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  1. #1
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    Prove that vectors ... are linearly dependent ...

    Prove that vectors
    v1, v2, ... ,vn of the same dimension are linearly dependent if and only if one of them can be written as a linear combination of the others.

    I'm not sure how exactly I would correctly go about proving this. Thanks!

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  2. #2
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    Quote Originally Posted by jlt1209 View Post
    [LEFT]Prove that vectors v1, v2, ... ,vn of the same dimension are linearly dependent if and only if one of them can be written as a linear combination of the others.
    Suppose that v_k  = \alpha _1 v_1  +  \cdots \alpha _{k - 1} v_{k - 1}  + \alpha _{k + 1} v_{k + 1}  \cdots  + \alpha _n v_n .
    But that means 0 = \alpha _1 v_1  +  \cdots \alpha _{k - 1} v_{k - 1}  + v_k  + \alpha _{k + 1}{\color{red}-1} v_{k + 1}  \cdots  + \alpha _n v_n.
    Do you know what it means for the set \left\{ {v_1 ,v_2 , \cdots v_{n - 1} ,v_n } \right\} to be linearly independent?
    If you do then you see the above proves that the set is not linearly independent.
    Moreover, you can reverse the argument.
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  3. #3
    Super Member Aryth's Avatar
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    Well, this can be proven directly:

    Theorem: \bold{v_1}, \bold{v_2}, ..., \bold{v_n} of the same dimension are linearly dependent iff one of them can be written as a linear combination of the others.

    Proof:

    Let the set \{\bold{v_1}, \bold{v_2}, ..., \bold{v_n}\} be linearly dependent. This means that there exists scalars s_1, s_2, ..., s_n, not all zero, such that:

    s_1\bold{v_1} + s_2\bold{v_2} + ... + s_n\bold{v_n} = 0

    So, assume that s_1 \neq 0. We can rewrite this as:

    \bold{v_1} = \left(-\frac{s_2}{s_1}\right)\bold{v_2} + \left(-\frac{s_3}{s_1}\right)\bold{v_3} + ... + \left(-\frac{s_n}{s_1}\right)\bold{v_n}

    Therefore, we get \bold{v_1} as a linear combination of the others.

    Now conversely, we assume that \bold{v_1} is a linear combination of \bold{v_2}, \bold{v_3}, ... , \bold{v_n} which means that there exists scalars c_2, c_3, ..., c_n such that:

    \bold{v_1} = (-c_2)\bold{v_2} + (-c_3)\bold{v_3} + ... + (-c_n)\bold{v_n}

    We can rewrite this to be:

    1\bold{v_1} + c_2\bold{v_2} + c_3\bold{v_3} + ... + c_n\bold{v_n} = 0

    Therefore, the set \{\bold{v_1}, \bold{v_2}, ... , \bold{v_n}\} is linearly dependent.
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  4. #4
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    awesome thank you that helps a lot!
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