A Problem of bounded linear operator in Hilbert Space Suppose $U$ is a bounded linear operator on a Hilbert Space $H$ which preserves the inner product, that is to say $(Ux, Uy) =(x,y)$. ($U$ not necessarily to be surjective). Then can we conclude that $ker(I-U) = ker(I-U^*)$, or give a counter-example?
 A: It is true: $"\subset"$:It has been stated in previous answers that $$
U^*U = I.$$
Therefore the following implications for an arbitrary $x$ hold true $$
x \in \operatorname{ker}(U-I) \Rightarrow U^*(U-I)x = 0 \Rightarrow (I-U^*)x = 0 \Rightarrow x \in \operatorname{ker}(U^*-I).$$
So we already have $\operatorname{ker}(U-I) \subset \operatorname{ker}(U^*-I)$.
$"\supset"$: Now let $x \in \operatorname{ker}(U^*-I)$. Then $U^*x = x$ and we conclude \begin{align}
||(U-I)x||^2 &= (Ux,Ux) - (Ux,x)-(x,Ux)+(x,x)\\
&= (x,x) - (x,U^*x) - (U^*x,x) + (x,x)\\
&= (x,x)-(x,x)-(x,x)+(x,x)\\
&= 0. 
\end{align}
Thus $x \in \operatorname{ker}(U-I)$.
A: Yes. I think $U$ is going to be surjective though since
$$ (x,y) = (Ux,Uy) = (x,U^*Uy)=(UU^*x,y)$$
for all $x,y \in H$. So $I=UU^*=U^*U$. Therefore, $U$ is invertible so it is injective and surjective. Then,
$$ I-U=UU^*-U = U(U^*-I)$$
Then, if $(I-U)x=0$, we have $U(U^*-I)x=0$ so $(U^*-I)x=0$.
A: This is true, although I'm not sure if my solution is an overkill. The result is true for normal operators because we have
$$ \| Tv \|^2 = \left< Tv, Tv \right> = \left< v, T^{*} T v \right> = \left< v, T T^{*} v \right> = \left< T^{*} v, T^{*}v \right> = \| T^{*} v \|^2 $$
for all $v \in H$. If $T$ is normal so is $T - I$ and then $v \in \ker(T - I)$ then
$$ 0 = \| (T - I)v \| = \| (T - I)^{*} v \| = \| (T^{*} - I)v \| $$
so $v \in \ker(T^{*} - I)$. Replacing $T$ with $T^{*}$ we see that $\ker(T - I) = \ker(T^{*} - I)$. In particular, this holds for unitary operators.
In addition, the result is true for the unilateral shift $U$ on $\ell_2(\mathbb{N})$ because both $\ker(U - I)$ and $\ker(U^{*} - I)$ are trivial as can be verified by a direct calculation. 
Since every isometry is either unitary, a direct sum of one or more copies of the unilateral shift or a direct sum of a unitary operator and some copies of the unilateral shift (problem 149 in Halmos's Hilbert Space Problem book) we get the required result.
