Vector weighting functions

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Orthogonal but not orthonormal basis sets

Suppose we have two vectors from an orthonormal system,  \vec \bold u and  \vec \bold v . Taking the inner product of these vectors, we get

 \vec \bold u \bullet \vec \bold v = \sum_{k=1}^3 u_k \vec \bold a_k \bullet \sum_{m=1}^3 v_m \vec \bold a_m  = \sum_{k=1}^3 u_k \sum_{m=1}^3  v_m \vec \bold a_k \bullet  \vec \bold 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

\vec \bold a_k \bullet \vec \bold a_n = w_k \delta_{k,n}

where the  w_k is the square of the length of  \vec \bold a_k and the symbol  \delta_{k,n} is one when k = n and zero otherwise? Well the general inner product of  \vec \bold u and  \vec \bold v becomes

 \vec \bold u \bullet \vec \bold v = \sum_{k=1}^3 u_k \vec \bold a_k \bullet \sum_{m=1}^3 v_m \vec \bold a_m  = \sum_{k=1}^3 u_k \sum_{m=1}^3  v_m \vec \bold a_k \bullet  \vec \bold 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 .

You can interpret the w_k as a weighting factor between the different directions so that different directions all end up in the units you would like. For example, suppose that the x and y directions were measured in meters, and the z direction was measured in centimeters, and you would like to use meters as your base unit. You could either convert the z dimensions to meters (probably simpler) or use a weighting function  w_x = 1, w_y = 1 and  w_z = 10^{-6} . In this sense, the system could be considered orthonormal with these units and this weighting arrangement.

This idea is often extended to functions.

Orthogonal functions

Principle author of this page: Rob Frohne