Entry
Math: Matrix: Eigenvector: Eigenvalue: Operation: Calculate: How to: 3 x 3 matrix
Jan 21st, 2006 16:56
Knud van Eeden,
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--- Knud van Eeden --- 15 January 2006 - 02:36 pm --------------------
Math: Matrix: Eigenvector: Eigenvalue: Operation: Calculate: How to: 3
x 3 matrix
===
Steps: Overview:
1. -Given the eigenvector equation
- - -
A . x = lambda . x
-
2. -Choose A to be a 3 x 3 matrix
3. -Writing this in components
[ a11 a12 a13 ]
[ ] - -
[ a21 a22 a23 ] . x = lambda . x
[ ]
[ a31 a32 a33 ]
4. -Writing this out further
[ a11 a12 a13 ] [ x1 ] [ lambda 0 lambda ] [ x1 ]
[ ] [ ] [ ] [ ]
[ a21 a22 a23 ] . [ x2 ] = [ 0 lambda 0 ] . [ x2 ]
[ ] [ ] [ ] [ ]
[ a31 a32 a33 ] [ x3 ] [ 0 0 0 ] [ x3 ]
5. -Working this out, according to the vector rules,
gives 3 equations
a11 . x1 + a12 . x2 + a13 . x3= lambda . x1 + 0 . x2 + 0 . x3
a21 . x1 + a22 . x2 + a23 . x3= 0 . x1 + lambda . x2 + 0 . x3
a31 . x1 + a32 . x2 + a33 . x3= 0 . x1 + 0 . x2 + lambda . x3
6. -Put everything on one side of the equation
(a11 - lambda) . x1 + a12 . x2 + a13 . x3 = 0
a21 . x1 + (a22 - lambda) . x2 + a23 . x3 = 0
a31 . x1 + a32 . x2 + (a33 - lambda) . x3 = 0
7. -Or thus
[ a11 - lambda a12 a13 ] [ x1 ] [ 0 ]
[ ] [ ] [ ]
[ a21 a22 - lambda a23 ] . [ x2 ] = [ 0 ]
[ ] [ ] [ ]
[ a31 a32 a33 - lambda ] [ x3 ] [ 0 ]
8. -This equation will only be generally true if the determinant is
zero
-
(as the vector x is in general not zero)
9. -Thus you have to work out the equation that the determinant is
zero.
| a11 - lambda a12 a13 |
| |
| a21 a22 - lambda a23 | = 0
| |
| a31 a32 a33 - lambda |
10. Working that equation out (according to the determinant rules)
shows that you will need to solve an 3th degree polynomial in the
unknown lambda.
+ ((a11 - lambda) . (((a22 - lambda) . (a33 - lambda)) - (a23 .
a32))) - ((a12) . ((a21 . (a33 - lambda)) - (a23 . a31))) + (a13 .
((a21 . a32) - ((a22 - lambda) . a31))) = 0
11. Working out the parenthesis (according to the rules of algebra) and
rewriting the order of the terms in this 3nd degree polynomial (of
a similar structure to e.g. 6 . x^3 + 3 . x^2 + 5 . x + 4 = 0) in
the variable lambda gives
-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
12. Making the sign of the highest degree positive, for esthetic
purposes, gives
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
13. Solving this 3rd degree polynomial
(similar as in a . x^3 + b . x^2 + c . x + d = 0)
in this special case here
a = 1
b = - (a11 + a22 + a33)
c = (a22 . a33 - a12 . a21 + a22 . a11 + a11 . a33 - a23 . a32 -
a31 . a13)
d = (- a23 . a12 . a31 - a32 . a21 . a13 + a12 . a21 . a33 -
a22 . a11 . a33 + a23 . a32 . a11 + a22 . a31 . a13)
for the unknown lambda always gives exactly 3 solutions (according
to the fundamental theorem of algebra).
---
This gives in general the solution for all 3 eigenvalues of the
3rd degree eigenvalue equation.
---
You can solve this equation exact using the Cardano formula, or
otherwise using numerical methods.
14. Solving for the eigenvectors
Steps: Overview:
1. -Given the original system of linear equations
[ a11 - lambda a12 a13 ] [ x1 ] [ 0 ]
[ ] [ ] [ ]
[ a21 a22 - lambda a23 ] . [ x2 ] = [ 0 ]
[ ] [ ] [ ]
[ a31 a32 a33 - lambda ] [ x3 ] [ 0 ]
2. Do the matrix multiplication to get the 3 linear equations
(a11 - lambda) . x1 + a12 . x2 + a13 . x3 = 0
a21 . x1 + (a22 - lambda) . x2 + a23 . x3 = 0
a31 . x1 + a32 . x2 + (a33 - lambda) . x3 = 0
3. -Remove (arbitrarily) one of the equations
(say the first equation)
a21 . x1 + (a22 - lambda) . x2 + a23 . x3 = 0
a31 . x1 + a32 . x2 + (a33 - lambda) . x3 = 0
4. -Select one of the variables (say the first variable, x1)
5. -Divide by that variable (say that that variable x1 is not zero)
a21 . x1 + (a22 - lambda) . x2 + a23 . x3 = 0
-- -- -- --
x1 x1 x1 x1
a31 . x1 + a32 . x2 + (a33 - lambda) . x3 = 0
-- -- -- --
x1 x1 x1 x1
6. Work this out
a21 + (a22 - lambda) . x2 + a23 . x3 = 0
-- --
x1 x1
a31 + a32 . x2 + (a33 - lambda) . x3 = 0
-- --
x1 x1
7. -Solve this 2 linear equations for this new variables
x2
--
x1
and
x3
--
x1
8. -This gives as a solution
x2 a31 . a23 - a33 . a21 + lambda . a21
-- = -------------------------------------------------------------
x1 2
a22 . a33 - a22 . lambda - lambda . a33 + lambda - a23 . a32
x3 a21 . a32 - a22 . a31 + lambda . a31
-- = -------------------------------------------------------------
x1 2
a22 . a33 - a22 . lambda - lambda . a33 + lambda - a23 . a32
9. -Possibly multiply nominator and denominator of
the solution quotient with some multiple m
(because a scaled version of the eigenvector is also
a solution). This will give the eigenspace for that
specific eigenvalue.
10. -Possibly normalize the magnitudes of the eigenvectors, by
making it a unit vector, as usual by dividing by its length
2 2 2
sqrt( x1 + x2 + x3 )
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Internet: see also:
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Math: Transformation: Eigenvector: Eigenvalue: Link: Can you give an
overview of links?
http://www.faqts.com/knowledge_base/view.phtml/aid/39001/fid/1856
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