Entry
Math: Matrix: Eigenvector: Eigenvalue: Operation: Calculate: How to: 2 x 2 matrix
Jan 16th, 2006 18:05
Knud van Eeden,
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--- Knud van Eeden --- 15 January 2006 - 04:29 am --------------------
Math: Matrix: Eigenvector: Eigenvalue: Operation: Calculate: How to: 2
x 2 matrix
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Steps: Overview:
1. -Given the eigenvector equation
- - -
A . x = lambda . x
-
2. -Choose A to be a 2 x 2 matrix
3. -Writing this in components
[ a11 a12 ] - -
[ ] . x = lambda . x
[ a21 a22 ]
4. -Writing this out further
[ a11 a12 ] [ x1 ] [ lambda 0 ] [ x1 ]
[ ] . [ ] = [ ] . [ ]
[ a21 a22 ] [ x2 ] [ 0 lambda ] [ x2 ]
5. -Working this out, according to the vector rules,
gives 2 equations
a11 . x1 + a12 . x2 = lambda . x1 + 0 . x2
a21 . x1 + a22 . x2 = 0 . x1 + lambda . x2
6. -Put everything on one side of the equation
( a11 - lambda ). x1 + a12 . x2 = 0
a21 . x1 + ( a21 - lambda ) . x2 = 0
7. -Or thus
[ a11 - lambda a12 ] [ x1 ] [ 0 ]
[ ] . [ ] = [ ]
[ a21 a22 - lambda ] [ x2 ] [ 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 |
| | = 0
| a21 a22 - lambda |
10. Working that equation out according to the determinant rules shows
that you will need to solve an Nth degree polynomial in the unknown
lambda.
(a11 - lambda) . (a22 - lambda) - a12 . a21 = 0
11. Working out the parenthesis, according to the rules of algebra
2
a11 . a22 - a11 . lambda - lambda . a22 + lambda - a12 . a21 = 0
12. Rewriting the order of the terms in this 2nd degree polynomial
(of a similar structure to e.g. 3 . x^2 + 5 . x + 4 = 0)
in the variable lambda gives
lambda^2 - ( a11 + a22 ) . lambda + a11 . a22 - a12 . a21 = 0
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Note:
Another way of seeing this equation is that this just equals
lambda^2 - ( trace of given matrix A ) . lambda + ( determinant
of given matrix A ) = 0
13. Solving this 2nd degree polynomial, using the root formula
(similar as in a . x^2 + b . x + c = 0 gives that
x = ( -b +/- sqrt( b^2 - 4 . a . c ) ) / ( 2 . a )
in this special case here
a = 1
b = - (a11 + a22)
c = a11 . a22 - a12 . a21
filling this in gives
lambda = ( (a11 + a22) +/- sqrt( (-(a22 + a11))^2 - 4 . 1 .
(a11 . a22 - a12 . a21) ) ) / ( 2 . 1 )
or thus
lambda = ( (a11 + a22) +/- sqrt( (a22 + a11)^2 - 4 . (a11 .
a22 - a12 . a21) ) ) / 2
or thus
lambda1 = ( (a11 + a22) + sqrt( (a11 + a22)^2 - 4 . (a11 . a22 -
a12 . a21) ) ) / 2
lambda2 = ( (a11 + a22) - sqrt( (a11 + a22)^2 - 4 . (a11 . a22 -
a12 . a21) ) ) / 2
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This gives in general the solution for the exactly 2 eigenvalues
of the 2nd degree eigenvalue equation.
---
You can solve this equation exact using the root formula, or
otherwise using numerical methods.
14. Solving for the eigenvectors
Steps: Overview:
1. -Given the original system of linear equations
[ a11 - lambda a12 ] [ x1 ] [ 0 ]
[ ] [ ] [ ]
[ a21 a22 - lambda ] . [ x2 ] = [ 0 ]
2. Do the matrix multiplication to get the 2 linear equations
(a11 - lambda) . x1 + a12 . x2 = 0
a21 . x1 + (a22 - lambda) . x2 = 0
3. -Remove (arbitrarily) one of the equations (say the first
equation)
a21 . x1 + (a22 - lambda) . x2 = 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 = 0
-- -- --
x1 x1 x1
6. Work this out
a21 + (a22 - lambda) . x2 = 0
--
x1
7. -Solve this 2 linear equations for this new variable
x2
--
x1
8. -This gives as a solution
x2 -a21
-- = --------------
x1 (a22 - lambda)
8. -Possibly multiply nominator and denominator of
the solution quotient with some multiple m
x2 -a21 . m1
-- = -------------------
x1 (a22 - lambda) . m1
9. -Possibly normalize the magnitudes of the eigenvectors
by dividing by its length
2 2
sqrt( x1 + x2 )
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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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