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Eigenvectors and Eigenvalues

  1. Nov 13, 2011 #1
    I have to be able to figure out eigenvalues and eigenvectors for 2x2 and 3x3 matrices for my physics course, but I have never taken linear algebra so I obviously have no idea what they even are. I need someone to basically teach me how to solve these problems because I have no knowledge of this material and cannot find any useful source to help me.

    First I have to find the eigenvalues and vectors for

    [a 0 1]
    [0 b 0]
    [0 0 c] where a =/= b =/= c.

    I looked around online and could only find that the answers for the eigenvalues are a, b, and c but I have no idea how that answer is derived. And I have no idea how to solve for eigenvectors except so any help would be great.

    Next I have to do the same thing for

    [a a 0]
    [2a 0 0]
    [0 a a] where a =/= 0.

    Again I know that the solution for the values is -a, a, and 2a but no clue on how to actually do the problem.

    I know nothing of this material except how to figure out the determinant of a matrix. So detailed steps on how to solve for eigenvalues and eigenvectors would be greatly appreciated.
     
  2. jcsd
  3. Nov 13, 2011 #2

    I like Serena

    User Avatar
    Homework Helper

    An eigenvalue λ and its eigenvector v are defined by:

    Av = λv

    That is, you try to find a vector that if you multiply A with it you find the same vector, except for length.

    To solve this, you do:
    Av - λv = 0
    (A - λI)v = 0

    This means (A - λI) must be singular, so its determinant must be zero.
    det(A - λI)=0 yields a polynomial in λ, that is called the "characteristic polynomial".
    Its roots are the eigenvalues of A.

    Once you have an eigenvalue, you solve:
    (A - λI)v = 0
    to find a vector v (or its multiple) that satisfies the equation.
     
  4. Nov 13, 2011 #3

    I like Serena

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    Homework Helper

    Applied to your first matrix you have:
    det(A - λI)=0
    (a-λ)(b-λ)(c-λ)=0
    Roots are a, b, and c.

    For the first eigenvalue "a", you get:
    A - aI =
    [0 0 1]
    [0 b 0]
    [0 0 c]

    The vector v=(1,0,0) satisfies (A - aI)v = 0, meaning that is the eigenvector.
     
  5. Nov 13, 2011 #4
    So i understand how you get the eigenvalues but i'm still confused with how you are arriving at the answer for the eigenvector.
     
  6. Nov 13, 2011 #5

    Mark44

    Staff: Mentor

    The matrix A - aI can be row reduced to
    [tex]\begin{bmatrix}0&0&1\\0&1&0\\0&0&0 \end{bmatrix}[/tex]

    This represents the equation (A - aI)x = 0, where x is a column vector with coordinates x1, x2 and x3.

    The reduced matrix represents the system
    x3 = 0
    x2 = 0
    with x1 being arbitrary. Since it's arbitrary, it's reasonable to set it to 1, giving the vector <1, 0, 0> as an eigenvector for the eigenvalue a.
     
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