For hermitian $x$ with $\|x\|\leq1$, its spectrum lies in $[-1,1]$. Continuous functional calculus therefore shows that $1-x^2$ is positive. Let
$$
v=(1-x^2)^{1/2}.
$$
It commutes with $x$, and
$$
u=x+iv
$$
satisfies $u^*u=uu^*=x^2+v^2=1$. Thus $u$ is <unitary operator>[unitary] and
$$
x=\frac12(u+u^*).
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 106 / 5 / 2 / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-106/5/2/solution