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 – a_n) = 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 (E_1,2) using the quadratic formula:

E_1,2 = 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:  E_1,2 = Ö{- 1 +/- [1² – 4 (-1)]} / 2(1)


E_1,2= 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)