Unitaries $u$ span $A$ linearly?

Question

I can't understand this paragraph in my book:

If $a$ is a self-adjoint element of the closed unit ball of a unital $C^\ast$-algebra $A$ then $1-a^2$ is positive and $u=a + i\sqrt{1-a^2}$ and $v = a - i\sqrt{1-a^2}$ are unitaries such that $a = {1\over 2}(u+v)$. Therefore the unitaries linearly span $A$.

I really don't see how it could be true that unitaries linearly span $A$. If $a\in A$ is any not self-adjoint element, then $u,v$ are not unitary. And I don't see how such $a$ can be written as a linear combination of unitaries. Please could someone shed some light?

Answer 1

Any $a \in A$ is canonically a (complex) linear combination of self-adjoint elements $\frac{a + a^{\ast}}{2} + i \frac{a - a^{\ast}}{2i}$, so to show that any element can be written as a linear combination of unitary elements it suffices to show that any self-adjoint element can. By rescaling such a thing we can assume WLOG that it has norm at most $1$. What is the issue?

Answer 2

As an example for better understanding, consider the case where $A=\mathbb C$. Then, for a fixed $z\in\mathbb C$, you write $$z=\sqrt{a^2+b^2}\left(\frac a{\sqrt{a^2+b^2}}+i\frac b{\sqrt{a^2+b^2}}\right),$$ with $a,b\in\mathbb R$. Since $$ -1\leq \frac a{\sqrt{a^2+b^2}}\leq1, $$ there exists $\theta$ such that $\frac a{\sqrt{a^2+b^2}}=\cos\theta$. Similarly, there exists $\phi$ such that $\frac b{\sqrt{a^2+b^2}}=\cos\phi$. So $$ z=\sqrt{a^2+b^2}\,\left(\cos\theta+i\cos\phi\right). $$ Now we construct the unitaries (i.e. complex numbers of absolute value $1$ in this case) $$ u_1=\cos\theta+i\sin\theta,\ \ u_2=\cos\theta-i\sin\theta,\ \ \ v_1=\cos\phi+i\sin\phi\ \ \ \ v_2=\cos\phi-i\sin\phi, $$ and $$ z=\frac{|z|}2\,(u_1+u_2+iv_1+iv_2) $$ is a linear combination of four unitaries.