= Solution
The four <Pauli matrices> form an orthogonal basis of $2\times2$ matrices for the <Hilbert-Schmidt inner product>, since $\operatorname{Tr}(\sigma_\mu\sigma_\nu)=2\delta_{\mu\nu}$. A Hermitian <density matrix> therefore has real coefficients in this basis, and its trace-one condition fixes the identity coefficient. Thus
$$
\boxed{\rho=\frac12(I+n_x\sigma_x+n_y\sigma_y+n_z\sigma_z),\qquad n_j=\operatorname{Tr}(\rho\sigma_j).}
$$
The multiplication rule for the <Pauli matrices> $\sigma_i\sigma_j=\delta_{ij}I+i\epsilon_{ijk}\sigma_k$ implies $(n\cdot\sigma)^2=\|n\|^2I$. Its trace is zero, so for $n\ne0$ its eigenvalues are $\pm\|n\|$; for $n=0$ it vanishes. Hence
$$
\lambda_\pm(\rho)=\frac{1\pm\|n\|}{2}.
$$
Positivity is equivalent to $\|n\|\le1$. Conversely every real vector in this unit ball defines a positive trace-one matrix by the boxed formula. Therefore $\Phi(\rho)=n$ is a one-to-one correspondence with the <Bloch ball>, with the displayed inverse. Each coordinate is linear in $\rho$, proving the affine property under any convex mixture.
A density matrix is pure precisely when its eigenvalues are $1,0$, equivalently when $\rho^2=\rho$. The eigenvalue formula makes this equivalent to $\|n\|=1$. Alternatively $\operatorname{Tr}\rho^2=(1+\|n\|^2)/2$. Thus \b[the pure states correspond exactly to the <Bloch sphere>].
For $n\ne0$, put $\widehat n=n/\|n\|$. The <spectral projectors> for the larger and smaller eigenvalues are
$$
P_+=\frac12(I+\widehat n\cdot\sigma),\qquad P_-=\frac12(I-\widehat n\cdot\sigma).
$$
They are orthogonal rank-one projections, have sum $I$, and give $\rho=\lambda_+P_++\lambda_-P_-$. For normalized eigenvectors in this nondegenerate case they equal $|\phi_1\rangle\langle\phi_1|$ and $|\phi_2\rangle\langle\phi_2|$. Their <Bloch vectors> are therefore \b[$\boxed{n_1=n/\|n\|,\quad n_2=-n/\|n\|}$], the <antipodal eigenprojectors of a qubit density matrix>.
At $n=0$, the state is $I/2$ and both eigenvalues coincide. There is no distinguished direction and the printed directional expression is undefined. Any orthonormal eigenbasis gives an antipodal pair of pure-state Bloch vectors, since its two projectors sum to $I$, but that pair can point in any direction. The nonzero-$n$ qualification is essential. The PDF denominators are $\|n\|$, not the converted TeX denominators involving $n_1,n_2$.
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