Positive operators on a Hilbert space

Question

I am reading about the positive operators, more generally about the positive elements of a $C^*$-algebra which are defined to be elements of the form $a^*a$ for some $a$ in the $C^*$-alegbra. My reference is the book “$C^*$-algebras and operator theory” written by Gerard J. Murphy.

It has been proved that any positive element $a$ has a unique square root i.e. $b$ such that $b^2=a$. I understand that a positive element can have two different representations of the form $a^*a$. My question is the following.

If $X$ is a $C^*$-algebra (prototype being the space of bounded linear operators acting on a Hilbert space) with multiplicative identity $a^*a=b^*b$ Then is there any unitary ($uu^*=id=u^*u$) such that $a=ub?$

-where $id$ is the multiplicative identity of the $C^*$-algebra.

If $a=ub$ then it is always true that $a^*a=b^*b.$ For the converse, I want to say that it need not be true.

Answer 1

No, it is not true. As an example let $X=K(H)\oplus \Bbb C1$, where $K(H)$ is the compact operators on some infinite dimensional hilbert space $H$. The only unitaries in this algebra are elements of the form $e^{i\lambda}1$ for some $\lambda\in\Bbb R$.

But if $u\in U(H)$ is a unitary operator on $H$ and $a\in K$ then clearly also $b:=ua\in K$. So $a^*a=b^*b$ but unless $u$ was central you do not have any unitary $\widetilde u$ in $X$ realising $b=\widetilde ua$.

Answer 2

Look at another counter-example:
Let $H:=\ell^2(\mathbb{N})$ (the Hillbert space of all sequnces of finite sum of squares)
Then take $a,b\in B(H)$ to be $a:=S_R$ (the right-shift operaor) and $b:=id$.
Then $$a^\ast a=(S_R)^\ast S_R=S_LS_R=id=b^\ast b$$ ($S_L$ is the left-shift operator which is the conjugate of $S_R$)
There is no unitary $u$ s.t $a=ub$ since then we'll have $a=u$. But $a$ is not unitary since $aa^\ast=S_RS_L\ne id$.