Question about resolvent set and spectrum











up vote
-2
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I'm trying to find the solution of this one:



Let $mathcal{H}$ a Hilbert space and $T in mathcal{B}(mathcal{H})$. Show that



1) $lambda in rho (T) Longleftrightarrow bar{lambda} in rho (T^*)$



2) $sigma (T^*) = {lambda in mathbb{C} : bar{lambda} in sigma (T)}$



3) $ lambda in rho (T) Longrightarrow R_{T^*} (bar{lambda}) = R_T(lambda)^*$.



I tried to write down the definition of resolvent set and spectrum as follows:



begin{equation}
rho(T) = {lambda in mathbb{C}: exists(T-lambda I)^{-1} in mathcal{B}(mathcal{H})}
end{equation}
and
begin{equation}
sigma(T) = mathbb{C}-rho(T)
end{equation}



But I can't figure out what to do..






functional-analysis spectral-theory







share|cite|improve this question






















  • what have you tried?
    – supinf
    Nov 23 at 10:59










  • just dealt with definitions..
    – James Arten
    Nov 23 at 11:08















up vote
-2
down vote

favorite












I'm trying to find the solution of this one:



Let $mathcal{H}$ a Hilbert space and $T in mathcal{B}(mathcal{H})$. Show that



1) $lambda in rho (T) Longleftrightarrow bar{lambda} in rho (T^*)$



2) $sigma (T^*) = {lambda in mathbb{C} : bar{lambda} in sigma (T)}$



3) $ lambda in rho (T) Longrightarrow R_{T^*} (bar{lambda}) = R_T(lambda)^*$.



I tried to write down the definition of resolvent set and spectrum as follows:



begin{equation}
rho(T) = {lambda in mathbb{C}: exists(T-lambda I)^{-1} in mathcal{B}(mathcal{H})}
end{equation}
and
begin{equation}
sigma(T) = mathbb{C}-rho(T)
end{equation}



But I can't figure out what to do..






functional-analysis spectral-theory







share|cite|improve this question






















  • what have you tried?
    – supinf
    Nov 23 at 10:59










  • just dealt with definitions..
    – James Arten
    Nov 23 at 11:08













up vote
-2
down vote

favorite









up vote
-2
down vote

favorite











I'm trying to find the solution of this one:



Let $mathcal{H}$ a Hilbert space and $T in mathcal{B}(mathcal{H})$. Show that



1) $lambda in rho (T) Longleftrightarrow bar{lambda} in rho (T^*)$



2) $sigma (T^*) = {lambda in mathbb{C} : bar{lambda} in sigma (T)}$



3) $ lambda in rho (T) Longrightarrow R_{T^*} (bar{lambda}) = R_T(lambda)^*$.



I tried to write down the definition of resolvent set and spectrum as follows:



begin{equation}
rho(T) = {lambda in mathbb{C}: exists(T-lambda I)^{-1} in mathcal{B}(mathcal{H})}
end{equation}
and
begin{equation}
sigma(T) = mathbb{C}-rho(T)
end{equation}



But I can't figure out what to do..






functional-analysis spectral-theory







share|cite|improve this question













I'm trying to find the solution of this one:



Let $mathcal{H}$ a Hilbert space and $T in mathcal{B}(mathcal{H})$. Show that



1) $lambda in rho (T) Longleftrightarrow bar{lambda} in rho (T^*)$



2) $sigma (T^*) = {lambda in mathbb{C} : bar{lambda} in sigma (T)}$



3) $ lambda in rho (T) Longrightarrow R_{T^*} (bar{lambda}) = R_T(lambda)^*$.



I tried to write down the definition of resolvent set and spectrum as follows:



begin{equation}
rho(T) = {lambda in mathbb{C}: exists(T-lambda I)^{-1} in mathcal{B}(mathcal{H})}
end{equation}
and
begin{equation}
sigma(T) = mathbb{C}-rho(T)
end{equation}



But I can't figure out what to do..






functional-analysis spectral-theory




functional-analysis spectral-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 23 at 10:56









James Arten

579




579












  • what have you tried?
    – supinf
    Nov 23 at 10:59










  • just dealt with definitions..
    – James Arten
    Nov 23 at 11:08


















  • what have you tried?
    – supinf
    Nov 23 at 10:59










  • just dealt with definitions..
    – James Arten
    Nov 23 at 11:08
















what have you tried?
– supinf
Nov 23 at 10:59




what have you tried?
– supinf
Nov 23 at 10:59












just dealt with definitions..
– James Arten
Nov 23 at 11:08




just dealt with definitions..
– James Arten
Nov 23 at 11:08










1 Answer
1






active

oldest

votes

















up vote
2
down vote



accepted










All what we need is: if $A in mathcal{B}(mathcal{H})$, then



$A$ is bijective $ iff A^*$ is bijective



and if $A=T - lambda I$, then $A^*=T^*- overline{lambda} I.$



Can you proceed ?






share|cite|improve this answer





















  • So if I suppose that there exists $(T-lambda I)^{-1} = A^{-1}$ then $A$ is bijective and this implies that also $A^*$ as well (and vice-versa). So the result should follows automatically from the definition, right?
    – James Arten
    Nov 23 at 11:32








  • 1




    Yes, the results follow easily from the definitions.
    – Fred
    Nov 23 at 11:35










  • thank you, I really need to grab my old book of linear algebra and check few things again. ^_^"
    – James Arten
    Nov 23 at 11:38











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote



accepted










All what we need is: if $A in mathcal{B}(mathcal{H})$, then



$A$ is bijective $ iff A^*$ is bijective



and if $A=T - lambda I$, then $A^*=T^*- overline{lambda} I.$



Can you proceed ?






share|cite|improve this answer





















  • So if I suppose that there exists $(T-lambda I)^{-1} = A^{-1}$ then $A$ is bijective and this implies that also $A^*$ as well (and vice-versa). So the result should follows automatically from the definition, right?
    – James Arten
    Nov 23 at 11:32








  • 1




    Yes, the results follow easily from the definitions.
    – Fred
    Nov 23 at 11:35










  • thank you, I really need to grab my old book of linear algebra and check few things again. ^_^"
    – James Arten
    Nov 23 at 11:38















up vote
2
down vote



accepted










All what we need is: if $A in mathcal{B}(mathcal{H})$, then



$A$ is bijective $ iff A^*$ is bijective



and if $A=T - lambda I$, then $A^*=T^*- overline{lambda} I.$



Can you proceed ?






share|cite|improve this answer





















  • So if I suppose that there exists $(T-lambda I)^{-1} = A^{-1}$ then $A$ is bijective and this implies that also $A^*$ as well (and vice-versa). So the result should follows automatically from the definition, right?
    – James Arten
    Nov 23 at 11:32








  • 1




    Yes, the results follow easily from the definitions.
    – Fred
    Nov 23 at 11:35










  • thank you, I really need to grab my old book of linear algebra and check few things again. ^_^"
    – James Arten
    Nov 23 at 11:38













up vote
2
down vote



accepted







up vote
2
down vote



accepted






All what we need is: if $A in mathcal{B}(mathcal{H})$, then



$A$ is bijective $ iff A^*$ is bijective



and if $A=T - lambda I$, then $A^*=T^*- overline{lambda} I.$



Can you proceed ?






share|cite|improve this answer












All what we need is: if $A in mathcal{B}(mathcal{H})$, then



$A$ is bijective $ iff A^*$ is bijective



and if $A=T - lambda I$, then $A^*=T^*- overline{lambda} I.$



Can you proceed ?







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 23 at 11:12









Fred

43.9k1644




43.9k1644












  • So if I suppose that there exists $(T-lambda I)^{-1} = A^{-1}$ then $A$ is bijective and this implies that also $A^*$ as well (and vice-versa). So the result should follows automatically from the definition, right?
    – James Arten
    Nov 23 at 11:32








  • 1




    Yes, the results follow easily from the definitions.
    – Fred
    Nov 23 at 11:35










  • thank you, I really need to grab my old book of linear algebra and check few things again. ^_^"
    – James Arten
    Nov 23 at 11:38


















  • So if I suppose that there exists $(T-lambda I)^{-1} = A^{-1}$ then $A$ is bijective and this implies that also $A^*$ as well (and vice-versa). So the result should follows automatically from the definition, right?
    – James Arten
    Nov 23 at 11:32








  • 1




    Yes, the results follow easily from the definitions.
    – Fred
    Nov 23 at 11:35










  • thank you, I really need to grab my old book of linear algebra and check few things again. ^_^"
    – James Arten
    Nov 23 at 11:38
















So if I suppose that there exists $(T-lambda I)^{-1} = A^{-1}$ then $A$ is bijective and this implies that also $A^*$ as well (and vice-versa). So the result should follows automatically from the definition, right?
– James Arten
Nov 23 at 11:32






So if I suppose that there exists $(T-lambda I)^{-1} = A^{-1}$ then $A$ is bijective and this implies that also $A^*$ as well (and vice-versa). So the result should follows automatically from the definition, right?
– James Arten
Nov 23 at 11:32






1




1




Yes, the results follow easily from the definitions.
– Fred
Nov 23 at 11:35




Yes, the results follow easily from the definitions.
– Fred
Nov 23 at 11:35












thank you, I really need to grab my old book of linear algebra and check few things again. ^_^"
– James Arten
Nov 23 at 11:38




thank you, I really need to grab my old book of linear algebra and check few things again. ^_^"
– James Arten
Nov 23 at 11:38


















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