Invertibility of the square root of an operator

Question

Let $\mathcal{B}(F)$ the algebra of all bounded linear operators on a complex Hilbert space $F$. Let $M\in \mathcal{B}(F)^+$ (i.e. $\langle Mx\;, \;x\rangle\geq 0$ for all $x\in F$).

• The square root of $M$ is defined to be the unique operator $N$ such that $$N^2=M.$$

• In this case we write $N=M^{1/2}$

If $M$ is an invertible operators, is $M^{1/2}$ invertible?

Answer 1

Note that $M$ is invertible if and only if there is $\epsilon>0$ such that $$\sigma(M)\subset [\epsilon, \infty).$$ Thus $\sigma(M^{1/2})\subset [\epsilon^{1/2},\infty)$ and $M^{1/2}$ is invertible.

Answer 2

$N(x)=N(y)$ implies $M(x)=N^2(x)=N^2(y)=M(y)$ since $M$ is injective, we deduce that $N$ is injective.

For every $y$, $y=M(x)=N(N(x))$ implies that $N$ is surjective and open (open mapping theorem)

This implies that the inverse of $N$ is continuous $N$ and $N$ is invertible.