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

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.

• Why doesn't this work for the infinite dimensional case? Isn't it still the case that if $B x = 0 \implies x = 0$ for some operator $B$ in an inner product space, then the map $B$ is invertible? May 6, 2020 at 1:57
• I am not aware if the invertibility result extends to infinite-dimensional spaces too. I have followed PR Halmos's "Finite-Dimensional Vector Spaces". Theorem 2 from S.36 (page 62) of the book states that "$A$ is invertible on a finite-dimensional vector space if and only if $Ax = 0 \implies x = 0$" with a proof. The book does not say that this result extends to infinite-dimensional spaces too. May 6, 2020 at 2:05
• Indeed, it is false in general. An operator may be injective but not invertible, e.g. the shift $$(x_1,x_2, \ldots) \mapsto (0,x_1,x_2, \ldots)$$ on $\ell^2$. May 6, 2020 at 2:18

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.
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.