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Vector sets and basis

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  1. Jan 8, 2017 #1
    1. The problem statement, all variables and given/known data

    1. Suppose {v1, . . . , vk} is a linearly independent set of vectors in Rn and suppose A is an m × n matrix such that Nul A = {0}.
    (a) Prove that {Av1, . . . , Avk} is linearly independent.
    (b) Suppose that {v1, . . . , vk} is actually a basis for Rn. Under what conditions on m and n will {Av1, . . . , Avk} be a basis for Rm?


    3. The attempt at a solution

    a)
    we know that c1V1+...+CnVn=0 as it is linearly independent

    suppose that

    C1AV1+...+CnAVn=0 and that A is an invertible matrix since the null space is 0 which means we can multiply both sides by A^-1 which gives c1V1+...+CnVn=0 which means that there is a trivial solution and that it is linearly independent

    Is this correct?

    b)
    Im unsure on how to get started on this question

    Thanks for any help
     
  2. jcsd
  3. Jan 8, 2017 #2

    fresh_42

    Staff: Mentor

    Is ##m=k##? And how can ##k## vectors be a basis of an ##n-##dimensional vector space ##\mathbb{R}^n##?
    The definition of linear independence goes as follows:
    ##\{v_1,\ldots , v_k\}## are linear independent if ##(c_1v_1+\ldots+c_kv_k=0 \Longrightarrow c_1=\ldots=c_k=0)##
    So you have to show this implication, the equation ##c_1v_1+\ldots+c_kv_k=0## alone doesn't mean anything.
    Why should ##A## be invertible? Is ##m=n##? Use the linearity of ##A## instead.
     
  4. Jan 8, 2017 #3
    what does the Linearity of A mean?
     
  5. Jan 8, 2017 #4

    fresh_42

    Staff: Mentor

    Linearity of a mapping, and ##A## is a linear mapping ##A\, : \,\mathbb{R}^n \longrightarrow \mathbb{R}^m##, that is ##A(v+w) = Av + Aw## and ##A(cv)=cA(v)## for ##c \in \mathbb{R}##.
     
  6. Jan 8, 2017 #5

    Mark44

    Staff: Mentor

    You can still write exactly the same equation if ##\{v_1, v_2, \dots, v_n\}## is a linearly dependent set. So what is it that distinguishes a linearly independent set from a linearly dependent set?

    It's given that A is m x n, so you can't assume that A is invertible.
     
  7. Jan 8, 2017 #6
    it is linearly independent if ##\{c_1, c_2, \dots, c_n=0\}##


    Then how would i go about showing that it is independent?
     
  8. Jan 8, 2017 #7

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    We cannot answer that without giving away the whole solution.
     
  9. Jan 8, 2017 #8

    Mark44

    Staff: Mentor

    Consider these vectors in ##\mathbb{R}^2##: ##v_1 = <1, 1>, v_2 = <2, 2>##
    Suppose that ##c_1v_1 + c_2v_2 = 0##. Clearly ##c_1 = c_2 = 0## is a solution. Does it follow that the two vectors are linearly independent?
     
  10. Jan 8, 2017 #9
    yes as there is only the trivial solution
     
  11. Jan 8, 2017 #10

    fresh_42

    Staff: Mentor

    And what about ##2 \cdot v_1 + (-1)\cdot v_2##?
     
  12. Jan 8, 2017 #11
    then adding the vectors would be 0 but this shouldn't effect the independence
     
  13. Jan 8, 2017 #12

    fresh_42

    Staff: Mentor

    You should really repeat the concept of linearity, of vectors as well as of mappings. Two vectors are linear independent, if they don't point to the same direction. Does this hold for ##(1,1)## and ##(2,2)##? Three vectors are linear independent, if they are not all on the same line or in the same plane. The formal notation of this is
    So the implication is the crucial point. Choosing ##c_i=0## is always a solution. What sense would it make to call vectors linear independent then? The point is, it has to be the only solution, for else we could always write a ##v_i= -\frac{c_1}{c_i}v_1 - \ldots -\frac{c_n}{c_i}v_n## which is clearly a linear dependence.
     
  14. Jan 8, 2017 #13

    Mark44

    Staff: Mentor

    No. ##c_1 = c_2 = 0## is a solution, but it is far from being the only solution. As fresh_42 explains, my two vectors are NOT linearly independent.
     
  15. Jan 10, 2017 #14
    If Null A = {0} that means that A would be Linearly independent as there are no free variables and it has a trivial solution

    Given that {v1,...;,vk} is linearly independent

    Then {AV1,...AVK} would also be linearly independent as the matrix A is linearly independent

    Would this be correct now or am I missing something?
     
  16. Jan 10, 2017 #15

    fresh_42

    Staff: Mentor

    You cannot say "##A## is linear independent" in this context. What does it mean? You might say that the column or row vectors are linear independent but not the matrix as a whole. Actually ##Null(A)=\{0\}## means, that ##A## is injective, or "into", or an embedding, in your case ##A\, : \,\mathbb{R}^n \hookrightarrow \mathbb{R}^m##. So the column vectors of ##A## are linear independent (whereas the row vectors are not!), but do you know why? Where are no free variables? What are the variables? A trivial solution ##A\cdot \vec{0} = \vec{0}## always exists, so this cannot be a criteria.

    Start at the beginning instead. You want to show, that ##\{Av_1,\ldots ,Av_k\}## are linear independent vectors, given ##\{v_1,\ldots ,v_k\}## are linear independent. What do you have to show then? Look up the definition of linear independence in #2.
    In the next steps, you should use the linearity of ##A## (see definition in #4), then the condition ##Null(A)=\{0\}## and at last the fact, that the ##v_i## are linear independent. Try to proceed along these four steps.
     
  17. Jan 10, 2017 #16
    Given that ##\{v_1,\ldots ,v_k\}## is linearly independent it means that ##c_1v_1+\ldots +c_kv_k=0## which implies that ##c_1=\ldots=c_k=0##

    to show that ##\{Av_1,\ldots ,Av_k\}## is linearly independent we need to show that ##c_1Av_1+\ldots +c_kAv_k=0##

    Given that ##Null(A)=\{0\}## it means that the column vectors are linearly independent and given that we know that the ##V_i## are linearly independent then ##c_1=\ldots=c_k=0## which means that ##\{Av_1,\ldots ,Av_k\}## is also linearly independent

    Is this better now?
     
  18. Jan 10, 2017 #17

    Mark44

    Staff: Mentor

    As already stated, even if ##\{v_1,\ldots ,v_k\}## is linearly dependent, the equation ##c_1v_1+\ldots +c_kv_k=0## still has a solution of ##c_1=\ldots=c_k=0##. There's a subtlety here that I don't think you are getting.
    No. You can always set up this equation, whether or not ##\{Av_1, \dots, Av_k \} is a linearly independent set.
    Look at the example I gave in post #8. You concluded incorrectly that <1, 1> and <2, 2> were linearly independent merely because I showed that if c1<1, 1> + c2<2, 2> = 0, then ##c_1 = c_2 = 0## is a solution.

    What is the definition of linear independence that you are using?
     
  19. Jan 10, 2017 #18

    fresh_42

    Staff: Mentor

    Almost. Cancel the "which". Linear independency means there is only the solution ##c_1= \ldots = c_k =0##, no others. To show it, we have to rule out the possibility of other solutions. In Mark's example ##2 \cdot (1,1) + (-1) \cdot (2,2) = 0## is another solution, so they cannot be linear independent. We could also write this equation as ##(2,2) = 2 \cdot (1,1)##, i.e. one vector is twice the other, which is a linear dependency.
    E.g. ##(1,2)## and ##(2,3)## are linear independent. They cannot be combined to the zero vector, other than multiplying them with zeros. You might try it.
    We need to show that ##c_1Av_1+\ldots +c_kAv_k=0## implies ##c_1= \ldots = c_k =0##, that it is forced. (see above)
    Why?
    Yes, but some techniques would be helpful here in order to be convincing. Which properties do you apply when?
    Don't misunderstand me. I don't want to torture you, nor shall you please me. I only think it is important to gain some certainty with these concepts as there might be more ahead.
     
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