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Math Help - Linearly independent set that spans same subspace as original set of vectors

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    Linearly independent set that spans same subspace as original set of vectors

    Problem:

    Determine a linearly independent set of vectors that spans the same subspace of V as that spanned by the original set of vectors.

    V=R3, {(1,1,1),(1,-1,1),(1,-3,1),(3,1,2)}

    Question:

    There is a theorem that says, "If one of the vectors in the set is a linear combination of the other vectors in the set, then that vector can be deleted from the given set of vectors".

    I found the following dependency relationship among these vectors to be:

    v1-2v2+v3=0

    Now, I can solve for any one of the vectors in terms of the other two. Does this mean I can remove one, two, or three vectors from the original set, or am I limited to just one "vector removal"?

    Thanks for any help!
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  2. #2
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    Quote Originally Posted by divinelogos View Post
    Problem:

    Determine a linearly independent set of vectors that spans the same subspace of V as that spanned by the original set of vectors.

    V=R3, {(1,1,1),(1,-1,1),(1,-3,1),(3,1,2)}

    Question:

    There is a theorem that says, "If one of the vectors in the set is a linear combination of the other vectors in the set, then that vector can be deleted from the given set of vectors".

    I found the following dependency relationship among these vectors to be:

    v1-2v2+v3=0

    Now, I can solve for any one of the vectors in terms of the other two. Does this mean I can remove one, two, or three vectors from the original set, or am I limited to just one "vector removal"?

    Thanks for any help!
    \begin{vmatrix} 1 & 1 & 1 \\ 1& -1 & 1 \\ 3& 1 & 2\end{vmatrix}=2

    So these vectors are linearly independent so they span all of \mathbb{R}^3

    Since they only have to span the same space why not just pick the standard basis vectors.
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  3. #3
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    Quote Originally Posted by TheEmptySet View Post
    \begin{vmatrix} 1 & 1 & 1 \\ 1& -1 & 1 \\ 3& 1 & 2\end{vmatrix}=2

    So these vectors are linearly independent so they span all of \mathbb{R}^3

    Since they only have to span the same space why not just pick the standard basis vectors.
    Can you elaborate a little? By standard basis vectors do you mean multiples of (1,0,0) (0,1,0) and (0,0,1)?
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    Quote Originally Posted by divinelogos View Post
    Can you elaborate a little? By standard basis vectors do you mean multiples of (1,0,0) (0,1,0) and (0,0,1)?
    That's correct.
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