Results 1 to 4 of 4

Math Help - [SOLVED] Problem about an inner product

  1. #1
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1

    [SOLVED] Problem about an inner product

    Let \alpha _1=(1,2) and \alpha _2 =(3,4). Let \text{( , )} be an inner product in \mathbb{C}^2 such that :
    (\alpha _1, \alpha _1)=1
    (\alpha _1, \alpha _2)=1+2i
    (\alpha _2, \alpha _2)=\frac{1}{2}
    Calculate (\alpha , \beta) \forall \alpha and \beta \in \mathbb{C}^2.
    My attempt : I tried to figure out how is definied the inner product but I didn't find. After all I'm not even sure I should start the problem this way. If I should, then it's not an obvious thing to do... I need help on this.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by arbolis View Post
    Let \alpha _1=(1,2) and \alpha _2 =(3,4). Let \text{( , )} be an inner product in \mathbb{C}^2 such that :
    (\alpha _1, \alpha _1)=1
    (\alpha _1, \alpha _2)=1+2i
    (\alpha _2, \alpha _2)=\frac{1}{2}
    Calculate (\alpha , \beta) \forall \alpha and \beta \in \mathbb{C}^2.
    My attempt : I tried to figure out how is definied the inner product but I didn't find. After all I'm not even sure I should start the problem this way. If I should, then it's not an obvious thing to do... I need help on this.
    let's show the elements of \mathbb{C}^2 by (x,y), \ x,y \in \mathbb{C}. and the inner product by <\alpha \mid \beta>, \ \alpha, \beta \in \mathbb{C}^2. for any

    a, b \in \mathbb{C} let f(x,y)=-2x+\frac{3y}{2}, \ \ g(x,y)=x-\frac{y}{2}. see that (x,y)=f(x,y)\alpha_1 + g(x,y)\alpha_2. thus for any

    x_1,y_1,x_2,y_2 \in \mathbb{C} we have:

    <(x_1,y_1) \mid (x_2,y_2)>=<f(x_1,y_1)\alpha_1+g(x_1,y_1)\alpha_2 \mid f(x_2,y_2)\alpha_1+g(x_2,y_2)\alpha_2>

    =f(x_1,y_1)\overline{f(x_2,y_2)}<\alpha_1 \mid \alpha_1>+f(x_1,y_1)\overline{g(x_2,y_2)}<\alpha_1 \mid \alpha_2> +

    g(x_1,y_1)\overline{f(x_2,y_2)} \ \overline{<\alpha_1 \mid \alpha_2>}+g(x_1,y_1)\overline{g(x_2,y_2)}<\alpha_  2 \mid \alpha_2>,

    where "overline" means the complex conjugate. you should be able to finish the solution very easily now.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1
    Thanks a lot for your reply!
    I must say it's hard for me to understand what you've done in some points.
    You said
    for any a,b \in \mathbb{C}
    but since then you don't use a and b. (but maybe I'll have to use them to finish the problem, so it's not a big deal).

    But I have no idea of
    see that (x,y)=f(x,y)\alpha_1 + g(x,y)\alpha_2
    . I've tried to see it writing it in my sheet but I didn't success to understand.
    We know that f(x,y)=-2x+\frac{3y}{2} and that \alpha _1=(1,2), so how do you multiply them? I then thought I'd have to substitute x for the first coordonate of \alpha_1 and y for the second one... but this makes no sense.
    So when I'll understand that part I'll try to go further and finish it alone.
    By the way I didn't see Calculus III (multivariable calculus) but I guess it doesn't matter.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1
    Today I could ask a helper at university... he told me a hint : \alpha _1 and \alpha _2 is a basis of \mathbb{C}^2.
    So I could solve the problem. (Well I think I did it correctly!)
    I wrote that any \alpha can be writen in the form x\alpha _1 +y \alpha _2 and wrote something similar for \beta. I ended with a not so beautiful expression (I had to deal with properties of inner product) as answer... but they don't ask it to be nice.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Inner Product
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: May 2nd 2010, 06:04 PM
  2. Replies: 3
    Last Post: April 4th 2010, 04:16 AM
  3. [SOLVED] Limit of a product...
    Posted in the Advanced Statistics Forum
    Replies: 5
    Last Post: April 20th 2009, 11:41 PM
  4. [SOLVED] Inner product question
    Posted in the Advanced Math Topics Forum
    Replies: 4
    Last Post: June 13th 2008, 12:29 PM
  5. [SOLVED] Semi-direct product
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: May 23rd 2008, 05:00 AM

Search Tags


/mathhelpforum @mathhelpforum