Aluthge transform

Let ${\displaystyle H}$ be a Hilbert space and let ${\displaystyle B(H)}$ be the algebra of linear operators from ${\displaystyle H}$ to ${\displaystyle H}$. By the polar decomposition theorem, there exists a unique partial isometry ${\displaystyle U}$ such that ${\displaystyle T=U|T|}$ and ${\displaystyle \ker(U)\supset \ker(T)}$, where ${\displaystyle |T|}$ is the square root of the operator ${\displaystyle T^{*}T}$. If ${\displaystyle T\in B(H)}$ and ${\displaystyle T=U|T|}$ is its polar decomposition, the Aluthge transform of ${\displaystyle T}$ is the operator ${\displaystyle \Delta (T)}$ defined as:

${\displaystyle \Delta (T)=|T|^{\frac {1}{2}}U|T|^{\frac {1}{2}}.}$

More generally, for any real number ${\displaystyle \lambda \in [0,1]}$, the ${\displaystyle \lambda }$-Aluthge transformation is defined as

${\displaystyle \Delta _{\lambda }(T):=|T|^{\lambda }U|T|^{1-\lambda }\in B(H).}$

For vectors ${\displaystyle x,y\in H}$, let ${\displaystyle x\otimes y}$ denote the operator defined as

${\displaystyle \forall z\in H\quad x\otimes y(z)=\langle z,y\rangle x.}$

An elementary calculation^[2] shows that if ${\displaystyle y\neq 0}$, then ${\displaystyle \Delta _{\lambda }(x\otimes y)=\Delta (x\otimes y)={\frac {\langle x,y\rangle }{\lVert y\rVert ^{2}}}y\otimes y.}$