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Math Help - Linear algebra

  1. #1
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    Linear algebra

    Can someone prove this claim?

    Claim:if w1,w2,.....,wl​ belongs to the span(u1,u2,....,uk) and l>k then w1,w2,.....,wl​ is linearly dependent.
    Last edited by davidciprut; January 11th 2014 at 06:16 AM.
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  2. #2
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    Re: Linear algebra

    Quote Originally Posted by davidciprut View Post
    Claim:if w1,w2,.....,wl​ belongs to the span(u1,u2,....,uk) and l>k then w1,w2,.....,wl​ is linearly dependent.
    dim span of u1...uk <= k
    if w1...wl linearly independent,
    dim span of u1...uk > k
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  3. #3
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    Re: Linear algebra

    A little more detail. The vectors \{w_1, w_2, ..., w_l\} are "dependent" if there exist number, a_1, a_2,..., a_l, not all 0, such that a_1w_1+ a_2w_2+ ...+ a_lw_l= 0. Since \{u_1, u_2, ..., u_k\} span the space, any vector, and in particular, the "w" vectors can be written as a linear combination of those: w_1= b_{11}u_1+ b_{12}u_2+ ...+ b_{1k}u_k, w_2= b_{21}u_1+ b_{22}u_2+ ...+ b_{2k}u_k}, etc. Replacing the w vectors with those,
    a_1(b_{11}u_1+ b_{12}u_2+ ...+ b_{1k}u_k)+ a_2(b_{21}u_1+ b_{22}u_2+ ...+ b_{2k}u_k))+ ...+ a_l(b_{l1}u_1+ b_{l2}u_2+ ...+b_{lk}u_k)= 0

    So (a_1b_{11}+ a_2b_{21}+ ...+ a_lb_{1l})u_1+ (a_1b_{12}+ a_2b_{22}+ ...+ a_lb_{2l})u_2+ ...+ (a_1b_{1l}+ a_2b_{2l}+ ... + a_lb_{kl})u_k= 0


    Now, we are NOT told that the u vectors are independent, only that they span the space. But certainly if the individual coefficients are 0 then certainly that sum will be the 0 vector. So one way that can happen is if
    a_1b_{11}+ a_2b_{21}+ ...+ a_lb_{1l}= 0
    a_1b_{12}+ a_2b_{22}+ ...+ a_lb_{2l}= 0
    .
    .
    .
    a_1b_{1l}+ a_2b_{2l}+ ... + a_lb_{kl}= 0

    We are given the "b" values, these are k equations for the l "a" values. If, as given, l>k, we have more unknown values than equations. We can take any values we want, in particular non- zero values, for l- k of them and have k equations to solve for the remaining k values.
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  4. #4
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    Re: Linear algebra

    Quote Originally Posted by HallsofIvy View Post
    A little more detail. ....
    That’s misleading. It implies you are elaborating on my proof, which is not the case. The detail applies to your proof, not mine. A better introduction to your post would be, “A different proof which requires more detail.”

    By the way, there is nothing wrong with a different proof. It makes things interesting. Thanks

    EDIT:
    A simple example might make your proof easier to follow:

    Let
    k=1 and l=2.

    Can we find a1 & a2 not both 0 st
    a1w1+a2w2=0

    Let
    w1=b1u1, w2=b2u1 (given),

    a1b1u1 + a2b2u1 = (a1b1+a2b2)u1=0 → a1b1+a2b2=0,
    which can be solved for a1 & a2 not both 0, and w1 & w2 are therefore linearly dependent.
    Last edited by Hartlw; January 13th 2014 at 09:24 AM. Reason: Add example
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  5. #5
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    Re: Linear algebra

    Quote Originally Posted by HallsofIvy View Post
    A little more detail. The vectors \{w_1, w_2, ..., w_l\} are "dependent" if there exist number, a_1, a_2,..., a_l, not all 0, such that a_1w_1+ a_2w_2+ ...+ a_lw_l= 0. Since \{u_1, u_2, ..., u_k\} span the space, any vector, and in particular, the "w" vectors can be written as a linear combination of those: w_1= b_{11}u_1+ b_{12}u_2+ ...+ b_{1k}u_k, w_2= b_{21}u_1+ b_{22}u_2+ ...+ b_{2k}u_k}, etc. Replacing the w vectors with those,
    a_1(b_{11}u_1+ b_{12}u_2+ ...+ b_{1k}u_k)+ a_2(b_{21}u_1+ b_{22}u_2+ ...+ b_{2k}u_k))+ ...+ a_l(b_{l1}u_1+ b_{l2}u_2+ ...+b_{lk}u_k)= 0

    So (a_1b_{11}+ a_2b_{21}+ ...+ a_lb_{1l})u_1+ (a_1b_{12}+ a_2b_{22}+ ...+ a_lb_{2l})u_2+ ...+ (a_1b_{1l}+ a_2b_{2l}+ ... + a_lb_{kl})u_k= 0


    Now, we are NOT told that the u vectors are independent, only that they span the space. But certainly if the individual coefficients are 0 then certainly that sum will be the 0 vector. So one way that can happen is if
    a_1b_{11}+ a_2b_{21}+ ...+ a_lb_{1l}= 0
    a_1b_{12}+ a_2b_{22}+ ...+ a_lb_{2l}= 0
    .
    .
    .
    a_1b_{1l}+ a_2b_{2l}+ ... + a_lb_{kl}= 0

    We are given the "b" values, these are k equations for the l "a" values. If, as given, l>k, we have more unknown values than equations. We can take any values we want, in particular non- zero values, for l- k of them and have k equations to solve for the remaining k values.
    I missed a crucial point in the above proof, which I pass on in case anyone else did.

    If
    1) (a1b11+…+albl1)u1+…..+ (a1b1k+…+alblk)uk=0,
    which it is if the coefficients of u1,…,uk=0 , whether or not u1,…,uk are independent,
    then
    a1(b11u1+….+b1kuk)+……+al(bl1u1+…+blkuk)=0, or
    2) a1w1+a2w2+…..alwl=0

    The point being, if you can solve 1) you can back-track to 2).

    1) is the second eq after the first paragraph above.
    Last edited by Hartlw; January 14th 2014 at 06:13 AM.
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