Let $N$ be a normal operator on a Hilbert space $H$. Show that there exists a normal operator $M$ on $H$ such that $M^2=N$.
How can I prove that. $N$ and $M$ are just normal operators.
My attempt:
Let $N∈B(H)$ normal operator then $N$ is self-adjoint and $σ(N)⊂[0,∞)$. If the function $f(t)=√t$ denotes the positive square root, then implies that $f∈C(σ(N)$. So there exists $M$ normal operator ($M∈C^∗(N)$ by construction) such that $M=f(N)$ . Hence $M^2=N$.
There is something wrong that $N$ does not positive. I do not have another idea. I need help to solve this problem
