Entry

Math: Matrix: Eigenvalue: Operation: Calculate:How to: 3 x 3 matrix: Example:[[1 2 3][4 5 6][7 8 9]]

Jan 15th, 2006 12:10
Knud van Eeden,

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--- Knud van Eeden --- 15 January 2021 - 04:39 pm --------------------
Math: Matrix: Eigenvalue: Operation: Calculate:How to: 3 x 3 matrix: 
Example:[[1 2 3][4 5 6][7 8 9]]
===
Steps: Overview:
 1. -Given the eigenvector equation
      -   -            -
      A . x = lambda . x
            -
 2. -Choose A to be a 3 x 3 matrix
 3. -Or thus written in components
     [ a11 a12 a13 ]
     [             ]     -             -
     [ a21 a22 a23 ]  .  x  = lambda . x
     [             ]
     [ a31 a32 a33 ]
                     -
 4. -Here the matrix A is given to be
     [ 1  2  3 ]
     [         ]
     [ 3  4  5 ]
     [         ]
     [ 6  7  8 ]
 5. -Or thus
     [ 1  2  3 ]
     [         ]     -             -
     [ 4  5  6 ]  .  x  = lambda . x
     [         ]
     [ 7  8  9 ]
 6. -Or thus you have to solve the equation, determinant equals zero
     | 1 - lambda   2            3          |
     |                                      |
     | 4            5 - lambda   6          | = 0
     |                                      |
     | 7            8            9 - lambda |
 7. To find lambda, the eigenvalue, or thus the scaling factor with
    which the eigenvector is multiply, you need to solve this 2nd
    degree polynomial where here thus given the components
     a11 = 1
     a12 = 2
     a13 = 3
     a21 = 4
     a22 = 5
     a23 = 6
     a31 = 7
     a32 = 8
     a33 = 9
 8. The characteristic polynomial
     lambda^3  - (a11 + a22 + a33) . lambda^2  - (-a22 . a33 + a12 . 
a21 - a22 . a11 - a11 . a33 + a23 . a32 + a31 . a13) . lambda - (a23 . 
a12 . a31 + a32 . a21 . a13 - a12 . a21 . a33 + a22 . a11 . a33 - 
a23 . a32 . a11 - a22 . a31 . a13) = 0
    becomes thus here
     lambda^3  - (1 + 5 + 9) . lambda^2  - (-5 . 9 + 2 . 4 - 5 . 1 - 
1 . 9 + 6 . 8 + 7 . 3) . lambda + (6 . 2 . 7 + 8 . 4 . 3 - 2 . 4 . 9 + 
5 . 1 . 9 - 6 . 8 . 1 - 5 . 7 . 3) = 0
    or thus worked out
     lambda^3 - 15 . lambda^2 - 18 . lambda = 0
 9. This 3rd degree polynomial has thus exactly 3 solutions
    1. -Using the exact Cardano formula or numeric methods to find the
        corresponding lambda gives the 3 solutions for lambda:
          lambda1 = 0
          lambda2 = 15/2 + 3/2 . sqrt(33) = 16.11684397...
          lambda3 = 15/2 - 3/2 . sqrt(33) = -1.116843971...
     The 3 eigenvalues, or scale factors of the eigenvectors, equal
     thus
      0, 16.11684397... and -1.116843971
===
To check this goto e.g.
http://www.arndt-bruenner.de/mathe/scripts/engl_eigenwert.htm
and fill in
1  2  3
3  4  5
6  7  8
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Internet: see also:
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Math: Transformation: Eigenvector: Link: Can you give an overview of 
links?
http://www.faqts.com/knowledge_base/view.phtml/aid/39001/fid/1856
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