Lets assume we have two "spaces" each characterised by a set of orthogonal basis functions. How can one "average" the spaces. That is to say find a third space, which is also orthogonal. I have considered the following:
vector space one with two basis.
$V^1_1, V^1_2$
These are orthonormal:
$V^1_1\cdot V^1_1 = 1$
$V^1_2\cdot V^1_2 = 1$
$V^1_2\cdot V^1_1 = V^1_1\cdot V^1_2 = 0$
We also consider a second vector space
$V^2_1, V^2_2$
These are orthonormal:
$V^2_1\cdot V^2_1 = 1$
$V^2_2\cdot V^2_2 = 1$
$V^2_2\cdot V^2_1 = V^2_1V^2_2 = 0$
Let us attempt to take the average of these vectors:
$V^3_1 = {1\over2}\left(V^1_1+V^2_1\right)$
$V^3_2 = {1\over2}\left(V^1_2+V^2_2\right)$
Then we consider the dot products of this new space:
$ \begin{align} V^3_1 \cdot V^3_1 & = {1\over2}\left(V^1_1+V^2_1\right) \cdot {1\over2}\left(V^1_1+V^2_1\right) \\ & = {1\over4} \left( V^1_1 \cdot V^1_1 + V^2_1 \cdot V^2_1 + V^1_1 \cdot V^2_1 + V^2_1 \cdot V^1_1\right) \\ & = {1\over2} \left(1+\text{cos}\theta_{V^2_1,V^1_1}\right) \\ \end{align} $ By using the above orthogonalities and the cosine angle rule. Now to retain orthogonality one could weight $V^3_1$ by the square root of this. Next consider
$ \begin{align} V^3_1 \cdot V^3_2 & = {1\over2}\left(V^1_1+V^2_1\right) \cdot {1\over2}\left(V^1_2+V^2_2\right) \\ & = {1\over4} \left( V^1_1 \cdot V^1_2 + V^2_1 \cdot V^2_2 + V^1_1 \cdot V^2_2 + V^2_1 \cdot V^1_2\right) \\ & = {1\over4} \left( \text{cos}\theta_{V^2_1,V^1_2}+\text{cos}\theta_{V^1_1,V^2_2}\right) \\ \end{align} $ This is however not zero unless $\theta_{V^2_1,V^1_2} = - \theta_{V^1_1,V^2_2}$. So how can I average two orthogonal spaces and retain orthonormality?