Infinite-dimensional inner product spaces: if $A$ is a skew operator, does it follow that $A-I$ is invertible?

Question

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.

Answer 1

I'm assuming that $H$ is a Hilbert space. For $x \in H$ we have $\langle Ax,x\rangle = -\langle x,Ax\rangle$ so $$\|(A-I)x\|^2 = \|Ax-x\|^2 = \langle Ax,x\rangle - \langle Ax,x\rangle - \langle x,Ax\rangle + \langle x,x\rangle = \|Ax\|^2 + \|x\|^2 \ge \|x\|^2.$$ Therefore, $A-I$ is bounded from below so in particular $A-I$ is injective and its range $R(A-I)$ is closed. Since $A-I$ is normal, we have $$H = N(A-I) \oplus \overline{R(A-I)} = R(A-I)$$ so $A-I$ is surjective. We conclude that $A-I$ is invertible.

Answer 2

If your space is complete (that is, a Hilbert space), the answer is yes. On a Hilbert space, the spectrum of a selfadjoint operator is real. As $iA$ is selfadjoint, ou have $\sigma(A)\subset i\mathbb R$, so $$\sigma(A-I)=\{-1+\lambda:\ \lambda\in\sigma(A)\}\subset -1+i\mathbb R,$$ so $0\not\in\sigma(A)$.

In general, the answer is no. Let $V=\mathbb C[x]$ with $\langle f,g\rangle=\int_0^1f(t)\,\overline{g(t)}\,dt$. Let $A$ be $(Af)(t)=i\,t^2\,f(t)$. We have $$\langle A^*f,g\rangle=\langle f,Ag\rangle=\int_0^1f(t)\,\overline{it^2g(t)}\,dt=\langle-Af,g\rangle. $$ So $A^*=-A$. But $A-I$ is multiplication by $it^2-1$, which is not surjective.