Purity is the boundary of the Bloch ball

Let $\mathrm{\Sigma }=\mathbf{r}\cdot \mathit{\sigma }$. From [F3] and [F4], $\mathrm{tr}({\rho }^2)=\frac{1}{4}\mathrm{tr}(I+2\mathrm{\Sigma }+{\mathrm{\Sigma }}^2)=\frac{1}{4}(2+2\mathrm{\parallel }\mathbf{r}{\mathrm{\parallel }}_2^2)=\frac{1}{2}(1+\mathrm{\parallel }\mathbf{r}{\mathrm{\parallel }}_2^2)$, because ${\mathrm{\Sigma }}^2=\mathrm{\parallel }\mathbf{r}{\mathrm{\parallel }}_2^{2}I$ by the Pauli anticommutation relation in [F4].

Let ${\lambda }_{+},{\lambda }_{-}$ be the eigenvalues from [F5]. They are nonnegative and sum to one by [F1]. Then $\mathrm{tr}({\rho }^2)={\lambda }_{+}^2+{\lambda }_{-}^2=1$ exactly when $2{\lambda }_{+}{\lambda }_{-}=0$, equivalently the eigenvalue pair is $(1,0)$. By [F2], $\rho$ has an orthonormal eigenbasis; in this case it is the orthogonal projection onto its one-dimensional $1$-eigenspace, so surjectivity in [F6] makes it the projector of a pure-state ray. Conversely, if $\rho$ is pure, [F7] gives a unit representative $u$ with $\rho =P_u$, and [F6] makes $P_u$ an orthogonal projection, so ${\rho }^2=\rho$. By [F1], $\mathrm{tr}(\rho )=1$, hence $\mathrm{tr}({\rho }^2)=1$. Thus purity is equivalent to $\mathrm{tr}({\rho }^2)=1$; step 1.1 makes this equivalent to $\mathrm{\parallel }\mathbf{r}{\mathrm{\parallel }}_2=1$.