= Solution
{c}
For any normalized vector $|u\rangle$, the <Born rule> assigns the <probability> $\langle u|\rho|u\rangle$ to the corresponding rank-one <orthogonal projection>. This must be nonnegative for every $u$, which is exactly <positive semidefiniteness>. For any complete <orthonormal basis> $\{|j\rangle\}$, the outcome <probabilities> must sum to one:
$$
\sum_j\langle j|\rho|j\rangle=\operatorname{Tr}\rho=1.
$$
Thus a <Hermitian matrix> represents a physical <density matrix> only if
$$
\boxed{\rho\geq0,\qquad\operatorname{Tr}\rho=1.}
$$
These conditions are also sufficient in finite dimension. The <spectral theorem> writes $\rho=\sum_jp_j|u_j\rangle\langle u_j|$ with $p_j\geq0$ and $\sum_jp_j=1$, describing a <quantum state ensemble> of <pure states> with valid mixing <probabilities>. A <Hermitian matrix> alone could have negative <eigenvalues>, hence negative measurement <probabilities>.
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