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T/F Question of linear independence

  1. Apr 1, 2017 #1
    1. The problem statement, all variables and given/known data
    T/F: Let ##T: V \rightarrow W##. If ##\{v_1,v_2,...,v_k \}## is a linearly independent set, then ##\{T(v_1), T(v_2),..., T(v_k) \}## is linearly independent.

    2. Relevant equations


    3. The attempt at a solution
    This seems to be true, because we know that ##a_1v_1 + a_2v_2 + \cdots + a_k v_k = 0## has only the trivial solution, since they are linearly independent. Then if we apply ##T## to both sides, we get that ##a_1T(v_1) + a_2T(v_2) + \cdots + a_kT(v_k) = 0##, which should mean that this only has the trivial solution as well, right? Meaning that they are linearly independent?
     
  2. jcsd
  3. Apr 1, 2017 #2

    fresh_42

    Staff: Mentor

    What if ##T=0##?
     
  4. Apr 1, 2017 #3
    So does a problem arise only when ##\{v_1,v_2,...,v_k \} \in \text{Ker}(T)##?
     
  5. Apr 2, 2017 #4

    Mark44

    Staff: Mentor

    No. Think about this in terms of simple examples, such as T(x, y) = <x, 0>; i.e. T is the projection onto the x-axis. The vectors <1, 0> and <0, 1> form a basis for R2. Do the vectors T(<1, 0>) and T(<0, 1>) also form a basis for the same space?
     
  6. Apr 2, 2017 #5

    fresh_42

    Staff: Mentor

    The point is: ##a_1=\ldots =a_k=0## is always a solution, but to be linearly independent, it has to be - as you correctly pointed out - the only possible solution. So every summand ##a_i T(v_i) = 0## allows two possibilities to become zero. And even worse, what if they are all unequal to zero, but add up to zero? Do you have an idea which property of ##T## would be needed, to conclude, that all ##a_i=0##?

    Your mistake was to start with ##a_1v_1+\ldots+a_kv_k=0##. But you want to find out, what ##x_1T(v_1)+\ldots+x_kT(v_k)=0## means for the ##x_i##, so better proceed the other way around.
     
  7. Apr 2, 2017 #6
    So, taking your advice, we start with ##a_1T(v_1)+\ldots+a_kT(v_k)=0##, which means that ##T(a_1v_1 + \cdots + a_nv_n) = 0##. We can only then conclude ##a_1v_1 + \cdots + a_nv_n = 0## from this if ##T## is injective, right?

    So is it true that you need ##T## to be injective to preserve linear independence?
     
  8. Apr 2, 2017 #7

    fresh_42

    Staff: Mentor

    That's right. Beside the missing ##0##. And if you have ##a_1v_1 + \cdots + a_nv_n = 0##, then you can apply the given fact, that the ##v_i## are linearly independent.

    Edit: One more important remark. If ##T## is not injective, the ##\{T(v_1),\ldots,T(v_k)\}## could still be linear independent, e.g. if ##k=1## and ##T(v_1) \neq 0##. But in general, we just don't know, if ##T## isn't injective. The statement, however, could still be true - or wrong. It all depends on what ##\left. T\right|_{span\{v_i\}}## does.
     
    Last edited: Apr 2, 2017
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