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Inner Product

  1. Jul 17, 2009 #1
    1. The problem statement, all variables and given/known data
    Prove that if u and v are given non-zero vectors in the arbitrary inner-product space V, and are such that <u,v>=0, then {u,v} is a linearly independent subset of V.

    2. Relevant equations


    3. The attempt at a solution
    I have no idea where to start. It's hard to prove because the inner-product could be defined in various ways. I can't just pick one and go with it, since another equally valid definition may not work.

    I am still working on this and I will post my attempt as soon as I have one, but until then I was hoping to get some ideas for a starting point.
     
    Last edited: Jul 17, 2009
  2. jcsd
  3. Jul 17, 2009 #2

    lanedance

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    Hi Daniel - I would try starting with the definition of linear independence

    u, v are linearly independent if the only solution to the equation
    c1.u + c2.v = 0, for scalars c1, c2, is c1=c2=0

    then you could try writing an arbitrary vector as a linear combination of u&v, and look at the inner product with u or v
     
  4. Jul 17, 2009 #3
    Thanks for the input! It was very helpful. Here's what I have so far. I'm not sure where the inner-product comes into play, though:

    Let [tex]\pmb{u}=\left(u_1,u_2,u_3\right)[/tex] and let [tex]\pmb{v}=\left(v_1,v_2,v_3\right)[/tex].
    If {u, v} is a linearly independent subset of V, then the only solution to [tex]c_1\pmb{u}+c_2\pmb{v}=0[/tex] is when [tex]c_1=c_2=0[/tex].

    Thus, [tex]c_1\pmb{u}+c_2\pmb{v}=c_1\left(u_1,u_2,u_3\right)+c_2\left(v_1,v_2,v_3\right)=0[/tex].

    Since [tex]c_1[/tex] and [tex]c_2[/tex] are coefficients for separate terms, and since it is given that u and v are non-zero vectors, the only way the above equation could be true is if [tex]c_1=c_2=0[/tex]
    QED

    Okay, this doesn't seem like much of a proof. Any ideas?
     
    Last edited: Jul 17, 2009
  5. Jul 17, 2009 #4

    Dick

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    It's not much of a proof. You don't need to split u and v into components. If c1*u+c2*v=0, what conclusions can you draw from <u,c1*u+c2*v>=0 and <v,c1*u+c2*v>=0, remembering <u,v>=0? That's what lanedance was suggesting. Use some properties of the inner product.
     
    Last edited: Jul 17, 2009
  6. Jul 17, 2009 #5
    Okay, this proof is actually easier than I thought. Tell me what you think. (I am using a and b for the constants to make it easier to type).

    If {u, v} is a linearly independent subset of V, then the only solution to au+bv=0 is when a=b=0. We seek to prove that this is in fact the case.

    Let u, v, and w be non-zero vectors.
    The properties of inner products require that <au+bv,w>=a<u,w>+b<v,w>=0.
    Since a and b are coefficients for separate terms, we know that a=b=0.
     
  7. Jul 17, 2009 #6

    Dick

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    Gack! No, that doesn't tell me anything since w could be anything. Put w=u. Try 0=<au+bv,u>=a<u,u>+b<v,u>. <v,u>=0. So 0=a<u,u>. What do the properties of the inner product tell you about <u,u> if u is nonzero?
     
  8. Jul 17, 2009 #7
    Let's try it again, then:

    If {u, v} is a linearly independent subset of V, then the only solution to au+bv=0 is when a=b=0. We seek to prove that this is in fact the case.

    The properties of inner products require that <au+bv,u>=a<u,u>+b<v,u>=0.
    The properties of inner products also require that <u,v>=<v,u>, so <v,u>=0.
    We are left with a<u,u>=0.
    However, the properties of inner products also require that <u,u> be greater than zero for a nonzero vector u.
    Thus, we know that <u,u> is greater than zero, which means a=0.

    Now we repeat this process, replacing u with v in the first argument:

    The properties of inner products require that <au+bv,v>=a<u,v>+b<v,v>=0.
    Since <u,v>=0, we are left with b<v,v>=0.
    However, the properties of inner products also require that <v,v> be greater than zero for a nonzero vector v.
    Thus, we know that <v,v> is greater than zero, which means b=0.

    Thus, both a and b must be equal to zero.

    QED
     
    Last edited: Jul 17, 2009
  9. Jul 17, 2009 #8

    Dick

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    Yes, yes, yes, yes. That's it. Exactly.
     
  10. Jul 17, 2009 #9
    Thank you so much!
     
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