It should be symmetric otherwise <u, v> won't be the same as <v, u>
0x484884
If we take the transpose of each side, we get v transpose times A transpose times u on the left and the same thing on the right (since its just a 1x1 matrix) and since <u,v> = <v,u> we get the same inner product when we use A transpose.
Osoii
So if we transpose u and v, u1v1 changes to v1u1, its value is not changed, same for u2v2. But u1v2 changes to v1u2, and u2v1 changes to v2u1. To make this inner product symmetric, the coefficients of u1v2 and u2v1 must be the same (in this case, they are both 1)
It should be symmetric otherwise <u, v> won't be the same as <v, u>
If we take the transpose of each side, we get v transpose times A transpose times u on the left and the same thing on the right (since its just a 1x1 matrix) and since <u,v> = <v,u> we get the same inner product when we use A transpose.
So if we transpose u and v, u1v1 changes to v1u1, its value is not changed, same for u2v2. But u1v2 changes to v1u2, and u2v1 changes to v2u1. To make this inner product symmetric, the coefficients of u1v2 and u2v1 must be the same (in this case, they are both 1)
In 2D, with a matrix [[ A B ][ C D ]] (row major)
u·v = Au1v1 + Bu1v2 + Cu2v1 + Du2v2
v·u = Au1v1 + Bu2v1 + Cu1v2 + Du2v2.
Since u·v = v·u and A, B, C, D are constants in terms of u and v, we have
Bu1v2 = Cu1v2
Cu2v1 = Bu2v1
so B = C
The inner product matrix is symmetric to maintain the requirement that <u, v> = <v, u>
Wow I never would have thought to represent dot and cross products with matrices instead of just vectors.