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 – a

_{1})(t – a

_{2}) (t – a

_{n})

The eigenvalues can be obtained via solving for a

_{1}, a

_{2}, a

_{n}, in the equation:

(t – a

_{1})(t – a

_{2}) (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= t

^{2}+ 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 (t

^{2}), 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 +/- [1

^{2}– 4 (-1)]} / 2(1)

E

_{1,2}=

**Ö**{-1 +/- [5] } / 2

So that:

E

_{1}=

**Ö**(-1 + [5] ) / 2 = 0.618

E

_{2}=

**Ö**(-1 - [5] ) / 2 = -1.618

_{}

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Practice Problems:

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

X =

(1.. …. .2)

(2.......-2)

Y =

(3.. ……2)

(-2...... 3)

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^{}

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