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Math Help - Vector Spaces

  1. #1
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    Question Vector Spaces

    Hi everyone, I have a question,
    I hope some two can help me.
    Here tjey are

    Let V be the set of all polynomials of degree 2 with the definitions of addidtion and scalar multiplication.
    1) Show that V is not a closed under addition.
    2) Show that V is closed under scalar multiplication.

    Thanks
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  2. #2
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    If you mean the set of all polynomials strictly of degree two then consider:
    \left( {x^2  - 3x} \right) + \left( { - x^2  - 3x} \right) .
    But if the problem is about polynomials of degree at most two, then part a is false.

    Part b should be very easy. What have you done?
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  3. #3
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    Quote Originally Posted by Plato View Post
    If you mean the set of all polynomials strictly of degree two then consider:
    \left( {x^2 - 3x} \right) + \left( { - x^2 - 3x} \right) .
    But if the problem is about polynomials of degree at most two, then part a is false.

    Part b should be very easy. What have you done?
    No I need to use the Real Vector Space definition, which is a set of V of elements and have two operations (+) and (*) and have 8 properties.
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  4. #4
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    Quote Originally Posted by MarianaA View Post
    No I need to use the Real Vector Space definition, which is a set of V of elements and have two operations (+) and (*) and have 8 properties.
    Did you understand any part of my first response?
    Part a asks you to show closure. Period.
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  5. #5
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    Quote Originally Posted by Plato View Post
    Did you understand any part of my first response?
    Part a asks you to show closure. Period.
    I know is asking to show clusure but I dont know how to do it,
    thats why I am asking for help.
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  6. #6
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    1 Says to show that V is NOT closed.
    If that is the set of all polynomials strictly of degree two then
    \left( {x^2  - 3x} \right) + \left( { - x^2  - 3x} \right)= -6x
    shows that it is not closed because -6x is not of degree 2.

    However, if the problem is about polynomials of degree at most two, then it is false.
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  7. #7
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    Quote Originally Posted by Plato View Post
    1 Says to show that V is NOT closed.
    If that is the set of all polynomials strictly of degree two then
    \left( {x^2 - 3x} \right) + \left( { - x^2 - 3x} \right)= -6x
    shows that it is not closed because -6x is not of degree 2.

    However, if the problem is about polynomials of degree at most two, then it is false.
    Ok, I got it
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