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Verification of the values in the continuous spectrum of an operator
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I am needing help with the continuous spectrum.
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 $$T(xi_j)=(lambda_j xi_j)$$
Questions: Find the point spectrum, residual spectrum, and continuous spectrum.
My thoughts...
I believe that the point spectrum is the set of the $lambda_j$. Since $$Te_j=lambda_j e_j$$ for $e_j$ having a 1 in the $jth$ position and zeroes elsewhere.
The operator is bounded and self adjoint since the $lambda_j$ are real (I will omit the details), so we know the residual spectrum is empty.
This is the part I am not sure about.
EDIT: I now think that the continuous spectrum only contains 1. I think this because that is the only value we can get arbitrarily close to for $lambda neq lambda_j)$. If someone could verify this that would be great. Thanks!
Since the spectrum is closed, we know that the closure of the set of $lambda_j subset $ the spectrum.
We can show that the set that is the closure of the set of $lambda_j$ without the $lambda_j$ is in the continuous spectrum since the resolvent would be unbounded. I think it is similar to this question here, (see the last paragraph). But I am not sure because we don't have that the $lambda_j$ are dense.
All other values are in the resolvent set.
functional-analysis proof-verification spectral-theory
asked Jul 30 at 5:51
MathIsHard
1,122415
add a comment |Â
up vote
1
down vote
favorite
I am needing help with the continuous spectrum.
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 $$T(xi_j)=(lambda_j xi_j)$$
Questions: Find the point spectrum, residual spectrum, and continuous spectrum.
My thoughts...
I believe that the point spectrum is the set of the $lambda_j$. Since $$Te_j=lambda_j e_j$$ for $e_j$ having a 1 in the $jth$ position and zeroes elsewhere.
The operator is bounded and self adjoint since the $lambda_j$ are real (I will omit the details), so we know the residual spectrum is empty.
This is the part I am not sure about.
EDIT: I now think that the continuous spectrum only contains 1. I think this because that is the only value we can get arbitrarily close to for $lambda neq lambda_j)$. If someone could verify this that would be great. Thanks!
Since the spectrum is closed, we know that the closure of the set of $lambda_j subset $ the spectrum.
We can show that the set that is the closure of the set of $lambda_j$ without the $lambda_j$ is in the continuous spectrum since the resolvent would be unbounded. I think it is similar to this question here, (see the last paragraph). But I am not sure because we don't have that the $lambda_j$ are dense.
All other values are in the resolvent set.
functional-analysis proof-verification spectral-theory
asked Jul 30 at 5:51
MathIsHard
1,122415
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am needing help with the continuous spectrum.
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 $$T(xi_j)=(lambda_j xi_j)$$
Questions: Find the point spectrum, residual spectrum, and continuous spectrum.
My thoughts...
I believe that the point spectrum is the set of the $lambda_j$. Since $$Te_j=lambda_j e_j$$ for $e_j$ having a 1 in the $jth$ position and zeroes elsewhere.
The operator is bounded and self adjoint since the $lambda_j$ are real (I will omit the details), so we know the residual spectrum is empty.
This is the part I am not sure about.
EDIT: I now think that the continuous spectrum only contains 1. I think this because that is the only value we can get arbitrarily close to for $lambda neq lambda_j)$. If someone could verify this that would be great. Thanks!
Since the spectrum is closed, we know that the closure of the set of $lambda_j subset $ the spectrum.
We can show that the set that is the closure of the set of $lambda_j$ without the $lambda_j$ is in the continuous spectrum since the resolvent would be unbounded. I think it is similar to this question here, (see the last paragraph). But I am not sure because we don't have that the $lambda_j$ are dense.
All other values are in the resolvent set.
functional-analysis proof-verification spectral-theory
asked Jul 30 at 5:51
MathIsHard
1,122415
I am needing help with the continuous spectrum.
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 $$T(xi_j)=(lambda_j xi_j)$$
Questions: Find the point spectrum, residual spectrum, and continuous spectrum.
My thoughts...
I believe that the point spectrum is the set of the $lambda_j$. Since $$Te_j=lambda_j e_j$$ for $e_j$ having a 1 in the $jth$ position and zeroes elsewhere.
The operator is bounded and self adjoint since the $lambda_j$ are real (I will omit the details), so we know the residual spectrum is empty.
This is the part I am not sure about.
EDIT: I now think that the continuous spectrum only contains 1. I think this because that is the only value we can get arbitrarily close to for $lambda neq lambda_j)$. If someone could verify this that would be great. Thanks!
Since the spectrum is closed, we know that the closure of the set of $lambda_j subset $ the spectrum.
We can show that the set that is the closure of the set of $lambda_j$ without the $lambda_j$ is in the continuous spectrum since the resolvent would be unbounded. I think it is similar to this question here, (see the last paragraph). But I am not sure because we don't have that the $lambda_j$ are dense.
All other values are in the resolvent set.
functional-analysis proof-verification spectral-theory
asked Jul 30 at 5:51
MathIsHard
1,122415
edited Jul 30 at 6:11
asked Jul 30 at 5:51
MathIsHard
1,122415
asked Jul 30 at 5:51
MathIsHard
1,122415
asked Jul 30 at 5:51
MathIsHard
1,122415
1,122415
add a comment |Â
add a comment |Â
2 Answers
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1
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In general, for any complex sequence $(lambda_n)_n$ the spectra for the diagonal operator $T$ are:
$$sigma(T) = overlinelambda_n : n in mathbbN$$
$$sigma_p(T) = lambda_n : n in mathbbN$$
$$sigma_r(T) = emptyset$$
$$sigma_c(T) = overlinelambda_n : n in mathbbN setminus lambda_n : n in mathbbN$$
So in your case we have $$sigma(T)= lambda_n : n in mathbbN cup 1$$ $$sigma_p(T) = lambda_n : n in mathbbN$$
$$sigma_r(T) = emptyset$$
$$sigma_c(T) = 1$$
answered Jul 30 at 9:27
mechanodroid
22.2k52041
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up vote
1
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If $lambda notin 1,lambda_1,lambda_2,...$ ther exists $a>0$ such that $|lambda -c|>a$ for all $c in 1,lambda_1,lambda_2,...$ (This is because the set here is compact). Now solve the equarion $Tx-lambda x=y$ for any $y in l^2$. You will get a unique solution. This implies that $T-lambda I$ is invertible. Hence $sigma (T) subset 1,lambda_1,lambda_2,...$. Since $lambda_1,lambda_2,...$ is contained in the spectrum and the spectrum is closed we get $sigma (T) = 1,lambda_1,lambda_2,...$.
answered Jul 30 at 6:25
Kavi Rama Murthy
19.7k2829
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
In general, for any complex sequence $(lambda_n)_n$ the spectra for the diagonal operator $T$ are:
$$sigma(T) = overlinelambda_n : n in mathbbN$$
$$sigma_p(T) = lambda_n : n in mathbbN$$
$$sigma_r(T) = emptyset$$
$$sigma_c(T) = overlinelambda_n : n in mathbbN setminus lambda_n : n in mathbbN$$
So in your case we have $$sigma(T)= lambda_n : n in mathbbN cup 1$$ $$sigma_p(T) = lambda_n : n in mathbbN$$
$$sigma_r(T) = emptyset$$
$$sigma_c(T) = 1$$
answered Jul 30 at 9:27
mechanodroid
22.2k52041
add a comment |Â
up vote
1
down vote
accepted
In general, for any complex sequence $(lambda_n)_n$ the spectra for the diagonal operator $T$ are:
$$sigma(T) = overlinelambda_n : n in mathbbN$$
$$sigma_p(T) = lambda_n : n in mathbbN$$
$$sigma_r(T) = emptyset$$
$$sigma_c(T) = overlinelambda_n : n in mathbbN setminus lambda_n : n in mathbbN$$
So in your case we have $$sigma(T)= lambda_n : n in mathbbN cup 1$$ $$sigma_p(T) = lambda_n : n in mathbbN$$
$$sigma_r(T) = emptyset$$
$$sigma_c(T) = 1$$
answered Jul 30 at 9:27
mechanodroid
22.2k52041
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
In general, for any complex sequence $(lambda_n)_n$ the spectra for the diagonal operator $T$ are:
$$sigma(T) = overlinelambda_n : n in mathbbN$$
$$sigma_p(T) = lambda_n : n in mathbbN$$
$$sigma_r(T) = emptyset$$
$$sigma_c(T) = overlinelambda_n : n in mathbbN setminus lambda_n : n in mathbbN$$
So in your case we have $$sigma(T)= lambda_n : n in mathbbN cup 1$$ $$sigma_p(T) = lambda_n : n in mathbbN$$
$$sigma_r(T) = emptyset$$
$$sigma_c(T) = 1$$
answered Jul 30 at 9:27
mechanodroid
22.2k52041
In general, for any complex sequence $(lambda_n)_n$ the spectra for the diagonal operator $T$ are:
$$sigma(T) = overlinelambda_n : n in mathbbN$$
$$sigma_p(T) = lambda_n : n in mathbbN$$
$$sigma_r(T) = emptyset$$
$$sigma_c(T) = overlinelambda_n : n in mathbbN setminus lambda_n : n in mathbbN$$
So in your case we have $$sigma(T)= lambda_n : n in mathbbN cup 1$$ $$sigma_p(T) = lambda_n : n in mathbbN$$
$$sigma_r(T) = emptyset$$
$$sigma_c(T) = 1$$
answered Jul 30 at 9:27
mechanodroid
22.2k52041
answered Jul 30 at 9:27
mechanodroid
22.2k52041
answered Jul 30 at 9:27
mechanodroid
22.2k52041
answered Jul 30 at 9:27
mechanodroid
22.2k52041
22.2k52041
add a comment |Â
add a comment |Â
up vote
1
down vote
If $lambda notin 1,lambda_1,lambda_2,...$ ther exists $a>0$ such that $|lambda -c|>a$ for all $c in 1,lambda_1,lambda_2,...$ (This is because the set here is compact). Now solve the equarion $Tx-lambda x=y$ for any $y in l^2$. You will get a unique solution. This implies that $T-lambda I$ is invertible. Hence $sigma (T) subset 1,lambda_1,lambda_2,...$. Since $lambda_1,lambda_2,...$ is contained in the spectrum and the spectrum is closed we get $sigma (T) = 1,lambda_1,lambda_2,...$.
answered Jul 30 at 6:25
Kavi Rama Murthy
19.7k2829
add a comment |Â
up vote
1
down vote
If $lambda notin 1,lambda_1,lambda_2,...$ ther exists $a>0$ such that $|lambda -c|>a$ for all $c in 1,lambda_1,lambda_2,...$ (This is because the set here is compact). Now solve the equarion $Tx-lambda x=y$ for any $y in l^2$. You will get a unique solution. This implies that $T-lambda I$ is invertible. Hence $sigma (T) subset 1,lambda_1,lambda_2,...$. Since $lambda_1,lambda_2,...$ is contained in the spectrum and the spectrum is closed we get $sigma (T) = 1,lambda_1,lambda_2,...$.
answered Jul 30 at 6:25
Kavi Rama Murthy
19.7k2829
add a comment |Â
up vote
1
down vote
up vote
1
down vote
If $lambda notin 1,lambda_1,lambda_2,...$ ther exists $a>0$ such that $|lambda -c|>a$ for all $c in 1,lambda_1,lambda_2,...$ (This is because the set here is compact). Now solve the equarion $Tx-lambda x=y$ for any $y in l^2$. You will get a unique solution. This implies that $T-lambda I$ is invertible. Hence $sigma (T) subset 1,lambda_1,lambda_2,...$. Since $lambda_1,lambda_2,...$ is contained in the spectrum and the spectrum is closed we get $sigma (T) = 1,lambda_1,lambda_2,...$.
answered Jul 30 at 6:25
Kavi Rama Murthy
19.7k2829
If $lambda notin 1,lambda_1,lambda_2,...$ ther exists $a>0$ such that $|lambda -c|>a$ for all $c in 1,lambda_1,lambda_2,...$ (This is because the set here is compact). Now solve the equarion $Tx-lambda x=y$ for any $y in l^2$. You will get a unique solution. This implies that $T-lambda I$ is invertible. Hence $sigma (T) subset 1,lambda_1,lambda_2,...$. Since $lambda_1,lambda_2,...$ is contained in the spectrum and the spectrum is closed we get $sigma (T) = 1,lambda_1,lambda_2,...$.
answered Jul 30 at 6:25
Kavi Rama Murthy
19.7k2829
answered Jul 30 at 6:25
Kavi Rama Murthy
19.7k2829
answered Jul 30 at 6:25
Kavi Rama Murthy
19.7k2829
answered Jul 30 at 6:25
Kavi Rama Murthy
19.7k2829
19.7k2829
add a comment |Â
add a comment |Â
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