Re: Defining Inner Products

Re: Defining Inner Products

Thanks so much. I can verify this is an inner product, but not sure how the length of each basis vector would be one.

Re: Defining Inner Products

The coordinates of a basis vector are (0, ..., 0, 1, 0, ..., 0), i.e., all zeros and one 1 at the th place.

Re: Defining Inner Products

define the length as √<u,u>, right?

since in this basis each vector has coordinates (0,0,....,1,....,0) (where the 1 is in the j-th place for the j-th basis vector)

we have <ej,ej>= √(0^2 + 0^2 +...+1^2 +....+ 0^2) = √1 = 1.

technically, what we are doing is making our given basis look like the standard basis for F^n (F is whatever field you're using, the reals, rationals or complex numbers probably).

if we call our basis B, and the standard basis for R^n S, what we are doing is replacing (for the inner product) the S-coordinates of our basis with their B-coordinates.

here is an in-depth example:

suppose our basis is {(1,0,1),(0,1,0),(1,0,-1)} = {u1,u2,u3}. in the USUAL inner product for R^3,

(1,0,1).(1,0,1) = 1+0+1 = 2, so this basis isn't orthonormal with respect to the usual inner product.

but (1,0,1) = u1 = 1u1 + 0u2 + 0u3 = [1,0,0]B.

so <u1,u1> , our "new" inner product, is the dot product of the B-coordinate vectors in R^3:

<u1,u1> = (1,0,0).(1,0,0) = 1+0+0 = 1.

the ability to squeeze a scalar from two vectors in a positive-definite bilinear way depends on us assigning an array of scalars to any vector. the way we do this is, in effect, tantamount to choosing a basis.

because, for any given vector space, we can choose a basis in many different ways, we can also come up with many different inner products. you can actually go either way with this:

start with a given inner product, and adjust the basis to be orthonormal (Gramm-Schmidt), or start with a basis, and adjust the inner product to be orthonormal.

the key to this whole process, is understanding that for any field F, we can make a finite-dimensional vector space V "look like F^n" (that is we can find an isomorphism to F^n, or in numerical terms: create a change-of-basis matrix).

well, in F^n, everything is simple, because the standard basis is so "nice". so all CaptainBlack did here is make our basis behave itself through the (implicit) transformation:

uj <--->ej (here, to avoid confusion i am using uj for our "given" basis, and ej for the standard basis in F^n)

so to calculate the (new) inner product, we pull our linear combinations in B back to linear combinations in the standard basis F^n, perform the standard inner product (dot product), and then assign that value back to the new inner product:

<u,v>: send u to [u]B, v to [v]B, calculate ([u]B.[v]B), assign to <u,v>.

Re: Defining Inner Products

Thank you so much, it makes so much sense now.