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Take any rank-one orthogonal projection $P$ and its generalized Bloch vector $s$. The antipodal vector $-s$ is on the same sphere, but its corresponding trace-one Hermitian matrix isOn the range of $P$, this matrix has eigenvalue $2/N-1$, which is negative for $N>2$. It is therefore not a positive semidefinite matrix and cannot be a density operator. Hence the pure states do not cover the sphere when $N>2$. This also shows directly why the purity bound for generalized Bloch vectors does not by itself guarantee a physical quantum state.

A density operator is Hermitian, positive semidefinite and has trace one. Part (a) therefore makes all its coordinates real, and the identity basis vector fixes the final coordinate:The remaining basis vectors are traceless by orthogonality to $I/\sqrt{N}$. The Generalized Bloch representation is consequentlyUsing orthonormality to expand the squared Hilbert-Schmidt norm gives the purity bound for generalized Bloch vectors:If $p_1,\dots ,p_N$ are its nonnegative eigenvalues, then ${\sum }_j{p}_j=1$ andIt follows thatEquality requires every pairwise product $p_j{p}_k$ to vanish, so exactly one eigenvalue equals one and all others are zero. Conversely, a rank-one projector has precisely that spectrum and attains equality. Thus equality holds exactly for a pure state. This normalization is different from the commonly used unit-radius qubit Bloch vector.