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Math Help - Dimension of span(Z)

  1. #1
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    Dimension of span(Z)

    thanks for the help!
    Last edited by MMAfanatic; January 18th 2010 at 06:23 AM. Reason: Changed post title
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  2. #2
    Senior Member Dinkydoe's Avatar
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    First: You can write these polynomials f_1,\cdots f_5 with respect to the basis \left\{1,x,x^2,x^3,x^4\right\}

    Thus we can write:

    f_1 = (1,0,1,1,0)
    f_2= (2,1,2,0,1)
    f_3 = (-4,-3,-4,2,-3)
    f_4=(1,-3,0,1,0)
    f_5= (13,-1,11,3,5)

    Put these vectors in a matrix and by performing elementary row-operations we get:

     \left[ \begin {array}{ccccc} 1&0&1&1&0\\\noalign{\medskip}2&1&2&0&1<br />
\\\noalign{\medskip}-4&-3&-4&2&-3\\\noalign{\medskip}1&-3&0&1&0<br />
\\\noalign{\medskip}13&-1&11&3&5\end {array} \right]  \to  \left[ \begin {array}{ccccc} 1&0&1&1&0\\\noalign{\medskip}0&1&0&-2&1<br />
\\\noalign{\medskip}0&0&-1&-6&3\\\noalign{\medskip}0&0&0&0&0<br />
\\\noalign{\medskip}0&0&0&0&0\end {array} \right]

    We see that span (Z)\subset \mathcal{P}_4(\mathbb{R}) is a 3-dimensional subset and:

    Now, letting:

    g_1 = 1+x^2+x^3
    g_2 = x-2x^3+x^4
     g_3=-x^2-6x^3+3x^4:

    Y: = \left\{g_1,g_2,g_3 \right\} is such that span( Y)= span( Z)
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  3. #3
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    hey thank you for your help.

    could u please send the operations you used to row reduce the matrix please.

    when i did it i just did row 2 - 2 x row 1, row 3 - 4 x row 1, row 4 - row 1, row 5 - 13 x row 1, row 3 + 3x row 2

    so im left with

    1 0 1 1 0
    0 1 0 -2 1
    0 -3 -1 0 0
    0 -1 -2 -10 5

    i know that i am wrong because im not sure how to row reduce properly i dont think :-(

    thanks for you help again.
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  4. #4
    Senior Member Dinkydoe's Avatar
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    Actually I let Maple calculate the reduced form. I admit it can be quite tedious work, but it's still probably the simplest method.
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  5. #5
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    oh right...fair enough lol.

    umm i know row 2 is row 2 - 2 x row 1, dont know how the other rows came about?

    can anyone help on the row reducing to get the final product.

    sorry guys for this.

    major thanks to the original helper and anyone else who may be able to help me out :-)
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  6. #6
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    i figured it


    :-)

    [/COLOR]
    Last edited by MMAfanatic; January 18th 2010 at 06:10 AM.
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  7. #7
    Senior Member Dinkydoe's Avatar
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    Very good. Now find even 2 more linear independant vectors: v_1,v_2\in\mathbb{R}^5 , write them on the basis \left\{1,x,x^2,x^3,x^4\right\} to find polynomials p_1,p_2

    Then together with Y you have a basis for \mathcal{P}_4(\mathbb{R})
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  8. #8
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    Quote Originally Posted by Dinkydoe View Post
    Very good. Now find even 2 more linear independant vectors: v_1,v_2\in\mathbb{R}^5 , write them on the basis \left\{1,x,x^2,x^3,x^4\right\} to find polynomials p_1,p_2

    Then together with Y you have a basis for \mathcal{P}_4(\mathbb{R})
    not sure about this.
    Last edited by MMAfanatic; January 18th 2010 at 08:18 AM.
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  9. #9
    Senior Member Dinkydoe's Avatar
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    No, it's the part where they ask to extend Y to find a basis for \mathcal{P}_4(\mathbb{R}).

    But they can be used to extend Y.

    Namely p_0(x) = 1 is independant of Y and p_1(x)= x is independant of Y

    We can even show that: Y\cup \left\{p_0(x),p_1(x)\right\} is a basis for \mathcal{P}_4(\mathbb{R})

    You can do this by showing that:

     \left[ \begin {array}{ccccc} 1&0&1&1&0\\\noalign{\medskip}0&1&0&-2&1<br />
\\\noalign{\medskip}0&0&-1&-6&3\\\noalign{\medskip}0&1&0&0&0<br />
\\\noalign{\medskip}1&0&0&0&0\end {array} \right]

    has rank 5.
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  10. #10
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    Quote Originally Posted by Dinkydoe View Post
    No, it's the part where they ask to extend Y to find a basis for \mathcal{P}_4(\mathbb{R}).

    But they can be used to extend Y.

    Namely p_0(x) = 1 is independant of Y and p_1(x)= x is independant of Y

    We can even show that: Y\cup \left\{p_0(x),p_1(x)\right\} is a basis for \mathcal{P}_4(\mathbb{R})

    You can do this by showing that:

     \left[ \begin {array}{ccccc} 1&0&1&1&0\\\noalign{\medskip}0&1&0&-2&1<br />
\\\noalign{\medskip}0&0&-1&-6&3\\\noalign{\medskip}0&1&0&0&0<br />
\\\noalign{\medskip}1&0&0&0&0\end {array} \right]

    has rank 5.

    Ok now im totally lost. I agree that matrix has rank 5 but where did you get that matrix from? Its different to our row operationed matrix we had before.
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  11. #11
    Senior Member Dinkydoe's Avatar
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    It's the same matrix with the 3 upper rows:

    with p_0 = (1,0,0,0,0) and p_1 = (0,1,0,0,0) instead of the zero-rows.

    Where p_0,p_1 are the vector representations of p_1(x), p_2(x) w.r.t. the basis \left\{1,x,x^2,x^3,x^4,\right\}

    You want to show that Y is linear independant of p_1(x),p_2(x). You can do that by showing that g_1,g_2,g_3,p_1,p_2 are linear independant, that is, the given matrix has rank 5. (meaning, having 5 linear independant rows).
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  12. #12
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    oh i see.

    i know why i was confused. i disagree with what your final row operated product looks like


    i have

    1 0 1 1 0

    0 1 0 -2 1

    0 0 1 6 -3

    0 0 0 0 0

    0 0 0 0 0
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  13. #13
    Senior Member Dinkydoe's Avatar
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    You're allways free to disagree with a computer output

    I didn't do the calculations, do whatever suits you.

    edit: Actually they're the same matrices.
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