the spectrum of normal operator
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In a Hilbert space if $T$ is a self-adjoint operator and $lambdain mathbb{R}$ is an eigenvalue of $T$ which is in a spectral gap $(r,s)$, then $spectrum(T)subseteq(-infty,r]cup[s,infty)$ and if $T$ is invertible we can say that $||T^{-1}||leq frac{1}{dist(lambda,spectrum(T))}$ ..
Now if we still work on $T$ but with $lambdain mathbb{{C|R}}$ then $T-lambda$ is a normal operator and hence we still have the relation with the distance above..
my question is how the spectrum of T with complex $lambda$ looks like in this case and do we still have the fact that $spectrum(T)subseteq(-infty,r]cup[s,infty)$ ??
functional-analysis eigenvalues-eigenvectors operator-theory
asked Nov 17 at 21:19
S.N.A
887
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In a Hilbert space if $T$ is a self-adjoint operator and $lambdain mathbb{R}$ is an eigenvalue of $T$ which is in a spectral gap $(r,s)$, then $spectrum(T)subseteq(-infty,r]cup[s,infty)$ and if $T$ is invertible we can say that $||T^{-1}||leq frac{1}{dist(lambda,spectrum(T))}$ ..
Now if we still work on $T$ but with $lambdain mathbb{{C|R}}$ then $T-lambda$ is a normal operator and hence we still have the relation with the distance above..
my question is how the spectrum of T with complex $lambda$ looks like in this case and do we still have the fact that $spectrum(T)subseteq(-infty,r]cup[s,infty)$ ??
functional-analysis eigenvalues-eigenvectors operator-theory
asked Nov 17 at 21:19
S.N.A
887
Adding a constant to an operator just moves the spectrum by that amount in the complex plane.
– Keith McClary
Nov 17 at 22:38
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
In a Hilbert space if $T$ is a self-adjoint operator and $lambdain mathbb{R}$ is an eigenvalue of $T$ which is in a spectral gap $(r,s)$, then $spectrum(T)subseteq(-infty,r]cup[s,infty)$ and if $T$ is invertible we can say that $||T^{-1}||leq frac{1}{dist(lambda,spectrum(T))}$ ..
Now if we still work on $T$ but with $lambdain mathbb{{C|R}}$ then $T-lambda$ is a normal operator and hence we still have the relation with the distance above..
my question is how the spectrum of T with complex $lambda$ looks like in this case and do we still have the fact that $spectrum(T)subseteq(-infty,r]cup[s,infty)$ ??
functional-analysis eigenvalues-eigenvectors operator-theory
asked Nov 17 at 21:19
S.N.A
887
In a Hilbert space if $T$ is a self-adjoint operator and $lambdain mathbb{R}$ is an eigenvalue of $T$ which is in a spectral gap $(r,s)$, then $spectrum(T)subseteq(-infty,r]cup[s,infty)$ and if $T$ is invertible we can say that $||T^{-1}||leq frac{1}{dist(lambda,spectrum(T))}$ ..
Now if we still work on $T$ but with $lambdain mathbb{{C|R}}$ then $T-lambda$ is a normal operator and hence we still have the relation with the distance above..
my question is how the spectrum of T with complex $lambda$ looks like in this case and do we still have the fact that $spectrum(T)subseteq(-infty,r]cup[s,infty)$ ??
functional-analysis eigenvalues-eigenvectors operator-theory
functional-analysis eigenvalues-eigenvectors operator-theory
asked Nov 17 at 21:19
S.N.A
887
asked Nov 17 at 21:19
S.N.A
887
asked Nov 17 at 21:19
S.N.A
887
asked Nov 17 at 21:19
S.N.A
887
asked Nov 17 at 21:19
S.N.A
887
887
Adding a constant to an operator just moves the spectrum by that amount in the complex plane.
– Keith McClary
Nov 17 at 22:38
add a comment |
Adding a constant to an operator just moves the spectrum by that amount in the complex plane.
– Keith McClary
Nov 17 at 22:38
Adding a constant to an operator just moves the spectrum by that amount in the complex plane.
– Keith McClary
Nov 17 at 22:38
Adding a constant to an operator just moves the spectrum by that amount in the complex plane.
– Keith McClary
Nov 17 at 22:38
add a comment |
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Adding a constant to an operator just moves the spectrum by that amount in the complex plane.
– Keith McClary
Nov 17 at 22:38