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