 Solution

The Pauli matrices are Hermitian matrices with zero trace, so a real Bloch vector makes $\varrho$ Hermitian and gives $\mathrm{tr}\varrho =1$. Their Pauli multiplication identity implies $(\mathbf{s}\cdot \mathbf{\sigma }{)}^2=|\mathbf{s}{|}^{2}I$. Consequently the two eigenvalues of $\mathbf{s}\cdot \mathbf{\sigma }$ are $\pm |\mathbf{s}|$, including the zero-vector case, andA density matrix must also be positive semidefinite. The displayed eigenvalues are nonnegative exactly when $\boxed{|\mathbf{s}|\le 1}$. Thus real coordinates alone are insufficient: the allowed vectors fill the Bloch ball. Its boundary consists of pure states, with eigenvalues one and zero; its center is the maximally mixed state $I/2$.