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Linear dependence and inner product space

  • Thread starter Defennder
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Defennder

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1. The problem statement, all variables and given/known data
The following is from the book Linear Algebra 3rd Edn by Stephen Friedberg, et al:
pg 327 said:
Let V be an inner product space, and let S be an orthogonal set of nonzero vectors. Then S is linearly independent. Proof:

Suppose that [tex]v_1, \ ... \,v_k \in S[/tex] and [tex]\sum_{i=1}^k a_i v_i = 0[/tex]

By theorem 6.3, aj = [tex]\langle 0,v_j \rangle / ||v_j||^2 = 0[/tex] for all j. So S is linearly independent.
Here aj are scalars of field F and vj are vectors of inner product space V.

2. Relevant equations
Theorem 6.3:
Let V be an inner product space, and let S = {v1, ... , vk} be an orthogonal set of non-zero vectors. If [tex]y = \sum^k_{i=1} a_i v_i[/tex] then [tex]a_j = \langle y, v_j \rangle / ||v_j||^2[/tex] for all j
3. The attempt at a solution
Now I don't understand why theorem 6.3 implies aj = 0 = [tex]\langle 0,v_j \rangle / ||v_j||^2 [/tex] for all j. I can see how this is zero if the numerator [tex]\langle 0,v_j \rangle[/tex] = 0, but the inner product isn't even defined yet and I don't see anywhere in the axioms that the inner product of the zero vector and an orthogonal vector would always be zero. So how does theorem 6.3 apply here?
 

CompuChip

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[...] the inner product isn't even defined yet
I find that hard to believe, since I don't see how you can talk about an inner product space without having defined an inner product :)
So I guess you should go back a bit and find that definition and you will see that it immediately follows from the properties of an inner product that <v, 0> = 0 for all v. For example:
  • Let x be any vector, then <v, 0> = <v, 0*x> (see definition of vector space) = 0 * <v, x> (by linearity of the inner product) = 0 (by multiplicative properties of the reals :smile:)
  • <v, 0> = <v, v - v> (by definition of the vector space) = <v, v> + <v, -v> (by additivity of the inner product) = <v, v> - <v, v> = 0. Of course, technically, v - v = v + (-v) with the existence of -v asserted by the definition of vector space, etc.
 

Defennder

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I find that hard to believe, since I don't see how you can talk about an inner product space without having defined an inner product :)
Well I meant the specific inner product function, not the general notion of an inner product on a vector space.
  • Let x be any vector, then <v, 0> = <v, 0*x> (see definition of vector space) = 0 * <v, x> (by linearity of the inner product) = 0 (by multiplicative properties of the reals :smile:)
This is just what I need. Thanks!
 

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