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Math Help - Is polynomial of degree 3 linearly independant

  1. #1
    Senior Member bugatti79's Avatar
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    Is polynomial of degree 3 linearly independant

    All,

    Let V be the vector space of all polynomials of degree 3 at most. Is the set

    A=\left({x^3+2x+3, x^2+4x-1, x+5,2x}\right) linearly independent? Is A a basis for V

    First of all i want to establish the difference between 'basis' and 'its dimension'. I am thinking they mean the same thing...ie, {\mathbb{R}}^3 is of dimension 3 because you need 3 linearly independant vectors to get to a point in 3D space. THis is the same as saying it has a basis involving 3 vectors.....correct me if im wrong...

    Thanks
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  2. #2
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    Re: Is polynomial of degree 3 linearly independant

    Quote Originally Posted by bugatti79 View Post
    All,

    Let V be the vector space of all polynomials of degree 3 at most. Is the set

    A=\left({x^3+2x+3, x^2+4x-1, x+5,2x}\right) linearly independent? Is A a basis for V

    First of all i want to establish the difference between 'basis' and 'its dimension'. I am thinking they mean the same thing...ie, {\mathbb{R}}^3 is of dimension 3 because you need 3 linearly independant vectors to get to a point in 3D space. THis is the same as saying it has a basis involving 3 vectors.....correct me if im wrong...

    Thanks
    You are correct that the dimention of a space tells you how many linearly independant vectors are required to span a space.

    In your example of \mathbb{R}^3 it says that the basis must have 3 vectors in it

    One basis could be

    v_1=(1,0,0) \quad v_2=(0,1,0) \quad v_3=(0,0,1) these are linearly independant and span \mathbb{R}^3

    but you could also have the basis

    v_1=(1,0,0) \quad v_2=(1,1,0) \quad v_3=(1,1,1) these are also linearly independant and span \mathbb{R}^3

    but they both still have 3 vectors.

    One way to solve the above problem is to express the polynomials in terms of the standard basis of P_{3}

    The basis is \{1, x, x^2, x^3 \} So for example the vector 2x^2-1=(-1,0,2,0). Now the vectors may look more familiar!
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  3. #3
    Senior Member bugatti79's Avatar
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    Re: Is polynomial of degree 3 linearly independant

    Quote Originally Posted by TheEmptySet View Post
    The basis is \{1, x, x^2, x^3 \}
    Before I attempt to solve this.

    Is the above basis a dimension of 4? Ie hypothetically in 4D space we need these 4 elements to define a point in 4D space?
    IF so...how did you know what to include in the basis? ie I would have left out the '1' but for no valid reason....

    Thanks
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  4. #4
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    Re: Is polynomial of degree 3 linearly independant

    one of the things a basis must be is a spanning set. note than any polynomial of degree 3 or less can be written:

    p(x) = a_0 + a_1x + a_2x^2 + a_3x^4, so \{1,x,x^2,x^3\} is a spanning set, and it should be clear

    that if p(x) = 0 for all x, that all 4 coefficients a_0,a_1,a_2,a_3 must be 0, so we have linear independence as well (the other thing a basis must be).

    (if we left out the element 1 from our basis, we couldn't have any constant polynomials).
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  5. #5
    Senior Member bugatti79's Avatar
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    Re: Is polynomial of degree 3 linearly independant

    Quote Originally Posted by TheEmptySet View Post
    You are correct that the dimention of a space tells you how many linearly independant vectors are required to span a space.

    In your example of \mathbb{R}^3 it says that the basis must have 3 vectors in it
    Where does it say the basis must have 3 vectors in it? The degree 3 is referring to the polynomial...

    The definition of a basis is that the set (x_1,x_2....x_n) is a finite basis for a vector space if

    a) it is linearly independent which we have established
    b) the set spans V ie span(x_1,x_2...x_n)

    How is b) shown mathematically?

    THanks
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  6. #6
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    Re: Is polynomial of degree 3 linearly independant

    in general, to prove that span(x1,...,xn) = V, you have to show that any v in V can be written:

    v = a1x1+...+anxn,

    that is you have to show you can find "coordinates" a1,...,an that make this true.

    in the case of R3, we can show {(1,0,0),(0,1,0),(0,0,1)} spans R3 rather easily:

    (x,y,z) = x(1,0,0) + y(0,1,0) + z(0,0,1).

    the same logic shows {1,x,x2,x3} spans P3(R), given a polynomial of degree 3 of less,

    say p(x) = a0 + a1x + a2x2 + a3x3,

    we see that p(x) is indeed a linear combination (in the span-set) of 1,x,x2 and x3

    (these play the role of the standard basis in R4, to which P3(R) is isomorphic).

    in fact, we have a "change-of-basis" linear transformation given by:

    p(x) = a0 + a1x + a2x2 + a3x3 ↔ (a0,a1,a2,a3),

    which maps "basis to basis":

    1 ↔ (1,0,0,0)
    x ↔ (0,1,0,0) (you can think of this as 0(1) + 1(x) + 0(x2) + 0(x3))
    x2 ↔ (0,0,1,0)
    x3 ↔ (0,0,0,1)
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