In earliers blogs (last year) we looked at examples of linear algebra in terms of the behavior of lines and planes, e.g.
http://brane-space.blogspot.com/2010/05/analyzing-lines-planes.html
Now, we look at this branch of math from the viewpoint of one of its main elements: matrices. One of the primary objectives here is, given a matrix, say 2 x 2 or 3 x 3 , one will want to obtain the 'characteristic matrix" from either form, and thence the eigenvalues. (Which we saw a lot of when we were looking at linear solutions of certain differential equations, i.e. http://brane-space.blogspot.com/2011/01/homogeneous-linear-de-systems-breaking.html
Now, in this linear algebra context, let a 3 x 3 matrix A =
(a1.....0.......0)
(0.......a2.....0)
(0.......0.......an)
Then its characteristic polynomial will be obtained from:
P_A(t) =
(t- a1.....0.......0)
(0.......t – a2.....0)
(0.......0....... t - an)
Or:
P_A(t) = (t – a1)(t – a2) (t – an)
The eigenvalues can be obtained via solving for a1, a2, an, in the equation:
(t – a1)(t – a2) (t – an) = 0
Example:
Given the matrix:
A =
(1.. ..i)
(i.......-2.)
Find the characteristic polynomial as well as the eigenvalues.
We have:
P_A(t) =
(t - 1…… i)
( i....... t + 2)
Whence: P_A(t) = (t – 1)(t + 2) – (i)^2
P_A(t) = t^2 –t + 2t -2 - (i)^2 = t^2 +t -2 + 1= t^2 + t – 1
Since this is a quadratic equation, so we can find the eigenvalues (E1,2) using the quadratic formula:
E1,2 = {- b +/- [b^2 – 4 ac]^ ½} / 2a
Where the a, b, c denote the coefficients for the quadratic, with a the numerical coefficient for the exponent 2 term (t^2), b for the exponent 1 term(t) and c the exponent 0 term. Thus: a = 1, b = 1, c = -1
Then: E1,2 = {- 1 +/- [1^2 – 4 (-1)]^ ½} / 2(1)
E1,2 = {-1 +/- [5] ^ ½} / 2
So that:
E1 = (-1 + [5] ^ ½) / 2 = 0.618
E2 = (-1 - [5] ^ ½) / 2 = -1.618
Problems for the Math adventurous:
Find the characteristic polynomials and the eigenvalues for each of the following matrices:
1) A =
(1.. ..i)
(-i.......1)
2) A =
(1.. …. .2)
(2.......-2)
3) A =
(3.. ……2)
(-2...... 3)
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