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Let $E$ be a complex Hilbert space. Let $M\in \mathcal{L}(E)^+$ (i.e. $M^*=M$ and $\langle Mx\;,x\;\rangle\geq 0,\,\forall x\in E$).

Why $$\overline{\operatorname{Im}(M)}=\operatorname{Im}(M)\Leftrightarrow \operatorname{Im}(M)=\operatorname{Im}(M^{1/2})\; ?$$ Thank you

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First off, note that $\text{ker}(M) = \text{ker}(M^{1/2})$. The inclusion $[\supseteq]$ is trivial. For the converse, if $x \in \text{ker}(M)$, then $ 0 = (Mx,x) = (M^{1/2}x,M^{1/2}x)$, so $M^{1/2}x = 0$.

$[\Rightarrow]$ Let $\text{Im}(M) = \overline{\text{Im}(M)}$. Then $$\text{Im}(M) = \overline{\text{Im}(M)} = \text{ker}(M)^{\perp} = \text{ker}(M^{1/2})^{\perp} = \overline{\text{Im}(M^{1/2})}$$ Now the restriction $M_0$ of $M^{1/2}$ to $\ker(M^{1/2})^{\perp}$ is bijective onto $\text{Im}(M)$ and bounded, hence invertible by the open mapping theorem. It follows that $M_0^{-1}$ is a homeomorphism from $\text{Im}(M)$ to $\text{Im}(M_0^{-1}) = \text{Im}(M^{1/2})$, which implies that $\text{Im}(M^{1/2})$ is closed.

$[\Leftarrow]$ Let $\text{Im}(M) = \text{Im}(M^{1/2})$. Then the restriction $M_{00}$ of $M^{1/2}$ to $\text{ker}(M^{1/2})^{\perp}$ as a map into $\text{ker}(M^{1/2})^{\perp}$ is a bijection, as $\text{Im}(M^{1/2})$ is dense in $\text{ker}(M^{1/2})^{\perp}$ and left invariant under $M_{00}$. Again, the open mapping theorem implies that $M_{00}$ is a homemorphism which shows that $\text{Im}(M_{00}) = \text{Im}(M)$ is closed.

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