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If the point spectrum of some operator is made up of 4 values, $1,2,3,4$ does that mean that all numbers in $[1,4]$ are in the spectrum?
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I am trying to understand what it means exactly for the spectrum of an operator to be closed. I am considering infinite dimensional operators.
Say we know, for example, that the point spectrum of some operator is made up of 4 values, $1,2,3,4$ does that mean that all numbers in $[1,4]$ are in the spectrum (in the residual or continuous spectrum) since by definition the spectrum is closed? Alternatively, is the $1,2,3,4$ set considered closed but just disjoint so that we don't necessarily have all of $[1,4]$ in the spectrum.
Can the spectrum be disjoint EDIT: not disjoint, disconnected?
I am afraid that I might be misunderstanding this idea.
I had a problem where the eigenvalues in the point spectrum were dense in $[0,1]$ and we concluded that the spectrum is made up of the closure of $[0,1]$. So I am wondering if the density was the important aspect of that, or if it works that way in general.
functional-analysis spectral-theory
asked Jul 30 at 3:41
MathIsHard
1,122415
add a comment |Â
up vote
1
down vote
favorite
I am trying to understand what it means exactly for the spectrum of an operator to be closed. I am considering infinite dimensional operators.
Say we know, for example, that the point spectrum of some operator is made up of 4 values, $1,2,3,4$ does that mean that all numbers in $[1,4]$ are in the spectrum (in the residual or continuous spectrum) since by definition the spectrum is closed? Alternatively, is the $1,2,3,4$ set considered closed but just disjoint so that we don't necessarily have all of $[1,4]$ in the spectrum.
Can the spectrum be disjoint EDIT: not disjoint, disconnected?
I am afraid that I might be misunderstanding this idea.
I had a problem where the eigenvalues in the point spectrum were dense in $[0,1]$ and we concluded that the spectrum is made up of the closure of $[0,1]$. So I am wondering if the density was the important aspect of that, or if it works that way in general.
functional-analysis spectral-theory
asked Jul 30 at 3:41
MathIsHard
1,122415
2
$1,2,3,4$ is already closed, so its closure is $1,2,3,4$ and not $[1, 4]$
– mercio
Jul 30 at 6:41
Thank you very much. I appreciate the clarification on that.
– MathIsHard
Jul 30 at 6:42
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am trying to understand what it means exactly for the spectrum of an operator to be closed. I am considering infinite dimensional operators.
Say we know, for example, that the point spectrum of some operator is made up of 4 values, $1,2,3,4$ does that mean that all numbers in $[1,4]$ are in the spectrum (in the residual or continuous spectrum) since by definition the spectrum is closed? Alternatively, is the $1,2,3,4$ set considered closed but just disjoint so that we don't necessarily have all of $[1,4]$ in the spectrum.
Can the spectrum be disjoint EDIT: not disjoint, disconnected?
I am afraid that I might be misunderstanding this idea.
I had a problem where the eigenvalues in the point spectrum were dense in $[0,1]$ and we concluded that the spectrum is made up of the closure of $[0,1]$. So I am wondering if the density was the important aspect of that, or if it works that way in general.
functional-analysis spectral-theory
asked Jul 30 at 3:41
MathIsHard
1,122415
I am trying to understand what it means exactly for the spectrum of an operator to be closed. I am considering infinite dimensional operators.
Say we know, for example, that the point spectrum of some operator is made up of 4 values, $1,2,3,4$ does that mean that all numbers in $[1,4]$ are in the spectrum (in the residual or continuous spectrum) since by definition the spectrum is closed? Alternatively, is the $1,2,3,4$ set considered closed but just disjoint so that we don't necessarily have all of $[1,4]$ in the spectrum.
Can the spectrum be disjoint EDIT: not disjoint, disconnected?
I am afraid that I might be misunderstanding this idea.
I had a problem where the eigenvalues in the point spectrum were dense in $[0,1]$ and we concluded that the spectrum is made up of the closure of $[0,1]$. So I am wondering if the density was the important aspect of that, or if it works that way in general.
functional-analysis spectral-theory
asked Jul 30 at 3:41
MathIsHard
1,122415
edited Jul 30 at 6:09
asked Jul 30 at 3:41
MathIsHard
1,122415
asked Jul 30 at 3:41
MathIsHard
1,122415
asked Jul 30 at 3:41
MathIsHard
1,122415
1,122415
2
$1,2,3,4$ is already closed, so its closure is $1,2,3,4$ and not $[1, 4]$
– mercio
Jul 30 at 6:41
Thank you very much. I appreciate the clarification on that.
– MathIsHard
Jul 30 at 6:42
add a comment |Â
2
$1,2,3,4$ is already closed, so its closure is $1,2,3,4$ and not $[1, 4]$
– mercio
Jul 30 at 6:41
Thank you very much. I appreciate the clarification on that.
– MathIsHard
Jul 30 at 6:42
2
2
$1,2,3,4$ is already closed, so its closure is $1,2,3,4$ and not $[1, 4]$
– mercio
Jul 30 at 6:41
$1,2,3,4$ is already closed, so its closure is $1,2,3,4$ and not $[1, 4]$
– mercio
Jul 30 at 6:41
Thank you very much. I appreciate the clarification on that.
– MathIsHard
Jul 30 at 6:42
Thank you very much. I appreciate the clarification on that.
– MathIsHard
Jul 30 at 6:42
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
Let $T(e_1)=e_1,T(e_2)=2e_2,T(e_3)=3e_3,T(e_4)=4e_4,Te_j=e_j$ for all $j>4$ where $e_j$ is the standard orthonormal basis for $l^2$. Then $1,2,3,4$ are eigen values of $T$. If $lambda notin 1,2,3,4$ you can solve the equation $Tx-lambda x=y$. You will get a unique solution for any $y in l^2$. Hence $T-lambda I$ is a bijection. Automatically the inverse is continuous (by OMT) so $lambda $ is n at in the spectrum. Thus $sigma (T)=1,2,3,4$. [The solution of $Tx-lambda x=y$ is given by $x_j=frac y_j 1-lambda $ if $j >4$ and $x_j=frac y_j j-lambda $ if $j leq 4$]. Incidentally, you are using the word disjoint for disconnected. The statement $A$ is disjoint has no meaning.
answered Jul 30 at 6:04
Kavi Rama Murthy
19.7k2829
Thank you very much for the help. If I am understanding correctly then, the set $1,2,3,4$ is closed and it is okay that the spectrum is the union closed sets?
– MathIsHard
Jul 30 at 6:07
You are right. I meant disconnected...
– MathIsHard
Jul 30 at 6:09
1
Yes, spectrum need not be a connected set.
– Kavi Rama Murthy
Jul 30 at 6:09
Thank you very much. I appreciate the help. :)
– MathIsHard
Jul 30 at 6:10
add a comment |Â
up vote
0
down vote
The answer of Kavi Rama is not O.K.
Example: Let $T: l^2 times mathbb C^4 to l^2 times mathbb C^4$ be defined by
$T((xi_1, xi_2, xi_3,...),(z_1,z_2,z_3,z_4))=(0, xi_1, xi_2, xi_3,...),(z_1,2z_2,3z_3,4z_4))$.
Then the point spectrum of $T$ is $1,2,3,4$, but the spectrum of $T$ is given by
$$lambda cup 1,2,3,4.$$
answered Jul 30 at 6:44
Fred
37k1237
Thank you for your answer. What part of Kavi Rama's answer and your's is different?
– MathIsHard
Jul 30 at 6:47
Oh, you mean that you used a shift so that your Tx isn't dense in $l^2$?
– MathIsHard
Jul 30 at 6:48
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Let $T(e_1)=e_1,T(e_2)=2e_2,T(e_3)=3e_3,T(e_4)=4e_4,Te_j=e_j$ for all $j>4$ where $e_j$ is the standard orthonormal basis for $l^2$. Then $1,2,3,4$ are eigen values of $T$. If $lambda notin 1,2,3,4$ you can solve the equation $Tx-lambda x=y$. You will get a unique solution for any $y in l^2$. Hence $T-lambda I$ is a bijection. Automatically the inverse is continuous (by OMT) so $lambda $ is n at in the spectrum. Thus $sigma (T)=1,2,3,4$. [The solution of $Tx-lambda x=y$ is given by $x_j=frac y_j 1-lambda $ if $j >4$ and $x_j=frac y_j j-lambda $ if $j leq 4$]. Incidentally, you are using the word disjoint for disconnected. The statement $A$ is disjoint has no meaning.
answered Jul 30 at 6:04
Kavi Rama Murthy
19.7k2829
Thank you very much for the help. If I am understanding correctly then, the set $1,2,3,4$ is closed and it is okay that the spectrum is the union closed sets?
– MathIsHard
Jul 30 at 6:07
You are right. I meant disconnected...
– MathIsHard
Jul 30 at 6:09
1
Yes, spectrum need not be a connected set.
– Kavi Rama Murthy
Jul 30 at 6:09
Thank you very much. I appreciate the help. :)
– MathIsHard
Jul 30 at 6:10
add a comment |Â
up vote
1
down vote
accepted
Let $T(e_1)=e_1,T(e_2)=2e_2,T(e_3)=3e_3,T(e_4)=4e_4,Te_j=e_j$ for all $j>4$ where $e_j$ is the standard orthonormal basis for $l^2$. Then $1,2,3,4$ are eigen values of $T$. If $lambda notin 1,2,3,4$ you can solve the equation $Tx-lambda x=y$. You will get a unique solution for any $y in l^2$. Hence $T-lambda I$ is a bijection. Automatically the inverse is continuous (by OMT) so $lambda $ is n at in the spectrum. Thus $sigma (T)=1,2,3,4$. [The solution of $Tx-lambda x=y$ is given by $x_j=frac y_j 1-lambda $ if $j >4$ and $x_j=frac y_j j-lambda $ if $j leq 4$]. Incidentally, you are using the word disjoint for disconnected. The statement $A$ is disjoint has no meaning.
answered Jul 30 at 6:04
Kavi Rama Murthy
19.7k2829
Thank you very much for the help. If I am understanding correctly then, the set $1,2,3,4$ is closed and it is okay that the spectrum is the union closed sets?
– MathIsHard
Jul 30 at 6:07
You are right. I meant disconnected...
– MathIsHard
Jul 30 at 6:09
1
Yes, spectrum need not be a connected set.
– Kavi Rama Murthy
Jul 30 at 6:09
Thank you very much. I appreciate the help. :)
– MathIsHard
Jul 30 at 6:10
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Let $T(e_1)=e_1,T(e_2)=2e_2,T(e_3)=3e_3,T(e_4)=4e_4,Te_j=e_j$ for all $j>4$ where $e_j$ is the standard orthonormal basis for $l^2$. Then $1,2,3,4$ are eigen values of $T$. If $lambda notin 1,2,3,4$ you can solve the equation $Tx-lambda x=y$. You will get a unique solution for any $y in l^2$. Hence $T-lambda I$ is a bijection. Automatically the inverse is continuous (by OMT) so $lambda $ is n at in the spectrum. Thus $sigma (T)=1,2,3,4$. [The solution of $Tx-lambda x=y$ is given by $x_j=frac y_j 1-lambda $ if $j >4$ and $x_j=frac y_j j-lambda $ if $j leq 4$]. Incidentally, you are using the word disjoint for disconnected. The statement $A$ is disjoint has no meaning.
answered Jul 30 at 6:04
Kavi Rama Murthy
19.7k2829
Let $T(e_1)=e_1,T(e_2)=2e_2,T(e_3)=3e_3,T(e_4)=4e_4,Te_j=e_j$ for all $j>4$ where $e_j$ is the standard orthonormal basis for $l^2$. Then $1,2,3,4$ are eigen values of $T$. If $lambda notin 1,2,3,4$ you can solve the equation $Tx-lambda x=y$. You will get a unique solution for any $y in l^2$. Hence $T-lambda I$ is a bijection. Automatically the inverse is continuous (by OMT) so $lambda $ is n at in the spectrum. Thus $sigma (T)=1,2,3,4$. [The solution of $Tx-lambda x=y$ is given by $x_j=frac y_j 1-lambda $ if $j >4$ and $x_j=frac y_j j-lambda $ if $j leq 4$]. Incidentally, you are using the word disjoint for disconnected. The statement $A$ is disjoint has no meaning.
answered Jul 30 at 6:04
Kavi Rama Murthy
19.7k2829
edited Jul 30 at 6:08
answered Jul 30 at 6:04
Kavi Rama Murthy
19.7k2829
answered Jul 30 at 6:04
Kavi Rama Murthy
19.7k2829
answered Jul 30 at 6:04
Kavi Rama Murthy
19.7k2829
19.7k2829
Thank you very much for the help. If I am understanding correctly then, the set $1,2,3,4$ is closed and it is okay that the spectrum is the union closed sets?
– MathIsHard
Jul 30 at 6:07
You are right. I meant disconnected...
– MathIsHard
Jul 30 at 6:09
1
Yes, spectrum need not be a connected set.
– Kavi Rama Murthy
Jul 30 at 6:09
Thank you very much. I appreciate the help. :)
– MathIsHard
Jul 30 at 6:10
add a comment |Â
Thank you very much for the help. If I am understanding correctly then, the set $1,2,3,4$ is closed and it is okay that the spectrum is the union closed sets?
– MathIsHard
Jul 30 at 6:07
You are right. I meant disconnected...
– MathIsHard
Jul 30 at 6:09
1
Yes, spectrum need not be a connected set.
– Kavi Rama Murthy
Jul 30 at 6:09
Thank you very much. I appreciate the help. :)
– MathIsHard
Jul 30 at 6:10
Thank you very much for the help. If I am understanding correctly then, the set $1,2,3,4$ is closed and it is okay that the spectrum is the union closed sets?
– MathIsHard
Jul 30 at 6:07
Thank you very much for the help. If I am understanding correctly then, the set $1,2,3,4$ is closed and it is okay that the spectrum is the union closed sets?
– MathIsHard
Jul 30 at 6:07
You are right. I meant disconnected...
– MathIsHard
Jul 30 at 6:09
You are right. I meant disconnected...
– MathIsHard
Jul 30 at 6:09
1
1
Yes, spectrum need not be a connected set.
– Kavi Rama Murthy
Jul 30 at 6:09
Yes, spectrum need not be a connected set.
– Kavi Rama Murthy
Jul 30 at 6:09
Thank you very much. I appreciate the help. :)
– MathIsHard
Jul 30 at 6:10
Thank you very much. I appreciate the help. :)
– MathIsHard
Jul 30 at 6:10
add a comment |Â
up vote
0
down vote
The answer of Kavi Rama is not O.K.
Example: Let $T: l^2 times mathbb C^4 to l^2 times mathbb C^4$ be defined by
$T((xi_1, xi_2, xi_3,...),(z_1,z_2,z_3,z_4))=(0, xi_1, xi_2, xi_3,...),(z_1,2z_2,3z_3,4z_4))$.
Then the point spectrum of $T$ is $1,2,3,4$, but the spectrum of $T$ is given by
$$lambda cup 1,2,3,4.$$
answered Jul 30 at 6:44
Fred
37k1237
Thank you for your answer. What part of Kavi Rama's answer and your's is different?
– MathIsHard
Jul 30 at 6:47
Oh, you mean that you used a shift so that your Tx isn't dense in $l^2$?
– MathIsHard
Jul 30 at 6:48
add a comment |Â
up vote
0
down vote
The answer of Kavi Rama is not O.K.
Example: Let $T: l^2 times mathbb C^4 to l^2 times mathbb C^4$ be defined by
$T((xi_1, xi_2, xi_3,...),(z_1,z_2,z_3,z_4))=(0, xi_1, xi_2, xi_3,...),(z_1,2z_2,3z_3,4z_4))$.
Then the point spectrum of $T$ is $1,2,3,4$, but the spectrum of $T$ is given by
$$lambda cup 1,2,3,4.$$
answered Jul 30 at 6:44
Fred
37k1237
Thank you for your answer. What part of Kavi Rama's answer and your's is different?
– MathIsHard
Jul 30 at 6:47
Oh, you mean that you used a shift so that your Tx isn't dense in $l^2$?
– MathIsHard
Jul 30 at 6:48
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The answer of Kavi Rama is not O.K.
Example: Let $T: l^2 times mathbb C^4 to l^2 times mathbb C^4$ be defined by
$T((xi_1, xi_2, xi_3,...),(z_1,z_2,z_3,z_4))=(0, xi_1, xi_2, xi_3,...),(z_1,2z_2,3z_3,4z_4))$.
Then the point spectrum of $T$ is $1,2,3,4$, but the spectrum of $T$ is given by
$$lambda cup 1,2,3,4.$$
answered Jul 30 at 6:44
Fred
37k1237
The answer of Kavi Rama is not O.K.
Example: Let $T: l^2 times mathbb C^4 to l^2 times mathbb C^4$ be defined by
$T((xi_1, xi_2, xi_3,...),(z_1,z_2,z_3,z_4))=(0, xi_1, xi_2, xi_3,...),(z_1,2z_2,3z_3,4z_4))$.
Then the point spectrum of $T$ is $1,2,3,4$, but the spectrum of $T$ is given by
$$lambda cup 1,2,3,4.$$
answered Jul 30 at 6:44
Fred
37k1237
answered Jul 30 at 6:44
Fred
37k1237
answered Jul 30 at 6:44
Fred
37k1237
answered Jul 30 at 6:44
Fred
37k1237
37k1237
Thank you for your answer. What part of Kavi Rama's answer and your's is different?
– MathIsHard
Jul 30 at 6:47
Oh, you mean that you used a shift so that your Tx isn't dense in $l^2$?
– MathIsHard
Jul 30 at 6:48
add a comment |Â
Thank you for your answer. What part of Kavi Rama's answer and your's is different?
– MathIsHard
Jul 30 at 6:47
Oh, you mean that you used a shift so that your Tx isn't dense in $l^2$?
– MathIsHard
Jul 30 at 6:48
Thank you for your answer. What part of Kavi Rama's answer and your's is different?
– MathIsHard
Jul 30 at 6:47
Thank you for your answer. What part of Kavi Rama's answer and your's is different?
– MathIsHard
Jul 30 at 6:47
Oh, you mean that you used a shift so that your Tx isn't dense in $l^2$?
– MathIsHard
Jul 30 at 6:48
Oh, you mean that you used a shift so that your Tx isn't dense in $l^2$?
– MathIsHard
Jul 30 at 6:48
add a comment |Â
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2
$1,2,3,4$ is already closed, so its closure is $1,2,3,4$ and not $[1, 4]$
– mercio
Jul 30 at 6:41
Thank you very much. I appreciate the clarification on that.
– MathIsHard
Jul 30 at 6:42