Problem related with the spectrum of a normal operator
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I am working on this problem:
Let $Ain L(H)$ be a normal operator on a nonzero complex Hilbert space $H$.
Prove that $sigma(A)cap imathbb{R}=emptysetiff A+A^* mathrm{ is bijective}$
I have already done one direction but get stuck now on $sigma(A)cap imathbb{R}=emptysetLongrightarrow A+A^* mathrm{ is bijective}$. The problem hints to show that if $sigma(A)cap imathbb{R}=emptyset$ we can find an $A$-invariant direct sum decomposition $H=H^+oplus H^-$ such that $pmmathrm{Re}lambda>0 $ for $lambdainsigma(A|_{H^{pm}})$. Prove that $H^{pm}$ is invariant under $A^*$ and use the result in a previous question which is $sup_{lambdainsigma(A)}mathrm{Re}lambda=sup_{|x|=1}mathrm{Re}left<x,Axright>,inf_{lambdainsigma(A)}mathrm{Re}lambda=inf_{|x|=1}mathrm{Re}left<x,Axright>$.
I know since $sigma(A)$ is closed we can use functional calculus to construct a projection $P$ such that its image and kernel are two closed subspace $H^{+}$ and $H^{-}$ which are invariant under $A$ and satisfy the properties described. I am not so clear about why $H^{pm}$ is also invariant under $A^*$(I feel it has something to do with the $C^*$ homomorphism but still get stuck).
Even I assume them to be valid, I still don't know how to use them to prove $A+A^* $ is bijective. The only method I know to prove the injectivity/surjectivity/bijectivity of an operator is the Closed Image Theorem (Closed Range Theorem). If I want to apply it here, since $A+A^*$ is self-adjoint, I just need to show $|(A+A^*)x|ge c|x|,forall xin H$ for some $c>0$. Expand explicitly to get $|(A+A^*)x|^2=|Ax|^2+|A^*x|^2+2mathrm{Re}left<Ax,A^*right>ge2mathrm{Re}left<Ax,A^*xright>$, so if I can prove $mathrm{Re}left<Ax,A^*xright>ge c|x|^2$ for some $c>0$ it's done. However, I have been frustrated for hours on how to exert the result $sup_{lambdainsigma(A)}mathrm{Re}lambda=sup_{|x|=1}mathrm{Re}left<x,Axright>,inf_{lambdainsigma(A)}mathrm{Re}lambda=inf_{|x|=1}mathrm{Re}left<x,Axright>$.
Can someone help me with this?
functional-analysis
asked Nov 16 at 3:03
Apocalypse
1206
add a comment |
up vote
0
down vote
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I am working on this problem:
Let $Ain L(H)$ be a normal operator on a nonzero complex Hilbert space $H$.
Prove that $sigma(A)cap imathbb{R}=emptysetiff A+A^* mathrm{ is bijective}$
I have already done one direction but get stuck now on $sigma(A)cap imathbb{R}=emptysetLongrightarrow A+A^* mathrm{ is bijective}$. The problem hints to show that if $sigma(A)cap imathbb{R}=emptyset$ we can find an $A$-invariant direct sum decomposition $H=H^+oplus H^-$ such that $pmmathrm{Re}lambda>0 $ for $lambdainsigma(A|_{H^{pm}})$. Prove that $H^{pm}$ is invariant under $A^*$ and use the result in a previous question which is $sup_{lambdainsigma(A)}mathrm{Re}lambda=sup_{|x|=1}mathrm{Re}left<x,Axright>,inf_{lambdainsigma(A)}mathrm{Re}lambda=inf_{|x|=1}mathrm{Re}left<x,Axright>$.
I know since $sigma(A)$ is closed we can use functional calculus to construct a projection $P$ such that its image and kernel are two closed subspace $H^{+}$ and $H^{-}$ which are invariant under $A$ and satisfy the properties described. I am not so clear about why $H^{pm}$ is also invariant under $A^*$(I feel it has something to do with the $C^*$ homomorphism but still get stuck).
Even I assume them to be valid, I still don't know how to use them to prove $A+A^* $ is bijective. The only method I know to prove the injectivity/surjectivity/bijectivity of an operator is the Closed Image Theorem (Closed Range Theorem). If I want to apply it here, since $A+A^*$ is self-adjoint, I just need to show $|(A+A^*)x|ge c|x|,forall xin H$ for some $c>0$. Expand explicitly to get $|(A+A^*)x|^2=|Ax|^2+|A^*x|^2+2mathrm{Re}left<Ax,A^*right>ge2mathrm{Re}left<Ax,A^*xright>$, so if I can prove $mathrm{Re}left<Ax,A^*xright>ge c|x|^2$ for some $c>0$ it's done. However, I have been frustrated for hours on how to exert the result $sup_{lambdainsigma(A)}mathrm{Re}lambda=sup_{|x|=1}mathrm{Re}left<x,Axright>,inf_{lambdainsigma(A)}mathrm{Re}lambda=inf_{|x|=1}mathrm{Re}left<x,Axright>$.
Can someone help me with this?
functional-analysis
asked Nov 16 at 3:03
Apocalypse
1206
Do you know the spectral mapping theorem for normal operators? If you do, then you know the spectrum of $A+A^*$ does not include $0$, and $A+A^*$ is selfadjoint.
– DisintegratingByParts
Nov 16 at 13:48
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am working on this problem:
Let $Ain L(H)$ be a normal operator on a nonzero complex Hilbert space $H$.
Prove that $sigma(A)cap imathbb{R}=emptysetiff A+A^* mathrm{ is bijective}$
I have already done one direction but get stuck now on $sigma(A)cap imathbb{R}=emptysetLongrightarrow A+A^* mathrm{ is bijective}$. The problem hints to show that if $sigma(A)cap imathbb{R}=emptyset$ we can find an $A$-invariant direct sum decomposition $H=H^+oplus H^-$ such that $pmmathrm{Re}lambda>0 $ for $lambdainsigma(A|_{H^{pm}})$. Prove that $H^{pm}$ is invariant under $A^*$ and use the result in a previous question which is $sup_{lambdainsigma(A)}mathrm{Re}lambda=sup_{|x|=1}mathrm{Re}left<x,Axright>,inf_{lambdainsigma(A)}mathrm{Re}lambda=inf_{|x|=1}mathrm{Re}left<x,Axright>$.
I know since $sigma(A)$ is closed we can use functional calculus to construct a projection $P$ such that its image and kernel are two closed subspace $H^{+}$ and $H^{-}$ which are invariant under $A$ and satisfy the properties described. I am not so clear about why $H^{pm}$ is also invariant under $A^*$(I feel it has something to do with the $C^*$ homomorphism but still get stuck).
Even I assume them to be valid, I still don't know how to use them to prove $A+A^* $ is bijective. The only method I know to prove the injectivity/surjectivity/bijectivity of an operator is the Closed Image Theorem (Closed Range Theorem). If I want to apply it here, since $A+A^*$ is self-adjoint, I just need to show $|(A+A^*)x|ge c|x|,forall xin H$ for some $c>0$. Expand explicitly to get $|(A+A^*)x|^2=|Ax|^2+|A^*x|^2+2mathrm{Re}left<Ax,A^*right>ge2mathrm{Re}left<Ax,A^*xright>$, so if I can prove $mathrm{Re}left<Ax,A^*xright>ge c|x|^2$ for some $c>0$ it's done. However, I have been frustrated for hours on how to exert the result $sup_{lambdainsigma(A)}mathrm{Re}lambda=sup_{|x|=1}mathrm{Re}left<x,Axright>,inf_{lambdainsigma(A)}mathrm{Re}lambda=inf_{|x|=1}mathrm{Re}left<x,Axright>$.
Can someone help me with this?
functional-analysis
asked Nov 16 at 3:03
Apocalypse
1206
I am working on this problem:
Let $Ain L(H)$ be a normal operator on a nonzero complex Hilbert space $H$.
Prove that $sigma(A)cap imathbb{R}=emptysetiff A+A^* mathrm{ is bijective}$
I have already done one direction but get stuck now on $sigma(A)cap imathbb{R}=emptysetLongrightarrow A+A^* mathrm{ is bijective}$. The problem hints to show that if $sigma(A)cap imathbb{R}=emptyset$ we can find an $A$-invariant direct sum decomposition $H=H^+oplus H^-$ such that $pmmathrm{Re}lambda>0 $ for $lambdainsigma(A|_{H^{pm}})$. Prove that $H^{pm}$ is invariant under $A^*$ and use the result in a previous question which is $sup_{lambdainsigma(A)}mathrm{Re}lambda=sup_{|x|=1}mathrm{Re}left<x,Axright>,inf_{lambdainsigma(A)}mathrm{Re}lambda=inf_{|x|=1}mathrm{Re}left<x,Axright>$.
I know since $sigma(A)$ is closed we can use functional calculus to construct a projection $P$ such that its image and kernel are two closed subspace $H^{+}$ and $H^{-}$ which are invariant under $A$ and satisfy the properties described. I am not so clear about why $H^{pm}$ is also invariant under $A^*$(I feel it has something to do with the $C^*$ homomorphism but still get stuck).
Even I assume them to be valid, I still don't know how to use them to prove $A+A^* $ is bijective. The only method I know to prove the injectivity/surjectivity/bijectivity of an operator is the Closed Image Theorem (Closed Range Theorem). If I want to apply it here, since $A+A^*$ is self-adjoint, I just need to show $|(A+A^*)x|ge c|x|,forall xin H$ for some $c>0$. Expand explicitly to get $|(A+A^*)x|^2=|Ax|^2+|A^*x|^2+2mathrm{Re}left<Ax,A^*right>ge2mathrm{Re}left<Ax,A^*xright>$, so if I can prove $mathrm{Re}left<Ax,A^*xright>ge c|x|^2$ for some $c>0$ it's done. However, I have been frustrated for hours on how to exert the result $sup_{lambdainsigma(A)}mathrm{Re}lambda=sup_{|x|=1}mathrm{Re}left<x,Axright>,inf_{lambdainsigma(A)}mathrm{Re}lambda=inf_{|x|=1}mathrm{Re}left<x,Axright>$.
Can someone help me with this?
functional-analysis
functional-analysis
asked Nov 16 at 3:03
Apocalypse
1206
asked Nov 16 at 3:03
Apocalypse
1206
asked Nov 16 at 3:03
Apocalypse
1206
asked Nov 16 at 3:03
Apocalypse
1206
asked Nov 16 at 3:03
Apocalypse
1206
1206
Do you know the spectral mapping theorem for normal operators? If you do, then you know the spectrum of $A+A^*$ does not include $0$, and $A+A^*$ is selfadjoint.
– DisintegratingByParts
Nov 16 at 13:48
add a comment |
Do you know the spectral mapping theorem for normal operators? If you do, then you know the spectrum of $A+A^*$ does not include $0$, and $A+A^*$ is selfadjoint.
– DisintegratingByParts
Nov 16 at 13:48
Do you know the spectral mapping theorem for normal operators? If you do, then you know the spectrum of $A+A^*$ does not include $0$, and $A+A^*$ is selfadjoint.
– DisintegratingByParts
Nov 16 at 13:48
Do you know the spectral mapping theorem for normal operators? If you do, then you know the spectrum of $A+A^*$ does not include $0$, and $A+A^*$ is selfadjoint.
– DisintegratingByParts
Nov 16 at 13:48
add a comment |
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Do you know the spectral mapping theorem for normal operators? If you do, then you know the spectrum of $A+A^*$ does not include $0$, and $A+A^*$ is selfadjoint.
– DisintegratingByParts
Nov 16 at 13:48