Linear dependence and inner product space

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  • #1
Defennder
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Homework Statement


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.

Homework 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

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?
 

Answers and Replies

  • #2
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.
 
  • #3
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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