1
$\begingroup$

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!

$\endgroup$
2
  • 2
    $\begingroup$ 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.) $\endgroup$ Mar 4, 2021 at 16:50
  • $\begingroup$ 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! $\endgroup$
    – Sky subO
    Mar 4, 2021 at 18:03

1 Answer 1

4
$\begingroup$

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.

$\endgroup$
3
  • 1
    $\begingroup$ Wow, Professor Axler, a big surprise to see you here! Thank you very much for the thoughtful answer! $\endgroup$
    – Sky subO
    Mar 5, 2021 at 13:41
  • $\begingroup$ +1 , although the second paragraph would be would be another Question for me. $\endgroup$ Mar 5, 2021 at 17:05
  • $\begingroup$ See the paper "Square Roots of Operators" by Halmos and Lumer and Schaffer in Proceedings of the AMS 4 (1953), 142-149 for details about a class of invertible operators on an infinite-dimensional Hilbert space that have no square root. $\endgroup$ Mar 5, 2021 at 17:37

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .