Isometric, self-adjoint linear operator is unitary

Question

$\newcommand{c}[1]{\langle #1 \rangle}$ Let $H$ be a Hilbert space, and let $T : H \to H$ be a linear operator that is both isometric and self-adjoint. Is it true that $T$ is surjective (and hence unitary)?

My proof is as follows: Let $x \in H$. Then: \begin{align*} \|T^2(x) - x\| &= \c{T^2(x) - x,T^2(x) - x} \\ &= \c{T^2(x),T^2(x)} - \c{T^2(x),x} - \c{x,T^2(x)} + \c{x,x} \\ &= \c{x,x} - \c{T(x),T(x)} - \c{T(x),T(x)} + \c{x,x} \\ &= \c{x,x} - \c{x,x} - \c{x,x} + \c{x,x} \\ &= 0 \end{align*}

Therefore, $T^2(x) = x$, and in particular $T$ is surjective.

Is this proof correct? Thanks in advance.

Answer 1

Yes, you are correct according to your definitions.

Unitary = Isometry + Surjectivity