Nico Patterson

2022-11-21

Is the invariant subspace problem open for invertible maps?
Let $T:H\to H$ be a bounded linear operator with bounded inverse on the separable complex Hilbert space. Does T preserve a closed proper non-trival invariant subspace?
I'm aware the question is (famously) open for bounded linear maps, and of partial results, but no survey (or Tao's blog, etc) seem to address the invertible case.
If it is open, does a positive or negative answer imply the answer in the non-invertible case?

Deanna Sweeney

The invertible case is indeed equivalent to the full (possibly non-invertible) problem.
Firstly, if T is invertible and has a proper non-trivial invariant subspace, then this also answers the full conjecture in the negative. So, a negative answer in the invertible case leads to a negative answer in the full case.
Now, suppose T has no proper non-trivial invariant subspace. Note that the kernel and range of T are both invariant subspaces, and hence must be {0} or H. If the range is {0} or the kernel is H, then T=0, in which case it leaves every subspace invariant! Hence, the range is H and the kernel is {0}, implying that T is invertible (boundedly so, by the bounded inverse theorem).
Therefore, a positive answer in the invertible case leads to a positive answer in the full problem. The invertible case is equivalent to the full problem.

Do you have a similar question?