Mixing
Does an infinite dimensional operator $T$ having an eigenvalue of $0$ imply that $T^-1$ doesn't exist?
Clash Royale CLAN TAG#URR8PPP
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
asked Jul 30 at 4:10
MathIsHard
1,122415
add a comment |Â
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
asked Jul 30 at 4:10
MathIsHard
1,122415
add a comment |Â
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
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
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
asked Jul 30 at 4:10
MathIsHard
1,122415
1,122415
add a comment |Â
add a comment |Â
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.
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
 |Â
show 2 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
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.
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
 |Â
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.
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
 |Â
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.
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.
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
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
 |Â
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
 |Â
show 2 more comments
Â
draft saved
draft discarded
Â
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2866651%2fdoes-an-infinite-dimensional-operator-t-having-an-eigenvalue-of-0-imply-that%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password