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Math Help - Show that if u,v,w form a basis.....

  1. #1
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    Show that if u,v,w form a basis.....

    Hi, i'm having trouble with the following question:

    Show that if u,v,w form a basis for a vector space V over the field of scalars K then so do, u+v, u+v+w, v+w.

    Any help would be great. Thanks.
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  2. #2
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    A sketch of the approach would be to notice that a sequence of vectors can form a basis if they're linearly independent. So you need to prove that the sums of vectors mentioned are linearly independent.

    Maybe this advances you a bit. If not, post back.
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Rafael Almeida View Post
    A sketch of the approach would be to notice that a sequence of vectors can form a basis if they're linearly independent and they span the vector space. So you need to prove that the sums of vectors mentioned are linearly independent.

    Maybe this advances you a bit. If not, post back.
    since u, v and w are linearly independent, it follows immediately that (u + v + w), (u + v) and (v + w) are linearly independent.

    now it leaves for us to show that the latter three vectors span the space. since we know u, v and w span the space, we can show this by showing that we can write u, v and w as linear combinations of (u + v + w), (u + v) and (v + w)
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  4. #4
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    Quote Originally Posted by Jhevon View Post
    since u, v and w are linearly independent, it follows immediately that (u + v + w), (u + v) and (v + w) are linearly independent.

    now it leaves for us to show that the latter three vectors span the space. since we know u, v and w span the space, we can show this by showing that we can write u, v and w as linear combinations of (u + v + w), (u + v) and (v + w)
    Indeed, i forgot to point out that they need to span the space, i.e., any point summed with a linear combination of the vectors must produce every point in the space.

    Anyway, it is somewhat easy to see that if three vectors are linearly independent and assuming the space as \mathbb{R}^3 for lack of other indication, they span the space.
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