For a Hilbert space $\mathcal{H}$, is every bounded linear operator on $\mathcal{H}$ a linear combination of unitary operators?

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.

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

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
• 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. Aug 14, 2014 at 14:03
• 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. Aug 15, 2014 at 18:19