9
$\begingroup$

Let $(\mathcal{H}, (\cdot, \cdot))$ be a Hilbert space, and let $B \in \mathcal{B}(H)$ be a bounded linear operator on $H$.

If $\mathcal{H}$ is a complex Hilbert space, then $B$ can be written as a linear combination of unitary operators. How do we do this? First, we write $B$ as a linear combination of self-adjoint operators:

$$B = \frac{1}{2}(B + B^*) - \frac{i}{2}(iB - iB^*)$$

So so it suffices to show that any self adjoint operator $A \in \mathcal{B}(H)$ is a linear combination of unitary operators. Even more, it is no problem to assume the operator norm $\|A\| \le 1$.

Then, a short computation shows that $(A \pm i\sqrt{I - A^2})$ is unitary, and furthermore we have: $$A = \frac{1}{2}(A + i\sqrt{I - A^2}) + \frac{1}{2}(A - i\sqrt{I - A^2}).$$

Note that, because $\|A\| \le 1$, it's easy to see that $I-A^2$ is a positive operator, hence it has a well-defined square root.

My concern is, can we still write $B$ as a combination of unitary operators, even if $\mathcal{H}$ is just a real Hilbert space? In the real case, we do not have the complex number $i$ to work with, and it seems to be crucial in the above argument.

Hints or solutions are greatly appreciated.

$\endgroup$
2
  • 1
    $\begingroup$ Why is it ok to assume $\| A \| \leq 1$? $\endgroup$
    – user319128
    Feb 20, 2017 at 22:14
  • 1
    $\begingroup$ @Elliot: If $\|A\| > 1$, just work with the operator $A/ \|A\|$, and then $\|A\|$ becomes part of the coefficients in the final linear combination. $\endgroup$
    – JZS
    Feb 22, 2017 at 14:38

1 Answer 1

5
$\begingroup$

Note that if $\mathcal{H}$ is real then the notion of a unitary operator does not make any sense. However, you can do that with five orthogonal operators instead. See e.g. Theorem 4.3 in

A. Böttcher, A. Pietsch, Orthogonal and Skew-Symmetric Operators in Real Hilbert Space, Integral Equations and Operator Theory 74 (2012), 497-511.

It should be added that a similar statement is valid for (complex) C*-algebras and is known as the Russo–Dye theorem.

$\endgroup$
4
  • $\begingroup$ Thank you. The definition of unitary that I have been using is the same as the definition of orthogonal that you made a link to. Could you explain what the different (not applicable in the real case) definition of unitary is? $\endgroup$
    – JZS
    Aug 14, 2014 at 14:00
  • 1
    $\begingroup$ To me an operator $U$ is unitary if $U^* = U^{-1}$; this definition involves the hermitian conjugation $\cdot^*$, which for matrices is just transposition followed by term-wise complex conjugation. $\endgroup$ Aug 14, 2014 at 14:03
  • $\begingroup$ @Tomek: it is not a good idea to define a property of an operator in terms of its entries in some basis (because you need to argue that the property works the same in every basis). Note that you define the adjoint of the operator in terms of its entries as a matrix in some basis. The most natural (and coordinate-less) definition of a unitary is surjective isometry. $\endgroup$ Aug 15, 2014 at 5:11
  • $\begingroup$ The definition $U^*=U^{-1}$ is independent of a basis; I just wanted to highlight the role played by complex scalars in the case of matrices. $\endgroup$ Aug 15, 2014 at 18:19

You must log in to answer this question.

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