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Math Help - show that the set is linerlly independent

  1. #1
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    show that the set is linerlly independent

    Suppose that {\mathbf{v}_1,\mathbf{v}_2...\mathbf{v}_k} is linearly independent oset of vectors in \mathbb{R}^n. Show that if A is an nxn nonsingular matrix then {A\mathbf{v}_1,A\mathbf{v}_2...A\mathbf{v}_k} is also linearly independent.

    So if the given set is linearlly independent that means that c_1 \mathbf{v}_1+c_2 \mathbf{v}_2+...+c_k \mathbf{v}_k=\mathbf{0}
    and that c_1=c_2=...=c_k=0
    doesn't that mean that A must be 0?

    can I use the theorem that if the larger set is linearally independent then it's subspace must be linearally independent? I'm not sure if I can use this becuse it's not like Av is part of the original set of vectors.
    Last edited by superdude; March 14th 2010 at 06:40 PM. Reason: made vectors bold
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  2. #2
    Senior Member Tinyboss's Avatar
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    I'm not sure what you're allowed to use at this point, but do you have any results about a linear map and the dimension of its domain, image, and kernel?
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  3. #3
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    Quote Originally Posted by superdude View Post
    Suppose that {\mathbf{v}_1,\mathbf{v}_2...\mathbf{v}_k} is linearly independent oset of vectors in \mathbb{R}^n. Show that if A is an nxn nonsingular matrix then {A\mathbf{v}_1,A\mathbf{v}_2...A\mathbf{v}_k} is also linearly independent.

    So if the given set is linearlly independent that means that c_1 \mathbf{v}_1+c_2 \mathbf{v}_2+...+c_k \mathbf{v}_k=\mathbf{0}
    and that c_1=c_2=...=c_k=0
    doesn't that mean that A must be 0?

    can I use the theorem that if the larger set is linearally independent then it's subspace must be linearally independent? I'm not sure if I can use this becuse it's not like Av is part of the original set of vectors.
    If a set \{v_1,...,v_k\} is linearly independent then c_1v_1 + c_2v_2 + ... + c_kv_k = 0 \Leftrightarrow c_1 = c_2 = ... = c_k = 0 where c_1, c_2,...,c_k are scalars.

    Now, note that for any scalars a_1,a_2,...,a_k, we have that a_1Av_1 + a_2Av_2 + ... + a_kAv_k = Aa_1v_1 + Aa_2v_2 + ... + Aa_kv_k = A(a_1v_1 + a_2v_2 + ... + a_kv_k)

    Now use the fact that A is nonsingular to finish.
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