Let $H$ be a Hilbert space and $B(H)$ be the space of Bounded operators on $H$. Let $x\in B(H)$ be self-adjoint, that is, $x=x^*$. Now consider the polar decomposition of $x$ as $x=ub$, where $u$ is the partial isometry and $b=\sqrt{x^*x}$. I want to show that $u$ is self-adjoint unitary.
Basically, we know that $\ker (b)=\ker (u)$ and also we have $x=x^*$. So for self-adjoint of $u$ if we show that $ub(\xi)=u^*b(\xi)$, for all $\xi \in H$, then we are done, because $b=b^*$ so $\overline{\text{Image} (b^*)}=\ker(b)^\perp =\overline{\text{Image} (b)}.$ Now notice that, $$b^2=bb^*=ub^2u^* \implies b=ubu^*, ~~~~\text{ by taking square root}.$$ So this follows that $ub=bu$. Also we have $u^*b=bu^*$. Since $x=x^*$ and $x=ub$ so they all are same and this gives $ub=u^*b$ and we proved that $u$ is self adjoint. Is my proof for $u=u^*$ correct? Also I am unable to show that $u$ is unitary, that is, $u^2=1$, $1 \in B(H)$ is the identity element. Please help me to show that $u$ is unitary. Thank you for your time and help.