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Linear Combination Proof of Orthonormal basis

  1. Oct 11, 2015 #1

    RJLiberator

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    1. The problem statement, all variables and given/known data
    Assume that [itex](|v_1>, |v_2>, |v_3>)[/itex] is an orthonormal basis for V. Show that any vector in V which is orthogonal to [itex]|v_3>[/itex] can be expressed as a linear combination of [itex]|v_1>[/itex] and [itex]|v_2>[/itex].


    2. Relevant equations
    Orthonormality conditions:
    |v_i>*|v_j> = 0 if i≠j OR 1 if i=j.

    3. The attempt at a solution

    I dont know how to mathematically prove this obvious one.
    I understand the definitions here.

    Orthonormal basis implies that the set of three vectors lives in dimension three and are all orthogonal to one another. This means the |v_1> *|v_2> = 0.

    If some vector is orthogonal to |v_3> that means |x> |v_3> = 0.

    I am pretty sure we have to use sum notation for an inner product and orthonormality conditions to prove this statement.

    Hints on starting out properly?
    Here's the only lead I have:

    |x> = c*|v_1> + t*|v_2>
    as it is a linear combination of v_1 and v_2. But we don't want to assume the proof.
     
  2. jcsd
  3. Oct 11, 2015 #2

    Mark44

    Staff: Mentor

    It helps to think of the geometry here. Since ##v_1, v_2## and ##v_3## form a basis for V, then they span V. If ##v \in V## is orthogonal to ##v_3##, then it must lie in the two-dimensional subspace of V spanned by ##v_1## and ##v_2##; i.e., the plane that is spanned by these two vectors. At this point, I would draw a sketch, showing the plane of ##v_1## and ##v_2##, with ##v_3## sticking straight out of the plane. Recognizing that ##v_1## and ##v_2## lie in this plane, and are a basis for that two-D subspace of V, then v must be a linear combination of ##v_1## and ##v_2##. When you have a handle on the geometry, it should be easier to write a proof that fleshes out the details.
     
  4. Oct 11, 2015 #3

    HallsofIvy

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    Since v1, v2, and v3 are a basis, any vector, v, can be written as v= av1+ bv2+ cv3. Now, take the inner product of both sides with v.
     
  5. Oct 11, 2015 #4

    RJLiberator

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    Thanks guys, what you have stated I understand.

    Taking the starting point that hallsofivy has gifted me, it's clear to see that c = 0, and a and b can be any numbers. This essentially is the proof.
    HallsofIvy suggests taking the inner product of both sides with v.

    LHS: <v|v> = |v|^2
    RHS: <v|av_1> + <v|bv_2> + <v|cv_3>
    a*<v|v_1>+b*<v|v_1>+c*<v|v_3> = |v|^2

    I'm not really sure what this proves, is this not assuming the proof?
     
  6. Oct 11, 2015 #5

    Mark44

    Staff: Mentor

    What else are you given in this problem? There is information given that you aren't using.

    Your last line should be an equation showing that v is a linear combination of ##v_1## and ##v_2##.
     
  7. Oct 11, 2015 #6

    RJLiberator

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    Is it really as simple as stating that c must = 0 by the definition of orthogonal, and so low and behold, v must be a linear combination of [itex]v_1[/itex] and [itex]v_2[/itex] ?

    Why do we need the inner product of both sides?

    Can we not just jump immediately to that step? C = 0, so linear combination is seen.
     
  8. Oct 12, 2015 #7

    Mark44

    Staff: Mentor

    No, there's nothing that says that c = 0.

    I asked this before, but you didn't answer.
    It's convenient and helpful for reaching the conclusion.
    No.
     
  9. Oct 12, 2015 #8

    RJLiberator

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    Is the piece of information that you are alluding to the fact that the set is an orthonormal basis?
     
  10. Oct 12, 2015 #9

    RJLiberator

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    Okay, so after taking the inner product, it is clear to see that the inner product of <v|v_3> = 0 because of the definition of orthogonality.

    And now we have
    <v|v> = a<v|v_1>+b<v|v_2>

    Which shows that it is a linear combination of the other two vectors.
     
  11. Oct 12, 2015 #10

    HallsofIvy

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    Actually, I miswrote. I meant to say "take the dot product of each side with v1, v2, and v3 in turn".
     
  12. Oct 12, 2015 #11

    RJLiberator

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    Ah, now this is the grove I needed.

    Taking the dot product in turn shows what |v_i> * |v> is clearly due to the orthogonal and orthonormal rules.

    It can be seen that |v_3>|v> must be 0 due to definition of orthogonality and so c=0.
    |v_2>|v> = b
    |v_1>|v> = a

    and so the answer seems clear. :D

    With c = 0, all you need to do is go back to basis, plug it in and we have a linear combination representation of v.
     
  13. Oct 12, 2015 #12

    Mark44

    Staff: Mentor

    More to the point, ##v_3 \cdot v = 0##, because this is given.
    A simpler proof:
    ##v = av_1 + bv_2 + cv_3##, since ##\{v_1, v_2, v_3\}## is a basis for V
    ##\Rightarrow cv_3 = v - av_1 - bv_2##
    ##\Rightarrow cv_3 \cdot v_3 = v \cdot v_3 - av_1 \cdot v_3 - bv_2 \cdot v_3##
    ##\Rightarrow c = 0 - 0 - 0 = 0## Can you give a justification for each of the four dot products?
    Hence ##v = av_1 + bv_2##, as required.
     
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