum) gives us that the inverse of an infinite dimensional operator doesn't exist.

functional-analysis spectral-theory

share|cite|improve this question

asked Jul 30 at 4:10

MathIsHard

1,122415

I can't seem to find anywhere a discussion on whether or not the idea of a zero eigenvalue (value in the point spectrum) implies the inverse doesn't exists for an infinite dimensional operator.

We know for a finite operator, this is the case (a 0 eigenvalue means a matrix is not invertible). I am wondering if this same condition (or some other condition on the eigenvalues or spectrum) gives us that the inverse of an infinite dimensional operator doesn't exist.

functional-analysis spectral-theory

share|cite|improve this question

asked Jul 30 at 4:10

MathIsHard

1,122415

share|cite|improve this question

share|cite|improve this question

asked Jul 30 at 4:10

MathIsHard

1,122415

asked Jul 30 at 4:10

MathIsHard

1,122415

asked Jul 30 at 4:10

MathIsHard

1,122415

1,122415

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
2
down vote



accepted

Yes. Having an eigenvalue of $0$ is equivalent to $operatornameker (T - 0I)$ is non-trivial, which is equivalent to $T$ not being injective. What breaks down is that $0$ not being an eigenvalue only implies $T$ is injective, not that $T^-1$ exists, as injectivity and bijectivity are no longer equivalent for linear operators on infinite-dimensional spaces.

share|cite|improve this answer

answered Jul 30 at 4:15

Theo Bendit

11.8k1841

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you. I appreciate the help. Is there a condition that will make the operator invertible? I am dealing with this problem particularly, Let $(lambda_j)$ be a sequence of real numbers with $lambda_j neq 1$ for all $j$ and $lambda_j rightarrow 1$. Consider $T: ell^2 rightarrow ell^2$ defined for $xi_j in ell^2$ by $$T(xi_j)=(lambda_j xi_j)$$ I am trying to find a condition on the $lambda_j$ so that $T^-1$ exists. The $lambda_j$ are the equivalent of eigenvalues since they are in the point spectrum of T
  – MathIsHard
  Jul 30 at 4:22
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Well, the obvious modified condition is that $0$ is not in the spectrum of the operator (by definition). My first potentially useful thought is that you might want to consider $|I - T|$, and show that it's strictly less than $1$. Then $sum_n=0^infty (I - T)^n$ converges to an inverse of $T$.
  – Theo Bendit
  Jul 30 at 4:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Wouldn't 0 not being in the spectrum not be enough from what you said above? I am confused what you mean by obvious modified condition... I was looking at that proof in my book that uses that sum, I will think about that some more, I am thinking that we need $||T||geq 1$ for the inverse to not exist. I think that is equivalent to one or more of the $lambda_j$ being larger than 1 in norm maybe?
  – MathIsHard
  Jul 30 at 4:34
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The spectrum is the set of all scalars $lambda$ such that $T - lambda I$ is not invertible. This is larger than the set of eigenvalues: the subset of the spectrum such that $T - lambda I$ is not injective. So $0$ might be in the spectrum ($T$ is not invertible), but not be an eigenvalue ($T$ is injective). I call the modified condition "obvious" because saying $0$ is not in the spectrum is literally just re-stating the condition that $T$ is not invertible, without hinting at any way forward (except possibly armed with a term to Google).
  – Theo Bendit
  Jul 30 at 4:38
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I just figured out why you said $||I-T||<1$ because $(I-(I-T))^-1$ in that series will give the inverse of $T$ :D
  – MathIsHard
  Jul 30 at 5:30
  

  
  
  
  

 | 
show 2 more comments

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1 Answer
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1 Answer
1

active

oldest

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active

oldest

votes

active

oldest

votes

up vote
2
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accepted

Yes. Having an eigenvalue of $0$ is equivalent to $operatornameker (T - 0I)$ is non-trivial, which is equivalent to $T$ not being injective. What breaks down is that $0$ not being an eigenvalue only implies $T$ is injective, not that $T^-1$ exists, as injectivity and bijectivity are no longer equivalent for linear operators on infinite-dimensional spaces.

share|cite|improve this answer

answered Jul 30 at 4:15

Theo Bendit

11.8k1841

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you. I appreciate the help. Is there a condition that will make the operator invertible? I am dealing with this problem particularly, Let $(lambda_j)$ be a sequence of real numbers with $lambda_j neq 1$ for all $j$ and $lambda_j rightarrow 1$. Consider $T: ell^2 rightarrow ell^2$ defined for $xi_j in ell^2$ by $$T(xi_j)=(lambda_j xi_j)$$ I am trying to find a condition on the $lambda_j$ so that $T^-1$ exists. The $lambda_j$ are the equivalent of eigenvalues since they are in the point spectrum of T
  – MathIsHard
  Jul 30 at 4:22
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Well, the obvious modified condition is that $0$ is not in the spectrum of the operator (by definition). My first potentially useful thought is that you might want to consider $|I - T|$, and show that it's strictly less than $1$. Then $sum_n=0^infty (I - T)^n$ converges to an inverse of $T$.
  – Theo Bendit
  Jul 30 at 4:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Wouldn't 0 not being in the spectrum not be enough from what you said above? I am confused what you mean by obvious modified condition... I was looking at that proof in my book that uses that sum, I will think about that some more, I am thinking that we need $||T||geq 1$ for the inverse to not exist. I think that is equivalent to one or more of the $lambda_j$ being larger than 1 in norm maybe?
  – MathIsHard
  Jul 30 at 4:34
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The spectrum is the set of all scalars $lambda$ such that $T - lambda I$ is not invertible. This is larger than the set of eigenvalues: the subset of the spectrum such that $T - lambda I$ is not injective. So $0$ might be in the spectrum ($T$ is not invertible), but not be an eigenvalue ($T$ is injective). I call the modified condition "obvious" because saying $0$ is not in the spectrum is literally just re-stating the condition that $T$ is not invertible, without hinting at any way forward (except possibly armed with a term to Google).
  – Theo Bendit
  Jul 30 at 4:38
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I just figured out why you said $||I-T||<1$ because $(I-(I-T))^-1$ in that series will give the inverse of $T$ :D
  – MathIsHard
  Jul 30 at 5:30
  

  
  
  
  

 | 
show 2 more comments

up vote
2
down vote



accepted

Yes. Having an eigenvalue of $0$ is equivalent to $operatornameker (T - 0I)$ is non-trivial, which is equivalent to $T$ not being injective. What breaks down is that $0$ not being an eigenvalue only implies $T$ is injective, not that $T^-1$ exists, as injectivity and bijectivity are no longer equivalent for linear operators on infinite-dimensional spaces.

share|cite|improve this answer

answered Jul 30 at 4:15

Theo Bendit

11.8k1841

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you. I appreciate the help. Is there a condition that will make the operator invertible? I am dealing with this problem particularly, Let $(lambda_j)$ be a sequence of real numbers with $lambda_j neq 1$ for all $j$ and $lambda_j rightarrow 1$. Consider $T: ell^2 rightarrow ell^2$ defined for $xi_j in ell^2$ by $$T(xi_j)=(lambda_j xi_j)$$ I am trying to find a condition on the $lambda_j$ so that $T^-1$ exists. The $lambda_j$ are the equivalent of eigenvalues since they are in the point spectrum of T
  – MathIsHard
  Jul 30 at 4:22
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Well, the obvious modified condition is that $0$ is not in the spectrum of the operator (by definition). My first potentially useful thought is that you might want to consider $|I - T|$, and show that it's strictly less than $1$. Then $sum_n=0^infty (I - T)^n$ converges to an inverse of $T$.
  – Theo Bendit
  Jul 30 at 4:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Wouldn't 0 not being in the spectrum not be enough from what you said above? I am confused what you mean by obvious modified condition... I was looking at that proof in my book that uses that sum, I will think about that some more, I am thinking that we need $||T||geq 1$ for the inverse to not exist. I think that is equivalent to one or more of the $lambda_j$ being larger than 1 in norm maybe?
  – MathIsHard
  Jul 30 at 4:34
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The spectrum is the set of all scalars $lambda$ such that $T - lambda I$ is not invertible. This is larger than the set of eigenvalues: the subset of the spectrum such that $T - lambda I$ is not injective. So $0$ might be in the spectrum ($T$ is not invertible), but not be an eigenvalue ($T$ is injective). I call the modified condition "obvious" because saying $0$ is not in the spectrum is literally just re-stating the condition that $T$ is not invertible, without hinting at any way forward (except possibly armed with a term to Google).
  – Theo Bendit
  Jul 30 at 4:38
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I just figured out why you said $||I-T||<1$ because $(I-(I-T))^-1$ in that series will give the inverse of $T$ :D
  – MathIsHard
  Jul 30 at 5:30
  

  
  
  
  

 | 
show 2 more comments

up vote
2
down vote



accepted

up vote
2
down vote



accepted

Yes. Having an eigenvalue of $0$ is equivalent to $operatornameker (T - 0I)$ is non-trivial, which is equivalent to $T$ not being injective. What breaks down is that $0$ not being an eigenvalue only implies $T$ is injective, not that $T^-1$ exists, as injectivity and bijectivity are no longer equivalent for linear operators on infinite-dimensional spaces.

share|cite|improve this answer

answered Jul 30 at 4:15

Theo Bendit

11.8k1841

Yes. Having an eigenvalue of $0$ is equivalent to $operatornameker (T - 0I)$ is non-trivial, which is equivalent to $T$ not being injective. What breaks down is that $0$ not being an eigenvalue only implies $T$ is injective, not that $T^-1$ exists, as injectivity and bijectivity are no longer equivalent for linear operators on infinite-dimensional spaces.

share|cite|improve this answer

answered Jul 30 at 4:15

Theo Bendit

11.8k1841

share|cite|improve this answer

share|cite|improve this answer

answered Jul 30 at 4:15

Theo Bendit

11.8k1841

answered Jul 30 at 4:15

Theo Bendit

11.8k1841

answered Jul 30 at 4:15

Theo Bendit

11.8k1841

11.8k1841

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you. I appreciate the help. Is there a condition that will make the operator invertible? I am dealing with this problem particularly, Let $(lambda_j)$ be a sequence of real numbers with $lambda_j neq 1$ for all $j$ and $lambda_j rightarrow 1$. Consider $T: ell^2 rightarrow ell^2$ defined for $xi_j in ell^2$ by $$T(xi_j)=(lambda_j xi_j)$$ I am trying to find a condition on the $lambda_j$ so that $T^-1$ exists. The $lambda_j$ are the equivalent of eigenvalues since they are in the point spectrum of T
  – MathIsHard
  Jul 30 at 4:22
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Well, the obvious modified condition is that $0$ is not in the spectrum of the operator (by definition). My first potentially useful thought is that you might want to consider $|I - T|$, and show that it's strictly less than $1$. Then $sum_n=0^infty (I - T)^n$ converges to an inverse of $T$.
  – Theo Bendit
  Jul 30 at 4:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Wouldn't 0 not being in the spectrum not be enough from what you said above? I am confused what you mean by obvious modified condition... I was looking at that proof in my book that uses that sum, I will think about that some more, I am thinking that we need $||T||geq 1$ for the inverse to not exist. I think that is equivalent to one or more of the $lambda_j$ being larger than 1 in norm maybe?
  – MathIsHard
  Jul 30 at 4:34
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The spectrum is the set of all scalars $lambda$ such that $T - lambda I$ is not invertible. This is larger than the set of eigenvalues: the subset of the spectrum such that $T - lambda I$ is not injective. So $0$ might be in the spectrum ($T$ is not invertible), but not be an eigenvalue ($T$ is injective). I call the modified condition "obvious" because saying $0$ is not in the spectrum is literally just re-stating the condition that $T$ is not invertible, without hinting at any way forward (except possibly armed with a term to Google).
  – Theo Bendit
  Jul 30 at 4:38
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I just figured out why you said $||I-T||<1$ because $(I-(I-T))^-1$ in that series will give the inverse of $T$ :D
  – MathIsHard
  Jul 30 at 5:30
  

  
  
  
  

 | 
show 2 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you. I appreciate the help. Is there a condition that will make the operator invertible? I am dealing with this problem particularly, Let $(lambda_j)$ be a sequence of real numbers with $lambda_j neq 1$ for all $j$ and $lambda_j rightarrow 1$. Consider $T: ell^2 rightarrow ell^2$ defined for $xi_j in ell^2$ by $$T(xi_j)=(lambda_j xi_j)$$ I am trying to find a condition on the $lambda_j$ so that $T^-1$ exists. The $lambda_j$ are the equivalent of eigenvalues since they are in the point spectrum of T
  – MathIsHard
  Jul 30 at 4:22
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Well, the obvious modified condition is that $0$ is not in the spectrum of the operator (by definition). My first potentially useful thought is that you might want to consider $|I - T|$, and show that it's strictly less than $1$. Then $sum_n=0^infty (I - T)^n$ converges to an inverse of $T$.
  – Theo Bendit
  Jul 30 at 4:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Wouldn't 0 not being in the spectrum not be enough from what you said above? I am confused what you mean by obvious modified condition... I was looking at that proof in my book that uses that sum, I will think about that some more, I am thinking that we need $||T||geq 1$ for the inverse to not exist. I think that is equivalent to one or more of the $lambda_j$ being larger than 1 in norm maybe?
  – MathIsHard
  Jul 30 at 4:34
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The spectrum is the set of all scalars $lambda$ such that $T - lambda I$ is not invertible. This is larger than the set of eigenvalues: the subset of the spectrum such that $T - lambda I$ is not injective. So $0$ might be in the spectrum ($T$ is not invertible), but not be an eigenvalue ($T$ is injective). I call the modified condition "obvious" because saying $0$ is not in the spectrum is literally just re-stating the condition that $T$ is not invertible, without hinting at any way forward (except possibly armed with a term to Google).
  – Theo Bendit
  Jul 30 at 4:38
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I just figured out why you said $||I-T||<1$ because $(I-(I-T))^-1$ in that series will give the inverse of $T$ :D
  – MathIsHard
  Jul 30 at 5:30
  

  
  
  
  

Thank you. I appreciate the help. Is there a condition that will make the operator invertible? I am dealing with this problem particularly, Let $(lambda_j)$ be a sequence of real numbers with $lambda_j neq 1$ for all $j$ and $lambda_j rightarrow 1$. Consider $T: ell^2 rightarrow ell^2$ defined for $xi_j in ell^2$ by $$T(xi_j)=(lambda_j xi_j)$$ I am trying to find a condition on the $lambda_j$ so that $T^-1$ exists. The $lambda_j$ are the equivalent of eigenvalues since they are in the point spectrum of T
– MathIsHard
Jul 30 at 4:22

Thank you. I appreciate the help. Is there a condition that will make the operator invertible? I am dealing with this problem particularly, Let $(lambda_j)$ be a sequence of real numbers with $lambda_j neq 1$ for all $j$ and $lambda_j rightarrow 1$. Consider $T: ell^2 rightarrow ell^2$ defined for $xi_j in ell^2$ by $$T(xi_j)=(lambda_j xi_j)$$ I am trying to find a condition on the $lambda_j$ so that $T^-1$ exists. The $lambda_j$ are the equivalent of eigenvalues since they are in the point spectrum of T
– MathIsHard
Jul 30 at 4:22

Well, the obvious modified condition is that $0$ is not in the spectrum of the operator (by definition). My first potentially useful thought is that you might want to consider $|I - T|$, and show that it's strictly less than $1$. Then $sum_n=0^infty (I - T)^n$ converges to an inverse of $T$.
– Theo Bendit
Jul 30 at 4:28

Well, the obvious modified condition is that $0$ is not in the spectrum of the operator (by definition). My first potentially useful thought is that you might want to consider $|I - T|$, and show that it's strictly less than $1$. Then $sum_n=0^infty (I - T)^n$ converges to an inverse of $T$.
– Theo Bendit
Jul 30 at 4:28

Wouldn't 0 not being in the spectrum not be enough from what you said above? I am confused what you mean by obvious modified condition... I was looking at that proof in my book that uses that sum, I will think about that some more, I am thinking that we need $||T||geq 1$ for the inverse to not exist. I think that is equivalent to one or more of the $lambda_j$ being larger than 1 in norm maybe?
– MathIsHard
Jul 30 at 4:34

Wouldn't 0 not being in the spectrum not be enough from what you said above? I am confused what you mean by obvious modified condition... I was looking at that proof in my book that uses that sum, I will think about that some more, I am thinking that we need $||T||geq 1$ for the inverse to not exist. I think that is equivalent to one or more of the $lambda_j$ being larger than 1 in norm maybe?
– MathIsHard
Jul 30 at 4:34

1

1

The spectrum is the set of all scalars $lambda$ such that $T - lambda I$ is not invertible. This is larger than the set of eigenvalues: the subset of the spectrum such that $T - lambda I$ is not injective. So $0$ might be in the spectrum ($T$ is not invertible), but not be an eigenvalue ($T$ is injective). I call the modified condition "obvious" because saying $0$ is not in the spectrum is literally just re-stating the condition that $T$ is not invertible, without hinting at any way forward (except possibly armed with a term to Google).
– Theo Bendit
Jul 30 at 4:38

The spectrum is the set of all scalars $lambda$ such that $T - lambda I$ is not invertible. This is larger than the set of eigenvalues: the subset of the spectrum such that $T - lambda I$ is not injective. So $0$ might be in the spectrum ($T$ is not invertible), but not be an eigenvalue ($T$ is injective). I call the modified condition "obvious" because saying $0$ is not in the spectrum is literally just re-stating the condition that $T$ is not invertible, without hinting at any way forward (except possibly armed with a term to Google).
– Theo Bendit
Jul 30 at 4:38

1

1

I just figured out why you said $||I-T||<1$ because $(I-(I-T))^-1$ in that series will give the inverse of $T$ :D
– MathIsHard
Jul 30 at 5:30

I just figured out why you said $||I-T||<1$ because $(I-(I-T))^-1$ in that series will give the inverse of $T$ :D
– MathIsHard
Jul 30 at 5:30

 | 
show 2 more comments

 

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