The argument I received from the lectures I attended on the same topic goes along the following lines. If $\|\cdot\|$ is a $C^*$-norm, then $\|A\|=\|A^*A\|^{1/2}$ for any $A \in \mathcal A$, and since $A^∗A$ is normal, $$ \|A\|^2 = \sup \{|λ| : λ \in σ(A^∗A) \} \,.$$ The conclusion (supposedly) follows from the fact that the right-hand side depends only on the algebraic structure of $\mathcal A.$
I am not able to make any sense out of this argument. Can anyone please explain to me what is going on? Or, perhaps provide another proof?
Let us denote by $\mathcal{A}^{-1}$ the set of the invertible elements of $\mathcal{A}$. This set is independent of the norm on $\mathcal{A}$ !!
Thus $ \sigma(A^*A)=\{\lambda: \lambda E-A^*A \notin \mathcal{A}^{-1}\}$ is independent of the norm on $\mathcal{A}$.
Hence $\sup \{|λ| : λ \in σ(A^∗A) \}$ is independent of the norm on $\mathcal{A}$.