Let $M$ be a **positive semidefinite** operator on a complex Hilbert space $(F,(\cdot,\cdot))$.


On the quotient space $F/\text{Ker}(M)$ we have the following inner product
$$\langle \overline{x},\overline{y}\rangle = (Mx,y),$$
for all $\overline{x},\overline{y}\in F/\text{Ker}(M)$.


According to some papers, the following theorem figures in [this book](https://rads.stackoverflow.com/amzn/click/com/B000VJD14C). However, I don't find it and I hope to get its proof.



> 
> **Theorem:** The completion of $F/\text{Ker}(M)$ denoted $\overline{F/\text{Ker}(M)}$ is isometrically isomorphic to the Hilbert space $\text{Im}(M^{1/2})$
>  with the inner product
>  $$( M^{1/2}x,M^{1/2}y)\_0=(P\_{\overline{\text{Im}(M)}}x, P\_{\overline{\text{Im}(M)}}y),\;\forall\, x,y \in F.$$
> $P\_{\overline{\text{Im}(M)}}$ is the orthogonal projection onto $\overline{\text{Im}(M)}$.
> 
> 
> 


I think that in order to prove the theorem it suffices to show that



> 
> The completion of $F/\text{Ker}(M)$ is isomorphic to $(\overline{\text{Im}(M)}, (\cdot,\cdot)\_0)$. Is this claim true?
> 
> 
> 



The completion of $F/\text{Ker}(M)$ is isomorphic to the closure of the range of $M$
I think what is going on is the following.


As Meisam Soleimani Malekan says, for any Hilbert space $H$ with inner product $(\cdot|\cdot)$, and $\newcommand{\mc}{\mathcal}T\in\mc B(H)$, we have that $\ker T^\*T=\ker T$, because $T^\*T\xi=0\implies (T^\*T\xi|\xi)=0 \implies \|T\xi\|^2=0\implies T\xi=0$. Also, $\newcommand{\im}{\operatorname{Im}}(\im T)^\perp = \ker T^\*$ by an easy calculation.


For your $F$ and $M\in\mc B(F)$ positive we hence have that $\ker M = \ker M^{1/2}$ and so on $F / \ker M$ we may define an inner product
$$ \newcommand{\ip}[2]{\langle{#1},{#2}\rangle} \ip{\overline{x}}{\overline{y}} = (Mx|y) = (M^{1/2}x|M^{1/2}y). $$
Here $\overline{x} = x+\ker M = x+\ker M^{1/2}$ is the equivalence class of $x\in F$. It is evident that $\overline{x} = \overline{x'} \implies x-x'\in\ker M^{1/2} \implies M^{1/2}x = M^{1/2}x'$, so the definition is well-defined. Notice that we have not established that $F/\ker M$ is complete for this inner-product.




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Alternatively, in any Hilbert space $H$, if $X\subseteq H$ is a closed subspace, then we can identify $H/X$ with $X^\perp$. Indeed, let $P:H\rightarrow X^\perp$ be the orthogonal projection, and define $\theta:H/X\rightarrow X^\perp; \overline{\xi}\mapsto P(\xi)$. This is well-defined, for $\overline{\xi}=\overline{\eta}\implies \xi-\eta\in X \implies P(\xi-\eta)=0$, and similarly $\theta$ is injective, as $P(\xi)=0$ exactly when $\xi\in X$. By construction $\theta$ is onto, and so $\theta$ is an isomorphism.


**However** this construction, applied to your setting, identifies $F/\ker M$ with $(\ker M)^\perp = (\im M)^{\perp\perp} = \overline{\im M}$ which is already a Hilbert space, so complete. The inner product we get on $F/\ker M$ is $(\overline x|\overline y) = (P(x)|P(y))$. Suppose we choose $x$ and $y$ already in $\overline{\im M}$. Then $(\overline x|\overline y) = (P(x)|P(y)) = (x|y)$. This **is not** the inner product you want.




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**Instead** let's define $U:\im(M^{1/2})\rightarrow F/\ker M$ as follows. Define $U(\xi) = \overline{x}$ if $\xi = M^{1/2}x$ for some $x\in F$. This is well-defined, for if also $\xi=M^{1/2}y$ then $x-y\in\ker M^{1/2}=\ker M$ so $\overline{x}=\overline{y}$. Furthermore, for $\xi=M^{1/2}x$ and $\eta=M^{1/2}y$,
$$ \ip{U(\xi)}{U(\eta)} = \ip{\overline x}{\overline y} = (Mx|y)
= (M^{1/2}x|M^{1/2}y) = (\xi|\eta). $$
Hence $U$ is an isometry. Clearly $U$ is surjective. So
$$ U^{-1} : \big( F/\ker M, \ip{\cdot}{\cdot} \big) \rightarrow
\big( \im(M^{1/2}), (\cdot|\cdot) \big) $$
is an isometric linear isomorphism. This identifies $F/\ker M$, given the inner-product you have defined, with the (in general not closed) subspace $\im(M^{1/2})$ of $F$. In particular, $F/\ker M$ might fail to be a Hilbert space.


However, if we take the *completion* of $F/\ker M$ then $U^{-1}$ extends to a unitary showing that $\overline{F/\ker M}$ is isomorphic to $\overline{\im(M^{1/2})} = \overline{\im(M)}$, the latter viewed as a subspace of $F$.




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Your theorem suggests that we give $\im(M^{1/2})$ a *different* inner-product,
$$ (M^{1/2}x|M^{1/2}y)\_0 = (Px|Py) $$
where I have written $(\cdot|\cdot)\_0$ to avoid confusion with $(\cdot|\cdot)$ which is the given inner-product on $F$. Define a *different* map $V:F/\ker M \rightarrow \im M^{1/2}$ by
$$ V(\overline x) = Mx = M^{1/2} M^{1/2} x \in \im M^{1/2}. $$
Again, this is well-defined. As $\overline{\im M} = (\ker M)^\perp = (\ker M^{1/2})^\perp = \overline{\im M^{1/2}}$, we have that $PM^{1/2}x=M^{1/2}x$, and so
$$ (Mx|My)\_0 = (M^{1/2} M^{1/2} x|M^{1/2} M^{1/2} y)\_0 = (PM^{1/2}x|PM^{1/2}y)
= (M^{1/2}x|M^{1/2}y) = \ip{\overline x}{\overline y}. $$
Thus
$$ V:\big( F/\ker M, \ip{\cdot}{\cdot} \big) \rightarrow \big( \im M^{1/2}, (\cdot|\cdot)\_0 \big) $$
is an isometry, as you want. Notice that $V$ is not (in general) onto. Finally, we can extend $V$ to the completion of $F/\ker M$, and then we will obtain a unitary transformation onto $\overline{\im M^{1/2}}$, completion with respect to $(\cdot|\cdot)\_0$.


