Question about resolvent set and spectrum
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
asked Nov 23 at 10:56
James Arten
579
add a comment |
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
asked Nov 23 at 10:56
James Arten
579
what have you tried?
– supinf
Nov 23 at 10:59
just dealt with definitions..
– James Arten
Nov 23 at 11:08
add a comment |
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
asked Nov 23 at 10:56
James Arten
579
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
asked Nov 23 at 10:56
James Arten
579
asked Nov 23 at 10:56
James Arten
579
asked Nov 23 at 10:56
James Arten
579
asked Nov 23 at 10:56
James Arten
579
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
add a comment |
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
add a comment |
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 ?
answered Nov 23 at 11:12
Fred
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
add a comment |
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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 ?
answered Nov 23 at 11:12
Fred
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
add a comment |
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 ?
answered Nov 23 at 11:12
Fred
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
add a comment |
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 ?
answered Nov 23 at 11:12
Fred
43.9k1644
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 ?
answered Nov 23 at 11:12
Fred
43.9k1644
answered Nov 23 at 11:12
Fred
43.9k1644
answered Nov 23 at 11:12
Fred
43.9k1644
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
add a comment |
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
add a comment |
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what have you tried?
– supinf
Nov 23 at 10:59
just dealt with definitions..
– James Arten
Nov 23 at 11:08