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Thread: Finding Eigenvalues for 3x3 matrices

  1. #1
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    Finding Eigenvalues for 3x3 matrices

    Hi all.

    Just wondering if I could get some tips on how to find eigenvalues for 3x3 matrices by row/column expansion. We've been taught to set all but one entries of a row or column to zero, then for that non-zero entry use it to find the other eigenvalues via the smaller 2x2 determinant. I'm having a hard time with this initial step though and often I rely on just guessing row/column operations until I get the values I need. This is too time consuming to do in an exam situation so I was wondering, are there any things I should be looking out for, or is this something that I will only be able to improve on with constant practice.

    Cheers.
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  2. #2
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    Re: Finding Eigenvalues for 3x3 matrices

    Eigenvalues? It sounds like you're trying to evaluate a determinant, which is not quite the same thing.

    If A is your 3x3 matrix, the first thing you do is to subtract [lambda]I, where I is the 3x3 identity matrix, and [lambda] is the Greek letter (you could use any variable, but [lambda] is used most often by convention) then come up with an expression for the determinant. After simplification, you'll get a polynomial in [lambda] that you must solve (i.e., set it equal to zero and find [lambda]). In textbook problems, that solution is usually not hard to find.

    Could you give us an example of a 3x3 matrix you're working on, and let us know where you're getting stuck?
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  3. #3
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    Re: Finding Eigenvalues for 3x3 matrices

    Surely you don't mean "set all but one entries of a row or column to zero". You can "row reduce" a matrix by row operations and that is what it seems you mean but row operations do not keep the determinant nor eigenvalues the same.

    Given the matrix \begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i\end{bmatrix}, the "characteristic equation" is \left|\begin{array}{ccc} a-\lambda & b & c \\ d & e-\lambda & f \\ g & h & i-\lambda \end{array}\right|= 0. That is a cubic equation to solve for \lambda.
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    Re: Finding Eigenvalues for 3x3 matrices

    I think he's talking about the process of finding a determinant by cofactor expansion. His description sort of makes sense. But he seems to think we find eigenvalues "one at a time," which is odd. We need more information from the OP.
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    Re: Finding Eigenvalues for 3x3 matrices

    Finding Eigenvalues for 3x3 matrices-eig-question.png
    Finding Eigenvalues for 3x3 matrices-eig-ans.png

    Here is a question and its answer. The step I'm referring to is highlighted. In this case the matrix is transformed(?) such that the first column has all zeroes except for the 1,1 entry. Getting to this point I find difficult or just very time consuming :<
    Last edited by nalo6451; Jun 18th 2017 at 07:16 PM.
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    Re: Finding Eigenvalues for 3x3 matrices

    I see what your book and/or teacher is trying to do. It's true that finding eigenvalues involves finding a determinant, and though determinants can change with row operations, it's usually in a fairly simple way. It didn't occur to me before that one could use that procedure to find eigenvalues.

    To get from the second given form of the A - I\lambda matrix to the third is easy enough, but I can't get from the first to the second.

    I decided to do what I'm used to. I wrote out the determinant using a cofactor expansion on the top row. Here's what I got:

    \begin{array}{l} \left( { - 3 - \lambda } \right)\left[ {\left( {5 - \lambda } \right)\left( { - 2 - \lambda } \right) - \left( { - 6} \right)} \right]\\  - \left( 1 \right)\left[ {\left( { - 7} \right)\left( { - 2 - \lambda } \right) - \left( { - 1} \right)\left( { - 6} \right)} \right]\\  + \left( { - 1} \right)\left[ {\left( { - 7} \right)\left( 6 \right) - \left( { - 6} \right)\left( {5 - \lambda } \right)} \right] \end{array}

    Now you must expand this and collect terms to get  - {\lambda ^3} + 12\lambda  + 16. This is a little tedious, but it's just high school stuff.

    I gather you're OK with the procedure for finding the eigenvectors. To my mind, it's better to use an augmented matrix with a column of 0's on the right, because you're looking for vectors in the null space of A - I\lambda .

    To get the roots, there's a cubic formula you can use, or you can do it by pattern recognition. Since it's a cubic, we expect three binomial factors. The first element in each will be  \pm \lambda . There must be either one or three minus signs. The second element in each binomial factor will be an integer factor of 16, that is, either 2 or 4.

    So we divide the cubic, using long division, by \lambda  + 2 to get  - {\lambda ^2} + 2\lambda  + 8, then divide that in turn by another \lambda  + 2 to get  - \lambda  + 4 and there's your eigenvalues.

    This is a defective matrix. Are you charged with finding the generalized eigenvector? Or the Jordan decomposition?
    Last edited by zhandele; Jun 20th 2017 at 03:25 PM.
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