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Thread: Conditions for Vector Space

  1. #1
    Senior Member AllanCuz's Avatar
    Apr 2010

    Conditions for Vector Space

    Hey guys,

    After a 3 year hiatus from taking any math courses i'm now taking an intermediate lin alg course and im a bit slow; I say this because I dont want you guys to hurt me for asking something so easy

    In my text it gives an example of,

    Let $\displaystyle S =\{ (a_1 , a_2): a_1 , a_2 \in R \} $ For $\displaystyle (a_1 , a_2), (b_1 , b_2) \in S $ and $\displaystyle c \in R $ define,

    $\displaystyle (a_1 , a_2) + (b_1 , b_2) = (a_1 + b_1, a_2 - b_2) $ and $\displaystyle c(a_1, a_2) = (ca_1, ca_2) $

    All is well, and the example goes on to say that the situation described above violates the commutatively of addition and the associativity of addition, which I agree. But it also says it violates VS8 (so it is therefore not a vector space), which is

    Quote Originally Posted by VS8
    For each pair of elements $\displaystyle a,b $ in $\displaystyle F$ and each element $\displaystyle x $ in $\displaystyle V$, $\displaystyle (a+b)x = ax + bx $
    I'm obviously missing something but I do not see how $\displaystyle c(a_1, a_2) = (ca_1, ca_2) $ violates the above condition. It actually makes sense that this is the case to me. If i have a physical vector and I multiply it by a scalar both its points should be multiplied by the scalar (I use a physical vector as an exmaple here I do know that vectors are not just physical).

    Also, the next example takes the same situation but with the definition of

    $\displaystyle c(a_1, a_2) = (ca_1, 0) $

    And this situation does not violate VS8. But i cannot see why, so clearly I am missing something with the VS8 condition!

    Thanks guys
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  2. #2
    Junior Member
    Mar 2011

    Re: Conditions for Vector Space

    In your example $\displaystyle V=S$ and $\displaystyle F=R$. So $\displaystyle x\in S$ means that $\displaystyle x=(x_1, x_2)$.

    Now, $\displaystyle (a+b)x=(a+b)(x_1, x_2)=((a+b)x_1, (a+b)x_2)=(ax_1 +bx_1, ax_2+bx_2)$

    and $\displaystyle ax+bx=a(x_1, x_2)+b(x_1, x_2)=(ax_1,ax_2)+(bx_1,bx_2)=$

    $\displaystyle =(ax_1+bx_1, ax_2-bx_2)\neq (ax_1 +bx_1, ax_2+bx_2)$.

    So, $\displaystyle (a+b)x \neq ax+bx$
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