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Spanning Sets

  1. Sep 21, 2005 #1
    Okay, so im doing this homework question, and its bothering me, so i thought perphaps somebody can help me out.

    " Let P4 denote the vector space of all polynomials with degree less than or equal to 4 and real coefficients. Describe percisely as you can the linear span of set {x^2 – x^4, 1 + x^2, 2x^4 – x^2 – 5} "

    To break it down in simple english, my prof said that you must subtract each part from each other until you get the span as simple as possible.

    Example.. ( x^2 – x^4 ) - ( 1 + x^2 ) = ( -1 + -x^4 )

    Now I don't understand WHY do I do that, and how it still keeps it as a Spanning Set? Could somebody please explain that.

    So anyway, I did what the prof said, and simplified the spanning set.

    I got.... P4 = span{ 1 + x^4, 1 + x^2, 6}

    Now the second part that I don't understand is, HOW do i explain what this linear span means?

    Can somebody please help.

  2. jcsd
  3. Sep 22, 2005 #2


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    The linear span means the set of all things that you can get by taking sums of (real) multiples. For example, if your basis set is {A,B,C}, then then linear span includes things like:


    Now I think your answer is good, but there is one change I would make. You've listed "6" as an element of the basis, but it would be a bit prettier to use "1" instead. The reason is that anything that is a multiple of 6 is also a multiple of 1.

    This all gets back to linear algebra and matrices.

    When you're working with complex numbers you might think about all the sums of complex multiples of the basis set.

  4. Sep 22, 2005 #3
    Thanks so much Carl, that helps!

    Now the part thats still troubling me is HOW do i describe the linear span set?

    Would I say "the spanning set is the set of all polynominals with degree less than or equal to 4 etc....."

    To be more percise, the part of the question that I'm not getting is " escribe percisely as you can the linear span of set {x^2 – x^4, 1 + x^2, 2x^4 – x^2 – 5} "

    Does that mean algebraically or in english?

  5. Sep 22, 2005 #4


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    " describe percisely as you can the linear span of set {x^2 – x^4, 1 + x^2, 2x^4 – x^2 – 5} "

    That just means to tell, either algebraically or in words what polynomials can be written as a linear combination of those three. Yes, you can "simplify" the set but since the three are independent, they form a basis themselves. From just looking at the set, I see that none of those contain "x" or "x3". Therefore, none of the polynomials in the span can. Since the span is 3 dimensional, you won't get any other limitations. Looks like all polynomials of the form ax4+ bx2+ c. If you want to be really fancy you could say "all even polynomials of degree 4 or less".
  6. Sep 22, 2005 #5
    I understand eveything you are saying, thanks!

    Just one thing, how did you know that it is linearly independent? Could it be linealy dependent?
  7. Sep 22, 2005 #6

    Tom Mattson

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    If you're studying spanning sets, then surely you have seen the definition of linear independence, right? You have to set up a linear combination of the basis vectors, set it equal to zero, and then........?
  8. Sep 22, 2005 #7
    Yes i have seen it, i was just having difficulites seeing it, but i figured it out!
  9. Sep 22, 2005 #8


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    I know they are linearly independent because I checked!

    A set of vectors are linearly dependent if and only if there exist constants, not all 0, such that a1v1+ ...+ anvn= 0.

    Here, that means we need to check if it is possible that
    a(x^2 – x^4)+ b(1 + x^2)+ c(2x^4 – x^2 – 5)= 0 (the 0 function: 0 for all x) with not all of a,b,c equal to 0.
    One way to check that is by looking at specific values of x. If x= 0 (chosen because it makes x2- x4= 0), that gives b+ c= 0. If x= 1 (also chosen because it makes x2- x4= 0), that gives 2b- 4c= 0. Multiply b+ c= 0 by 2 and subtract from that: (2b-2b)-(4c-2c)= (0-0) or -2c= 0 so c= 0. Since 2b-4c= 0, 2b-0= 0 so b= 0. Now, taking x= 2, a(22- 24)+ 0(1+ 22)+ 0(2(24)- 22-5)= a(4-16)= 0 so a= 0. Since all of a,b, c must be 0, the three vectors are independent.
  10. Sep 22, 2005 #9
    for linear independence, the linear combination has to = o and also, the scalars all have to eqal zero also.... yes, i know what linear independence and dependence is.

    now to solve or show that something is linear independent or dependent... gets me sometimes.... can somebody check my work and see if i did it properly?

    Say for example, this question comes right out of the text book, " Let f and g be functions on [a,b], and assume that f(a) = 1 = g(b) and f(b) = 0 = g(a). Show that {f,g} is linearly independent"

    This is how I did it.

    { f(x), g(x) }

    the linear combination is.... w( f(x) ) + m( g(x) ) = 0

    at X = a

    w( f(a) ) + m( g(a) ) = 0
    w( 1 ) + m( 0 ) = 0
    w( 1) = 0
    w = 0

    at X = b

    w( f(b) ) + m( g(b) ) = 0
    w( 0 ) + m( 1 ) = 0
    m( 1 ) = 0
    m = 0

    Therefore, w=m=0 and therefore, it is the set { f, g } is linearly independent

    Can somebody please tell me if that sound right?
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