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Can't solve a problem about subspaces

  1. Jan 3, 2004 #1
    I can't solve a problem about subspaces. Help would be great!

    U and V are subspaces in the vector space R^4[x] given with:

    U={p(x)=a0+a1*x+a2*x^2+a3*x^3+a4*x^4; a1+a2+a3+a4=0, a1+a2+2a3+2a4=0, a0+a1=a3+a4}

    V=L{x^3-x^2+x, x^4+1}

    Find the dimensions and basis for U, U+V and U?V. Is the expresion 1-x+x^2-x^3+x^4 an element of U+V or U?V?

    Thanks for your answers.
     
  2. jcsd
  3. Jan 3, 2004 #2
    Let's first look at the conditions:

    (1) ... a_1 + a_2 + a_3 + a_4 = 0
    (2) ... a_1 + a_2 + 2a_3 + 2a_4 = 0
    (3) ... a_0 + a_1 = a_3 + a_4

    We solve for a_1 in (1):

    (4) ... a_1 = -(a_2 + a_3 + a_4)

    And put that into (2):

    a_1 + a_2 + 2a_3 + 2a_4 = 0
    <=>
    -(a_2 + a_3 + a_4) + a_2 + 2a_3 + 2a_4 = 0
    <=>
    -a_2 - a_3 - a_4 + a_2 + 2a_3 + 2a_4 = 0
    <=>
    a_3 + a_4 = 0
    <=>
    a_3 = -a_4

    Now we put that back into (4):

    a_1 = -(a_2 - a_4 + a_4)
    <=>
    a_1 = -a_2

    Now, replace some stuff in (3):

    a_0 + a_1 = a_3 + a_4
    <=>
    a_0 = a_3 + a_4 - a_1
    <=>
    a_0 = -a_4 + a_4 - (-a_2)
    <=>
    a_0 = a_2

    So, we have that a_2 and a_4 can be chosen arbitrarily and that:

    a_0 = a_2
    a_1 = -a_2
    a_3 = -a_4

    The set U can then be rewritten as U = { a_2 - a_2 * x + a_2 * x^2 - a_4 * x^3 + a_4 * x^4 | a_2, a_4 in R }, or equivalently U = { a_2(1 - x + x^2) + a_4(-x^3 + x^4) | a_2, a_4 in R }. This suggests the basis (1 - x + x^2, -x^3 + x^4), and hence, dim U = 2. Hope I didn't go wrong in my calculations, I'm a bit new at this ;)

    I'm not really sure what you mean by "U + V" or "U questionmark V"?

    *edit* Messed up a single minus sign, tried to fix it, hope this didn't break anything. :P
     
    Last edited: Jan 3, 2004
  4. Jan 3, 2004 #3
    Thanks Muzza, I think i got it now.
     
  5. Jan 3, 2004 #4

    HallsofIvy

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    Staff Emeritus
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    Another, and I think simpler way, to find a basis for
    U={p(x)=a0+a1*x+a2*x^2+a3*x^3+a4*x^4; a1+a2+a3+a4=0, a1+a2+2a3+2a4=0, a0+a1=a3+a4} is to set up the equations in matrix form:

    [ 1 1 0 -1 -1] (a0+ a1- a3- a4= 0)
    [ 0 1 1 1 1] (a1+ a2+ a3+ a4= 0)
    [ 0 1 1 2 2] (a1+ a2+2a3+2a4= 0)

    and row reduce to
    [1 0 -1 0 0]
    [0 1 1 0 0]
    [0 0 0 1 1]

    which shows that we must have a0= a2, a1= -a2 and a3= -a4.

    Notice that the 5 coefficients actually depend on 2 values: a2 and a4. That's reasonable, we had 3 equations connecting them so the "degrees of freedom" are reduced from 5 to 5-3= 2.

    If we take a2= 1, a4= 0, a0= 1, a1= -1, a3= 0 and one basis vector is 1- x+ x<sup>2</sup>.
    If we take a2= 0, a4= 1, a0= 0, a1= 0, a3= -1 so another basis vector is -x3+ x4.
    A basis for U is {1- x+ x2,-x3 x4}.

    For V=L{x^3-x^2+x, x^4+1}, if I interpret this correctly, this is the suspace generated by x3- x2+ x and x4+ 1. Since those two "vectors" are independent (they have different order so one is not a multiple of the other), they constitute a basis for V.
    A basis for V is {x3- x2+ x,x4+ 1}.

    If by "U+ V" you mean the direct sum of the two subspaces, then a basis for U+ V is "generated by" the union of bases for U and V separately. "Generated by" meaning that the union may not be independent so you have to drop any "redundant" vectors.

    The way I would do that is set up the matrix
    [1 -1 1 0 0] (1- x+ x2)
    [0 0 0 -1 1] (- x3+ x4)
    [0 1 -1 1 0] (-x2+ x3)
    [1 0 0 0 1] (1+ x4)
    and row-reduce to
    [1 0 0 0 0]
    [0 1 -1 0 0]
    [0 0 0 1 -1]
    [0 0 0 0 0]

    That last "[0 0 0 0 0]" row tells us that the four vectors were not independent and so a basis for the direct sum of U and V is { 1, x- x2, x3- x4}.


    I don't know what you mean by U?V. I guess you used a character that did not "translate" properly. The union of two suspaces is not, in general, a subspace. The intersection of two subspaces is a subspace. In this case, since the four vectors from the 2 separate bases are not independent, the intersection of the two subspaces is a one dimensional subspace so a basis will consist of a single vector. I'll leave it to you to find a vector that is in both U and V.
     
  6. Jan 3, 2004 #5
    Thanks HallsofIvy!
     
  7. Jan 3, 2004 #6
    What's the basis for this subspace?

    V={p an element of R^4[x]; p'(1)=p(1)=0}
     
  8. Jan 8, 2004 #7

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Well, p contained in R^4[x] can be written as
    A+ Bx+ Cx^2+ Dx^3+ Ex^4 so p(1)= A+ B+ C+ D+ E.
    p'(x)= B+ 2Cx+ 3Dx^2+ 4Ex^3 so p'(1)= B+ 2C+ 3D+ 4E

    p(x)= 0 and p'(x)= 0 means that A+ B+ C+ D+ E= 0 and
    B+ 2C+ 3D+ 4E= 0. R^4[x] has dimension 5. Since there are 2 conditions, this subspace has dimension 3 and so you need to find 3 vectors. Try this: Let A= 1, B= 0, C= 0 and solve for D and E, then let A= 0, B= 1, C= 0 and solve for D and E, finally, let A= 0, B= 0, C= 1 and solve for D and E. That gives you three independent vector and so a basis.
     
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