Mixing
A self-adjoint operator without eigenvalues and with spectrum equal to {0}
$begingroup$
Let $A$ be a self-adjoint operator on a Hilbert space $H$. We know that the spectrum of $A$ ( $sigma(A)$) can be decomposed into an essential spectrum ($sigma_{ess}(A)$) and a set of eigenvalues ($sigma_e(A)$) : $sigma(A)=sigma_{ess}(A)cupsigma_e(A)$.
The question is: if we know that $sigma(A)=sigma_{ess}={0}$ can we prove that $A=0$?
spectral-theory self-adjoint-operators
asked Feb 1 at 10:53
MahranMahran
1235
$endgroup$
add a comment |
$begingroup$
Let $A$ be a self-adjoint operator on a Hilbert space $H$. We know that the spectrum of $A$ ( $sigma(A)$) can be decomposed into an essential spectrum ($sigma_{ess}(A)$) and a set of eigenvalues ($sigma_e(A)$) : $sigma(A)=sigma_{ess}(A)cupsigma_e(A)$.
The question is: if we know that $sigma(A)=sigma_{ess}={0}$ can we prove that $A=0$?
spectral-theory self-adjoint-operators
asked Feb 1 at 10:53
MahranMahran
1235
$endgroup$
add a comment |
$begingroup$
Let $A$ be a self-adjoint operator on a Hilbert space $H$. We know that the spectrum of $A$ ( $sigma(A)$) can be decomposed into an essential spectrum ($sigma_{ess}(A)$) and a set of eigenvalues ($sigma_e(A)$) : $sigma(A)=sigma_{ess}(A)cupsigma_e(A)$.
The question is: if we know that $sigma(A)=sigma_{ess}={0}$ can we prove that $A=0$?
spectral-theory self-adjoint-operators
asked Feb 1 at 10:53
MahranMahran
1235
$endgroup$
Let $A$ be a self-adjoint operator on a Hilbert space $H$. We know that the spectrum of $A$ ( $sigma(A)$) can be decomposed into an essential spectrum ($sigma_{ess}(A)$) and a set of eigenvalues ($sigma_e(A)$) : $sigma(A)=sigma_{ess}(A)cupsigma_e(A)$.
The question is: if we know that $sigma(A)=sigma_{ess}={0}$ can we prove that $A=0$?
spectral-theory self-adjoint-operators
spectral-theory self-adjoint-operators
asked Feb 1 at 10:53
MahranMahran
1235
asked Feb 1 at 10:53
MahranMahran
1235
asked Feb 1 at 10:53
MahranMahran
1235
asked Feb 1 at 10:53
MahranMahran
1235
asked Feb 1 at 10:53
MahranMahran
1235
1235
add a comment |
add a comment |
1 Answer
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$begingroup$
If $A: H to H$ is self-adjoint, then $A$ is bounded (Hellinger Toeplitz !).
If $r(A)= max{|lambda|: lambda in sigma(A)}$ denotes the spectral radius of $A$, then it is well-known that $r(A)=||A||$, if $A$ is self-adjoint.
Hence, if $sigma(A)={0}$, then $r(A)=0$, hence $A=0.$
answered Feb 1 at 10:57
FredFred
48.5k11849
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1 Answer
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1 Answer
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$begingroup$
If $A: H to H$ is self-adjoint, then $A$ is bounded (Hellinger Toeplitz !).
If $r(A)= max{|lambda|: lambda in sigma(A)}$ denotes the spectral radius of $A$, then it is well-known that $r(A)=||A||$, if $A$ is self-adjoint.
Hence, if $sigma(A)={0}$, then $r(A)=0$, hence $A=0.$
answered Feb 1 at 10:57
FredFred
48.5k11849
$endgroup$
add a comment |
$begingroup$
If $A: H to H$ is self-adjoint, then $A$ is bounded (Hellinger Toeplitz !).
If $r(A)= max{|lambda|: lambda in sigma(A)}$ denotes the spectral radius of $A$, then it is well-known that $r(A)=||A||$, if $A$ is self-adjoint.
Hence, if $sigma(A)={0}$, then $r(A)=0$, hence $A=0.$
answered Feb 1 at 10:57
FredFred
48.5k11849
$endgroup$
add a comment |
$begingroup$
If $A: H to H$ is self-adjoint, then $A$ is bounded (Hellinger Toeplitz !).
If $r(A)= max{|lambda|: lambda in sigma(A)}$ denotes the spectral radius of $A$, then it is well-known that $r(A)=||A||$, if $A$ is self-adjoint.
Hence, if $sigma(A)={0}$, then $r(A)=0$, hence $A=0.$
answered Feb 1 at 10:57
FredFred
48.5k11849
$endgroup$
If $A: H to H$ is self-adjoint, then $A$ is bounded (Hellinger Toeplitz !).
If $r(A)= max{|lambda|: lambda in sigma(A)}$ denotes the spectral radius of $A$, then it is well-known that $r(A)=||A||$, if $A$ is self-adjoint.
Hence, if $sigma(A)={0}$, then $r(A)=0$, hence $A=0.$
answered Feb 1 at 10:57
FredFred
48.5k11849
answered Feb 1 at 10:57
FredFred
48.5k11849
answered Feb 1 at 10:57
FredFred
48.5k11849
answered Feb 1 at 10:57
FredFred
48.5k11849
48.5k11849
add a comment |
add a comment |
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