Mixing
Self-adjoint implies symmetry
$begingroup$
I have found little statement that I'd like to prove. It's about: self-adjoin operator implies that this operator is symmetric.
Definitions:
Operator T: H-> H (where H-Hilbert's) is symmetric, if = for every x,t in D(T). <-,-> is the scalar product.
Operator is self-adjoint if D(T)=D(T*) and $Tf=T$*$f$ for $f in D(T)$, where
$D(T*)= { y in H: |<Tx,y>| leq C_{y} ||x||, $ for every $ x in D(T)$.
Then proof of the following: If T is self-adjoint, then it's symmetric.
Let us take $x,t in D(T)$.
$<Tx,y> = <x, T$*$y>$, because T is self-adjoint.
And then:
$<x, T$*$y> = <x, Ty>$.
Can somebody explain why the last equation occurs? I can't explain it from which property it comes. And does the D(T) means domain in this example? Could this equation comes from equality D(T)=D(T*)?
self-adjoint-operators
asked Jan 24 at 8:07
OlgaOlga
176
$endgroup$
add a comment |
$begingroup$
I have found little statement that I'd like to prove. It's about: self-adjoin operator implies that this operator is symmetric.
Definitions:
Operator T: H-> H (where H-Hilbert's) is symmetric, if = for every x,t in D(T). <-,-> is the scalar product.
Operator is self-adjoint if D(T)=D(T*) and $Tf=T$*$f$ for $f in D(T)$, where
$D(T*)= { y in H: |<Tx,y>| leq C_{y} ||x||, $ for every $ x in D(T)$.
Then proof of the following: If T is self-adjoint, then it's symmetric.
Let us take $x,t in D(T)$.
$<Tx,y> = <x, T$*$y>$, because T is self-adjoint.
And then:
$<x, T$*$y> = <x, Ty>$.
Can somebody explain why the last equation occurs? I can't explain it from which property it comes. And does the D(T) means domain in this example? Could this equation comes from equality D(T)=D(T*)?
self-adjoint-operators
asked Jan 24 at 8:07
OlgaOlga
176
$endgroup$
$begingroup$
on LHS use the definition of $T^{*}$ and on RHS use the hypothesis that $T^{*}=T$.
$endgroup$
– Kavi Rama Murthy
Jan 24 at 8:14
add a comment |
$begingroup$
I have found little statement that I'd like to prove. It's about: self-adjoin operator implies that this operator is symmetric.
Definitions:
Operator T: H-> H (where H-Hilbert's) is symmetric, if = for every x,t in D(T). <-,-> is the scalar product.
Operator is self-adjoint if D(T)=D(T*) and $Tf=T$*$f$ for $f in D(T)$, where
$D(T*)= { y in H: |<Tx,y>| leq C_{y} ||x||, $ for every $ x in D(T)$.
Then proof of the following: If T is self-adjoint, then it's symmetric.
Let us take $x,t in D(T)$.
$<Tx,y> = <x, T$*$y>$, because T is self-adjoint.
And then:
$<x, T$*$y> = <x, Ty>$.
Can somebody explain why the last equation occurs? I can't explain it from which property it comes. And does the D(T) means domain in this example? Could this equation comes from equality D(T)=D(T*)?
self-adjoint-operators
asked Jan 24 at 8:07
OlgaOlga
176
$endgroup$
I have found little statement that I'd like to prove. It's about: self-adjoin operator implies that this operator is symmetric.
Definitions:
Operator T: H-> H (where H-Hilbert's) is symmetric, if = for every x,t in D(T). <-,-> is the scalar product.
Operator is self-adjoint if D(T)=D(T*) and $Tf=T$*$f$ for $f in D(T)$, where
$D(T*)= { y in H: |<Tx,y>| leq C_{y} ||x||, $ for every $ x in D(T)$.
Then proof of the following: If T is self-adjoint, then it's symmetric.
Let us take $x,t in D(T)$.
$<Tx,y> = <x, T$*$y>$, because T is self-adjoint.
And then:
$<x, T$*$y> = <x, Ty>$.
Can somebody explain why the last equation occurs? I can't explain it from which property it comes. And does the D(T) means domain in this example? Could this equation comes from equality D(T)=D(T*)?
self-adjoint-operators
self-adjoint-operators
asked Jan 24 at 8:07
OlgaOlga
176
asked Jan 24 at 8:07
OlgaOlga
176
asked Jan 24 at 8:07
OlgaOlga
176
asked Jan 24 at 8:07
OlgaOlga
176
asked Jan 24 at 8:07
OlgaOlga
176
176
$begingroup$
on LHS use the definition of $T^{*}$ and on RHS use the hypothesis that $T^{*}=T$.
$endgroup$
– Kavi Rama Murthy
Jan 24 at 8:14
add a comment |
$begingroup$
on LHS use the definition of $T^{*}$ and on RHS use the hypothesis that $T^{*}=T$.
$endgroup$
– Kavi Rama Murthy
Jan 24 at 8:14
$begingroup$
on LHS use the definition of $T^{*}$ and on RHS use the hypothesis that $T^{*}=T$.
$endgroup$
– Kavi Rama Murthy
Jan 24 at 8:14
$begingroup$
on LHS use the definition of $T^{*}$ and on RHS use the hypothesis that $T^{*}=T$.
$endgroup$
– Kavi Rama Murthy
Jan 24 at 8:14
add a comment |
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$begingroup$
on LHS use the definition of $T^{*}$ and on RHS use the hypothesis that $T^{*}=T$.
$endgroup$
– Kavi Rama Murthy
Jan 24 at 8:14