# Square-root for nonnegative (and not necessarily self-adjoint) operators

Let $H$ be a $\mathbb R$-Hilbert space and $A$ be a bounded linear operator on $H$. If $A$ is nonnegative and self-adjoint, then there is a unique nonnegative and self-adjoint bounded linear operator $B$ on $H$ with $B^2=A$. This can be proved by an elementary version of the spectral theorem.

Now, suppose that $A$ is only nonnegative (and not necessarily self-adjoint). I’ve read that we can still find a unique nonnegative bounded linear operator $B$ on $H$ with $B^2=A$. How can we prove this statement?

My knowledge in operator theory is rather limited. So, I hope we can provide an elementary proof.