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Homogenous linear system

  1. Jan 11, 2013 #1
    1. The problem statement, all variables and given/known data
    In the proof of Lemma 3.6, what happens if there are no non-‘0 = 0’ equations?


    2. Relevant equations
    3.6 Lemma For any homogeneous linear system there exist vectors B1,...,Bksuch that the solution set of the system is
    {c1 B1+..+ckBk|c1,..ck εR}
    where k is the number of free variables in an echelon form version of the system.
    This theorem is from Jim Heffermin's book page 32.

    3. The attempt at a solution
    I just don't get the part when they say "there are no non-‘0 = 0’ equations"? Do they mean every single equation in the linear system is just 0=0?
     
  2. jcsd
  3. Jan 11, 2013 #2

    mfb

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    I don't have the book, but I think I see what you mean:
    If there are no "non-'0=0'-solutions, every single solution to the system has 0 for all coefficients ci on the left side. This is a trivial solution (it exists independent of Bi), solutions with non-zero coefficients are interesting.
     
  4. Jan 11, 2013 #3
    Oh so if that is the case it the solution set only generates the zero vector?
     
  5. Jan 11, 2013 #4

    HallsofIvy

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    If A is non-singular, then Ax= b has a unique solution. If A is singular then it maps Rn (A is n by n) to a subspace of Rn. If b is is in that subspace, the Ax= b has an infinite number of solutions. If b is not in that subspace, then there is no solution.

    Since b= 0 is in every supspace, If A is non-singular, x= 0 is the only solution to Ax= 0. If A is singular, Ax=0 has an infinite number of solutions, all in the "null-space" of A.
     
  6. Jan 11, 2013 #5
    In the proof in the book they state
    "Apply Gauss’s Method to get to echelon form. We may get some 0 = 0 equations (if the entire system consists of such equations then the result Linear Systems is trivially true) but because the system is homogeneous we cannot get any contradictory equations. We will use induction to show this statement: each leading variable can be expressed in terms of free variables. That will finish the proof because we can then use the free variables as the parameters and the B1,...,Bk are the vectors of coefficients of those free variables..."

    This problem is worded a bit confusing, but I think this is what its trying to say. For example-

    0x1+0x2+...+0xn=0
    :
    :
    0x1+0x2+...+0xn=0

    Then wouldn't that mean x1...xn are free variables. And it spans all of R^n since you could plug in any solution and it will work. So the solution set would look like

    {x1(1,0,0...0)+...+xn(0,0,...,1)|xi εR}
     
    Last edited: Jan 11, 2013
  7. Jan 11, 2013 #6
    yeah that looks right to me.
     
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