Results 1 to 3 of 3

Thread: Mystery Product Space

  1. #1

    Mystery Product Space

    P2 has a mysterious inner product * for which {x+4, 3x-5} is an orthonormal basis. IS there enough information to tell what is the length of 1? If so, compute it. If not say why not.


    I"m confused by the length of 1 portion at the end. Do we input 1 for x, solve and get{5,-2} and then compute the length?

  2. #2
    Full Member
    Join Date
    Aug 2011
    Posts
    919
    Quote Originally Posted by h2oskidude View Post
    P2 has a mysterious inner product * for which {x+4, 3x-5} is an orthonormal basis. IS there enough information to tell what is the length of 1? If so, compute it. If not say why not.


    I"m confused by the length of 1 portion at the end. Do we input 1 for x, solve and get{5,-2} and then compute the length?
    Verify the polynomials are linearly independent (one is not a multiple of the other). Write the constant polynomial 1 as a linear combination of these and calculate <1|1>, which is the length of 1 squared. Use the fact that both of these polynomials have norm 1 and are orthogonal to each other.

  3. #3
    Senior Member
    Join Date
    Jan 2012
    Posts
    2,316
    Quote Originally Posted by h2oskidude View Post
    P2 has a mysterious inner product * for which {x+4, 3x-5} is an orthonormal basis. IS there enough information to tell what is the length of 1? If so, compute it. If not say why not.


    I"m confused by the length of 1 portion at the end. Do we input 1 for x, solve and get{5,-2} and then compute the length?
    What you have written makes no sense. "P2" is the space of quadratic polynomials which has dimension 3 and so requires three basis vectors. You must mean "P1", the set of linear (polynomial of order 1) functions. Since {x+4, 3x- 5} is a basis, any such function can be written, in a unique way, as a linear combination of them. In particular, 1= A(x+4)+ B(3x- 5) for numbers A and B. That is the same as A(x+ 4)+ B(3x- 5)= (A+ 3B)x+ (4A- 5B)= 1 for all x. So we must have A+ 3B= 0 and 4A- 5B= 1. But 5(x+ 4)- 2(3x- 5)= 5x+ 20- 6x+ 10= -x+ 30, not 1.

    IF you find the right coefficients, that is find A and B so that A(x+4)+ B(3x- 5)= 1, then you can say that
    <1, 1>= <A(x+4)+ B(3x-5), A(x+4)+ B(3x-5)>= A<x+4, A(x+4)+ B(3x- 5)>+ B<3x- 5, A(x+4)+ B(3x-5)>
    =A<x+ 4, A(x+4)>+ A<x+4, B(3x- 5)>+ B<3x-5, A(x+4)>+ B<3x- 5, B(3x- 5)>
    = A^2<x+4, x+4>+ AB<x+ 4, 3x- 5>+ BA<3x- 5, x+ 4>= B^2<3x- 5, 3x- 5>
    = A^2<x+4, x+4>+ 2AB<x+4, 3x-5>+ B^2<3x-5, 3x-5>

    And, since x+ 4 and 3x- 5 are "orthonormal", you know what each of those is.

Bookmarks

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •