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System of vectors, linear dependence

  1. Apr 19, 2015 #1
    1. The problem statement, all variables and given/known data
    Prove that if in a system of vectors: [itex]S_a =\{a_1, a_2, ..., a_n\} [/itex] every vector [itex]a_i[/itex] is a linear combination of a system of vectors: [itex]S_b = \{b_1, b_2, ..., b_m\}[/itex], then [itex]\mathrm{span}(S_a)\subseteq \mathrm{span}(S_b)[/itex]

    2. Relevant equations


    3. The attempt at a solution
    We know due to [itex]a_j[/itex] being a linear combination, that every [itex]a_j\in S_a = \sum\limits_{j=1}^m c_j\cdot b_j[/itex] where [itex]b_j\in S_b, c_j\in\mathbb{R}\setminus\{0\}[/itex]
    But where should I go from here? Suggestions?
     
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  3. Apr 19, 2015 #2

    Orodruin

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    I suggest taking a vector in ##span(S_a)## and show that it is necessarily in ##span(S_b)##.
     
  4. Apr 19, 2015 #3

    Fredrik

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    The ##c_j## can be zero. The span of a set S is the set of all linear combinations of elements of S, including linear combinations where one or more (maybe all) of the coefficients are zero.

    I would do what Orodruin said, and avoid notations like
    every [itex]a_j\in S_a = \sum\limits_{j=1}^m c_j\cdot b_j[/itex] where...​
    It's ##a_j## that's equal to a linear combination, not ##S_a##. Oddly enough, the phrase
    every ##a_j\in S_a## is equal to ##\sum\limits_{j=1}^m c_j\cdot b_j## where...​
    would be considered acceptable.
     
  5. Apr 20, 2015 #4
    Alright. Let's denote the systems:
    [itex]A = \{a_1, a_2, ..., a_n\}\\B = \{b_1, b_2, ..., b_m\}[/itex]
    Let's denote the linear span of a system [itex]L(A), L(B)[/itex]. Then the respective linear spans would be:
    [itex]L(A) = \left\{a\ |\ a = \sum\limits_{k=1}^n \lambda _k\cdot a_k, \lambda _k\in\mathbb{R}, a_k\in A \right\}\\
    L(B) = \left\{b\ |\ b = \sum\limits_{k=1}^m \lambda _k\cdot b_k, \lambda _k\in\mathbb{R}, b_k\in B \right\}[/itex]
    We know that every vector [itex]a\in A[/itex] is a linear combination of the vectors in system [itex]B[/itex], that is:
    [itex]a = \sum\limits_{k=1}^m\lambda _k\cdot b_k[/itex] where [itex] \lambda _k\in\mathbb{R}, b_k\in B[/itex].
    Considering that a linear span is a vector space, then it is closed under multiplication with a scalar. Therefore, every [itex]a\in L(A)[/itex] implies [itex]a\in L(B)\Leftrightarrow L(A)\subseteq L(B)_{\square}[/itex]
     
  6. Apr 20, 2015 #5

    Fredrik

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    Writing this down is a good start, but I don't follow your argument here:

    I would just start with a simple statement like "Let ##x\in L(A)##." Then you can use the definition of ##L(A)## to say something about ##x##. This statement will involve the ##a_k##. Then you can use what you know about the ##a_k## to say something else. And so on. At some point you should be able to conclude that ##x\in L(B)##. Then you will have proved that ##L(A)\subseteq L(B)##.

    Be careful with your statements. The quoted statement above is saying that every vector in the subspace ##L(A)## implies some statement. Statements are implied by other statements, not by vectors.
     
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