 Solution

A Unitary element of a C-star algebra satisfies $u^{\ast }u=u{u}^{\ast }=1$. A projection in a C-star algebra satisfies $p=p^{\ast }=p^2$.

If $q=u^{\ast }u$ is a projection in a C-star algebra, the C-star identity gives $\Vert u(1-q){\Vert }^2=\Vert (1-q)q(1-q)\Vert =0$. Hence $u=uq=u{u}^{\ast }u$. Conversely $u{u}^{\ast }u=u$ implies $(u^{\ast }u{)}^2=u^{\ast }u$, and $u^{\ast }u$ 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 $u^{\ast }$, provingFor nonzero $u$, let $p=u{u}^{\ast }$ and $q=u^{\ast }u$. These are projections, $u=pu=uq$, and $\Vert u\Vert =1$. 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 $\Theta (a)=u^{\ast }a$. It lands in the corner, and its inverse is $b\mapsto ub$. The identities $u{u}^{\ast }a=pa=a$ and $u^{\ast }ub=qb=b$ verify both compositions. Both maps are contractive because $\Vert u\Vert =\Vert u^{\ast }\Vert =1$, so $\Theta$ is an isometry. For the proposed multiplication and involution,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 the unit of $A_u$ is $u$. If $u$ is unitary, $p=q=1$, and the map $a\mapsto u^{\ast }a$ is an onto isometric *-isomorphism from $A_u$ to $A$.