[proofplan] We construct the polar partial isometry directly on the [Hilbert space](/page/Hilbert%20Space) $H$ by sending $|x|\xi$ to $x\xi$ and extending by continuity. The identity $\||x|\xi\|_H=\|x\xi\|_H$ makes this construction well-defined and isometric on $\overline{|x|H}$, while the operator is set to zero on $\ker |x|$. To prove that the resulting partial isometry lies in $M$, we show that it commutes with every operator in the commutant of $M$ and then use the bicommutant theorem for von Neumann algebras. Finally, we identify the initial and final projections with the support projections of $|x|$ and $xx^*$, and prove uniqueness from the defining equation and the kernel condition. [/proofplan] [step:Construct the Hilbert space partial isometry from $|x|H$ to $xH$] Set \begin{align*} a:=|x|=(x^*x)^{1/2}. \end{align*} Since $x^*x\in M$ is positive, the continuous functional calculus for positive operators in a von Neumann algebra gives $a\in M$. For every $\xi\in H$, \begin{align*} \|a\xi\|_H^2=(a\xi,a\xi)_H=(a^2\xi,\xi)_H=(x^*x\xi,\xi)_H=\|x\xi\|_H^2. \end{align*} Hence $\ker a=\ker x$. Define \begin{align*} W_0:aH\to H,\qquad a\xi\mapsto x\xi. \end{align*} This map is well-defined: if $a\xi=a\eta$, then $a(\xi-\eta)=0$, so $x(\xi-\eta)=0$, and therefore $x\xi=x\eta$. The displayed norm identity also shows \begin{align*} \|W_0(a\xi)\|_H=\|a\xi\|_H \end{align*} for every $\xi\in H$. Thus $W_0$ extends uniquely to an isometry \begin{align*} W:\overline{aH}\to \overline{xH}. \end{align*} The range of $W$ is $\overline{xH}$ because $W(a\xi)=x\xi$ for every $\xi\in H$. Now define $v\in\mathcal{L}(H)$ by letting $v=W$ on $\overline{aH}$ and $v=0$ on $(\overline{aH})^\perp$. Since $a=a^*$, we have \begin{align*} (\overline{aH})^\perp=\ker a^*=\ker a. \end{align*} Thus $v$ is an isometry on $\overline{aH}$ and vanishes on its orthogonal complement, so $v$ is a partial isometry. For every $\xi\in H$, \begin{align*} v a\xi=W(a\xi)=x\xi. \end{align*} Therefore $x=va$. Also, \begin{align*} \ker v=(\overline{aH})^\perp=\ker a=\ker x. \end{align*} [guided] First note that $a=|x|=(x^*x)^{1/2}$ belongs to $M$: the operator $x^*x$ is a positive element of $M$, and the continuous functional calculus for positive operators in a von Neumann algebra keeps $(x^*x)^{1/2}$ inside $M$. The construction starts from the observation that $a$ and $x$ have the same size on every vector. Indeed, since $a^2=x^*x$, for each $\xi\in H$ we compute \begin{align*} \|a\xi\|_H^2=(a\xi,a\xi)_H=(a^2\xi,\xi)_H=(x^*x\xi,\xi)_H=\|x\xi\|_H^2. \end{align*} This identity has two consequences. First, $a\xi=0$ if and only if $x\xi=0$, so $\ker a=\ker x$. Second, it allows us to define an isometry on the range of $a$ by sending $a\xi$ to $x\xi$. We define \begin{align*} W_0:aH\to H,\qquad a\xi\mapsto x\xi. \end{align*} The only possible issue is whether this rule depends on the choice of $\xi$. Suppose $a\xi=a\eta$. Then $a(\xi-\eta)=0$. Since $\ker a=\ker x$, we get $x(\xi-\eta)=0$, hence $x\xi=x\eta$. Therefore $W_0$ is well-defined. The same norm identity gives \begin{align*} \|W_0(a\xi)\|_H=\|x\xi\|_H=\|a\xi\|_H. \end{align*} Thus $W_0$ is an isometry on the subspace $aH$, and it extends uniquely by continuity to an isometry \begin{align*} W:\overline{aH}\to \overline{xH}. \end{align*} We now make this isometry into an operator on all of $H$. Since $a=a^*$, the orthogonal complement of $\overline{aH}$ is \begin{align*} (\overline{aH})^\perp=\ker a^*=\ker a. \end{align*} Define $v$ to equal $W$ on $\overline{aH}$ and to equal $0$ on $\ker a$. This makes $v$ a partial isometry: its initial space is $\overline{aH}$ and its final space is $\overline{xH}$. Finally, for every $\xi\in H$, \begin{align*} v a\xi=W(a\xi)=x\xi. \end{align*} Hence $x=va$. Since $v$ vanishes exactly on $(\overline{aH})^\perp=\ker a$, and $\ker a=\ker x$, we also have $\ker v=\ker x$. [/guided] [/step] [step:Show the partial isometry commutes with the commutant] Define the commutant of $M$ by \begin{align*} M':=\{T\in\mathcal{L}(H):Ty=yT\text{ for every }y\in M\}. \end{align*} Let $y\in M'$. Since $x\in M$ and $x^*\in M$, the operator $y$ commutes with both $x$ and $x^*$, hence with $x^*x=a^2$. By the commutation property of the continuous functional calculus, if a bounded operator commutes with a self-adjoint operator then it commutes with every [continuous function](/page/Continuous%20Function) of that self-adjoint operator. Applying this to the function $t\mapsto t^{1/2}$ on the spectrum of $a^2$, the operator $y$ commutes with $a=(a^2)^{1/2}$. For every $\xi\in H$, \begin{align*} y v(a\xi)=y x\xi=x y\xi=v a y\xi=v y a\xi. \end{align*} Thus $yv=vy$ on $aH$, and by continuity on $\overline{aH}$. If $\eta\in(\overline{aH})^\perp=\ker a$, then \begin{align*} a y\eta=y a\eta=0. \end{align*} Hence $y\eta\in\ker a=(\overline{aH})^\perp$, and therefore \begin{align*} v y\eta=0=y v\eta. \end{align*} So $yv=vy$ on $(\overline{aH})^\perp$ as well. Since \begin{align*} H=\overline{aH}\oplus(\overline{aH})^\perp, \end{align*} we have $yv=vy$ on all of $H$. [/step] [step:Conclude that the partial isometry belongs to $M$] Define the double commutant by $M'':=(M')'$. The previous step shows that $v$ commutes with every $y\in M'$, so $v\in M''$. Since $M$ is a von Neumann algebra, the Bicommutant Theorem gives $M=M''$. Therefore $v\in M$. [/step] [step:Identify the initial and final projections] Since $v$ is the isometry $W$ on $\overline{aH}$ and vanishes on $(\overline{aH})^\perp$, the projection $v^*v$ is the [orthogonal projection](/theorems/437) onto $\overline{aH}$. For a positive operator, the support projection is the orthogonal projection onto the closure of its range, so \begin{align*} v^*v=s(a)=s(|x|). \end{align*} Similarly, $vv^*$ is the orthogonal projection onto the final space $\overline{xH}$. We identify this with $s(xx^*)$. For every $\xi\in H$, \begin{align*} \|x^*\xi\|_H^2=(xx^*\xi,\xi)_H. \end{align*} Thus $\ker(xx^*)=\ker x^*$. For any bounded operator $T\in\mathcal{L}(H)$, the Hilbert-space identity $\overline{TH}=(\ker T^*)^\perp$ follows by taking orthogonal complements of $TH$. Applying this identity to $T=x$, the support projection of $xx^*$ is the orthogonal projection onto \begin{align*} (\ker x^*)^\perp=\overline{xH}. \end{align*} Hence \begin{align*} vv^*=s(xx^*). \end{align*} [/step] [step:Prove uniqueness from the equation and the kernel condition] Let $w\in\mathcal{L}(H)$ be a partial isometry such that \begin{align*} x=w|x| \end{align*} and \begin{align*} \ker w=\ker x. \end{align*} Since $\ker x=\ker a=(\overline{aH})^\perp$, both $v$ and $w$ vanish on $(\overline{aH})^\perp$. For every $\xi\in H$, \begin{align*} w(a\xi)=x\xi=v(a\xi). \end{align*} Thus $w$ and $v$ agree on $aH$. Since both operators are bounded, they agree on $\overline{aH}$. Combining this agreement on $\overline{aH}$ with their common vanishing on $(\overline{aH})^\perp$, we obtain $w=v$ on all of $H$. Therefore $v$ is uniquely determined by $x=v|x|$ together with $\ker v=\ker x$. [/step]