# Modification of a pure state by a unitary on a C*-algebra

Let $$A$$ be a $$C^*$$-algebra and $$\rho$$ be a pure state on $$A$$ i.e. $$\rho\in PS(A)$$. Let $$\tilde{A}$$ be the unitisation of $$A$$ and $$u$$ be a unitary element in $$\tilde{A}$$. If we define $$\rho^u:A\rightarrow \mathbb{C}$$ such that $$\rho^u(a)=\rho(uau^*)$$. I have to show that $$\rho^u$$ is also a pure state on $$A$$.

I know $$A$$ is a hereditary $$C^*$$-algebra in it's unitisation $$\tilde{A}$$, hence the map is well defined. Now for proving that it's a state I have to show somehow that $$\operatorname{lim}_{\lambda}\rho^u(u_\lambda)=1$$ where $$\{u_{\lambda}\}_{\lambda\in \Lambda}$$ is an approximate unit of $$A$$. I do not know how to proceed further.

Any form of help is highly appreciated. Thanks.

• $uu_\lambda u^*$ is another approximate unit.
– Ruy
Oct 29, 2020 at 15:42
• @Ruy Can you please give some hint on how to prove this fact? Oct 29, 2020 at 16:20
• $\|(uu_\lambda u^*)a - a\| = \|u^*(uu_\lambda u^*a - a)u\| = \|u_\lambda (u^*au) - u^*au\|.$
– Ruy
Oct 29, 2020 at 16:27