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Why the completion of $F/\text{Ker}(M)$ is isometrically isomorphic to $\text{Im}(M^{1/2})$?

Let $\mathcal{B}(F)$ the algebra of all bounded linear operators on an infinite dimensionel complex Hilbert space $(F,\langle\cdot,\cdot\rangle)$. Let $M\in \mathcal{B}(F)^+$ (i.e. $\langle Mx\;, \;x\rangle\geq 0$ for all $x\in F$). Then $$\langle\cdot,\cdot\rangle_{M}:F\times F\longrightarrow\mathbb{C},\;(x,y)\longmapsto\langle Mx, y\rangle,$$ is a semi-inner product.

$\langle\cdot,\cdot\rangle_M$ induces an inner product on the quotient space $F/\text{Ker}(M)$.

I want to know why the completion of $F/\text{Ker}(M)$ is isometrically isomorphic to the Hilbert space $\text{Im}(M^{1/2})$ with the inner product $$(M^{1/2}x,M^{1/2}y)_{\text{Im}(M^{1/2})}:=\langle Px, Py\rangle,\;\forall\, x,y \in F,$$ where $P$ denotes the orthogonal projection of $F$ onto the closure of $\text{Im}(M)$.

Schüler
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