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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 $T:H\to H$ is bounded, linear map, with $\langle Tx,x\rangle >0$ for $x\neq0$ then prove that $T$ is surjective. My way of thinking was to use the adjoint map $T^*$ like that:

$\langle Tx,x\rangle = \langle x,T^* x\rangle >0$ , for $x\neq0$. 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.

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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.)

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  • $\begingroup$ 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 '19 at 17:37
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    $\begingroup$ 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 '19 at 18:04
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    $\begingroup$ 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 '19 at 18:05
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    $\begingroup$ 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 '19 at 9:27
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    $\begingroup$ 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 '19 at 17:53

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