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Tips on finding the eigenvalues of a 3x3 matrix

  1. Apr 16, 2009 #1
    I find it rather tedious to calculate the eigenvalues of a 3x3 matrix. For example
    [tex]The \emph{characteristic polynomial} $\chi(\lambda)$ of the
    3$3 \times 3$~matrix
    \[ \left( \begin{array}{ccc}
    1 & -1 & -1 \\
    -1 & 1 & -1 \\
    -1 & -1 & 1 \end{array} \right)\]
    is given by the formula
    \[ \chi(\lambda) = \left| \begin{array}{ccc}
    1-\lambda & -1 & -1 \\
    -1 & 1-\lambda & -1 \\
    -1 & -1 & 1-\lambda \end{array} \right|.\] [/tex]

    Now if I do this by develloping the minors I get a cubic equation and I can't solve it without at least 30 minutes. I find it time consuming, especially during an exam.
     
  2. jcsd
  3. Apr 16, 2009 #2

    Pengwuino

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    Gold Member

    One tip is that row operations can help you solve for the characteristic polynomial. Remember, given the matrix [tex] A [/tex], however, perform row operations on the matrix [tex]A - \lambda I[/tex], and not just [tex] A [/tex] in order for it to be valid.
     
  4. Apr 17, 2009 #3

    matt grime

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    Science Advisor
    Homework Helper

    The first thing to remember is that you got this question from a teacher or a book (possibly both) in all likelihood. They're not trying to make you really good at solving cubic equations (for which there is a formula by the way) or test your abilities to do something not particularly interesting like searching hard for solutions - they're testing that you understand what an eigenvalue is.

    Thus one can deduce that there will almost always be simple roots of the equations in the examples they set you. Have you tried plugging in some integers near to zero to the characteristic polynomial? Just by looking at the matrix one can see that -1 is an eigenvalue (there is an obvious candidate for an eigenvector, the column vector of 1s, equivalently the sum of the entries in each row is -1).
     
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