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Determine whether the set is a vector space

  1. Oct 7, 2007 #1
    15. Determine wheter the set is a vector space.
    The set of all fifth-degree polynomials with the standard operations.
    1.u+v is in V
    6. cu is in V

    the back of my book says that axioms 1,4,5, and 6 fail. I dont know why 4,5,and 6 fail. Can anyone help me?
    Last edited: Oct 7, 2007
  2. jcsd
  3. Oct 7, 2007 #2
    hm..if it fails 1 and 6 shouldn't it also fail 7? o.o...Sorry I can't see a counterexample that 1/4/5/6 fails.

    say you have
    where a,b,...,l are arbitrary constants where at least 1 of a,b,...,f and g,h,...,l doesn't equal 0.


    you can write it them out and I think they do hold, or can you give 1 counter-example that proves it doesn't?
    Last edited: Oct 8, 2007
  4. Oct 8, 2007 #3

    matt grime

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    Oh boy is that too complicated. 4,5,6 simply fail because 0 is not a 5th degree polynomial, thus there is no 0 to add to u in 4, no 0 for u+(-u) to equal, and 0*u is not a 5th degree poly
  5. Oct 8, 2007 #4


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    Notice that the set of all polynomials with degree less than or equal to 5 is a vector space. But here, you must have polynomials of precisely degree 5. The "0" polynomial is not in that set.
  6. Oct 8, 2007 #5
    to matts response:
    im still getting use to this abstract way of thinking and am not sure if this is stupid or not. For axiom 4 to pass, does the zero vector have to be a 5th degree polynomial? cant the zero vector just be 0? b/c lets say (x^5+x)+(0)=(x^5+x). Whats wrong with the way im approaching this?
  7. Oct 8, 2007 #6


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    Well, first of all, you wrote the axioms in such an unprecise manner, that some of them represent nothing. For example, axiom 4 sould be something like: "There exists 0 in V, such that for every x from V, 0 + x = x + 0 = x", etc, which means that merely by reading the axioms you can get the answers to your questions.

    You can't add a polynomial and a number. Of course 0 has to be a polynomial. And since it is not a polynomial of degree 4 (or whatever you defined the space to be, I don't remember right now), it fails to be in the vector space at all.
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