= Solution
\b[<Orthogonality> means zero <inner product>, not merely that two objects look different.] For complex <vectors>, $\langle u,v\rangle=\sum_i u_i^*v_i$; for <functions> a common choice is $\langle u,v\rangle=\int w(x)u(x)^*v(x)dx$ with positive weight $w$. An <orthonormal set> has $\langle u_m,u_n\rangle=\delta_{mn}$.
For an <orthogonal> set, the component of $f$ along $u_n$ has coefficient
$$
\boxed{c_n=\frac{\langle u_n,f\rangle}{\langle u_n,u_n\rangle}.}
$$
A complete <orthonormal set> gives $f=\sum_nc_nu_n$ in the appropriate <norm> and <Parseval's identity> $\|f\|^2=\sum_n|c_n|^2$. Without <completeness> there can be a nonzero residual <orthogonal> to the entire chosen set. <Orthogonal> nonzero <vectors> are <linearly independent>, but independence alone does not imply <orthogonality>.
<Fourier coefficients> are <projections> because
$$
\int_0^T e^{-2\pi imt/T}e^{2\pi int/T}\,dt=T\delta_{mn}.
$$
Likewise, for a <self-adjoint operator>, $Au_n=\lambda_nu_n$ gives $(\lambda_n-\lambda_m)\langle u_m,u_n\rangle=0$, so distinct <eigenvalues> have <orthogonal> <eigenvectors>. Degeneracy does not itself force a chosen pair to be <orthogonal>, but an <orthonormal basis> can be selected within each <eigenspace>. The differential version uses the correct Sturm-Liouville weight and <boundary conditions>.
For mechanical <normal modes> with unequal masses the relevant <inner product> is $\langle u,v\rangle_{\mathsf M}=u^\dagger\mathsf Mv$. This makes the modes in D14 <orthogonal> even when an ordinary unweighted dot product does not. Expanding in such coordinates diagonalizes the kinetic and potential <quadratic forms>, explaining why modal motions evolve independently and their <energies> add.
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