Results 1 to 9 of 9

Math Help - Problem: Inner Product Space and Linear dependence

  1. #1
    Super Member
    Joined
    Apr 2009
    Posts
    677

    Problem: Inner Product Space and Linear dependence

    Let V be an inner product space
    Let u,v belong to V

    Is it true that if (u,v).(v,u) = (u,u).(v,v), then u, v are linearly dependent? If yes then why?

    Thanks,
    Aman
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Bruno J.'s Avatar
    Joined
    Jun 2009
    From
    Canada
    Posts
    1,266
    Thanks
    1
    Awards
    1
    Edit : misread question.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Aug 2009
    From
    Israel
    Posts
    976
    Do you mean <u, v> \cdot <v,u> = <u,u> \cdot <v,v>?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Apr 2009
    Posts
    677
    yes
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor Swlabr's Avatar
    Joined
    May 2009
    Posts
    1,176
    Quote Originally Posted by Defunkt View Post
    Do you mean <u, v> \cdot <v,u> = <u,u> \cdot <v,v>?
    I'm confused now - by the brackets are we meaning the vector spaces generated by u and v or the inner product operation thingy? Or neither of these things? My first thought was the latter, but your question, and the reply, makes me wonder if the latter is what was meant...
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Aug 2009
    From
    Israel
    Posts
    976
    Quote Originally Posted by Swlabr View Post
    I'm confused now - by the brackets are we meaning the vector spaces generated by u and v or the inner product operation thingy? Or neither of these things? My first thought was the latter, but your question, and the reply, makes me wonder if the latter is what was meant...
    Inner product.

    @aman: I tend to think the statement is false.

    For example, take V = R^2, x,y \in V, <x, y> \ = \ <(x_1, x_2),(y_1, y_2)> \ = \ x_1y_1 + x_2y_2 (This is indeed an inner product).

    Then, you can take x = (0,1) and y = (1, 0) which are clearly not linearly dependent, yet you still get:

    <(0,1) \ , \ (1,0)> \cdot <(1,0) \ , \ (0,1)> = (0,0) = <(0,1) \ , \ (0,1)> \cdot <(1,0) \ , \ (1,0)>

    Which contradicts the statement.
    Last edited by Defunkt; August 31st 2009 at 11:25 AM.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member
    Joined
    Apr 2009
    From
    México
    Posts
    721
    Quote Originally Posted by Defunkt View Post

    <(0,1) \ , \ (1,0)> \cdot <(1,0) \ , \ (0,1)> = (0,0) = <(0,1) \ , \ (0,1)> \cdot <(1,0) \ , \ (1,0)>
    This doesn't really make sense. It should be: <(1,0),(0,1)><(0,1),(1,0)>=0 and <(0,1),(0,1)>=<(1,0),(1,0)>=1

    I think it's true: If the underlying field of V is \mathbb{C} then <u,v><v,u>= \vert <u,v> \vert ^2 (because <u,v>= \overline{<v,u>} ) and <u,u>=\Vert u \Vert ^2 (where this last means the norm induced by the scalar product) doing the same for v, by the Cauchy-Schwarz inequality we have \vert <u,v> \vert ^2 \leq \Vert u \Vert ^2 \Vert v \Vert ^2 with equality iff u,v are l.d. which proves the statement. The real case is easier
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member
    Joined
    Apr 2009
    Posts
    677
    Thanks Jose27. Yes - I get your point when you say it is a form of Cauchy-Schwarz inequality. But I would appreciate if someone can help me prove that

    "the equality holds IFF u,v are linearly dependent"

    So basically I want to start from <u, v> <v,u> = <u,u> <v,v> and prove u,v and lineraly dependent

    Thanks
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member
    Joined
    Apr 2009
    Posts
    677
    @Swlabr - It means inner product. Sorry if I am using non-std notations
    Last edited by aman_cc; September 1st 2009 at 06:19 AM. Reason: typo error
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Linear Dependence Problem
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: September 16th 2011, 11:43 AM
  2. another linear dependence problem
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: December 5th 2009, 12:20 PM
  3. trouble with inner product space problem...
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: November 18th 2009, 07:27 PM
  4. Another linear dependence problem
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: April 15th 2009, 08:03 PM
  5. Inner Product Space and Linear Transformation
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: November 2nd 2008, 02:55 AM

Search Tags


/mathhelpforum @mathhelpforum