 Solution

By orthonormality in the Generalized Bloch representation,A pure state has density operator $\rho =|\psi ⟩⟨\psi |$ with $⟨\psi |\psi ⟩=1$, so ${\rho }^2=\rho$ and its purity of a density operator is one. ConsequentlyEvery pure state therefore maps to the specified radius-$\sqrt{1-1/N}$ sphere in ${\mathbb{R}}^{N^2-1}$, of dimension $N^2-2$. The expansion is unique, so it distinguishes pure-state density operators, while vectors differing only by global phase describe the same point. This proves that pure states lie on the sphere, without asserting that they fill it.