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Math Help - Proving vector space properties

  1. #1
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    Proving vector space properties

    The question:

    Prove that the following properties are true for every vector space.

    a) 2v = v + v
    b) nv = v + ... + v, where there are n terms on the right

    My attempt:

    a) Let v be an element of vector space V^n.

    2v = 2\begin{pmatrix} x_1 \\ x_2 \\ ...\\ x_n \end{pmatrix} = \begin{pmatrix} 2x_1 \\ 2x_2 \\ ...\\ 2x_n \end{pmatrix}

    v + v = \begin{pmatrix} x_1 \\ x_2 \\ ...\\ x_n \end{pmatrix} + \begin{pmatrix} x_1 \\ x_2 \\ ...\\ x_n \end{pmatrix} = \begin{pmatrix} 2x_1 \\ 2x_2 \\ ...\\ 2x_n \end{pmatrix}

    Therefore 2v = v + v

    Is this correct, or am I approaching this the wrong way? Thanks!
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  2. #2
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    Why do you have doubts about your proof?

    Can you continue with the use of induction?
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  3. #3
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    Not sure, I guess it's different to the answer in my text which is:
    2v = (1 + 1)v = 1v + 1v = v + v

    As for part b, there's no answer in the text, but I got this:

    P(n) = nv = v + ... + v
    P(1) = 1v = v (by axiom 8)
    P(n+1) = (n+1)v = nv + v (by axiom 9)

    Not sure if that's a proper proof (I've never really done induction proofs before).
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  4. #4
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    Is the above correct? Thanks.
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  5. #5
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    In your initial problem nothing is said about the vector space being of finite dimension yet when you introduce " V^n" you are assuming that. Typically "Linear Algebra" classes assume finite dimension but if that is not explicetely said, you should not assume it- and for this proof you don't need to.

    The proof your book gives is directly from the "axioms" for a vector space:
    For all scalars, a, b, and vectors, u, v, a(u+ v)= au+ av and (a+ b)v= av+ bv.
    Also, 1v= v. That is why you can write 2v= (1+ 1)v= 1v+ 1v= v+ v.

    Your induction looks good. More formally stated:

    For n= 1, we just have v= v which is true.

    Assume that, for some k, kv= v+ v+ v+ ...+ v (k times).
    Then (k+ 1)v= kv+ v= (v+ v+ ...+ v)+ v= (v+ v+ ...+ v) (k+1 times).
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