Absolute value of a bounded operator

Assume AC and let $H$ be a nonzero complex Hilbert space with $T\in \mathcal{B}(H)$ (The operator norm as the least bound and as the unit-sphere or unit-ball supremum). The absolute value of $T$ is the bounded positive operator

$\mathrm{\mid }T\mathrm{\mid }:=(T^{\ast }T{)}^{1\mathrm{/}2},$

the unique bounded positive square root of $T^{\ast }T$ supplied by the positive-square-root theorem (Positive square root).

Well-definedness. $T^{\ast }T$ is self-adjoint and positive: $(T^{\ast }T{)}^{\ast }=T^{\ast }T$ by the involution rule, and $⟨T^{\ast }Tx,x⟩=⟨Tx,Tx⟩=\mathrm{\parallel }Tx{\mathrm{\parallel }}^2\ge 0$ for every $x$ by the adjoint identity (Hilbert-adjoint identities, Self-adjoint, positive, unitary and normal operators). The square root theorem therefore applies and its root is unique, so $\mathrm{\mid }T\mathrm{\mid }$ is a well-defined bounded positive operator; indeed $\mathrm{\mid }T\mathrm{\mid }\ge 0$ and $\mathrm{\mid }T{\mathrm{\mid }}^2=T^{\ast }T$. Moreover $\mathrm{\mid }T\mathrm{\mid }$ is self-adjoint: positivity makes $⟨\mathrm{\mid }T\mathrm{\mid }x,x⟩$ real for every $x$, hence $⟨(\mathrm{\mid }T\mathrm{\mid }-\mathrm{\mid }T{\mathrm{\mid }}^{\ast })x,x⟩=0$, and the four-term polarization identity applied to the sesquilinear form $(x,y)\mapsto ⟨(\mathrm{\mid }T\mathrm{\mid }-\mathrm{\mid }T{\mathrm{\mid }}^{\ast })x,y⟩$ gives $\mathrm{\mid }T\mathrm{\mid }=\mathrm{\mid }T{\mathrm{\mid }}^{\ast }$.

The two identities used later. For every $x$,

$\mathrm{\parallel }\mathrm{\mid }T\mathrm{\mid }x{\mathrm{\parallel }}^2=⟨\mathrm{\mid }T\mathrm{\mid }x,\mathrm{\mid }T\mathrm{\mid }x⟩=⟨\mathrm{\mid }T{\mathrm{\mid }}^{\ast }\mathrm{\mid }T\mathrm{\mid }x,x⟩=⟨\mathrm{\mid }T{\mathrm{\mid }}^{2}x,x⟩=⟨T^{\ast }Tx,x⟩=\mathrm{\parallel }Tx{\mathrm{\parallel }}^2,$

so $\mathrm{\parallel }\mathrm{\mid }T\mathrm{\mid }x\mathrm{\parallel }=\mathrm{\parallel }Tx\mathrm{\parallel }$; consequently $\mathrm{\mid }T\mathrm{\mid }x=0$ exactly when $Tx=0$, that is $\ker \mathrm{\mid }T\mathrm{\mid }=\ker T$, and $\mathrm{\parallel }\mathrm{\mid }T\mathrm{\mid }\mathrm{\parallel }=\mathrm{\parallel }T\mathrm{\parallel }$ because the two operators have the same unit-ball images of norms. The absolute value depends on $T$ through the self-adjoint operator $T^{\ast }T$, and the square root lies in $C^{\ast }(I,T^{\ast }T)$; no polar decomposition or Borel calculus is used in its definition.