Normal operator with real spectrum is hermitian.
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Consider a normal operator on a complex Hilbert space with its spectrum contained in the real line. Show that the operator is hermitian without using the spectral theorem.
What I have tried- Since the spectrum is inside the real line, the boundary of the spectrum as a subset of the complex numbers is the spectrum itself. Hence the spectrum is equal to the approximate point spectrum. I have been unable to proceed further.
functional-analysis operator-theory self-adjoint-operators
asked Dec 2 '18 at 13:12
MananManan
1,085611
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add a comment |
$begingroup$
Consider a normal operator on a complex Hilbert space with its spectrum contained in the real line. Show that the operator is hermitian without using the spectral theorem.
What I have tried- Since the spectrum is inside the real line, the boundary of the spectrum as a subset of the complex numbers is the spectrum itself. Hence the spectrum is equal to the approximate point spectrum. I have been unable to proceed further.
functional-analysis operator-theory self-adjoint-operators
asked Dec 2 '18 at 13:12
MananManan
1,085611
$endgroup$
add a comment |
$begingroup$
Consider a normal operator on a complex Hilbert space with its spectrum contained in the real line. Show that the operator is hermitian without using the spectral theorem.
What I have tried- Since the spectrum is inside the real line, the boundary of the spectrum as a subset of the complex numbers is the spectrum itself. Hence the spectrum is equal to the approximate point spectrum. I have been unable to proceed further.
functional-analysis operator-theory self-adjoint-operators
asked Dec 2 '18 at 13:12
MananManan
1,085611
$endgroup$
Consider a normal operator on a complex Hilbert space with its spectrum contained in the real line. Show that the operator is hermitian without using the spectral theorem.
What I have tried- Since the spectrum is inside the real line, the boundary of the spectrum as a subset of the complex numbers is the spectrum itself. Hence the spectrum is equal to the approximate point spectrum. I have been unable to proceed further.
functional-analysis operator-theory self-adjoint-operators
functional-analysis operator-theory self-adjoint-operators
asked Dec 2 '18 at 13:12
MananManan
1,085611
asked Dec 2 '18 at 13:12
MananManan
1,085611
edited Dec 2 '18 at 13:19
Manan
asked Dec 2 '18 at 13:12
MananManan
1,085611
asked Dec 2 '18 at 13:12
MananManan
1,085611
asked Dec 2 '18 at 13:12
MananManan
1,085611
1,085611
add a comment |
add a comment |
1 Answer
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If the spectrum $sigma(N)$ of $N$ is contained in $mathbb{R}$, then you want to show that $N=N^*$. So assume $sigma(N)subseteqmathbb{R}$ and consider $M=N-N^*$. The spectrum of $M$ lies on the real axis by the spectral mapping theorem. But $iM$ is selfadjoint, which means that the spectrum of $M$ lies on the imaginary axis. So $sigma(M)={0}$. Because the norm and spectral radius are the same for a normal operator, then $M=0$, which proves that $N=N^*$.
answered 2 days ago
DisintegratingByPartsDisintegratingByParts
58.9k42580
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1 Answer
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1 Answer
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$begingroup$
If the spectrum $sigma(N)$ of $N$ is contained in $mathbb{R}$, then you want to show that $N=N^*$. So assume $sigma(N)subseteqmathbb{R}$ and consider $M=N-N^*$. The spectrum of $M$ lies on the real axis by the spectral mapping theorem. But $iM$ is selfadjoint, which means that the spectrum of $M$ lies on the imaginary axis. So $sigma(M)={0}$. Because the norm and spectral radius are the same for a normal operator, then $M=0$, which proves that $N=N^*$.
answered 2 days ago
DisintegratingByPartsDisintegratingByParts
58.9k42580
$endgroup$
add a comment |
$begingroup$
If the spectrum $sigma(N)$ of $N$ is contained in $mathbb{R}$, then you want to show that $N=N^*$. So assume $sigma(N)subseteqmathbb{R}$ and consider $M=N-N^*$. The spectrum of $M$ lies on the real axis by the spectral mapping theorem. But $iM$ is selfadjoint, which means that the spectrum of $M$ lies on the imaginary axis. So $sigma(M)={0}$. Because the norm and spectral radius are the same for a normal operator, then $M=0$, which proves that $N=N^*$.
answered 2 days ago
DisintegratingByPartsDisintegratingByParts
58.9k42580
$endgroup$
add a comment |
$begingroup$
If the spectrum $sigma(N)$ of $N$ is contained in $mathbb{R}$, then you want to show that $N=N^*$. So assume $sigma(N)subseteqmathbb{R}$ and consider $M=N-N^*$. The spectrum of $M$ lies on the real axis by the spectral mapping theorem. But $iM$ is selfadjoint, which means that the spectrum of $M$ lies on the imaginary axis. So $sigma(M)={0}$. Because the norm and spectral radius are the same for a normal operator, then $M=0$, which proves that $N=N^*$.
answered 2 days ago
DisintegratingByPartsDisintegratingByParts
58.9k42580
$endgroup$
If the spectrum $sigma(N)$ of $N$ is contained in $mathbb{R}$, then you want to show that $N=N^*$. So assume $sigma(N)subseteqmathbb{R}$ and consider $M=N-N^*$. The spectrum of $M$ lies on the real axis by the spectral mapping theorem. But $iM$ is selfadjoint, which means that the spectrum of $M$ lies on the imaginary axis. So $sigma(M)={0}$. Because the norm and spectral radius are the same for a normal operator, then $M=0$, which proves that $N=N^*$.
answered 2 days ago
DisintegratingByPartsDisintegratingByParts
58.9k42580
answered 2 days ago
DisintegratingByPartsDisintegratingByParts
58.9k42580
answered 2 days ago
DisintegratingByPartsDisintegratingByParts
58.9k42580
answered 2 days ago
DisintegratingByPartsDisintegratingByParts
58.9k42580
58.9k42580
add a comment |
add a comment |
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