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Eigenvalues homework problem

  1. Jan 4, 2010 #1
    1. The problem statement, all variables and given/known data

    Show that lambda = 1 is an eigenvalue of the matrix

    2,-1, 6
    3,-3, 27
    1,-1, 7

    and find the eigenvalues and the corresponding eigenvectors


    2. Relevant equations



    3. The attempt at a solution

    I dont understand how to actually get eigenvalues and eigenvectors
    could someone help me out please
    =]
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 4, 2010 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Re: Eigenvalues

    Do you understand the definition of "eigenvalue"? "[itex]\alpha[/itex]" is an eigenvaue of matrix A if and only if there is some non-zero vector, v, such that [itex]Av= \alpha v[/itex].

    In particular, "1" is an eigenvalue of this matrix if and only if
    [tex]\begin{bmatrix}2 &-1 & 6 \\ 3 & -3 & 27 \\ 1 & -1 & 7\end{bmatrix}\begin{bmatrix} x \\ y \\ z\end{bmatrix}= 1\begin{bmatrix}x \\ y \\ z\end{bmatrix}[/tex]
    for some not all of x, y, z zero.

    Multiplying that left side gives
    [tex]\begin{bmatrix}2x- y+ 6z \\3x- 3y+ 27z \\ x- y+ 7z\end{bmatrix}= \begin{bmatrix}x \\ y \\ z\end{bmatrix}[/tex]

    Now, solve that using whatever method (row reduction, determinants, etc.) you choose. The most basic is to write that as the system of equations
    2x- y+ 6z= x, 3x- 3y+ 27z= y, and x- y+ 7z= z. Those are the same as x- y+ 6z= 0, 3x- 2y+ 27z= 0, and x- y+ 6z= 0.

    Notice that the first and second equations are the same- we really have only two equations, not 3 and so there is no unique answer. That had to happen. Obviously, v= 0 is a solution to [itex]Av= \alpha v[/itex] for any A and [itex]\alpha[/itex]. Saying there is a non-zero vector such that that is true means there is NO "unique" answer.

    Getting back to our equations: if we multiply x- y+ 6z= 0 by 2 we get 2x- 2y+ 12z= 0 while the second equation above was 3x- 2y+ 27z= 0. Subtracting 2x- 2y+ 12z= 0 from that, x+ 15z= 0 so x= -15z. That's as far as we can go, we cannot solve for a specific 5value of x. But putting x= -15z into x- y+ 6z= 0 gives -15x- y+ 6z= 0 or y= -7z.

    What does that mean? That means that any vector of the form v= <15z, -7z, z> satisfies Av= v, no matter what z is. Any such vector, for non-zero z, is an eigenvector corresponding to eigenvalue [itex]\lambda[/itex].

    More generally, if [itex]Av= \lambda v[/itex], then [itex]Av- \lambda v= (A- \lamba I)v= 0[/itex]. The point of writing it that way is that equation always has the solution v= 0 and will have a non-zero answer only if v= 0 is not unique. If [itex]A- \lambda I[/itex] had an inverse matrix, then we could multiply on both sides by that inverse and get the unique answer [itex]v= (A- \lambda I)^{-1}0= 0[/itex]. That is, [itex]Av= \lambda v[/itex] has a non-zero solution if and only if [itex]A- \lambda I[/itex] is not invertible which is the same as saying it has determinant 0.

    The "eigenvalue equation" or "characteristic equation" for matrix A is "[itex]det(A- \lambda I)= 0[/itex]". For your example, that is
    [tex]\left|\begin{array}{ccc}2-\lambda & -1 & 6 \\ 3 & -3-\lambda & 27 \\ 1 & -1 & 7-\lambda \end{array}\right|= 0[/tex]
    Working out that determinant on the left will give you a cubic polynomial equation (the "eigenvalue equation" or "characteristic equation" for an n by n matrix is typically an [itex]n^{th}[/itex] degree polynomial). Since you already know that [itex]\lambda= 1[/itex] is a solution, you can reduce it to a quadratic equation for the other two eigenvalues.
     
    Last edited: Jan 4, 2010
  4. Jan 4, 2010 #3
    Re: Eigenvalues

    ok i got to the cubic polynomial and just want to check that i have the right polynomial

    im rubbish with latex so let x = lambda

    x3+6x2+20x=0

    i then simplified to get

    -x(x2-6x-20)=0
     
  5. Jan 4, 2010 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Re: Eigenvalues

    Unfortunately, there is one serious problem with that! Remember that you were asked to show that "1" is an eigenvalue? (It is.) But [itex]1^3+ 5(1)+ 10= 26[/itex], not 0.
     
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