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Homework Help: Homogeneous system

  1. Oct 28, 2008 #1
    1. The problem statement, all variables and given/known data

    Let A * x= 0 be a homogeneous system of n linear equations in n unknowns, that has only the trivial solution. Show that if k is any postive integer, than the system A^k * X = 0 also has only the trivial solution.


    3. The attempt at a solution


    I'm so lost plz help and what is trivial solution?
     
  2. jcsd
  3. Oct 28, 2008 #2

    Mark44

    Staff: Mentor

    The trivial solution is the vector x = 0. All n entries in this vector are zero.
     
  4. Oct 29, 2008 #3

    Mark44

    Staff: Mentor

    If A is an n x n matrix, and Ax = 0 has only the trivial solution, what things does that tell you about A?
     
  5. Oct 29, 2008 #4
    I am just picking up on what Mark44 said. If the only sol to the matri eq is the trivial one Ax=0, that is for =0,(x-vector, A matrix (nxn), this means that A is nonsingular. Then since A is nonsingular, we know that A has an inverse, a unique one. SO:

    [tex] A^kx=0=>A*A*A*A*A*...*Ax=0[/tex] Now multiplying by A^-1 , k-1 times from the left side we get to Ax=0, which as we know has only the trivial sol. so we are set.
     
  6. Oct 29, 2008 #5

    Mark44

    Staff: Mentor

    But that gets you to Ax = 0. There is one more thing that needs to be done to arrive at the conclusion you want.
     
  7. Oct 30, 2008 #6
    Which is?
     
  8. Oct 31, 2008 #7

    HallsofIvy

    User Avatar
    Science Advisor

    First, any equation like Ax= 0 has the "trivial solution": it is always true that A(0)= 0. The whole question is whether that is the only solution or whether other solutions exist.

    Do you know that Ax= 0 has only the trivial solution if and only if A is invertible? And so det(A)= 0? If you do and also know that det(An)= (det(A))n then the problem is trivial.

    If not then proof by induction may be simplest. Since we are given that Ax= 0 has only the trivial solution, the base case, n= 1 is trivial. Suppose Akx= 0 has only the trivial solution. Then Ak+1x= A(Akx)= 0. What can Akx be? And what does that tell you?
     
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