show that $S^{-1}:S(H)\to H $ exist. where is $H$ is Hilbert space and $T^*$ is Adjoint of $T$ on $H$. let $S=I+T^*T:H\to H$, where $T$ is linear and bounded .
show that $S^{-1}:S(H)\to H $ exist. where is $H$ is Hilbert space and $T^*$ is Adjoint of $T$ on $H$.
by definition of $S$ it is linear map from $S(H)\to H$.so we are only required to prove $S$ is one one.but how do we prove that . any hint??
 A: $$(I+T^*T)x=0 \implies x+T^*Tx=0 \implies T^*Tx=-x$$
Then $$\langle-x,x\rangle=\langle T^*Tx,x\rangle=\|Tx\|^2$$
This is a contradiction unless $x=0$.
A: One can also argue like this:
$\langle x, Sx \rangle = \langle x, (I + T^\ast T)x \rangle = \langle x, Ix \rangle + \langle x, T^\ast Tx \rangle$
$= \langle x, x \rangle + \langle Tx, Tx \rangle = \Vert x \Vert^2 + \Vert Tx \Vert^2 \ge \Vert x \Vert^2, \tag 1$
which shows that
$\langle x, Sx \rangle \ge \Vert x \Vert^2; \tag 2$
by (2),
$Sx = 0 \Longleftrightarrow x = 0; \tag 3$
(3) means that $S$ is injective, i.e., one-to-one from $H \to S(H)$; thus, for any $w \in S(H)$, there is a unique $v \in H$ such that $Sv = w$, and we may set $v = S^{-1}w$; $S^{-1}$ is clearly linear, and in fact, we may show it is continuous, i.e., bounded: from (2) we have, using Cauchy-Schwarz, 
$\Vert x \Vert^2 \le \langle x, Sx \rangle = \vert \langle x, Sx \rangle \vert \le \Vert x \Vert \Vert Sx \Vert, \tag 4$, whence for $\Vert x \Vert \ne 0$, 
$\Vert x \Vert \le \Vert Sx \Vert; \tag 5$
then taking $x = S^{-1}y$ we find
$\Vert S^{-1} y \Vert \le \Vert S(S^{-1}y) \Vert = \Vert y \Vert, \tag 6$
so we see that
$\Vert S^{-1} \Vert \le 1. \tag 7$
Note Added in Edit, Friday 20 October 2017 10:05 AM PST:  Since $S = I + T^\ast T$ is injective, it follows that for $\dim(H) < \infty$, $S$ is also surjective; whether it is surjective in the infinite-dimensional case I do not fully know.  It is clear that if the self-adjoint operator $T^\ast T$ satisfies certain additional conditions, surjectivity may be attained; for example, if $T^\ast T$ is compact on a separable $H$, then there is an orthonormal basis of $H$ consisting of eigevectors $e_n$ of $T^\ast T$, with 
$T^\ast Te_n = \lambda_n e_n \tag 8$
and $\lambda_n \to 0$; we note that
$\lambda_n = \lambda_n \langle e_n, e_n \rangle = \langle e_n, \lambda_n e_n \rangle = \langle e_n, T^\ast T e_n \rangle = \langle Te_n, Te_n \rangle \ge 0; \tag 9$
these things being the case, if $y \in H$ is written
$y = \displaystyle \sum_1^\infty y_i e_i, \tag{10}$
and we set
$z_n = \dfrac{y_n}{1 + \lambda_n}, \tag{11}$
then 
$\displaystyle \sum_1^\infty \bar z_i z_i = \sum_1^\infty \dfrac{\bar y_i}{1 + \lambda_i}\dfrac{y_i}{1 + \lambda_i} = \sum_1^\infty \dfrac{\bar y_i y_i}{(1 + \lambda_i)^2} \le \sum_1^\infty \bar y_i y_i < \infty; \tag{12}$
so if we set
$z = \displaystyle  \sum_1^\infty \dfrac{y_i}{1 + \lambda_i} e_i, \tag{13}$
we find that
$Sz = (I + T^\ast T)z = \displaystyle  \sum_1^\infty z_i (I + T^\ast T)e_i = \sum_1^\infty z_i (e_i + T^\ast Te_i)$
$= \sum_1^\infty z_i (e_i + \lambda_i e_i) =  \sum_1^\infty z_i (1 + \lambda_i) e_i = \sum_1^\infty y_i e_i = y, \tag{14}$
using (8) and (11).  So in this special case $T^\ast T$ compact, we see that $S = I + T^\ast T$ is surjective.  But I know neither how to prove $S = I + T^\ast T$ is surjective in general, nor even if this is true.
Finally, I think it is worth noting that $S(H)$ is a closed subspace of $H$; for if $y_n \in S(H)$ is Cauchy, then by the continuity of $S^{-1}$ (i.e., by (7)), the sequence $x_n = S^{-1}y_n$ is Cauchy as well; thus there exists $x \in H$ with $x_n \to x$; then 
$y_n = Sx_n \to Sx, \tag{15}$
that is, the limit of the Cauchy sequence $y_n \in S(H)$ is $Sx \in S(H)$. End of Note.
