Orthogonality- Gram-Schmidt Process for Complex Sequences

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SUMMARY

The discussion centers on the application of the Gram-Schmidt process to determine orthogonality in the inner product space L2 of complex sequences. The vectors x = (1,0,1,0,...), y = (1,i,0,i,...), and z = (-1,1,i,-1,...) are analyzed, revealing that they are not orthogonal since = 1. The user seeks clarification on the Gram-Schmidt process, specifically regarding the removal of the zero vector and the correct application of the orthogonalization formula. The confusion arises from misinterpretation of instructions and the handling of zero entries within the vectors.

PREREQUISITES
  • Understanding of inner product spaces, specifically L2 for complex sequences.
  • Familiarity with the Gram-Schmidt orthogonalization process.
  • Knowledge of complex numbers and their properties.
  • Ability to compute inner products and norms in vector spaces.
NEXT STEPS
  • Study the properties of inner product spaces, focusing on L2 convergence criteria.
  • Review the Gram-Schmidt process with examples involving complex vectors.
  • Learn about orthonormal sets and the conditions for orthogonality in vector spaces.
  • Practice computing inner products and norms for complex sequences to reinforce understanding.
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Mathematicians, students studying linear algebra, and anyone interested in the application of the Gram-Schmidt process to complex vector spaces.

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Homework Statement



Consider L2, the inner product space of the complex sequences x = (xn) such that \sum xi converges,
with the inner product given by
<x,y> = (sum of) xi yi(complex conjugate)

Now let
x = (1,0,1,0,1,0,0,0...)
y = (1,i,0,i,0,i,0,0,0...)
z = (-1,1,i,-1,1,i,0,0..)

(xn = yn = zn for all n>7 = 0)

a) Is the set {x,y,z} an orthogonal set in L2?

b) If not use the Gram-Schmidth orthogonalization process to get an orthogonal set with the same span

Sol.

A) well i know it can't be orthogonal because if it was there wouldn't be a part b but i can't give that as an answer so for them to be orthogonal <x,y> = 0 but i get <x,y> = 1 so they are not orthogonal(but are orthonormal) so the answer is no, the set is not an orthogonal set in L2

i think I've got that right?

b) to apply Gram-schmidth you first have to remove 0 from the list

which i did giving
x = 1,1,1,0,0,0,0...
y = 1,i,i,i,0,0,0,0,..
z = -1,1,i,-1,1,i,0,0...

and then fill into the formula, but when i do this its not working out, is there something I am missing?

Thanks a mill for reading and sorry some of the code didn't work,i hope you can understand the question.
 
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First of all...in part (a) you write that the set of vectors is orthonormal. This is not true: if a set of vectors are not orthogonal, then they are certainly not orthonormal.

Secondly, you are confusing the instuction about removing 0 from the set: it means you should remove the 0 vector if it appears in the set, NOT remove any zero entries in the vectors.
 
thanks for the reply,your right they obviously aren't ON either but i think i have shown that they are not orthogonal, right?

in second part, so it only the 0 vector i remove but leave the 0 entries in the vectors, thanks a million.
It still doesn't seem to be working out.is there something else I am missing.
im try to find 3 new sequences say u1,u2 and u3 i start with letting
x = u1
then
u2 = y- [<y,u1>/||u1||2] u1

u3 = z- [<z,u1>/||u1||2] u1 - [<z,u2>/||u2||2] u2

am i doing this right,its not working out?
Thanks for the help
 

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