= Solution
{c}
For a <Hermitian matrix> $A$, the <spectral theorem> gives $A=U\operatorname{diag}(\lambda_1,\ldots,\lambda_d)U^\dagger$, with real <eigenvalues> and a <unitary matrix> $U$. Hence
$$
A^\dagger A=A^2=U\operatorname{diag}(\lambda_1^2,\ldots,\lambda_d^2)U^\dagger,
\qquad
\sqrt{A^\dagger A}=U\operatorname{diag}(|\lambda_1|,\ldots,|\lambda_d|)U^\dagger.
$$
The square root here is the unique <positive square root of an operator>. Taking the <trace>, which is unchanged by <unitary similarity>, proves
$$
\boxed{\|A\|_{\rm tr}=\sum_{j=1}^d|\lambda_j|.}
$$
For a <density matrix>, all <eigenvalues> are nonnegative and sum to one. Its <trace norm> is therefore
$$
\boxed{\|\rho\|_{\rm tr}=\operatorname{Tr}\rho=1.}
$$
The absolute values cannot be omitted for a general <Hermitian matrix>, since positive and negative <eigenvalues> can cancel in its <trace> but never in its <trace norm>.
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