# 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.

• What does nonnegative mean for an operator which isn't self-adjoint? – Qiaochu Yuan May 31 '17 at 5:15
• @Qiaochu: I can't speak for the OP, but I guess it is supposed to mean $\langle Ax, x\rangle \geq 0$ for all $x$. – PhoemueX May 31 '17 at 11:15
• @0xbadf00d: Is $A$ at least normal? If so, then there is the continuous functional calculus available that does what you want. – Alex M. Jun 1 '17 at 8:33
• @0xbadf00d: Alternatively, consider a branch of the complex square root (it will be holomorphic on its domain of definition): if $A$ is not normal but its spectrum is contained in the domain of this complex square root, then you may use the holomorphic functional calculus. – Alex M. Jun 1 '17 at 8:44
• @san: This is a good remark, but the links considers complex Hilbert spaces. In contrast, the OP considers real Hilbert spaces. In those spaces, it is not true that every nonnegative (in the sense of $\langle Ax, x\rangle \geq 0$ operator is self-adjoint. A counterexample is any operator with $A^\ast = -A$, i.e., any skew-symmetric matrix in the finite dimensional case. – PhoemueX Jun 3 '17 at 14:32