Thursday, July 9, 2009

Eigenvalues, part 2

Continuing with the previous post, let's do a bigger problem. Suppose we have a 3 x 3 matrix A:

A = [ 2  1  0 ]
[ 1 4 0 ]
[ 2 5 2 ]


The characteristic polynomial of A is given by:

det|A - kI|

kI = [ k 0 0 ]
[ 0 k 0 ]
[ 0 0 k ]

A - kI = [ 2-k 1 0 ]
[ 1 4-k 0 ]
[ 2 5 2-k ]


Remembering that for a 3 x 3 matrix

M = [ a1  a2  a3 ]
[ b1 b2 b3 ]
[ c1 c2 c3 ]


the determinant det|M| is the sum of three terms. The first is:

  a1 * det | [ b2  b3 ] |
[ c2 c3 ]


The other two are similar, except that second term, starting with a2 * ..., is multiplied by - 1. So, going back to our problem:

A - kI = [ 2-k    1    0 ]
[ 1 4-k 0 ]
[ 2 5 2-k ]


det|A = kI| is:

[ (2 - k) * ((4 - k)(2 - k) - 0) ] 
- 1 * ( (1)(2 - k) - 0 )
+ 0

= (2 - k)(8 - 6k + k^2) - (2 - k)
= (2 - k)(k^2 - 6k + 7)


Setting this equal to zero, one solution is

k = 2


We use the quadratic formula to obtain the
other two roots:

( -b +/- sqrt(b^2 - 4ac) ) / 2a
( 6 +/- sqrt (36 - 28) ) / 2
( 6 +/- sqrt(8) ) / 2
3 +/- sqrt(2)


The solutions are: 4.414, 2, 1.586

To find the first eigenvector (for k = 3 + sqrt(2) ),
we have:

Av = kv


where v is a vector with values x,y,z:

v = [ x ]
[ y ]
[ z ]


For the first eigenvalue we have:

A v1 = k1 v1

[ 2 1 0 ] [ x ] = (3 + sqrt(2)) [ x ]
[ 1 4 0 ] [ y ] = (3 + sqrt(2)) [ y ]
[ 2 5 2 ] [ z ] = (3 + sqrt(2)) [ z ]


That is:

2x + y =  (3 + sqrt(2)) x
x + 4y = (3 + sqrt(2)) y
2x + 5y + 2z = (3 + sqrt(2)) z


From the first equation:

y =  (1 + sqrt(2)) x


If we choose

x = 1
y = 1 + 1.414 = 2.414


2x + 5y + 2z = 4.414z
z = (2 + 5 * 2.414) / 2.414
z = 5.82


To normalize, we divide each by:

sqrt(1 + 2.414^2 + 5.82^2)
= 6.38

x = 0.157
y = 0.378
z = 0.912


Check this is a unit vector:

0.157^2 + 0.378^2 + 0.912^2 = 0.9993


In R:

v = c(2,1,0,1,4,0,2,5,2)
A = matrix(v,nrow=3,byrow=T)
A

[,1] [,2] [,3]
[1,] 2 1 0
[2,] 1 4 0
[3,] 2 5 2

eigen(A)

$values
[1] 4.414214 2.000000 1.585786

$vectors
[,1] [,2] [,3]
[1,] 0.1565580 0 0.9124870
[2,] 0.3779645 0 -0.3779645
[3,] 0.9124870 1 0.1565580

1 comment:

Thomas Olson said...


Umm, this is tedious. How bout instead I put a u in front of another letter for the unit vector. I'm leaving out the square root and other mess.

uv=[1, 1+sqrt(2), (1+sqrt(2))^2]

vw= [(1/2-(1/2)*sqrt(5))^2, -(1/2-(1/2)*sqrt(5)), 1]
uv=vw

Basically it's [1, a, a^2] for a quadratic solution. You can get the other value with [a^2, -a, 1]

Notice I reversed the order and snuck in a negative sign? Btw, I found this by doing three consecutive Pell numbers. This works with Fibonacci number and any of the recursive number things.