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Homework Help: Systems of Linear Equations

  1. May 3, 2009 #1
    1. The problem statement, all variables and given/known data

    For a few problems dealing with eigenvectors, I substituted my eigenvalues into the characteristic equations. I got systems of linear equations. I need to find the general solution to the systems in order to find the corresponding eigenvectors.
    For example in one problem I have to solve:

    [tex]A = \left[\begin{array}{ccccc} -3&-3 \\ -4&-4 \end{array}\right][/tex] [tex]\left[\begin{array}{ccccc} x\\ y \end{array}\right][/tex] [tex]= \left[\begin{array}{ccccc} 0\\ 0 \end{array}\right][/tex]


    It looks as if x,y are equal. I think I might need to write one in terms of the other but I'm no sure which.

    or for example the system of equations:


    We have 3 variables and 3 equations which are exactly the same. How do we decide which variable to use as the "free" variable?

  2. jcsd
  3. May 3, 2009 #2


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    Both these are easy to solve by inspection. For example, in the first one, set x = 1 and y = -1.
  4. May 3, 2009 #3


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    It doesn't matter. In the 2nd example, you can pick x = -y-z or y = -x-z or z = -x-y. You will still have the same solution space even though the eigenvectors don't look the same.
  5. May 4, 2009 #4


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    Remember that the set of all eigenvectors corresponding to a given eigenvalue for a vector space- there is not one single solution. In fact, the condition that [itex]A-\lambda I[/itex] NOT have an inverse guarentees that your set of equation not have a single solution.

    For the first, -3x-3y= 0, -4x- 4y= 0, it is not the case that "x,y are equal". Both equations are equivalent to -x-y= 0 and then y= -x, not x. Every eigenvector is of the form <x, -x> = x<1, -1>. The vector space of all eigenvectors is spanned by the single vector <1, -1>.

    For the second set where every equation is of the form -x-y-z= 0 you have one equation in three variables. That means you can solve for any one in terms of the other two- it doesn't matter which you choose. For example, z= -x- y. Taking x=1, y= 0, z= -1 so <1, 0, -1> is in the space. Taking x= 0, y= 1, z= -1 so <0, 1, -1> is also in the space. Those two vectors form a basis for the space of eigenvectors.

    If you had chosen instead to solve for y, y= -x- z. Now taking x=1, z= 0 gives <1, -1, 0> and taking x=0, z= 1 gives <0, -1, 1> . Those form a different basis for the same space.

    If you had chosen to solve for x, x= -y- z giving <-1, 1, 0> and <-1, 0, 1>, yet another basis for the same space.
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