Invertibility in infinite dimensional HIlbert spaces

Question

Suppose the operator $A$ acts on an infinite dimensional Hilbert space and has the property that $A^\dagger A = 1$. Can such an operator not be invertible? For a finite dimensional Hilbert space, applying the determinate to $A^\dagger A = 1$ easily reveals invertibility, but I do not think I can apply the determinant logic in the infinite-dimensional case. If there are such operators that are not invertible, what are some examples?

Answer 1

Yes, it can happen. The canonical example is the unilateral shift. For instance take $H=\ell^2(\mathbb N)$, and put $$ Ax=(0,x_1,x_2,x_3,\ldots). $$ It is straightforward to see that $A$ is an isometry, which immediately implies that $A^\dagger A=1$. But $A$ is not surjective. With just a tiny more work, one can see that $$ AA^\dagger x=(0,x_2,x_3,\ldots). $$ Such an operator can be defined in any separable infinite-dimensional Hilbert space, by fixing an orthonormal basis $\{e_n\}$ and defining $A$ as the linear operator that satisfies $Ae_n=e_{n+1}$ for all $n$.

As mentioned by David, the Wold Decomposition guarantees that the only counterexamples are versions of this one.