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Vector Space example

  1. Jun 24, 2007 #1
    1. The problem statement, all variables and given/known data
    I have been going through some past exam papers and have come across this vector space question that I cannot find relevant examples for.

    Consider the vector space V of n-th order polynomials
    p(x) = a0 + a1x + a2x^2 +· · ·+anx^n,
    where a0,a1,a2, ...,an are real numbers, and n is a fixed positive integer.

    Show that the vector space V is closed under addition, and also under multiplication with a real scalar.
    Also, what is the dimension of V, and determine a set of basis functions for V.
    Determine an inner product for V.

    3. The attempt at a solution
    I have looked through a couple of relevant textbooks that covered the theory vaguely but did not show how a question such as this should be approached.
    Any assistance with this question would be appreciated.

    Is the dimension simply n+1?
    Also, is the basis simply an?
     
    Last edited: Jun 24, 2007
  2. jcsd
  3. Jun 24, 2007 #2

    radou

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    First of all, the "vector space" of all polynomials of degree n is no vector space. You were probably referring to the vector space of all polynomials of degree less or equal to n.

    What is the most obvious basis for this vector space?
     
  4. Jun 24, 2007 #3

    HallsofIvy

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    Also, a "basis" is a set of vectors so the answer to "Also, is the basis simply an?" is clearly NO- those are numbers, not vectors (i.e. not polynomials). A possible basis for this space is 1, x, x2, ..., xn.

    The "dimension" of a vector space is the number of vectors in every basis (although there are many bases for any vector space, they all have the same number of vectors).

    What course are you reviewing for? These are very basic questions about Linear Algebra which is generally not considered "precalcululus".
     
  5. Jun 24, 2007 #4
    So you are saying that the question itself doesn't make sense?
    I copied and pasted the question from the paper itself, so there shouldn't be any crucial words missing, however I always thought a vector space was a collection of vectors, rather then simply a polynomial which is just one vector.
    As far as the most obvious basis goes I thought an but obviously that's not correct.
    Thanks for the input nonetheless.
     
  6. Jun 24, 2007 #5

    CompuChip

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    Remember, the dimension is the minimum number of basis elements you need to express any element as a linear combination of them.

    The best way to find a basis is in this case: just write down a set of functions which can describe any element of V by taking a linear combination for them (usually, there is an obvious choice, as is the case here). Then check if it is a basis (what is the requirement for a basis?). Then check if you really need all of them (can you do with less?)
     
  7. Jun 24, 2007 #6
    Yes, I was clearly confused about this point.
     
  8. Jun 24, 2007 #7

    radou

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    As I already mentioned, they were probably referring to the vector space of all polynomials of degree less or equal to n. It is easy to convince yourself that the "vector space" of all polynomials of degree n is no vector space - take the polynmoials x^n and -x^n. What do you get when you add them together?
     
  9. Jun 24, 2007 #8
    Only the first term of the polynomial doesn't cancel, therefore a0 remains.
     
  10. Jun 24, 2007 #9

    radou

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    Think about what you're writing.

    Take p1(x) = x^n, and p2(x) = -x^n (no a's here, it really doesn't matter!). p1(x) + p2(x) = x^n + (-x^n) = 0, for every x. And the zero polynomial doesn't have degree n! Hence it is no vector space.

    Sorry if I caused confusion, but I just thought this little digression would be useful.
     
  11. Jun 24, 2007 #10

    CompuChip

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    You don't even need to do it that "complicated". Any vector space just contains the null vectors, period :)
    It follows from the basic properties that for any [tex]\lambda \in F[/tex] (with F the real or complex numbers, whatever it's a vector space over) and [tex]v \in V[/tex], [tex]\lambda v \in V[/tex]. Take [tex]\lambda = 0[/tex] => done.

    Anyway, mtb4x4mad, how is the solution coming?
     
  12. Jun 24, 2007 #11
    Thanks for your help with this, I originally decided to post this query on here for a simple answer, but it has turned into more of a discussion. As such, I haven't understood everything that has been raised here. Is there a website that simply defines what has been asked in the question?
     
  13. Jun 25, 2007 #12
    These are all just definitions. The proofs are straightforward, however, the inner product one might be a bit tricky.
     
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