- #1

- 201

- 2

- Thread starter matqkks
- Start date

- #1

- 201

- 2

- #2

chiro

Science Advisor

- 4,790

- 132

It's easier to think of the inner product in an abstract space as a projection and the "weighting" or coeffecient that an object has with respect to the thing you are projecting on.

The concept of an angle is to help you make the leap from the geometric idea of relating one vector to another (through the inner product) with the "angle" between them to going to a generalized way of having a coeffecient that simply relates two vectors together with respect to its orientation.

If you recall your 3D geometry, you know that the inner product of two vectors is zero when they are independent, positive when they are relatively in the same direction and negative when they are not.

If you have a positive value it means the projection has a positive directional component with respect to the other vector and if negative it means the opposite. A zero value means that the basis vector contributes absolutely nothing to the other vector and vice-versa.

It's the idea of contribution between the two vectors and the relation of that contribution (the sign and the weighting) that is important.

So now to answer your question.

What you are basically doing in this analysis is you are taking some vector, your decomposing it by finding its projections with respect to a set of orthonormal basis vectors (which are polynomials themselves) and then just like the geometric examples, you re-construct the polynomial using a linear combination of the basis vectors (which in this case are polynomials not direction vectors in some geometric R^n space) by weighting each vector by the coeffecient given by taking the inner product of the polynomial with each individual basis vector.

So keep in mind that this idea of angle is just meant to be used as a way to numerically relate any two arbitrary vectors together and when you are able to relate one vector with all basis vectors that form a basis, then you are able to relate a vector to the entire basis (i.e. the entire space).

- Last Post

- Replies
- 2

- Views
- 14K

- Last Post

- Replies
- 2

- Views
- 2K

- Last Post

- Replies
- 2

- Views
- 2K

- Last Post

- Replies
- 4

- Views
- 2K

- Last Post

- Replies
- 3

- Views
- 2K

- Last Post

- Replies
- 2

- Views
- 836

- Last Post

- Replies
- 4

- Views
- 769

- Replies
- 6

- Views
- 4K

- Replies
- 8

- Views
- 961

- Last Post

- Replies
- 5

- Views
- 3K