Vector weighting functions

Orthogonal but not Orthonormal Basis Sets

Suppose we have two vectors from an orthonormal system, $\overrightarrow{\mathbf{u}}$ and $\overrightarrow{\mathbf{v}}$. Taking the inner product of these vectors, we get

$\overrightarrow{\mathbf{u}}\bullet \overrightarrow{\mathbf{v}}=\sum _{k=1}^3{u}_k{\overrightarrow{\mathbf{a}}}_k\bullet \sum _{m=1}^3{v}_m{\overrightarrow{\mathbf{a}}}_m=\sum _{k=1}^3{u}_k\sum _{m=1}^3{v}_m{\overrightarrow{\mathbf{a}}}_k\bullet {\overrightarrow{\mathbf{a}}}_m=\sum _{k=1}^3{u}_k\sum _{m=1}^3{v}_m{\delta }_{k,m}=\sum _{k=1}^3{v}_k{u}_k$

What if they aren't from a normalized system, so that

${\overrightarrow{\mathbf{a}}}_k\bullet {\overrightarrow{\mathbf{a}}}_n=w_k{\delta }_{k,n}$

where the $w_k$ is the square of the length of ${\overrightarrow{\mathbf{a}}}_k$ and the symbol ${\delta }_{k,n}$ is one when k = n and zero otherwise? Well the general inner product of $\overrightarrow{\mathbf{u}}$ and $\overrightarrow{\mathbf{v}}$ becomes

$\overrightarrow{\mathbf{u}}\bullet \overrightarrow{\mathbf{v}}=\sum _{k=1}^3{u}_k{\overrightarrow{\mathbf{a}}}_k\bullet \sum _{m=1}^3{v}_m{\overrightarrow{\mathbf{a}}}_m=\sum _{k=1}^3{u}_k\sum _{m=1}^3{v}_m{\overrightarrow{\mathbf{a}}}_k\bullet {\overrightarrow{\mathbf{a}}}_m=\sum _{k=1}^3{u}_k\sum _{m=1}^3{v}_m{w}_k{\delta }_{k,m}=\sum _{k=1}^3{w}_k{v}_k{u}_k$.