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Homework Help: Proving a Linear System

  1. Apr 14, 2010 #1
    1. The problem statement, all variables and given/known data

    let ui , ci [tex]\in[/tex] Rn

    and

    A [tex]\in[/tex] Mn(R) be such that AciT = uiT , i=1,2,...,n

    Suppose u1,u2,....,un are linearly independent. Show that

    c1,c2,....,cn are linearly independent

    2. Relevant equations

    N/A

    3. The attempt at a solution

    i was just learning the eigenvalue, eigenvector, but i didn't have any clue how to relate to this or this really related to eigen?

    can someone help me
     
  2. jcsd
  3. Apr 14, 2010 #2

    Mark44

    Staff: Mentor

    I hope you don't mind, but I changed the names of your c vectors to v. c is almost always used for constants, while u, v, w, etc are used for vectors.
    Since u1, u2, ..., un are linearly independent, the equation c1u1 + c2u2 + ... + cnun = 0 has only one solution for the constants c1, c2, ... , cn. (What is that solution?)

    Now, what can you say about the equation c1Av1 + c2Av2 + ... + cnAvn = 0?
     
  4. Apr 14, 2010 #3
    one solution which ci= 0 , i = 1,2,....,n

    right? i'm still not sure how to state that because, i remembered there are only one solution, but how to state that it is "only one solution"

    this one, since

    AviT = uiT

    implies

    ui = viAT

    implies

    c1v1AT + c2v2AT + ... + cnvnAT = 0

    has only one solution ci = 0 , i=1,2,...,n

    implies

    v1AT , v2AT ,...., vnAT are linear independent

    am i right?? did i over complicating things?

    if i right,

    i only show v1AT , v2AT ,...., vnAT are independent

    how to show

    v1 , v2 ,...., vn are independent
     
    Last edited: Apr 14, 2010
  5. Apr 14, 2010 #4
    can i make like this?

    c1v1AT + c2v2AT + ... + cnvnAT = 0

    has only one solution ci = 0 , i=1,2,...,n

    implies

    (c1v1 + c2v2 + ... + cnvn)AT = 0

    has only one solution ci = 0 , i=1,2,...,n

    implies

    c1v1 + c2v2 + ... + cnvn = 0

    has only one solution ci = 0 , i=1,2,...,n

    implies

    v1 , v2 ,...., vn are independent


    im sorry if i do over complicating things and annoy you much. T_T
     
    Last edited: Apr 14, 2010
  6. Apr 15, 2010 #5

    Mark44

    Staff: Mentor

    Right. Just say that this solution is the only solution because it is given that the vectors are linearly independent.
    I guess you took the transpose of each side. Not sure that this is worth doing. I would say that you can leave off the T's for transpose.
    Yes, a little, but you have the main idea. Instead of saying this:
    c1v1AT + c2v2AT + ... + cnvnAT = 0


    you can instead say this:
    c1Av1 + c2Av2 + ... + cnAvn = 0

    All you have done is replace ui in the first equation with Avi in the second equation. Since the first equation has only one solution (the trivial solution c1=c2=...=cn=0), then so does the second equation.

     
  7. Apr 15, 2010 #6
    hoho, thankyou very much mark44, hmm now only bugging me is the other question .. Let me try my best.

    before that, can you tell whether A is invertible or not?
     
  8. Apr 15, 2010 #7

    Mark44

    Staff: Mentor

    A is invertible.

    If A were not invertible, then the dimension of the nullspace of A would have to be at least 1. That means that for at least one vector vi, Avi = 0.

    Since you are given that Avi = ui, and the u vectors are linearly independent, none of them can be zero, hence Avi [itex]\neq[/itex] 0 for i = 1, 2, ..., n.

    That's not the complete proof, but it should give you the idea.
     
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