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Cross Product

  1. Jul 16, 2009 #1
    1. The problem statement, all variables and given/known data
    Which of the following functions is not an inner product in [tex]\mathbb{R}^2[/tex], where u=(a, b), v=(c,d)?

    a. <u,v> = ac+bd
    b. <u,v> = 4ad+7bc
    c. <u,v> = ac/2+bd/3
    d. <u,v> = 2ac+9bd

    2. Relevant equations

    3. The attempt at a solution
    u x v = det(u v) = ad-bc

    I don't see how ANY of these answers could be right. Any ideas?
  2. jcsd
  3. Jul 16, 2009 #2


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    Inner product, not cross product. Look up the axioms for what a function has to satisfy to be an inner product
  4. Jul 16, 2009 #3
    Oops! I forgot the definition of inner product - for some reason I was thinking of cross products.

    Okay, so <u,v> = ac+bd, which means b is the correct answer because it doesn't have this form (the others have this form but are simply multiplied by constants).

  5. Jul 16, 2009 #4


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    For your final answer you'll actually need to prove either all three parts of the definition are satisfied, or demonstrate that one of them isn't for each one

    For example

    ac-bd is not an inner product since <(1,1), (1,1)> = 0 even though it's of the 'form' ac+bd with constants
  6. Jul 16, 2009 #5
    I don't quite follow you. Where does the <(1,1), (1,1)> = 0 come from? And doesn't <(1,1), (1,1)> = 2?
  7. Jul 16, 2009 #6
    Isn't it enough just to say that 4ad+7bc is impossible because there is no way that a and b could be multiplied by d and c (respectively)?
    Last edited: Jul 16, 2009
  8. Jul 16, 2009 #7
    You have the standard inner product stuck in your head. It's true that in Euclidean-space (the usual space we're used to working with), the inner product is defined as:
    but this is just one possible definition of the inner product. It turns out that you can define lots of different inner products (in terms of the vector's components) as long as you follow three rules:
    1) Symmetry: [itex]<\mathbf{x},\mathbf{y}>=<\mathbf{y},\mathbf{x}> [/itex]
    2) Linearity: [itex]<a\mathbf{x}+b\mathbf{y},\mathbf{z}>=a<\mathbf{x},\mathbf{z}>+b<\mathbf{y},\mathbf{z}> [/itex]
    3) Positive-Definiteness: [itex]<\mathbf{x},\mathbf{x}>\geq 0, \mathbf{x}\neq 0[/itex]

    You need to check to see if those different inner products satisfy these conditions.
  9. Jul 17, 2009 #8

    Thanks for your input. It was very helpful. Tell me if my logic is correct:

    The correct answer is b since it fails the symmetry test below.

    [tex]\langle u,v\rangle = 4ad+7bc = 4da+7cb = \langle (d,c),(b,a)\rangle \neq \langle v,u\rangle[/tex]

    Is this correct?
    Last edited: Jul 17, 2009
  10. Jul 17, 2009 #9


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    You didn't actually show that symmetry fails, you just said it did. You have to find two vectors u and v, and show that <u,v> is not equal to <v,u>
  11. Jul 17, 2009 #10
    What about this. My two vectors are u=(a,b) and v=(c,d), assuming a, b, c, and d are unique real numbers.

    <u,v> = 4ad+7bc
    <v,u> = 4cb+7da

    Since 4ad+7bc is not equal to 4cb+7da, symmetry fails.

    P.S. I still think I was right in the first place. The only reason why this fails is because it doesn't follow the form (first*first)+(last*last).
  12. Jul 17, 2009 #11


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    You still haven't shown anything. How do you know 4ad+7bc=/=4cb+7da? Pick two ACTUAL vectors, with ACTUAL numbers. It doesn't need to be that complicated.

    And I demonstrated how something of the form (first*first) + (last*last) can not be an inner product: ac-bd. Then the inner product of (1,1) with itself comes out to

    1*1-1*1 = 1-1=0. So while your intuition was correct, you need a more formal proof because the way you described was, in its entirety, incorrect
  13. Jul 17, 2009 #12
    Let u=(1,2) and v=(3,4):

    [tex]<u,v>=4 (1*4)+7 (2*3)=58[/tex]

    [tex]<v,u>=4 (3*2)+7 (4*1)=52[/tex]

    [tex]58\neq 52[/tex]
  14. Jul 17, 2009 #13
    There you go! Now that looks like a compelling reason why that is not a well-defined inner product.
  15. Jul 17, 2009 #14
    Okay, I finally get it. Thanks for all your help!
  16. Jul 17, 2009 #15


    Staff: Mentor

    Your conclusion shouldn't be that [itex]58\neq 52[/itex], which is obvious, but that [itex]<u,v> \neq <v,u>[/itex]
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