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