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Linear algebra: Determine if this set is a vector space.

  1. Feb 15, 2012 #1
    1. The problem statement, all variables and given/known data

    Determine if this set is a vector space. Either show that the necessary properties are satisfied, or give an example showing that at least one of them is not.

    -The set P of all polynomials, with the usual definition of scalar multiplication, but with addition defined as polynomial multiplication (so f + g (warped addition) is equal to the product of f and g).

    2. Relevant equations



    3. The attempt at a solution

    I know that scalar multiplication holds up, but I don't understand what to do with the polynomial multiplication part.
     
  2. jcsd
  3. Feb 15, 2012 #2
    Hopefully this gets you started. For associativity of addition:

    Let V be the set of all polynomials, and let f, g, and h be polynomials in V such that f + g = fg, f+h = fh, g+h = gh.

    Associativity of addition says that if V is a vector space, then it satisfies the property that f + (g+h) = (f+g) + h:

    f + (g + h) = f + gh = f(gh) = (fg)h = (f + g)h = (f+g) + h. Thus, V holds for the associativity property of addition.

    Try using the same procedure for the remaining 3 properties.
     
  4. Feb 15, 2012 #3

    Deveno

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    well, for one thing, you have to define the coefficient set for the polynomials (or, tell us what underlying field you are considering for your proposed vector space).

    since you don't say, my guess is you mean the field of real numbers, but this is just that: a guess.

    if you suspect one of the axioms does not hold, it suffices to give a single counter-example for that axiom.
     
  5. Feb 15, 2012 #4

    Dick

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    Put f and g both equal to the polynomial x. What is 2f + 2g in your warped addition? What should it be in a vector space?
     
  6. Feb 15, 2012 #5
    Yes, it is a vector space. Sorry about that.
     
  7. Feb 15, 2012 #6
    Sorry, I REALLY don't get vector spaces. But I'll take a guess at your question.

    Wouldn't you factor out the 2 and go from there?

    2(f+g) = 2(f*g), they don't equal each other and therefore polynomial addition doesn't hold.

    That is just a guess. So I'm thinking this would show that it's not a vector space.
     
  8. Feb 15, 2012 #7

    Dick

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    Ooops. I really picked the wrong number to scalar multiply there. Sorry. Try 3f+3g. What is 3f+3g (warped addition) and what is 3(f+g) (warped addition). They are supposed to be equal in a vector space. Just answer those and I think you'll have it. This is a bit more confusing than usual because of (warped addition) thing. It does make it hard to follow.
     
    Last edited: Feb 15, 2012
  9. Feb 15, 2012 #8
    So what you're saying is that 3f+3g does not equal 3(f*g), right?

    For example, let's say f=2 and g=5

    (3)2+(3)5 does not equal 3(2*5) ?

    Forgive me for my ignorance, this is a weird subject for me.
     
  10. Feb 15, 2012 #9

    Dick

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    It's a weird (but legitimate) question. 3f+3g (warped addition) is (3f)*(3g). 3(f+g) (warped addition) is 3(f*g). They are just substituting the '*' operation for the '+' operation. Are they equal?
     
  11. Feb 15, 2012 #10

    Dick

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    To stick with your original example, it's because (3)2*3(5) is not the same as 3(2*5).
     
  12. Feb 15, 2012 #11
    I see where you're going and since I'm so unsure, I was just trying to confirm it :-)

    Thank you again. I am so lost that even my questions don't make sense!
     
  13. Feb 15, 2012 #12
    At this point I'm hoping to get through the class by the skin of my teeth :-/
     
  14. Feb 15, 2012 #13

    Dick

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    Most questions shouldn't be this confusing. (warped addition) is odd. You probably won't see it again. Have a little more confidence.
     
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