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How to start this Vector Space Property Proof?

  1. Oct 10, 2015 #1

    RJLiberator

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    1. The problem statement, all variables and given/known data
    Let |v> ∈ V with |v> ≠ |0>, and let λ, μ ∈ ℂ.
    Prove that if λ|v> = μ|v>, then λ = μ

    2. Relevant equations
    Vector Space Axioms

    3. The attempt at a solution
    I am struggling to begin with this one.
    I can think of tons of different ways to begin, but all seem to get into a hazy area where I am unsure.
    I could start by adding or "subtracting" from both sides λ|v> or by adding 0 to both sides, or any other traditional proof ways.

    Is there any hint to help me go in the right direction?
     
  2. jcsd
  3. Oct 10, 2015 #2

    Ray Vickson

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    PF rules: show us your attempt first, then we are allowed to give hints.
     
  4. Oct 10, 2015 #3

    RJLiberator

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    Thanks for the reply.

    I can give you a first attempt, I have many disaster steps.
    It should be a simple proof, I am just seemingly taking the wrong first steps.

    λ|v> = μ|v> (this is what we have)
    First step is the challenging part: I don't know, let's try adding λ|v> to both sides.
    λ|v> + λ|v> = μ|v> + λ|v>
    (λ+λ)|v> = (μ+λ) |v>
    Now we see that λ+λ = μ+λ
    And so with simple subtraction we get λ = μ

    Is that a correct way of doing it? I feel like I am skidding by something here...
     
  5. Oct 10, 2015 #4

    Mark44

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    I would suggest subtracting μ|v> from both sides (or adding -μ|v> to both sides).
    This would give you λ|v> - μ|v> = |0>
    (I have to admit, I'm not very familiar with this notation. It seems to be used more by physicists than by mathematicians, AFAIK.)
     
  6. Oct 10, 2015 #5

    RJLiberator

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    If I were to subtract, the proof looks as follows:
    λ|v> = μ|v>
    λ|v> - μ|v> = μ|v> - μ|v>
    λ|v> - μ|v> = 0

    But isn't this essentially what we started with?
    How do we go from here to prove that λ = μ ?
    Do we say:

    λ|v> - μ|v> = 0
    (λ-μ)|v> = 0 and since v ≠ 0 then λ-μ must = 0, and so it follows that λ = μ.

    That seems like a correct path! Excellent.

    It's as if my proof writing abilities are increased tenfold when I write a post on Physics Forums.
     
  7. Oct 10, 2015 #6

    Ray Vickson

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    No: if you have not yet proven that ##\lambda |v \rangle = \mu |v \rangle \Longrightarrow \lambda = \mu##, then you have no reason to suppose that
    ## (\lambda + \lambda) |v \rangle = (\lambda + \mu) |v \rangle \Longrightarrow \lambda + \lambda = \lambda + \mu##. Basically, you are just assuming the result you want to prove.
     
  8. Oct 10, 2015 #7

    RJLiberator

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    @Ray Vickson , How about this last part that I concluded off of Mark's hint

    If we have (λ-μ) |v> = 0
    and since |v> ≠0, we can say (λ-μ) must equal 0, correct?

    Or is this still assuming the result?
     
  9. Oct 10, 2015 #8

    Mark44

    Staff: Mentor

    Correct. You that λ|v> - μ|v> = (λ-μ) |v> because of one of the vector space axioms that pertain to scalar multiplication.
    Assuming the result would be assuming that λ = μ.
     
  10. Oct 10, 2015 #9

    Ray Vickson

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    No, probably not, provided you have already proven that ##c |v\rangle = |0 \rangle## implies ##c = 0##, or else have assumed it as an axiom.
     
  11. Oct 10, 2015 #10

    RJLiberator

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    Ah, I see the difference.

    In my first proof we had λ+λ = λ+μ
    Where I was assuming the result.

    In this last proof we have λ-μ = 0
    Where we are not assuming anything here, we are showing what is needed.
     
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