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Math Help - linearly independent?

  1. #1
    jht
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    linearly independent?

    u=
    (5)
    (1)
    (2)
    v=
    (1)
    (2)
    (1)
    w=
    (3)
    (1)
    (-2)

    is it u+v+w = 0 if they are linearly independent?
    And how do i find out if its form a basis of R^3
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  2. #2
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    Quote Originally Posted by jht View Post
    u=
    (5)
    (1)
    (2)
    v=
    (1)
    (2)
    (1)
    w=
    (3)
    (1)
    (-2)

    is it u+v+w = 0 if they are linearly independent?
    And how do i find out if its form a basis of R^3

    What you have to check is whether au+bv+cw=0 is possible with at least one of the scalars a,b,c not zero. In our case you've to check whether:

    au+bv+cw=\begin{pmatrix}5a\\a\\2a\end{pmatrix}+\be  gin{pmatrix}b\\2b\\b\end{pmatrix}+\begin{pmatrix}3  c\\c\\-2c\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix  }\iff a=b=c=0.

    If the above homogeneous linear system only has the trivial solution then the three vectors are lin. ind., otherwise they're lin. dep.

    Tonio
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  3. #3
    jht
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    Thanks, i also have a question other than part (b) above,
    say if u,v,w is lin,indep
    q=w-v+lu and l E R
    show u,v,q is lin,indep.

    I can see u,v,q is lin,indep if q=w-v+u
    But what if the equaition at above with a l real number multiply by u in the equation?
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  4. #4
    jht
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    so i found a=20/63 ; b = -5/9; c= 5/7
    i am wondering is it right?
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  5. #5
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    Quote Originally Posted by jht View Post
    u=
    (5)
    (1)
    (2)
    v=
    (1)
    (2)
    (1)
    w=
    (3)
    (1)
    (-2)

    is it u+v+w = 0 if they are linearly independent?
    No, if there exist any numbers, a, b, c, not all 0, so that au+ bv+ cw= 0 then they are dependent if u+ v+ w= 0 then a= b= c= 1 are not all 0 so they are dependent. Of course, with the u, v, w you give here, u+ v+ w is NOT 0.
    And how do i find out if its form a basis of R^3
    If 3 vectors are independent, they they form a basis for a 3 dimensional space.
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  6. #6
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    Quote Originally Posted by jht View Post
    so i found a=20/63 ; b = -5/9; c= 5/7
    i am wondering is it right?
    Tonio told you that you needed
    au+bv+cw=\begin{pmatrix}5a\\a\\2a\end{pmatrix}+\be  gin{pmatrix}b\\2b\\b\end{pmatrix}+\begin{pmatrix}3  c\\c\\-2c\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix  }
    So you must have 5a+ b+ 3c= 0, a+ 2b+ c= 0, and 2a+ b- 2c= 0.

    with a= 20/63, b= -5/9, c= 5/7, the second equation is
    a+ 2b+ c= (20/63)+ 2(-5/9)+ 5/7= 20/63- 10/9+ 5/7= 20/63- 70/63+ 45/63= -5/63, not 0.
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  7. #7
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    Quote Originally Posted by jht View Post
    Thanks, i also have a question other than part (b) above,
    say if u,v,w is lin,indep
    q=w-v+lu and l E R
    show u,v,q is lin,indep.

    I can see u,v,q is lin,indep if q=w-v+u
    But what if the equaition at above with a l real number multiply by u in the equation?
    Suppose u, v, q were NOT linearly independent. The there exist a, b, c, not all 0, such that au+ bv+ cq= 0. But q= w- v+ lu so that is au+ bv+ c(w- v+ lu)= (a+ lc)u+ (b- c)v+ cw= 0. Since u, v, and w are independent, we must have a+ lc= 0, b- c= 0, and c= 0. Those reduce to a= 0, b= 0, and c= 0 contradicting the assertion that they are not all 0. Since c= 0, the "l" in lc does not matter.
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