Mixing
Relations between Trace and Spectrum of an Operator
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As the trace of a matrix equals to the sum of all eigenvalues, do we have an analogous result that the trace of an operator (if it is well-defined) equals to an integral on the spectrum space?
reference-request operator-theory trace
edited Jan 10 at 9:40
José Carlos Santos
159k22126230
asked Aug 18 '17 at 14:36
FanFan
780313
$endgroup$
add a comment |
$begingroup$
As the trace of a matrix equals to the sum of all eigenvalues, do we have an analogous result that the trace of an operator (if it is well-defined) equals to an integral on the spectrum space?
reference-request operator-theory trace
edited Jan 10 at 9:40
José Carlos Santos
159k22126230
asked Aug 18 '17 at 14:36
FanFan
780313
$endgroup$
add a comment |
$begingroup$
As the trace of a matrix equals to the sum of all eigenvalues, do we have an analogous result that the trace of an operator (if it is well-defined) equals to an integral on the spectrum space?
reference-request operator-theory trace
edited Jan 10 at 9:40
José Carlos Santos
159k22126230
asked Aug 18 '17 at 14:36
FanFan
780313
$endgroup$
As the trace of a matrix equals to the sum of all eigenvalues, do we have an analogous result that the trace of an operator (if it is well-defined) equals to an integral on the spectrum space?
reference-request operator-theory trace
reference-request operator-theory trace
edited Jan 10 at 9:40
José Carlos Santos
159k22126230
asked Aug 18 '17 at 14:36
FanFan
780313
edited Jan 10 at 9:40
José Carlos Santos
159k22126230
asked Aug 18 '17 at 14:36
FanFan
780313
edited Jan 10 at 9:40
José Carlos Santos
159k22126230
edited Jan 10 at 9:40
José Carlos Santos
159k22126230
edited Jan 10 at 9:40
José Carlos Santos
159k22126230
159k22126230
asked Aug 18 '17 at 14:36
FanFan
780313
asked Aug 18 '17 at 14:36
FanFan
780313
asked Aug 18 '17 at 14:36
FanFan
780313
780313
add a comment |
add a comment |
1 Answer
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It seems to me that you're after Lidskii's theorem. It states that if $H$ is a separable Hilbert space, if $Acolon Hlongrightarrow H$ is an operator of trace class, and if ${lambda_1,lambda_2,ldots}$ are the eigenvalues of $A$ (enumerated with algebraic multiplicities taken into account), then$$sum_{n=1}^inftylambda_n=operatorname{Tr}(A).$$Note that the previous sum may well be just a finite sum.
answered Aug 18 '17 at 17:31
José Carlos SantosJosé Carlos Santos
159k22126230
$endgroup$
$begingroup$
Thanks for your answer. I think operators of trace class are compact operators and thus have countable spectrum if I'm right? So this result is not extremely surprising. What also interests me is, for example, the irrational rotation operator $Vf(t)=f(t-theta)$ on $L^2(mathbb{R}/mathbb{Z})$, whose spectrum is the whole unit circle and the trace is exactly $int e^{2pi it}dt=0.$ Is there any deeper explanation for this?
$endgroup$
– Fan
Aug 18 '17 at 18:01
$begingroup$
@Fan No. The definition is: a trace class operator is a compact operator $A$ such that, for some orthonormal basis $(e_n)_{ninmathbb N}$ of $H$, the sum $sum_nlangle (A^*A)^{1/2} e_n, e_n rangle$ converges. Yes, the set of eigenvalues is countable, but that is not part of the definition. I don't know the answer to your question.
$endgroup$
– José Carlos Santos
Aug 18 '17 at 18:07
add a comment |
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1 Answer
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1 Answer
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$begingroup$
It seems to me that you're after Lidskii's theorem. It states that if $H$ is a separable Hilbert space, if $Acolon Hlongrightarrow H$ is an operator of trace class, and if ${lambda_1,lambda_2,ldots}$ are the eigenvalues of $A$ (enumerated with algebraic multiplicities taken into account), then$$sum_{n=1}^inftylambda_n=operatorname{Tr}(A).$$Note that the previous sum may well be just a finite sum.
answered Aug 18 '17 at 17:31
José Carlos SantosJosé Carlos Santos
159k22126230
$endgroup$
$begingroup$
Thanks for your answer. I think operators of trace class are compact operators and thus have countable spectrum if I'm right? So this result is not extremely surprising. What also interests me is, for example, the irrational rotation operator $Vf(t)=f(t-theta)$ on $L^2(mathbb{R}/mathbb{Z})$, whose spectrum is the whole unit circle and the trace is exactly $int e^{2pi it}dt=0.$ Is there any deeper explanation for this?
$endgroup$
– Fan
Aug 18 '17 at 18:01
$begingroup$
@Fan No. The definition is: a trace class operator is a compact operator $A$ such that, for some orthonormal basis $(e_n)_{ninmathbb N}$ of $H$, the sum $sum_nlangle (A^*A)^{1/2} e_n, e_n rangle$ converges. Yes, the set of eigenvalues is countable, but that is not part of the definition. I don't know the answer to your question.
$endgroup$
– José Carlos Santos
Aug 18 '17 at 18:07
add a comment |
$begingroup$
It seems to me that you're after Lidskii's theorem. It states that if $H$ is a separable Hilbert space, if $Acolon Hlongrightarrow H$ is an operator of trace class, and if ${lambda_1,lambda_2,ldots}$ are the eigenvalues of $A$ (enumerated with algebraic multiplicities taken into account), then$$sum_{n=1}^inftylambda_n=operatorname{Tr}(A).$$Note that the previous sum may well be just a finite sum.
answered Aug 18 '17 at 17:31
José Carlos SantosJosé Carlos Santos
159k22126230
$endgroup$
$begingroup$
Thanks for your answer. I think operators of trace class are compact operators and thus have countable spectrum if I'm right? So this result is not extremely surprising. What also interests me is, for example, the irrational rotation operator $Vf(t)=f(t-theta)$ on $L^2(mathbb{R}/mathbb{Z})$, whose spectrum is the whole unit circle and the trace is exactly $int e^{2pi it}dt=0.$ Is there any deeper explanation for this?
$endgroup$
– Fan
Aug 18 '17 at 18:01
$begingroup$
@Fan No. The definition is: a trace class operator is a compact operator $A$ such that, for some orthonormal basis $(e_n)_{ninmathbb N}$ of $H$, the sum $sum_nlangle (A^*A)^{1/2} e_n, e_n rangle$ converges. Yes, the set of eigenvalues is countable, but that is not part of the definition. I don't know the answer to your question.
$endgroup$
– José Carlos Santos
Aug 18 '17 at 18:07
add a comment |
$begingroup$
It seems to me that you're after Lidskii's theorem. It states that if $H$ is a separable Hilbert space, if $Acolon Hlongrightarrow H$ is an operator of trace class, and if ${lambda_1,lambda_2,ldots}$ are the eigenvalues of $A$ (enumerated with algebraic multiplicities taken into account), then$$sum_{n=1}^inftylambda_n=operatorname{Tr}(A).$$Note that the previous sum may well be just a finite sum.
answered Aug 18 '17 at 17:31
José Carlos SantosJosé Carlos Santos
159k22126230
$endgroup$
It seems to me that you're after Lidskii's theorem. It states that if $H$ is a separable Hilbert space, if $Acolon Hlongrightarrow H$ is an operator of trace class, and if ${lambda_1,lambda_2,ldots}$ are the eigenvalues of $A$ (enumerated with algebraic multiplicities taken into account), then$$sum_{n=1}^inftylambda_n=operatorname{Tr}(A).$$Note that the previous sum may well be just a finite sum.
answered Aug 18 '17 at 17:31
José Carlos SantosJosé Carlos Santos
159k22126230
answered Aug 18 '17 at 17:31
José Carlos SantosJosé Carlos Santos
159k22126230
answered Aug 18 '17 at 17:31
José Carlos SantosJosé Carlos Santos
159k22126230
answered Aug 18 '17 at 17:31
José Carlos SantosJosé Carlos Santos
159k22126230
159k22126230
$begingroup$
Thanks for your answer. I think operators of trace class are compact operators and thus have countable spectrum if I'm right? So this result is not extremely surprising. What also interests me is, for example, the irrational rotation operator $Vf(t)=f(t-theta)$ on $L^2(mathbb{R}/mathbb{Z})$, whose spectrum is the whole unit circle and the trace is exactly $int e^{2pi it}dt=0.$ Is there any deeper explanation for this?
$endgroup$
– Fan
Aug 18 '17 at 18:01
$begingroup$
@Fan No. The definition is: a trace class operator is a compact operator $A$ such that, for some orthonormal basis $(e_n)_{ninmathbb N}$ of $H$, the sum $sum_nlangle (A^*A)^{1/2} e_n, e_n rangle$ converges. Yes, the set of eigenvalues is countable, but that is not part of the definition. I don't know the answer to your question.
$endgroup$
– José Carlos Santos
Aug 18 '17 at 18:07
add a comment |
$begingroup$
Thanks for your answer. I think operators of trace class are compact operators and thus have countable spectrum if I'm right? So this result is not extremely surprising. What also interests me is, for example, the irrational rotation operator $Vf(t)=f(t-theta)$ on $L^2(mathbb{R}/mathbb{Z})$, whose spectrum is the whole unit circle and the trace is exactly $int e^{2pi it}dt=0.$ Is there any deeper explanation for this?
$endgroup$
– Fan
Aug 18 '17 at 18:01
$begingroup$
@Fan No. The definition is: a trace class operator is a compact operator $A$ such that, for some orthonormal basis $(e_n)_{ninmathbb N}$ of $H$, the sum $sum_nlangle (A^*A)^{1/2} e_n, e_n rangle$ converges. Yes, the set of eigenvalues is countable, but that is not part of the definition. I don't know the answer to your question.
$endgroup$
– José Carlos Santos
Aug 18 '17 at 18:07
$begingroup$
Thanks for your answer. I think operators of trace class are compact operators and thus have countable spectrum if I'm right? So this result is not extremely surprising. What also interests me is, for example, the irrational rotation operator $Vf(t)=f(t-theta)$ on $L^2(mathbb{R}/mathbb{Z})$, whose spectrum is the whole unit circle and the trace is exactly $int e^{2pi it}dt=0.$ Is there any deeper explanation for this?
$endgroup$
– Fan
Aug 18 '17 at 18:01
$begingroup$
Thanks for your answer. I think operators of trace class are compact operators and thus have countable spectrum if I'm right? So this result is not extremely surprising. What also interests me is, for example, the irrational rotation operator $Vf(t)=f(t-theta)$ on $L^2(mathbb{R}/mathbb{Z})$, whose spectrum is the whole unit circle and the trace is exactly $int e^{2pi it}dt=0.$ Is there any deeper explanation for this?
$endgroup$
– Fan
Aug 18 '17 at 18:01
$begingroup$
@Fan No. The definition is: a trace class operator is a compact operator $A$ such that, for some orthonormal basis $(e_n)_{ninmathbb N}$ of $H$, the sum $sum_nlangle (A^*A)^{1/2} e_n, e_n rangle$ converges. Yes, the set of eigenvalues is countable, but that is not part of the definition. I don't know the answer to your question.
$endgroup$
– José Carlos Santos
Aug 18 '17 at 18:07
$begingroup$
@Fan No. The definition is: a trace class operator is a compact operator $A$ such that, for some orthonormal basis $(e_n)_{ninmathbb N}$ of $H$, the sum $sum_nlangle (A^*A)^{1/2} e_n, e_n rangle$ converges. Yes, the set of eigenvalues is countable, but that is not part of the definition. I don't know the answer to your question.
$endgroup$
– José Carlos Santos
Aug 18 '17 at 18:07
add a comment |
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