# compact and self adjoint square root of an operator

Let H a Hilbert space and $T:H\rightarrow H$ a linear bounded, self-adjoint, positive and compact operator. How can i prove that the square root of T, $\ T^{1/2}:H\rightarrow H$ is also compact and self-adjoint; thanks.

• As $T$ is a self-adjoint operator, we can apply the Continuous Functional Calculus (C.F.C.) to $T$.

• As $T$ is a positive operator, i.e., ${\sigma_{B(\mathcal{H})}}(T) \subseteq [0,\infty)$, we can apply the $\sqrt{\bullet}$-function to $T$ to obtain $\sqrt{T}$.

• The range of the C.F.C. of $T$ is the $\| \cdot \|_{B(\mathcal{H})}$-closed $^{\ast}$-subalgebra of $B(\mathcal{H})$ generated by $T$.

• As involution is a continuous operation on $B(\mathcal{H})$, it follows that $\sqrt{T}$ is a self-adjoint operator.

• As $T \in K(\mathcal{H})$ and $K(\mathcal{H})$ is a $\| \cdot \|_{B(\mathcal{H})}$-closed ideal of $B(\mathcal{H})$, it follows that the range of the C.F.C. of $T$ is contained in $K(\mathcal{H})$.

• Therefore, $\sqrt{T} \in K(\mathcal{H})$.

• i was looking for an answer whithout using $C^*-$algebra tools. the compactness can be derived from spectral representation but what about self-adjointness; – nikosp Feb 23 '13 at 9:39
• The range of C.F.C is not in the ideal of compacts. Take 1+T for example. – user207135 Jan 11 '15 at 14:56

Hint: the diagonalization of compact self-adjoint operators makes this easy. See here for instance.