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Dimension of a subspace

  1. Oct 24, 2004 #1
    Hi everyone, I was hoping someone could help me with something. Could someone explain to me exactly what this expression means:

    H = {(a+b) + (a - 2b)t + bt^2 | a E R, b E R}

    the purpose is to find the dimension of the given subspace, which I know how to do, I have just never seen this notation, so I'm not exactly sure what this expresssion is telling me about the subspace. since a and e are real numbers, does that mean t is not necessarily a real number? and is the first part of the expression a single vector? or is it some sort of conglomeration of multiple vectors? Thanks for your help
    Davy
     
  2. jcsd
  3. Oct 24, 2004 #2
    How on earth did you find its dimension without knowing what it is?

    It appears to be a subspace of the vector space of all polynomials with real coefficients. The notation simply tells you that H is the set of all polynomials with a particular kind of coefficients.
     
    Last edited: Oct 24, 2004
  4. Oct 24, 2004 #3
    thats what i don't understand, how is that expression telling me that it is the set of all polynomials with real coefficients. what is the definition of a, b and t?
    thanks
     
  5. Oct 24, 2004 #4

    matt grime

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    a,b, are defined as Real numbers in the definition of H. t is an indeterminate, this is assumed, apparently, and should be clear from the context of the statement in wherever you found it.

    It reads as: H is the set of all expressions of the form (a+b) + (a-2b)t^2 + bt^2 where a and b are elements of R.

    But it should be read in context
     
  6. Oct 24, 2004 #5
    So if it has 3 variables, then it has 3 dimensions?
     
  7. Oct 24, 2004 #6

    matt grime

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    No, that certainly isn't true. t isn't a variable in the sense you're using. It is an indeterminate.
     
  8. Oct 24, 2004 #7

    James R

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    The set of all 2nd degree polynomials in t consists of all polynomials of the form

    [tex]a + bt + ct^2[/tex]

    where a,b and c are arbitrary constants.

    Your H is a subset of this set, where the coefficients are limited to certain real values.
     
  9. Oct 24, 2004 #8
    Here is what I have come up with: someone please let me know if this is correct, or even close to correct.

    if H is the set of all polynomials of the form: (a+b) + (a - 2b)t + bt^2 where a and b are real numbers, then the matrix should sort look like this, but infinite:

    ...
    | b a-2b a+b |
    | b a-2b a+b |
    | b a-2b a+b |
    | b a-2b a+b |
    ...

    and (-1)((a-2b) + (a+b)) = b, therefore this is a linearly dependent set, since b is a linear combination of a-2b and a+b. (is this correct?)

    Therefore, the dimension is less than 3and since a-2b and a+b are linearly independent, the dimension would be 2

    am I even close?

    thanks for all help
     
  10. Oct 24, 2004 #9

    HallsofIvy

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    What matrix?? The problem didn't say anything about a matrix.

    [/quote]
    and (-1)((a-2b) + (a+b)) = b, therefore this is a linearly dependent set, since b is a linear combination of a-2b and a+b. (is this correct?)

    Therefore, the dimension is less than 3and since a-2b and a+b are linearly independent, the dimension would be 2

    am I even close?

    thanks for all help[/QUOTE]

    What you are saying is that the three coefficients of the polynomial are dependent. Effectively, what is happening is that the polynomials in this set all depend upon the choice of two numbers, a and b. That's why this subspace is 2 dimensional.
     
  11. Oct 24, 2004 #10
    thank you so much, thats all i needed to hear
     
  12. Oct 26, 2004 #11

    James R

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    You can write any of polynomial in H as a linear combination of two polynomials which form a basis for the subspace H.

    A general polynomial in H takes the form:

    [tex]a(1 + t + t^2) + b(1 - 2t + t^2)[/tex]

    The polynomials in brackets are the basis vectors of H. Two basis vectors are needed, so the dimension of the space H is 2.
     
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