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How do I determine whether a set of polynomials form a basis?

  1. Oct 4, 2011 #1
    1. The problem statement, all variables and given/known data

    Are the following statements true or false? Explain your answers carefully, giving all necessary working.

    (1) p[itex]_{1}[/itex](t) = 3 + t[itex]^{2}[/itex] and p[itex]_{2}[/itex](t) = -1 +5t +7t[itex]^{2}[/itex] form a basis for [itex]P_{2}[/itex]

    (2) p[itex]_{1}[/itex](t) = 1 + 2t + t[itex]^{2}[/itex], p[itex]_{2}[/itex](t) = -1 + t[itex]^{2}[/itex] and p[itex]_{3}[/itex](t) = 7 + 5t -6t[itex]^{2}[/itex] form a basis for [itex]P_{2}[/itex]



    3. The attempt at a solution

    So I rendered both (1) and (2) into matrices and did row reduction on them:

    [PLAIN]http://img411.imageshack.us/img411/9639/43512286.jpg [Broken]

    I believe (1) does not form a basis for [itex]P_{2}[/itex] because there is no solution even though the vectors are linearly independent. Where as (2) does have a solution and the vectors are linearly independent so therefore it should form a basis.

    Thoughts: To form a basis in [itex]P_{2}[/itex] wouldn't you need at least 3 vectors always? In my book it states that to form a basis the vectors need to be linearly independent (which is established) and also must be a spanning set, what does this exactly mean?

    It also does an example of 3 vectors just like (2): here's what the end result of their row-reduction looked like:

    [PLAIN]http://img443.imageshack.us/img443/489/17419228.jpg [Broken]

    Their comment was

    So does this mean if b3 - 2b2 + b1 = 0, then it wouldn't be a spanning set and hence not a basis? Why is this?
     
    Last edited by a moderator: May 5, 2017
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  3. Oct 4, 2011 #2

    vela

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    To be a bit more precise, (1) has a solution only when
    [tex]-3b_1 + \frac{22}{5}b_2 + b_3 = 0[/tex]so it's not that there's never a solution but that there isn't always a solution.
    Yes, that's right. P2 is a three-dimensional vector space, so any basis for it will have exactly 3 vectors.
    When you say a set of vectors {v1, v2, …, vn} spans a space V, that means if you take any element b in V, you can find some linear combination of v1, v2, …, and vn that's equal to b. In other words, you can find a solution to the equation
    [tex]c_1\vec{v}_1 + c_2\vec{v}_2 + \cdots + c_n\vec{v}_n = \vec{b}[/tex]
    In (1), what you found was that you could find solutions only some of the time, not all of the time. Consequently, even though p1(x) and p2(x) are independent, they do not span P2 and therefore do not form a basis for P2.
    No, that's not what they mean.
     
  4. Oct 4, 2011 #3
    So if [tex]-3b_1 + \frac{22}{5}b_2 + b_3 = 3[/tex] for e.g. would that mean it would be a spanning set? I'm not sure I understand what the distinction is if the equation was equal to zero as opposed to non-zero.

    You said:
    How is this so? What is the major distinguishing feature in the the equation that's going to tell me whether it's some of the time or all of the time?

    So if b had elements (x,x,x) just as an example. I'd get a system of equations to solve for c[itex]_{1}[/itex], c[itex]_{2}[/itex], c[itex]_{3}[/itex] - wouldn't you always be able to find solutions for c[itex]_{x}[/itex]? Hence any vector b would always be in the span?

    Thank you for the very detailed response by the way, it's helped in understanding this.
     
  5. Oct 4, 2011 #4

    vela

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    What's the reasoning behind forming the matrix and then row-reducing it? Can you explain that?
     
  6. Oct 4, 2011 #5
    So that we can get the values of b[itex]_{1..x}[/itex]? Effectively the scalar multipliers for the vectors in a set to determine if a given vector is in a spanning set?

    So if we got b[itex]_{1}[/itex], b[itex]_{2}[/itex], b[itex]_{3}[/itex] all equal to 3.

    Then 3v[itex]_{1}[/itex] + 3v[itex]_{2}[/itex] + 3v[itex]_{3}[/itex] = b ?
     
  7. Oct 4, 2011 #6

    vela

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    Not exactly. In (1), you're trying to find c1 and c2 such that [tex]c_1 p_1(t) + c_2 p_2(t) = f(t)[/tex]where f(t)=b1t2+b2t+b3 is an element of P2. This is the whole point of setting up the matrix and reducing it.

    Now if you plug everything in, you get
    [tex]c_1 (t^2+3) + c_2(7t^2+5t-1) = b_1t^2+b_2t+b_3[/tex]or
    [tex](c_1 + 7c_2)t^2 + (5c_2)t + (3c_1 - c_2) = b_1t^2+b_2t+b_3[/tex]Matching coefficients from the two sides of the equations, you get
    \begin{align*}
    c_1 + 7c_2 &= b_1 \\
    5c_2 &= b_2 \\
    3c_1-c_2 &= b_3
    \end{align*}To solve this system of equations, you set up the augmented matrix
    [tex]\left(\begin{array}{cc|c} 1 & 7 & b_1 \\ 0 & 5 & b_2 \\ 3 & -1 & b_3 \end{array}\right)[/tex]This is the matrix you formed. Solving this system of equations is equivalent to solving the top equation. And remember you're solving for c1 and c2.


    So now what you're doing is using what you learned before about solving systems of equations to see if you can always find a solution, or if there are no solutions, or if there are infinite solutions.

    1. If you have a basis, you should find you get a unique solution for any possible values of the b's.
    2. If you find you get no solution for some values of the b's, that means that some vectors can not be expressed as a linear combination of the vectors. In other words, the vectors don't span the space. This is what you found for problem (1).
    3. If you find you can get an infinite number of solutions, that means the vectors are linearly dependent.
     
  8. Oct 4, 2011 #7
    Ah, I'm starting to see the picture now. I guess I never really thought about where b[itex]_{1}[/itex], b[itex]_{2}[/itex], and b[itex]_{3}[/itex] came about, but now I do. Thanks so much for your help, this was very interesting.
     
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