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Finding Bases of Subspaces

  1. Jan 21, 2012 #1
    1. The problem statement, all variables and given/known data

    http://img856.imageshack.us/img856/5586/screenshot20120121at328.png [Broken]

    3. The attempt at a solution
    I propose the vectors [itex]x,x^2,x^3[/itex] form a basis of V. To test for linear independence, let [itex]0 = a_1 x + a_2 x^2 + a_3 x^3[/itex], where [itex]a \in R[/itex]. A polynomial is 0 iff all of its coefficients are zero. Thus, linear independence is proved. That is, [itex]a_1 = a_2 = a_3 = 0[/itex]

    To prove that [itex]x,x^2,x^3[/itex] spans V, let [itex]p(x) = b_1 x + b_2 x^2 + b_3 x^3 \in V[/itex], where [itex]b \in R[/itex]. We need numbers [itex]c_1,c_2,c_3[/itex] such that [itex]b_1 x + b_2 x^2 + b_3 x^3 = c_1 x + c_2 x^2 + c_3 x^3[/itex]. This implies [itex]b_1 = c_1[/itex], [itex]b_2 = c_2[/itex], [itex]b_3 = c_3[/itex]. Thus, [itex]p(x) = c_1 x + c_2 x^2 + c_3 x^3[/itex] and clearly [itex]p(0) = c_1 (0) + c_2 (0) + c_3 (0) = 0[/itex]. Therefore, [itex]x,x^2,x^3[/itex] span V.

    Am I doing this right?
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jan 21, 2012 #2
    This is the second part of the question:

    http://img819.imageshack.us/img819/9069/screenshot20120121at350.png [Broken]

    I began with [itex]\int \frac{d}{dx} p(x) dx = 0x + c_1[/itex], where [itex]c_1 \in R[/itex]. Thus, I reasoned that p(x) must be of the form: [itex]c_1 + 0(x) + 0(x^2) + 0(x^3)[/itex]. A polynomial is zero iff all of its coefficients are zero. [itex]c_1[/itex] must necessarily equal zero in order for the zero vector to result. Thus, linear independence is proven. To show that [itex]c_1[/itex] spans S, let [itex]p(x) = b_1 + 0(x) + 0(x^2) + 0(x^3) \in S[/itex], where [itex]b_1 \in R[/itex]. We need [itex]p(x) = b_1 = c_1[/itex]. This implies [itex]b_1 = c_1[/itex]. So [itex]p(x) = c_1[/itex] and clearly [itex]\frac{d}{dx} p(x) =\frac{d}{dx} c_1 = 0[/itex]. Thus, [itex]c_1[/itex] spans S.

    Man, I don't think I'm doing these right....
     
    Last edited by a moderator: May 5, 2017
  4. Jan 21, 2012 #3

    vela

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    In the second part, you can't assume p(x) has the form ##b_1 x + b_2 x^2 + b_3 x^3## right off, otherwise you're assuming what you're trying to prove. When you say ##p(x) \in V##, you know that p is a polynomial and that p(0)=0. From what you know about polynomials, you should be able deduce that p(x) has the form required, from which it follows that p(x) is in the span of {x, x2, x3}.
     
    Last edited by a moderator: May 5, 2017
  5. Jan 21, 2012 #4
    Okay. Can I assume that p(x) has the form [itex]b_0 +b_1 x + b_2 x^2 + b_3 x^3[/itex]? But, clearly p(0) cannot equal zero unless [itex]b_0 = 0[/itex], which immediately gets us to the required form.

    Is that better?
     
  6. Jan 21, 2012 #5

    vela

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    Yes, that's fine. Because you know V is a subspace of P3 and presumably you know that {1, x, x2, x3} is a basis for P3, you can write p(x) in that form.
     
  7. Jan 21, 2012 #6
    This is the last problem. I need only find a basis.

    http://img861.imageshack.us/img861/7183/screenshot20120121at517.png [Broken]

    I propose that the vectors [itex]1,x^2[/itex] form a basis for W. To prove linear independence, we write, [itex]0 = a_0 + a_2 x^2[/itex] where [itex]a \in R[/itex]. A polynomial is zero iff all coefficients equal zero. To prove that [itex]1,x^2[/itex] span W let [itex]p(x) = b_0 + b_1 x + b_2 x^2 + b_3 x^3[/itex], but [itex] b_1 = b_3 = 0[/itex] otherwise condition fails. We need numbers [itex]c_0, c_2[/itex] such that [itex]b_0 +b_2 x^2 = c_0 +c_2 x^2[/itex]. This implies that [itex]b_0 = c_0[/itex] and [itex]b_2 = c_2[/itex]. Thus [itex]p(x) = c_0 + c_2 x^2[/itex] in which case [itex]\forall x[/itex] [itex]p(x) = p(-x)[/itex], as desired. Thus [itex]1,x^2[/itex] form a basis for W.
     
    Last edited by a moderator: May 5, 2017
  8. Jan 21, 2012 #7

    vela

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    You should go into more detail about why you must have b1=b3=0.

    This part is unnecessary. After you show that b1=b3=0, you know that p(x)=b0+b2x2. Since it's a linear combination of 1 and x2, it's in the span of {1, x2}.
     
    Last edited by a moderator: May 5, 2017
  9. Jan 21, 2012 #8
    Should I say something like:

    If [itex]p(x) = p(-x)[/itex] then it follows that all of the n-degree components of p(x) should be equal to their respective n-degree components of p(-x). However, clearly this can't be so since x ≠ -x (in the first degree) and (x)3 ≠ (-x)3 (in the third degree). Thus, in order to preserve the equality of the condition, it is necessary to zero these terms.

    ?
     
  10. Jan 21, 2012 #9

    vela

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    You made a slight error. You showed that p(x)∈S has the form p(x)=c1. The basis vector is 1, not c1, i.e. p(x) = c11. You want to show the vector 1 spans S, not c1 as you said.

    Since you've already shown that p(x)∈S implies p(x)=c1=c11, you're done because you've shown that p(x) can be written as a linear combination of the set of vectors {1}.
     
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  11. Jan 21, 2012 #10

    vela

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    It would be more straightforward to let ##p(x) = b_0 + b_1 x + b_2 x^2 + b_3 x^3## and calculate g(x)=p(x)-p(-x). Then show that g(x)=0 implies b1=b3=0.
     
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