Proving that $|T|=max{sqrt{lambda};;lambdain sigma(T^*T)}$.












3














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 $A$.











share|cite|improve this question
























  • 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
















3














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 $A$.











share|cite|improve this question
























  • 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














3












3








3







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 $A$.











share|cite|improve this question















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






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share|cite|improve this question













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edited Nov 26 at 15:02

























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


















  • 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










1 Answer
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3















  1. for each $A in mathcal{B}(F)$ we have $||A^*A||=||A||^2$.


  2. if $A in mathcal{B}(F)$ is self-adjoint, then $||A|| =max {| mu|: mu in sigma(A)}$.


  3. 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 $|T|=max{sqrt{lambda};;lambdain sigma(T^*T)}$.






share|cite|improve this answer





















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    1 Answer
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    active

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    3















    1. for each $A in mathcal{B}(F)$ we have $||A^*A||=||A||^2$.


    2. if $A in mathcal{B}(F)$ is self-adjoint, then $||A|| =max {| mu|: mu in sigma(A)}$.


    3. 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 $|T|=max{sqrt{lambda};;lambdain sigma(T^*T)}$.






    share|cite|improve this answer


























      3















      1. for each $A in mathcal{B}(F)$ we have $||A^*A||=||A||^2$.


      2. if $A in mathcal{B}(F)$ is self-adjoint, then $||A|| =max {| mu|: mu in sigma(A)}$.


      3. 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 $|T|=max{sqrt{lambda};;lambdain sigma(T^*T)}$.






      share|cite|improve this answer
























        3












        3








        3







        1. for each $A in mathcal{B}(F)$ we have $||A^*A||=||A||^2$.


        2. if $A in mathcal{B}(F)$ is self-adjoint, then $||A|| =max {| mu|: mu in sigma(A)}$.


        3. 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 $|T|=max{sqrt{lambda};;lambdain sigma(T^*T)}$.






        share|cite|improve this answer













        1. for each $A in mathcal{B}(F)$ we have $||A^*A||=||A||^2$.


        2. if $A in mathcal{B}(F)$ is self-adjoint, then $||A|| =max {| mu|: mu in sigma(A)}$.


        3. 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 $|T|=max{sqrt{lambda};;lambdain sigma(T^*T)}$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 26 at 15:07









        Fred

        44.2k1645




        44.2k1645






























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