= Solution
For $N=2$, take the normalized <Pauli operators> as the traceless basis. The matrix associated with a vector $s$ is $\rho=I/2+s\cdot\sigma$, where these basis matrices are the usual Pauli matrices divided by $\sqrt2$. Their multiplication relations give
$$
(s\cdot\sigma)^2=\frac{\|s\|^2}{2}I,
\qquad
\operatorname{spec}\rho=\left\{\frac12+\frac{\|s\|}{\sqrt2},\
\frac12-\frac{\|s\|}{\sqrt2}\right\}.
$$
At radius $1/\sqrt2$, these <eigenvalues> are one and zero. Every sphere point therefore gives a positive rank-one projector. Any other orthonormal Hermitian basis merely changes the real coordinates by an <orthogonal transformation>, so the result is basis independent.
For every $N>2$, choose a rank-one projector $P$ and consider the Hermitian trace-one matrix
$$
A=\frac2N I-P.
$$
Its <eigenvalues> are $2/N-1$ once and $2/N$ with multiplicity $N-1$. They sum to one and satisfy
$$
\operatorname{Tr}(A^2)=\left(\frac2N-1\right)^2+(N-1)\frac4{N^2}=1.
$$
By orthonormality its traceless coordinate vector has squared length $1-1/N$, so it is on the specified sphere. But $2/N-1<0$, so $A$ is not a <density operator>. Thus \b[for $N>2$, the <pure states> occupy a proper subset of the sphere]; positivity excludes some of its points. In fact, this example is the antipodal <Bloch vector> to that of $P$.
The usual assertion presumes $N\geq2$. If the trivial dimension $N=1$ is admitted, the radius-zero sphere consists of the single coordinate point, also corresponding to the single <pure state>, so that dimension is an additional trivial exception.
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