An application of Fredholm Alternative Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Fredholm and Compact OperatorsLinear operators with no adjointQuestions about the proof of the Riesz representation theoremFinding the Hilbert Adjoint in this caseexamples of using the Fredholm alternative explicitlyA possible Corollary of the Fredholm alternative?On Fredholm operator on Hilbert spacesSelf adjoint map are bounded(bounded linear) orthogonal projections on Hilbert spacesFredholmness of formal selfadjoint operator $AA^*$ and Fredholmenss of $A$.
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An application of Fredholm Alternative
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Fredholm and Compact OperatorsLinear operators with no adjointQuestions about the proof of the Riesz representation theoremFinding the Hilbert Adjoint in this caseexamples of using the Fredholm alternative explicitlyA possible Corollary of the Fredholm alternative?On Fredholm operator on Hilbert spacesSelf adjoint map are bounded(bounded linear) orthogonal projections on Hilbert spacesFredholmness of formal selfadjoint operator $AA^*$ and Fredholmenss of $A$.
$begingroup$
I have just started reading the Fredholm Alternative for finite dimensional spaces and I came towards an excersise that it reads as follows:
If $H$ is a Hilbert space and $T:Hto H$ is bounded, linear map, with
$langle Tx,xrangle >0$ for $xneq0$ then prove that $T$ is surjective. My way of thinking was to use the adjoint map $T^*$ like that:
$langle Tx,xrangle = langle x,T^* xrangle >0$ , for $xneq0$.
The last thing means that $KerT^* =0$ and therefor , $ (KerT^*)^bot =H$, which means that $T$ is surjective.
But I do have some considerations in case the dimension of $H$ is not finite:
1) Is the adjoint $T^*$ always defined ?
2) Why we have $KerT^*oplus(KerT^*)^bot =H$ ?
Propably the above hold as $H$ is Hilbert and $T$ is bounded, but I cannot understand how I can deduce those...
Any clarification or hint is really appreciated.
functional-analysis vector-spaces
edited Mar 26 at 1:23
Saad
20.7k92452
asked Mar 25 at 18:23
dmtridmtri
1,7612521
$endgroup$
add a comment |
$begingroup$
I have just started reading the Fredholm Alternative for finite dimensional spaces and I came towards an excersise that it reads as follows:
If $H$ is a Hilbert space and $T:Hto H$ is bounded, linear map, with
$langle Tx,xrangle >0$ for $xneq0$ then prove that $T$ is surjective. My way of thinking was to use the adjoint map $T^*$ like that:
$langle Tx,xrangle = langle x,T^* xrangle >0$ , for $xneq0$.
The last thing means that $KerT^* =0$ and therefor , $ (KerT^*)^bot =H$, which means that $T$ is surjective.
But I do have some considerations in case the dimension of $H$ is not finite:
1) Is the adjoint $T^*$ always defined ?
2) Why we have $KerT^*oplus(KerT^*)^bot =H$ ?
Propably the above hold as $H$ is Hilbert and $T$ is bounded, but I cannot understand how I can deduce those...
Any clarification or hint is really appreciated.
functional-analysis vector-spaces
edited Mar 26 at 1:23
Saad
20.7k92452
asked Mar 25 at 18:23
dmtridmtri
1,7612521
$endgroup$
add a comment |
$begingroup$
I have just started reading the Fredholm Alternative for finite dimensional spaces and I came towards an excersise that it reads as follows:
If $H$ is a Hilbert space and $T:Hto H$ is bounded, linear map, with
$langle Tx,xrangle >0$ for $xneq0$ then prove that $T$ is surjective. My way of thinking was to use the adjoint map $T^*$ like that:
$langle Tx,xrangle = langle x,T^* xrangle >0$ , for $xneq0$.
The last thing means that $KerT^* =0$ and therefor , $ (KerT^*)^bot =H$, which means that $T$ is surjective.
But I do have some considerations in case the dimension of $H$ is not finite:
1) Is the adjoint $T^*$ always defined ?
2) Why we have $KerT^*oplus(KerT^*)^bot =H$ ?
Propably the above hold as $H$ is Hilbert and $T$ is bounded, but I cannot understand how I can deduce those...
Any clarification or hint is really appreciated.
functional-analysis vector-spaces
edited Mar 26 at 1:23
Saad
20.7k92452
asked Mar 25 at 18:23
dmtridmtri
1,7612521
$endgroup$
I have just started reading the Fredholm Alternative for finite dimensional spaces and I came towards an excersise that it reads as follows:
If $H$ is a Hilbert space and $T:Hto H$ is bounded, linear map, with
$langle Tx,xrangle >0$ for $xneq0$ then prove that $T$ is surjective. My way of thinking was to use the adjoint map $T^*$ like that:
$langle Tx,xrangle = langle x,T^* xrangle >0$ , for $xneq0$.
The last thing means that $KerT^* =0$ and therefor , $ (KerT^*)^bot =H$, which means that $T$ is surjective.
But I do have some considerations in case the dimension of $H$ is not finite:
1) Is the adjoint $T^*$ always defined ?
2) Why we have $KerT^*oplus(KerT^*)^bot =H$ ?
Propably the above hold as $H$ is Hilbert and $T$ is bounded, but I cannot understand how I can deduce those...
Any clarification or hint is really appreciated.
functional-analysis vector-spaces
functional-analysis vector-spaces
edited Mar 26 at 1:23
Saad
20.7k92452
asked Mar 25 at 18:23
dmtridmtri
1,7612521
edited Mar 26 at 1:23
Saad
20.7k92452
asked Mar 25 at 18:23
dmtridmtri
1,7612521
edited Mar 26 at 1:23
Saad
20.7k92452
edited Mar 26 at 1:23
Saad
20.7k92452
edited Mar 26 at 1:23
Saad
20.7k92452
20.7k92452
asked Mar 25 at 18:23
dmtridmtri
1,7612521
asked Mar 25 at 18:23
dmtridmtri
1,7612521
asked Mar 25 at 18:23
dmtridmtri
1,7612521
1,7612521
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
1) Yes. It will be helpful to know the other terms for adjoint: "dual" and "transpose" (even for operators that are not matrices). See Transpose of a linear map. For a full, excellent exposition (which addresses infinite-dimensional linear algebra despite the title), see FDVS. I don't recommend learning linear algebra from any other textbook (Halmos also wrote a problem book on the subject), nor going forward to functional analysis before studying linear algebra.
2) Because (see the FDVS book cited above for every italicized term) if $U$ is any subspace of an inner product space $V$, then $V$ is the direct sum of $U$ and the orthogonal complement of $U$. (For now, see the last bulleted item under "Inner Product Spaces, Properties" in the article Orthogonal complement.)
answered Mar 25 at 18:39
avsavs
4,197515
$endgroup$
$begingroup$
Thanks a lot for your nice answer! As far as I can see that the linear operator $T$ is bounded, is irrelevant here to prove that it is surjeective ....Am I wrong?
$endgroup$
– dmtri
Mar 26 at 17:37
1
$begingroup$
I think boundedness is important. The exercise says to prove that a bounded, positive-definite operator on a Hilbert space is surjective. The closest result I could find was able to find is the more general Browder-Minty Theorem, but they require that the operator be coercive, which is a stronger condition that positive definiteness. On the other hand, they don't require the operator to be linear. This may be an overkill. The question you probably want is,
$endgroup$
– avs
Mar 26 at 18:04
1
$begingroup$
Can an unbounded, positive-definite linear operator on a Hilbert space fail to be surjective? (Of course, for finite dimensions, boundedness is guaranteed.)
$endgroup$
– avs
Mar 26 at 18:05
1
$begingroup$
After watching around some books in the net, at least the pages they let you see, about Hilbert spaces, I see that the boundedness is required for the existence of the $T^*$ , so you are right and thanks.
$endgroup$
– dmtri
Mar 27 at 9:27
1
$begingroup$
Thanks for doing the research and confirming: I wasn't sure and didn't have the time to look into it properly.
$endgroup$
– avs
Mar 27 at 17:53
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
1) Yes. It will be helpful to know the other terms for adjoint: "dual" and "transpose" (even for operators that are not matrices). See Transpose of a linear map. For a full, excellent exposition (which addresses infinite-dimensional linear algebra despite the title), see FDVS. I don't recommend learning linear algebra from any other textbook (Halmos also wrote a problem book on the subject), nor going forward to functional analysis before studying linear algebra.
2) Because (see the FDVS book cited above for every italicized term) if $U$ is any subspace of an inner product space $V$, then $V$ is the direct sum of $U$ and the orthogonal complement of $U$. (For now, see the last bulleted item under "Inner Product Spaces, Properties" in the article Orthogonal complement.)
answered Mar 25 at 18:39
avsavs
4,197515
$endgroup$
$begingroup$
Thanks a lot for your nice answer! As far as I can see that the linear operator $T$ is bounded, is irrelevant here to prove that it is surjeective ....Am I wrong?
$endgroup$
– dmtri
Mar 26 at 17:37
1
$begingroup$
I think boundedness is important. The exercise says to prove that a bounded, positive-definite operator on a Hilbert space is surjective. The closest result I could find was able to find is the more general Browder-Minty Theorem, but they require that the operator be coercive, which is a stronger condition that positive definiteness. On the other hand, they don't require the operator to be linear. This may be an overkill. The question you probably want is,
$endgroup$
– avs
Mar 26 at 18:04
1
$begingroup$
Can an unbounded, positive-definite linear operator on a Hilbert space fail to be surjective? (Of course, for finite dimensions, boundedness is guaranteed.)
$endgroup$
– avs
Mar 26 at 18:05
1
$begingroup$
After watching around some books in the net, at least the pages they let you see, about Hilbert spaces, I see that the boundedness is required for the existence of the $T^*$ , so you are right and thanks.
$endgroup$
– dmtri
Mar 27 at 9:27
1
$begingroup$
Thanks for doing the research and confirming: I wasn't sure and didn't have the time to look into it properly.
$endgroup$
– avs
Mar 27 at 17:53
add a comment |
$begingroup$
1) Yes. It will be helpful to know the other terms for adjoint: "dual" and "transpose" (even for operators that are not matrices). See Transpose of a linear map. For a full, excellent exposition (which addresses infinite-dimensional linear algebra despite the title), see FDVS. I don't recommend learning linear algebra from any other textbook (Halmos also wrote a problem book on the subject), nor going forward to functional analysis before studying linear algebra.
2) Because (see the FDVS book cited above for every italicized term) if $U$ is any subspace of an inner product space $V$, then $V$ is the direct sum of $U$ and the orthogonal complement of $U$. (For now, see the last bulleted item under "Inner Product Spaces, Properties" in the article Orthogonal complement.)
answered Mar 25 at 18:39
avsavs
4,197515
$endgroup$
$begingroup$
Thanks a lot for your nice answer! As far as I can see that the linear operator $T$ is bounded, is irrelevant here to prove that it is surjeective ....Am I wrong?
$endgroup$
– dmtri
Mar 26 at 17:37
1
$begingroup$
I think boundedness is important. The exercise says to prove that a bounded, positive-definite operator on a Hilbert space is surjective. The closest result I could find was able to find is the more general Browder-Minty Theorem, but they require that the operator be coercive, which is a stronger condition that positive definiteness. On the other hand, they don't require the operator to be linear. This may be an overkill. The question you probably want is,
$endgroup$
– avs
Mar 26 at 18:04
1
$begingroup$
Can an unbounded, positive-definite linear operator on a Hilbert space fail to be surjective? (Of course, for finite dimensions, boundedness is guaranteed.)
$endgroup$
– avs
Mar 26 at 18:05
1
$begingroup$
After watching around some books in the net, at least the pages they let you see, about Hilbert spaces, I see that the boundedness is required for the existence of the $T^*$ , so you are right and thanks.
$endgroup$
– dmtri
Mar 27 at 9:27
1
$begingroup$
Thanks for doing the research and confirming: I wasn't sure and didn't have the time to look into it properly.
$endgroup$
– avs
Mar 27 at 17:53
add a comment |
$begingroup$
1) Yes. It will be helpful to know the other terms for adjoint: "dual" and "transpose" (even for operators that are not matrices). See Transpose of a linear map. For a full, excellent exposition (which addresses infinite-dimensional linear algebra despite the title), see FDVS. I don't recommend learning linear algebra from any other textbook (Halmos also wrote a problem book on the subject), nor going forward to functional analysis before studying linear algebra.
2) Because (see the FDVS book cited above for every italicized term) if $U$ is any subspace of an inner product space $V$, then $V$ is the direct sum of $U$ and the orthogonal complement of $U$. (For now, see the last bulleted item under "Inner Product Spaces, Properties" in the article Orthogonal complement.)
answered Mar 25 at 18:39
avsavs
4,197515
$endgroup$
1) Yes. It will be helpful to know the other terms for adjoint: "dual" and "transpose" (even for operators that are not matrices). See Transpose of a linear map. For a full, excellent exposition (which addresses infinite-dimensional linear algebra despite the title), see FDVS. I don't recommend learning linear algebra from any other textbook (Halmos also wrote a problem book on the subject), nor going forward to functional analysis before studying linear algebra.
2) Because (see the FDVS book cited above for every italicized term) if $U$ is any subspace of an inner product space $V$, then $V$ is the direct sum of $U$ and the orthogonal complement of $U$. (For now, see the last bulleted item under "Inner Product Spaces, Properties" in the article Orthogonal complement.)
answered Mar 25 at 18:39
avsavs
4,197515
answered Mar 25 at 18:39
avsavs
4,197515
answered Mar 25 at 18:39
avsavs
4,197515
answered Mar 25 at 18:39
avsavs
4,197515
4,197515
$begingroup$
Thanks a lot for your nice answer! As far as I can see that the linear operator $T$ is bounded, is irrelevant here to prove that it is surjeective ....Am I wrong?
$endgroup$
– dmtri
Mar 26 at 17:37
1
$begingroup$
I think boundedness is important. The exercise says to prove that a bounded, positive-definite operator on a Hilbert space is surjective. The closest result I could find was able to find is the more general Browder-Minty Theorem, but they require that the operator be coercive, which is a stronger condition that positive definiteness. On the other hand, they don't require the operator to be linear. This may be an overkill. The question you probably want is,
$endgroup$
– avs
Mar 26 at 18:04
1
$begingroup$
Can an unbounded, positive-definite linear operator on a Hilbert space fail to be surjective? (Of course, for finite dimensions, boundedness is guaranteed.)
$endgroup$
– avs
Mar 26 at 18:05
1
$begingroup$
After watching around some books in the net, at least the pages they let you see, about Hilbert spaces, I see that the boundedness is required for the existence of the $T^*$ , so you are right and thanks.
$endgroup$
– dmtri
Mar 27 at 9:27
1
$begingroup$
Thanks for doing the research and confirming: I wasn't sure and didn't have the time to look into it properly.
$endgroup$
– avs
Mar 27 at 17:53
add a comment |
$begingroup$
Thanks a lot for your nice answer! As far as I can see that the linear operator $T$ is bounded, is irrelevant here to prove that it is surjeective ....Am I wrong?
$endgroup$
– dmtri
Mar 26 at 17:37
1
$begingroup$
I think boundedness is important. The exercise says to prove that a bounded, positive-definite operator on a Hilbert space is surjective. The closest result I could find was able to find is the more general Browder-Minty Theorem, but they require that the operator be coercive, which is a stronger condition that positive definiteness. On the other hand, they don't require the operator to be linear. This may be an overkill. The question you probably want is,
$endgroup$
– avs
Mar 26 at 18:04
1
$begingroup$
Can an unbounded, positive-definite linear operator on a Hilbert space fail to be surjective? (Of course, for finite dimensions, boundedness is guaranteed.)
$endgroup$
– avs
Mar 26 at 18:05
1
$begingroup$
After watching around some books in the net, at least the pages they let you see, about Hilbert spaces, I see that the boundedness is required for the existence of the $T^*$ , so you are right and thanks.
$endgroup$
– dmtri
Mar 27 at 9:27
1
$begingroup$
Thanks for doing the research and confirming: I wasn't sure and didn't have the time to look into it properly.
$endgroup$
– avs
Mar 27 at 17:53
$begingroup$
Thanks a lot for your nice answer! As far as I can see that the linear operator $T$ is bounded, is irrelevant here to prove that it is surjeective ....Am I wrong?
$endgroup$
– dmtri
Mar 26 at 17:37
$begingroup$
Thanks a lot for your nice answer! As far as I can see that the linear operator $T$ is bounded, is irrelevant here to prove that it is surjeective ....Am I wrong?
$endgroup$
– dmtri
Mar 26 at 17:37
1
1
$begingroup$
I think boundedness is important. The exercise says to prove that a bounded, positive-definite operator on a Hilbert space is surjective. The closest result I could find was able to find is the more general Browder-Minty Theorem, but they require that the operator be coercive, which is a stronger condition that positive definiteness. On the other hand, they don't require the operator to be linear. This may be an overkill. The question you probably want is,
$endgroup$
– avs
Mar 26 at 18:04
$begingroup$
I think boundedness is important. The exercise says to prove that a bounded, positive-definite operator on a Hilbert space is surjective. The closest result I could find was able to find is the more general Browder-Minty Theorem, but they require that the operator be coercive, which is a stronger condition that positive definiteness. On the other hand, they don't require the operator to be linear. This may be an overkill. The question you probably want is,
$endgroup$
– avs
Mar 26 at 18:04
1
1
$begingroup$
Can an unbounded, positive-definite linear operator on a Hilbert space fail to be surjective? (Of course, for finite dimensions, boundedness is guaranteed.)
$endgroup$
– avs
Mar 26 at 18:05
$begingroup$
Can an unbounded, positive-definite linear operator on a Hilbert space fail to be surjective? (Of course, for finite dimensions, boundedness is guaranteed.)
$endgroup$
– avs
Mar 26 at 18:05
1
1
$begingroup$
After watching around some books in the net, at least the pages they let you see, about Hilbert spaces, I see that the boundedness is required for the existence of the $T^*$ , so you are right and thanks.
$endgroup$
– dmtri
Mar 27 at 9:27
$begingroup$
After watching around some books in the net, at least the pages they let you see, about Hilbert spaces, I see that the boundedness is required for the existence of the $T^*$ , so you are right and thanks.
$endgroup$
– dmtri
Mar 27 at 9:27
1
1
$begingroup$
Thanks for doing the research and confirming: I wasn't sure and didn't have the time to look into it properly.
$endgroup$
– avs
Mar 27 at 17:53
$begingroup$
Thanks for doing the research and confirming: I wasn't sure and didn't have the time to look into it properly.
$endgroup$
– avs
Mar 27 at 17:53
add a comment |
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