= Solution
A <Unitary element of a C-star algebra> satisfies $u^*u=uu^*=1$. A <projection in a C-star algebra> satisfies $p=p^*=p^2$.
If $q=u^*u$ is a <projection in a C-star algebra>, the <C-star identity> gives $\|u(1-q)\|^2=\|(1-q)q(1-q)\|=0$. Hence $u=uq=uu^*u$. Conversely $uu^*u=u$ implies $(u^*u)^2=u^*u$, and $u^*u$ is a <Hermitian C-star element>, so it is a <projection in a C-star algebra>. Taking adjoints of the triple-product equality gives the corresponding equality for $u^*$, proving
$$
\boxed{u\text{ partial isometry}\iff uu^*u=u\iff u^*\text{ partial isometry}.}
$$
For nonzero $u$, let $p=uu^*$ and $q=u^*u$. These are projections, $u=pu=uq$, and $\|u\|=1$. The rectangular corner $pAq$ is closed because it is the range of the bounded idempotent map $a\mapsto paq$.
Define $\Theta:pAq\to qAq$ by $\Theta(a)=u^*a$. It lands in the corner, and its inverse is $b\mapsto ub$. The identities $uu^*a=pa=a$ and $u^*ub=qb=b$ verify both compositions. Both maps are contractive because $\|u\|=\|u^*\|=1$, so $\Theta$ is an <isometry>. For the proposed multiplication and <involution>,
$$
\Theta(au^*b)=\Theta(a)\Theta(b),\qquad\Theta(ua^*u)=q a^*u=a^*u=\Theta(a)^*.
$$
Thus the operations correspond exactly to the ordinary operations in the closed <C-star algebra> $qAq$. This proves associativity, <closure>, all <involution> identities and the C\*-<norm> identity at once. Moreover $\Theta(u)=q$, so \b[the unit of $A_u$ is $u$]. If $u$ is unitary, $p=q=1$, and the map $a\mapsto u^*a$ is an onto isometric \*-isomorphism from $A_u$ to $A$.
Back to article page
