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Linear Algebra: Linear Independence

  1. Apr 4, 2012 #1
    1. The problem statement, all variables and given/known data

    Let S be a basis for an n-dimensional vector space V. Show that if v1,v2,...,vr form a linearly independent set of the vectors in V, then the coordinate vectors (v1)s, (v2)s,...,(vr)s form a linearly independent set in the Rn, and conversely.


    2. Relevant equations



    3. The attempt at a solution

    I tried working this problem but i got stuck almost at the end. i know that to show that the coordinate vectors form a linearly independent set that the following equation

    k1((v1)s)+ k2((v2)s) +...+ kr((v)s)=0 has to have only the trivial solution. Could i please get some help. I wrote v1, v2,..vn as a linear combination of the set S which i defined as S={w1,w2,...,wn}. Help please.
     
  2. jcsd
  3. Apr 4, 2012 #2

    Dick

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    If v1 is a vector, then what is (v1)s supposed to mean?
     
  4. Apr 4, 2012 #3
    It is notation. It is called the coordinate vector of v1 relative to S.

    for example v1 can be written as a linear combination of the basis S

    v1= c1(w1)+ c2(w2)+...+ cn(wn)

    thus

    (v1)s= {c1,c2,...,cn}
     
  5. Apr 5, 2012 #4

    Fredrik

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    You want to prove that the two implications
    $$\sum_i k_i v_i=0\ \Rightarrow\ \forall i~~k_i=0$$ and
    $$\sum_i k_i (v_i)_S=0\ \Rightarrow\ \forall i~~k_i=0.$$ are either both true or both false. You can do this by proving that ##\sum_i k_i (v_i)_S=\big(\sum_i k_i v_i)_S##. This is a matrix equation, so it holds if and only if the jth components of the left-hand side and the right-hand side are equal for all j.

    The following observation is useful. For all vectors x, we have
    $$
    \begin{align}
    x &=\sum_j x_j w_j\\
    x_S &=\sum_j (x_S)_j e_j=\sum_j x_j e_j,
    \end{align}
    $$ where the ##e_j## are the standard basis vectors for ##\mathbb R^n##. The important detail here is that ##(x_S)_j=x_j##, by definition of the "S" notation.
     
  6. Apr 5, 2012 #5
    Thank You!
     
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