I am trying to find an answer to this question: if $A$ is a skew-Hermitian operator (i.e., $A^* = -A$) on an infinite-dimensional inner product space, does it follow that $A-I$ is invertible? The question appears as exercise 7(a) after S.74 on page 145 of PR Halmos's "Finite-Dimensional Vector Spaces" - Second Edition.
So far, I have managed to establish the result in the finite-dimensional inner product spaces alone. Proof: if $(A-I)x = 0$ for any vector $x$, then $Ax = x$. Thus, we have the inner product $(x, x) = (Ax, x) = (x, A^*x) = (x, -Ax) = (x, -x) = -(x, x) \implies (x, x) = 0$. It follows that $x = 0$ due to the inner product property. In summary, $(A-I)x = 0 \implies x = 0$, and therefore $A-I$ is invertible (since the space is finite-dimensional).
Haven't been able to prove the assertion in infinite-dimensional inner product spaces. Would appreciate a guidance. Thanks.