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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