= Solution
For $Bx_i=\lambda_ix_i$, Hermitian symmetry gives
$$
\lambda_jx_i^\dagger x_j=x_i^\dagger Bx_j=(Bx_i)^\dagger x_j=\lambda_ix_i^\dagger x_j.
$$
Distinct <eigenvalues> therefore imply $x_i^\dagger x_j=0$ for $i\ne j$. Together with the unit normalizations,
$$
\boxed{x_i^\dagger x_j=\delta_{ij}.}
$$
The $n$ mutually orthogonal nonzero vectors are linearly independent and span the $n$-dimensional space, so they form an <orthonormal basis>. Here $\delta_{ij}$ is the <Kronecker delta>.
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