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Linear Algebra: Linearity of inner product

  1. May 8, 2012 #1
    1. The problem statement, all variables and given/known data
    An inner product is linear in both components.

    2. Relevant equations
    <x,y> = <conjugate(y),conjugate(x)>
    <x+y,z> = <x,z> +<y,z>


    3. The attempt at a solution

    I thought it was true . It is obvious that it is linear for the first component by definition
    Attempt to show it is for second component:

    <x,y+z> = <conjugate(y+z),conjugate(x)>
    =<conjugate(y),conjugate(x)> + <conjugate(z),conjugate(x)>
    = <x,y> + <x,y>


    But the answer is false. I am having trouble understanding why it is not linear in both components. The answer key says that the second component is conjugate- linear.
     
  2. jcsd
  3. May 8, 2012 #2

    Ray Vickson

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    What is the vector space you are working in? What is the relevant definition of inner product?

    RGV
     
    Last edited by a moderator: May 9, 2012
  4. May 9, 2012 #3
    It doesn't say which space. It was just a true or false statement, but I know that the entries of the vectors can be from the complex space. And it doesn't tell me which definition of the inner product but I believe it is the standard inner product.
     
    Last edited by a moderator: May 9, 2012
  5. May 9, 2012 #4

    Ray Vickson

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    In your OP YOU said "It is obvious that it is linear for the first component by definition", implying that you had some definition in mind--otherwise, why use the word 'definition'? So, are you now saying that is not true?

    If we assume you are speaking of [itex]C^n[/itex] with [itex] <x,y> = \sum_{i=1}^n \bar{x_i} y_i, [/itex] it is definitely linear in both x and y. Your "equation"
    [tex] <\text{conjugate}(y),\text{conjugate}(x)> = <x,y>[/tex] is false, if by "conjugate" you mean "complex conjugate". It is not even true in one dimension, with [itex] <x,y> = \bar{y}x.[/itex]

    RGV
     
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