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Math Help - Vector Spaces

  1. #1
    Jav
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    Question Vector Spaces

    Let W be the subspace of P3(all third degree polynomials) such that p(0)=0, and let U be the subspace of all polynomials such that p(1)=0. Find a basis for W, a basis for U, and a basis for their intersection W and U.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: Vector Spaces

    I suppose you mean by P_3 the vector space of all polynomials of degree \leq 3. Choose a generic polynomial p(x)=ax^3+bx^2+cx+d then, p\in W\Leftrightarrow p(0)=0\Leftrightarrow d=0 and p\in U\Leftrightarrow p(1)=0\Leftrightarrow a+b+c+d=0 . Could you continue?
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    Re: Vector Spaces

    hint #2. suppose a+b+c+d = 0. if we choose a,b and c, does that tell us what d has to be? what does that tell you about dim(U)?

    for the intersection, we have to have a+b+c = 0 and d = 0 (why?).
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  4. #4
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    Re: Vector Spaces

    To Fernando:
    I'm not sure, how to continue. Should I set this arbitrary polynomial in a matrix?
    If "d" is always zero for W, then the basis has to include only the variables x, x^2, x^3?

    As for U, I have no clue, how to set a basis. There can be a large amount of combinations of a+b+c+d that yield 0.
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    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: Vector Spaces

    Quote Originally Posted by Jav View Post
    To Fernando:
    I'm not sure, how to continue. Should I set this arbitrary polynomial in a matrix?
    If "d" is always zero for W, then the basis has to include only the variables x, x^2, x^3?

    As for U, I have no clue, how to set a basis. There can be a large amount of combinations of a+b+c+d that yield 0.
    But not every (a,b,c,d)\in\mathbb{R}^4 satisfies a+b+c+d=0
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  6. #6
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    Re: Vector Spaces

    Quote Originally Posted by Also sprach Zarathustra View Post
    But not every (a,b,c,d)\in\mathbb{R}^4 satisfies a+b+c+d=0
    Yeah, if "a" is the coefficient of the variable with the leading exponent, it can happen that by adding two vectors the result is not in the space R^4.

    -ax^3+bx^2+cx+d +
    ax^3+ex^2+gx+d= (b+e)x^2...

    How would I set up a basis then?
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  7. #7
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    Re: Vector Spaces

    here is my claim: dim(U) = 3. how can i prove this? i shall exhibit a basis. so i need to find 3 linearly independent polynomials p,q,r such that p(1) = q(1) = r(1) = 0.

    for p, i choose p(x) = x-1. for q, i choose q(x) = (x-1)^2. for r, i choose (x-1)^3. are these linearly independent?

    suppose ap + bq + cr = 0. that means: a(x-1) + b(x^2 - 2x + 1) + c(x^3 - 3x^2 + 3x - 1) is the 0-polynomial. now,

    a(x-1) + b(x^2 - 2x + 1) + c(x^3 - 3x^2 + 3x - 1) = ax - a + bx^2 - 2bx + b + cx^3 - 3cx^2 + 3cx - c

    = cx^3 + (b - 3c)x^2 + (a - 2bc + 3c)x - (a - b + c). if this is the 0-polynomial, we have to have c = 0. so

    bx^2 + (a - 2b)x - (a - b) = 0. from this, we see that b = 0, as well. so:

    ax - a = 0, which is only identically the 0-polynomial when a = 0.

    how does this relate to my hint? if a+b+c+d = 0, if we choose any 3 numbers for a,b and c, then we are forced to set d = -a - b - c.

    this suggests that picking 3 coefficients determines an element of U. the three basis vectors i chose correspond the the following choices for a,b,c,d:

    x - 1 (a = 0, b = 0, c = 1, d = -1)
    (x-1)^2 = x^2 - 2x + 1 (a = 0, b = 1, c = -2, d = 1).
    (x-1)^3 = x^3 - 3x^2 + 3x - 1 (a = 1, b = -3, c = 3, d = -1).

    i could have chosen p,q, and r quite differently. suppose i just worked from d = -a - b - c.

    one "natural" choice might be: a = 1, b = c = 0, d = -1 ---> r(x) = x^3 - 1; a = c = 0, b = 1, d = -1 --> q(x) = x^2 - 1;

    a = b = 0, c = 1, d = -1 --> p(x) = x - 1. you may verify for yourself that {x - 1, x^2 - 1, x^3 - 1} is also a basis for U.

    now for W, it is clear that {x,x^2,x^3} is a linearly independent set, and that span({x, x^2, x^3}) is clearly a subspace of W.

    suppose p(x) is in W. since p(0) = 0, p has a 0 constant term. hence p(x) = ax^3 + bx^2 + cx, which is in span({x, x^2, x^3}),

    so {x, x^2, x^3} is a linearly independent spanning set for W. what do we call a linearly independent spanning set?

    it would be nice if the bases we had exhibited so far had a nice intersection. but, oh well. to find a basis for U∩W, my advice is

    to think about how you can factor polynomials for which p(0) = 0, and p(1) = 0. another approach:

    dim(U) = dim(W) = 3. since neither U nor W is contained in the other, dim(U∩W) < 3, which leaves 0,1, or 2.

    x^2 - x is in U∩W, so dim(U∩W) > 0. can you find another element of U∩W linearly independent from x^2 - x?
    Last edited by Deveno; July 6th 2011 at 12:47 AM.
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  8. #8
    MHF Contributor FernandoRevilla's Avatar
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    Re: Vector Spaces

    Quote Originally Posted by Jav View Post
    To Fernando: I'm not sure, how to continue.
    When we have the implicit equations of a subspace, it is just routine to find a basis of the subspace, you only need to solve the system . Working in coordinates with respect to B=\{x^3,x^2,x,1\} we have

    Basis of W

    W \equiv  (a,b,c,d)=a(1,0,0,0)+b(0,1,0,0)+c(0,0,1,0)

    so, \{(1,0,0,0),(0,1,0,0),(0,0,1,0)\} are linearly independent an span W . As a consequence a basis of W is B_W=\{x^3,x^2,x\} .

    Basis of U

    U \equiv (a,b,c,d)=a(1,0,0,-1)+b(0,1,0,-1)+c(0,0,1,-1)

    so, \{(1,0,0,-1),(0,1,0,-1),(0,0,1,-1)\} are linearly independent an span U . As a consequence a basis of U is B_U=\{x^3-1,x^2-1,x-1\}

    Basis of U\cap W

    U\cap W \equiv \begin{Bmatrix} a +b+c+d=0\\d=0\end{matrix}\Leftrightarrow \ldots
    \Leftrightarrow (a,b,c,d)=b(-1,1,0,0)+c(-1,0,1,0)

    so, \{(-1,1,0,0),(-1,0,1,0)\} are linearly independent an span U\cap W . As a consequence a basis of U\cap W is B_{U\cap W}=\{-x^3+x^2,-x^2+x\}
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  9. #9
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    Re: Vector Spaces

    Quote Originally Posted by Deveno View Post
    here is my claim: dim(U) = 3. how can i prove this? ?
    Thank you. It's a good insight, but I'm not well rounded in Linear Algebra to come up with claims yet. Perhaps in the future I can come up with this on my own.
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  10. #10
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    Question Re: Vector Spaces

    Quote Originally Posted by FernandoRevilla View Post
    When we have the implicit equations of a subspace, it is just routine to find a basis of the subspace, you only need to solve the system . Working in coordinates with respect to B=\{x^3,x^2,x,1\} we have

    Basis of U

    U \equiv (a,b,c,d)=a(1,0,0,-1)+b(0,1,0,-1)+c(0,0,1,-1)

    so, \{(1,0,0,-1),(0,1,0,-1),(0,0,1,-1)\} are linearly independent an span U . As a consequence a basis of U is B_U=\{x^3-1,x^2-1,x-1\}
    I understand that U can be expressed as {(1,0,0,-1),(0,1,0,-1),(0,0,1,-1)} and this vectors are linearly independent and span the polynomial.
    I've been taught to put this vectors in a matrix and solve in row echelon or reduced row-echelon to find a basis, but I don't understand your methods to get the basis of U.
    Could you develop further?
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  11. #11
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    Re: Vector Spaces

    You have a+ b+ c+ d= 0 and can solve that for any one component- for example, d= -a- b- c. Then any vector is of the form <a, b, c, d>= <a, b, c, -a- b- c>= <a, 0, 0, -a>+ <0, b, 0, -b>+ <0, 0, c, -c>= a<1, 0, 0, -1>+ b<0, 1, 0, 1>+ c<0, 0, 1, -1>.
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  12. #12
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    Re: Vector Spaces

    Quote Originally Posted by HallsofIvy View Post
    You have a+ b+ c+ d= 0 and can solve that for any one component- for example, d= -a- b- c. Then any vector is of the form <a, b, c, d>= <a, b, c, -a- b- c>= <a, 0, 0, -a>+ <0, b, 0, -b>+ <0, 0, c, -c>= a<1, 0, 0, -1>+ b<0, 1, 0, 1>+ c<0, 0, 1, -1>.
    I understand, but why is the basis x^(3) -1, x^(2) -1, x-1 ?
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  13. #13
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    Re: Vector Spaces

    in P3, x^3 - 1 is the vector (1,0,0,-1) relative to the basis {x^3,x^2,x,1}: x^3 - 1 = (1)x^3 + (0)x^2 + (0)x + (-1)1.

    similarly x^2 - 1 is the vector (0,1,0,-1) relative to that same basis, and x - 1 is (0,0,1,-1).

    ax^3 + bx^2 + cx + d <--> (a,b,c,d).
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  14. #14
    Jav
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    Re: Vector Spaces

    Thank you Deveno and most of you. Particularly, to Fernando. You are very clear.
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    Re: Vector Spaces

    The new question has been moved here. The posts here must be related to the original problem only.


    Jav,
    Two suggestions for a healthy stay at MHF:

    1) Start a new thread for a new question.
    2) Mark the thread as [SOLVED] by clicking on 'Thread Tools' in the bar above your first post in the thread
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