An application of Fredholm Alternative 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 therefore, $ ({KerT^*})^\bot =H$, which means that $T$ is surjective.
But I do have some considerations in case the dimension of $H$ is not finite:

*

*Is the adjoint $T^*$ always defined ?


*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.

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.)
