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Math Help - eigenvalues

  1. #1
    Newbie euler42's Avatar
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    eigenvalues

    Could anyone tell me how to find the eigenvalue of a 4x4 matrix. I don't think the Rule of Sarrus applies.
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  2. #2
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    Quote Originally Posted by euler42 View Post
    Could anyone tell me how to find the eigenvalue of a 4x4 matrix. I don't think the Rule of Sarrus applies.
    There isn't really any general way to find a matrix's eigenvalues, other than explicitly calculating \Delta_A(\lambda) = det(\lambda I-A) (that is, assuming A is our matrix) and factoring. The roots of \Delta_A(\lambda) are the eigenvalues of A.

    Also, I don't really know what you're referring to by the Rule of Sarrus.

    Why don't you post some of your attempts and we will help you.

    P.S.

    There are some theorems that help finding eigenvalues, such as:

    (*) If the sum of elements of each row in the matrix A is equal to n for some constant n, then n is an eigenvalue with corresponding eigenvector (1 1 ... 1)^T

    Similarly, if the sum of elements of each column is equal to k for some constant k, then k is also an eigenvalue, however we do not know if (1 1 ... 1)^T will be an eigenvector.

    (*) If A is of order nxn and rank(A) < n, then 0 is an eigenvalue of A.

    Similarly, if rank(A-\lambda_0 I) < n then \lambda_0 is an eigenvalue of A, with geometric multiplicity n-rank(A - \lambda_0 I) = dim (Ker (A-\lambda_0 I))
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  3. #3
    Newbie euler42's Avatar
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    I know how to find the eigenvalue for a 2x2 and 3x3 matrix. That's easy. Essentially you're finding the roots for a quadratic and cubic equation respectively. The Rule of Sarrus is used to find the eigenvalue for a 3x3 matrix as follows:

    Rule of Sarrus - Wikipedia, the free encyclopedia

    I just wanted to know what method you would use to come up with a quartic equation so you can find its roots, assuming of course that is how you find the eigenvalue of a 4x4 matrix.
    Last edited by euler42; December 3rd 2009 at 09:22 PM.
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  4. #4
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    Quote Originally Posted by euler42 View Post
    I know how to find the eigenvalue for a 2x2 and 3x3 matrix. That's easy. Essentially you're finding the roots for a quadratic and cubic equation respectively. The Rule of Sarrus is used to find the eigenvalue for a 3x3 matrix as follows:

    Rule of Sarrus - Wikipedia, the free encyclopedia

    I just wanted to know what method you would use to come up with a quartic equation so you can find its roots, assuming of course that is how you find the eigenvalue of a 4x4 matrix.
    Well, as I said in my first post, the eigenvalues of a matrix A are the roots of the equation det(\lambda I - A) = 0.
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  5. #5
    Newbie euler42's Avatar
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    Ok, so how do you find the determinant of a 4x4 matrix?
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  6. #6
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    I had to look up "rule of Sarrus"! I've been using that for years! Unfortunately, it only applies to 3 by 3 determinants.

    There are two different ways to find the determinant of a general matrix:

    1) Expand by minors. That is choose on row or column to expand on. The determinant is a value in that row, times "+" or "_", depending on whether the "row+ column" is even or odd, times the "minor" which is the determinant formed by dropping that row and column. For an "n by n" determinant, that gives a sum of the numbers on the chosen row or column times "n-1 by n-1" determinants.

    For example, if the determinant is \left|\begin{array}{ccc}1 & 0 & 2\\ 1 & 2 & 1 \\ 2 & 0 & 1\end{array}\right| (I'm using a 3 by 3 determinant so you can check by using the "rule of Sarrus"), if I expand on the first column, I have 1\left|\begin{array}{cc}2 & 1 \\ 0 & 1\end{array}\right| - 1\left|\begin{array}{cc}0 & 2\\ 0 & 1\end{array}\right| + 2\left|\begin{array}{cc}0& 2 \\ 2 & 1\end{array}\right|= (1)(2)- (1)(0)+ 2(-4)= -6.[/tex]

    If I were clever (and weren't trying to show the general procedure), I would see those two "0"s in the second column and decide to expand on that column instead: I would get 2\left|\begin{array}{cc}1 & 2 \\ 2 & 1\end{array}\right|. That is equal to 2(1- 4)= -6.

    2) Use row reduction to a triangular matrix. There are three "row operations"
    a) Multiply an entire row by a number. That multiplies the determinant by that number.
    b) Swap two rows. That multiplies the determinant by -1.
    c) Add a multiple of one row to another. That does not change the determinant.

    Of course, once we have the triangular matrix, its determinant is just the product of the numbers on the diagonal. After we have that, we need to "correct" for (a) and (b)- dividing by any multiple used in (a) and multiplying by -1 if we used (b) an odd number of times.

    With the same example as before, to find the determinant of the matrix \begin{bmatrix}1 & 0 & 2\\ 1 & 2 & 1 \\ 2 & 0 & 1\end{bmatrix}, I would subtract the second row from the second and subtract 2 times the first row from the third to get \begin{bmatrix}1 & 0 & 2 \\ 0 & 2 & -1 \\ 0 & 0 & -3\end{bmatrix}. That is now in triangular form so its determinant is just (1)(2)(-3)= -6. Because I only used row operation (c), I have not changed the determinant. The determinant of the original matrix is, as we saw before, -6.

    Frankly, you should not be working with eignvalues and eigenvectors until after you have learned "expansion by minors" and/or "row reduction".
    Last edited by HallsofIvy; December 5th 2009 at 06:49 AM.
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