## Monday, April 24, 2017

### Math Revisited: Looking At Linear Algebra

In earlier posts 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 examine this fascinating branch of advanced math at deeper levels. Some aspects will resonate from when we were looked linear solutions of certain differential equations, i.e. http://brane-space.blogspot.com/2011/01/homogeneous-linear-de-systems-breaking.html

In this linear algebra context, let a 3 x 3 matrix A =

(a 1.....0.......0)
(0.......a 2.....0)
(0.......0.......a n)

We're first  interested in obtaining its characteristic polynomial from:

P_A(t) =

(t- a 1.....0.......0)
(0.......t – a 2.....0)
(0.......0....... t - a n)

Or:

P_A(t) = (t – a1)(t – a2) (t – a n)

The eigenvalues can be obtained via solving for a1, a 2, a n, 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= t2 + t – 1

Since this is a quadratic equation, so we can find the eigenvalues (E1,2) using the quadratic formula:

E1,2 = Ö{- b +  [b 2 – 4 ac]} / 2a

Where  a, b, c denote the coefficients for the quadratic, with a the numerical coefficient for the exponent 2 term (t2), b for the exponent 1 term(t) and c the exponent 0 term. Thus: a = 1, b = 1, c = -1

Then:  E1,2 = Ö{- 1 +/- [12 – 4 (-1)]} / 2(1)

E1,2= Ö{-1 +/- [5] } / 2

So that:

E1 = Ö(-1 + [5] ) / 2 = 0.618

E2=Ö (-1 - [5] ) / 2 = -1.618

Practice Problems:

Find the characteristic polynomials and eigenvalues for each of the following matrices:

X =

(1.. …. .2)
(2.......-2)

Y =

(3.. ……2)
(-2...... 3)