Orthonormal Basis Homework: Gram-Schmidt Process w/ Inner Product

Click For Summary
SUMMARY

The discussion focuses on applying the Gram-Schmidt process to the basis {1, 2x, 3x²} using the inner product defined as = ∫ p(x)q(x) dx from 0 to 1. The user encountered difficulties in ensuring orthogonality between the basis vectors, specifically when calculating the scalar multiplier α for the second vector b2. The correct approach involves setting the inner product to zero, leading to the equation ∫_0^1 (2x + α) dx = 0, which provides the necessary condition for α to maintain orthogonality.

PREREQUISITES
  • Understanding of the Gram-Schmidt process
  • Familiarity with inner product spaces
  • Knowledge of definite integrals
  • Basic algebra for solving equations
NEXT STEPS
  • Study the properties of inner products in vector spaces
  • Learn about the Gram-Schmidt process in detail
  • Practice calculating definite integrals of polynomial functions
  • Explore examples of orthonormal bases in functional analysis
USEFUL FOR

Students and educators in linear algebra, mathematicians working with orthonormal bases, and anyone interested in the Gram-Schmidt process and inner product spaces.

maggie56
Messages
26
Reaction score
0

Homework Statement


Hi, i am applying the gram-schmidt procedure to a basis of {1,2x,3x^2} with inner product <p,q> = \int p(x)q(x) from 0 to 1.

i am unsure what to do with the inner product

Homework Equations





The Attempt at a Solution



I have followed the procedure i have for converting this basis to an orthonormal basis, where v1=1, v2=2x, v3=3x^2 are the initial vectors

i let b1=v1=1

so b2 = v2 + \alpha v1 = 2x + \alpha
then <b1,b2> = 0 = <1, 2x + \alpha >
which gives me \alpha = -2x but then b2=0
I get a result of 0 for b3 also, and think i must have something wrong because i haven't used the inner product <p,q> = \int p(x)q(x) from 0 to 1. but can't see what to do with this.


thanks
 
Physics news on Phys.org
ok so you want the inner product to be zero, so the vectors orthogonal
&lt;b1,b2&gt; = 0 = &lt;1, 2x + \alpha &gt; =\int_0^1 dx 2x+\alpha = (x^2 + \alpha.x)|_0^1

so this is the condition for alpha

notes alpha is a scalar multiplier of the basis vectors and cannot be a function of x.
 

Similar threads

  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 5 ·
Replies
5
Views
6K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
8K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 14 ·
Replies
14
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 1 ·
Replies
1
Views
6K