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Math Help - Scalar by vector multiplication

  1. #1
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    Scalar by vector multiplication

    Show why scalar multiplicative inverse property (∀a ≠ 0) (∃ b) ab=1 has no analogue using vector-by-scalar multiplication.

    I'm not sure how to go about this. Help would be appreciated I know what the two symbols ∀ ∃ represent, but not sure what to do for this question.
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  2. #2
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    I don't know what the question is even asking. There is no notion of the "multiplicative inverse" of a vector (well... not in the sense of the vector space structure anyway).
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  3. #3
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    Quote Originally Posted by topspin1617 View Post
    I don't know what the question is even asking. There is no notion of the "multiplicative inverse" of a vector (well... not in the sense of the vector space structure anyway).
    Yup, I don't understand either. Anyone else know? Really don't know how to do this.
    Last edited by brumby_3; March 9th 2011 at 06:50 PM.
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  4. #4
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    Quote Originally Posted by brumby_3 View Post
    Show why scalar multiplicative inverse property (∀a ≠ 0) (∃ b) ab=1 has no analogue using vector-by-scalar multiplication.

    I'm not sure how to go about this. Help would be appreciated I know what the two symbols ∀ ∃ represent, but not sure what to do for this question.
    1. If you have real numbers you know that for every number a \in \mathbb{R} \setminus \{0\} exists a number \frac1a, called the reciprocal of a, such that a \cdot \frac1a = 1.

    2. If you have the scalar product of two vectors the result is not a vector but a single real number. This relation 2\ vectors \longrightarrow real\ number is not invertible.

    As a practical consequence you are not allowed to calculate

    \vec a \cdot \vec b = 1~\implies~\vec a = \dfrac1{\vec b}
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  5. #5
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    But this question was about "vector by scalar multiplication" (which I take to mean scalar multiplication), not the dot product.

    In order to have a multiplicative inverse, you must first have a multiplicative identity- but there is no vector, v, such that, for every scalar, a, av= a because the product is a vector, not a scalar. There is a scalar multiplicative identity, 1, so that 1v= 1 for every vector, v, but we cannot get that as the result of a multiplication because, again, the product is a vector, not a scalar.
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