# Over C, invertible operators have square roots?

We know that on finite-dimensional vector spaces, it is true. This is 8.33 in Sheldon Axler's Linear Algebra Done Right:

Suppose $$V$$ is a complex vector space and $$\, T\in L(V) \,$$ is invertible. Then $$T$$ has a square root.

The question is that, if V is infinite-dimensional, does this result still hold?

I'd be much obliged if anyone could give a counterexample or prove that this result is true in general. Thanks!

• If the vector space is the space of functions $\mathbb{Z}\to \mathbb{C}$, and $T : (\ldots, x_{-2}, x_{-1}, x_0, x_1, x_2, \ldots) \mapsto (\ldots, x_{-1}, x_0, x_1, x_2, x_3, \ldots)$, it's not immediately obvious what a square root of that operator would be. (But it's also not immediately obvious that no square root exists at all.) Mar 4, 2021 at 16:50
• Although difficult for me, this counterexample looks nice. The important thing is that it tells us this result is wrong under infinite-dimensional conditions. Thank you very much! Mar 4, 2021 at 18:03

Be careful about the first example above. On the space $$\ell^2$$, the shift operator $$T$$ defined in the first comment above is a bounded invertible operator. Intuitively, a square root of $$T$$ should be a shift by one-half slot, which of course does not make sense. Thus one can easily believe that this invertible operator $$T$$ does not have a square root. However, this operator does have square root!! To see this, identify $$\ell^2$$ with $$L^2$$ of the unit circle and identify the shift with the operator of multiplication by $$z$$. Then the operator of multiplication by the square root of $$z$$ (choose any branch of the square root function) is a square root of $$T$$.
For an invertible operator that does not have a square root, consider the operator of multiplication by $$z$$ on the Hardy space of an annulus centered at the origin. This operator is invertible (its inverse is multiplication by $$1/z$$) but it does not have a square root.