Proving that $|T|=max{sqrt{lambda};;lambdain sigma(T^*T)}$.
Let $mathcal{B}(F)$ the algebra of all bounded linear operators on an infinite-dimensional complex Hilbert space $F$.
It is well known that if $Tin mathcal{B}(F)$, then
$$|T|=displaystylesup_{|x|=1}|Tx|.$$
I want to prove that for $Tin mathcal{B}(F)$, we have
$$|T|=max{sqrt{lambda};;lambdain sigma(T^*T)=sigma(TT^*)},$$
where $sigma(A)$ denotes the spectrum of an operator $A$.
functional-analysis reference-request operator-theory
asked Nov 26 at 14:57
Schüler
1,5181421
add a comment |
Let $mathcal{B}(F)$ the algebra of all bounded linear operators on an infinite-dimensional complex Hilbert space $F$.
It is well known that if $Tin mathcal{B}(F)$, then
$$|T|=displaystylesup_{|x|=1}|Tx|.$$
I want to prove that for $Tin mathcal{B}(F)$, we have
$$|T|=max{sqrt{lambda};;lambdain sigma(T^*T)=sigma(TT^*)},$$
where $sigma(A)$ denotes the spectrum of an operator $A$.
functional-analysis reference-request operator-theory
asked Nov 26 at 14:57
Schüler
1,5181421
The operators $T^ast T$ and $TT^ast$ do not always have the same spectrum, but $sigma(T^ast T)setminus{0}=sigma(T T^ast)setminus{0}$.
– MaoWao
Nov 26 at 18:12
@MaoWao Thank you. So the formula is only true for $sigma(T^*T)$?
– Schüler
Nov 26 at 19:22
No, it works either way. The maxima of $sigma(T^ast T)$ and $sigma(T T^ast)$ do coincide, as the formula $sigma(T^ast T)setminus{0}=sigma(TT^ast)setminus{0}$ shows.
– MaoWao
Nov 26 at 20:28
add a comment |
Let $mathcal{B}(F)$ the algebra of all bounded linear operators on an infinite-dimensional complex Hilbert space $F$.
It is well known that if $Tin mathcal{B}(F)$, then
$$|T|=displaystylesup_{|x|=1}|Tx|.$$
I want to prove that for $Tin mathcal{B}(F)$, we have
$$|T|=max{sqrt{lambda};;lambdain sigma(T^*T)=sigma(TT^*)},$$
where $sigma(A)$ denotes the spectrum of an operator $A$.
functional-analysis reference-request operator-theory
asked Nov 26 at 14:57
Schüler
1,5181421
Let $mathcal{B}(F)$ the algebra of all bounded linear operators on an infinite-dimensional complex Hilbert space $F$.
It is well known that if $Tin mathcal{B}(F)$, then
$$|T|=displaystylesup_{|x|=1}|Tx|.$$
I want to prove that for $Tin mathcal{B}(F)$, we have
$$|T|=max{sqrt{lambda};;lambdain sigma(T^*T)=sigma(TT^*)},$$
where $sigma(A)$ denotes the spectrum of an operator $A$.
functional-analysis reference-request operator-theory
functional-analysis reference-request operator-theory
asked Nov 26 at 14:57
Schüler
1,5181421
asked Nov 26 at 14:57
Schüler
1,5181421
edited Nov 26 at 15:02
asked Nov 26 at 14:57
Schüler
1,5181421
asked Nov 26 at 14:57
Schüler
1,5181421
asked Nov 26 at 14:57
Schüler
1,5181421
1,5181421
The operators $T^ast T$ and $TT^ast$ do not always have the same spectrum, but $sigma(T^ast T)setminus{0}=sigma(T T^ast)setminus{0}$.
– MaoWao
Nov 26 at 18:12
@MaoWao Thank you. So the formula is only true for $sigma(T^*T)$?
– Schüler
Nov 26 at 19:22
No, it works either way. The maxima of $sigma(T^ast T)$ and $sigma(T T^ast)$ do coincide, as the formula $sigma(T^ast T)setminus{0}=sigma(TT^ast)setminus{0}$ shows.
– MaoWao
Nov 26 at 20:28
add a comment |
The operators $T^ast T$ and $TT^ast$ do not always have the same spectrum, but $sigma(T^ast T)setminus{0}=sigma(T T^ast)setminus{0}$.
– MaoWao
Nov 26 at 18:12
@MaoWao Thank you. So the formula is only true for $sigma(T^*T)$?
– Schüler
Nov 26 at 19:22
No, it works either way. The maxima of $sigma(T^ast T)$ and $sigma(T T^ast)$ do coincide, as the formula $sigma(T^ast T)setminus{0}=sigma(TT^ast)setminus{0}$ shows.
– MaoWao
Nov 26 at 20:28
The operators $T^ast T$ and $TT^ast$ do not always have the same spectrum, but $sigma(T^ast T)setminus{0}=sigma(T T^ast)setminus{0}$.
– MaoWao
Nov 26 at 18:12
The operators $T^ast T$ and $TT^ast$ do not always have the same spectrum, but $sigma(T^ast T)setminus{0}=sigma(T T^ast)setminus{0}$.
– MaoWao
Nov 26 at 18:12
@MaoWao Thank you. So the formula is only true for $sigma(T^*T)$?
– Schüler
Nov 26 at 19:22
@MaoWao Thank you. So the formula is only true for $sigma(T^*T)$?
– Schüler
Nov 26 at 19:22
No, it works either way. The maxima of $sigma(T^ast T)$ and $sigma(T T^ast)$ do coincide, as the formula $sigma(T^ast T)setminus{0}=sigma(TT^ast)setminus{0}$ shows.
– MaoWao
Nov 26 at 20:28
No, it works either way. The maxima of $sigma(T^ast T)$ and $sigma(T T^ast)$ do coincide, as the formula $sigma(T^ast T)setminus{0}=sigma(TT^ast)setminus{0}$ shows.
– MaoWao
Nov 26 at 20:28
add a comment |
1 Answer
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for each $A in mathcal{B}(F)$ we have $||A^*A||=||A||^2$.
if $A in mathcal{B}(F)$ is self-adjoint, then $||A|| =max {| mu|: mu in sigma(A)}$.
if $T in mathcal{B}(F)$, then $T^*T$ is self-adjoint and $ sigma(T^*T) subseteq [0, infty)$.
Now you should be in a position to prove $|T|=max{sqrt{lambda};;lambdain sigma(T^*T)}$.
answered Nov 26 at 15:07
Fred
44.2k1645
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for each $A in mathcal{B}(F)$ we have $||A^*A||=||A||^2$.
if $A in mathcal{B}(F)$ is self-adjoint, then $||A|| =max {| mu|: mu in sigma(A)}$.
if $T in mathcal{B}(F)$, then $T^*T$ is self-adjoint and $ sigma(T^*T) subseteq [0, infty)$.
Now you should be in a position to prove $|T|=max{sqrt{lambda};;lambdain sigma(T^*T)}$.
answered Nov 26 at 15:07
Fred
44.2k1645
add a comment |
for each $A in mathcal{B}(F)$ we have $||A^*A||=||A||^2$.
if $A in mathcal{B}(F)$ is self-adjoint, then $||A|| =max {| mu|: mu in sigma(A)}$.
if $T in mathcal{B}(F)$, then $T^*T$ is self-adjoint and $ sigma(T^*T) subseteq [0, infty)$.
Now you should be in a position to prove $|T|=max{sqrt{lambda};;lambdain sigma(T^*T)}$.
answered Nov 26 at 15:07
Fred
44.2k1645
add a comment |
for each $A in mathcal{B}(F)$ we have $||A^*A||=||A||^2$.
if $A in mathcal{B}(F)$ is self-adjoint, then $||A|| =max {| mu|: mu in sigma(A)}$.
if $T in mathcal{B}(F)$, then $T^*T$ is self-adjoint and $ sigma(T^*T) subseteq [0, infty)$.
Now you should be in a position to prove $|T|=max{sqrt{lambda};;lambdain sigma(T^*T)}$.
answered Nov 26 at 15:07
Fred
44.2k1645
for each $A in mathcal{B}(F)$ we have $||A^*A||=||A||^2$.
if $A in mathcal{B}(F)$ is self-adjoint, then $||A|| =max {| mu|: mu in sigma(A)}$.
if $T in mathcal{B}(F)$, then $T^*T$ is self-adjoint and $ sigma(T^*T) subseteq [0, infty)$.
Now you should be in a position to prove $|T|=max{sqrt{lambda};;lambdain sigma(T^*T)}$.
answered Nov 26 at 15:07
Fred
44.2k1645
answered Nov 26 at 15:07
Fred
44.2k1645
answered Nov 26 at 15:07
Fred
44.2k1645
answered Nov 26 at 15:07
Fred
44.2k1645
44.2k1645
add a comment |
add a comment |
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The operators $T^ast T$ and $TT^ast$ do not always have the same spectrum, but $sigma(T^ast T)setminus{0}=sigma(T T^ast)setminus{0}$.
– MaoWao
Nov 26 at 18:12
@MaoWao Thank you. So the formula is only true for $sigma(T^*T)$?
– Schüler
Nov 26 at 19:22
No, it works either way. The maxima of $sigma(T^ast T)$ and $sigma(T T^ast)$ do coincide, as the formula $sigma(T^ast T)setminus{0}=sigma(TT^ast)setminus{0}$ shows.
– MaoWao
Nov 26 at 20:28