Math 344 Notes Fall '11   This page  requires Internet Explorer 6+ MathPlayer or Mozilla/Firefox/Netscape 7+.

Martin Flashman's Courses
MATH 344 Linear Algebra Fall, 2011
Class Notes and Summaries
 WEEK Monday Wednesday Friday week1 8-22 8-24 8-26 week 2 8-29 8-31 9-2 week 3 9-5 9-7 9-9 week 4 9-12 9-11 9-13 week 5 9-19 9-21 9-23 week 6 9-26 9-28 9-30 week 7 10-3 10-5 10-7 week 8 10-10 10-12 10-14 week 9 10-17 10-19 10-21 week 10 10-24 10-24 10-26 week 11 10-31 11-2 11-4 week 12 11-7 11-9 11-11 week 13 11-14 11-16 11-18 week 14 NoClasses 11-21 11-23 11-27 week 15 11-28 11-30 12-2 week 16 12-5 12-7 11-9 week 17 Final Exam 12-12 Exam

8-22 discussion of course organization. [See syllabus on line.]
• Other texts in Linear Algebra:

• Finite Dimensional Vector Spaces by Paul Halmos
Linear algebra by Kenneth Hoffman and Ray Kunze.
Linear Algebra by Friedberg, Insel, and Spence
Linear Algebra by  Lang
Applied linear algebra by Ben Noble.
Linear Algebra and Its Applications by G. Strang
Topics in Abstract Algebra by I. Herstein
HSU Library Holdings
• On Proofs:

• How to Read and Do Proofs by D.Solow
The Keys to Advanced Mathematics by D. Solow
How to Solve It by G. Polya

• Motivational Question I:
• What can we say about powers of a square 2 by 2 matrix A and AX , where X is `(x,y)^{tr}`.
• What is the geometric interpretation of AX when X is  `(1,0)^{tr}` or `(0,1)^{tr}`?
• Examples:

A= (
 0 1 1 0

)
`A^n = I `  if `n` is even
`A^n =A` if `n` is odd.

A= (
 1 1 0 1

)

`A^n `  =
(
 1 n 0 1

)
Prove? [Use induction]

A  =
(
 r 0 0 s

)

1. r = 1/2;  s= 1/2   Discussion:  Same if  0< |r|, |s| < 1

`A^n`  `->`
(
 0 0 0 0

)
2. r= 2, s = 1/2  Discussion: ....
3. r= 1, s = -1   Discussion: `A^n = I `  if `n` is even; `A^n =A` if  `n `is odd

• Warm up:  Find the eigenvalues for

A= (

 0 1 1 0

)
• Complex numbers: C = { z = a+bi, where a and b are real numbers}

• The arithmetic of C. `||z|| = sqrt(a^2 + b^2) `
z = ||z||(cos(t) +i sin(t))    [called the "polar form" of the complex number]

The geometry of complex arithmetic:

If z = a+bi = ||z||(cos(t) +i sin(t)) and w  = c+di = ||w||(cos(s) +i sin(s)) then

z+w = (a+c)+(b+d)i which corresponds geometrically to the "vector " sum  of z and w in the plane, and

zw = ||z||(cos(t) +i sin(t)) ||w||(cos(s) +i sin(s))=  ||z|| ||w|| (cos(t) +i sin(t))(cos(s) +i sin(s))
= ||z|| ||w|| (cos(t) cos(s) - sin(t)sin(s) + (sin(t) cos(s) + sin(s)cos(t)) i)
= ||z|| ||w|| (cos(t+s) + sin(t+s) i)

So you use the product of the magnitudes of z and w to determine the magnitude of the product and use the sum of the angles to determine the angle of the product.

Powers of complex numbers.
Based on the previous result on products,
`z^n =  ||z||^n  (cos(nt) + sin(nt) i) `
If ||z||<1, then the powers of z will spiral into 0. If ||z||>1, then the powers of z will spiral outward without bound. If ||z||=1 and z is not 1, the the powers of Z will oscillate around the circle of radius 1 without having any limit.

Notation: cos(t) + i sin(t) is somtimes written as cis(t).
Note: If we consider the series for ex = 1 + x + x2/2! +x3/3! + ...
then eix = 1 + ix + (ix)2/2! +(ix)3/3! + ... = 1 + ix -  x2/2! - ix3/3! + ...
... = cos(x) + i sin(x)
Thus `e^{i*p} = cos(pi) + i sin(pi)= -1`.  So `ln(-1) = i *p`.
Furthermore: `e^{a+bi} = e^a*e^{bi} = e^a ( cos (b) + sin(b) i) `

Matrices with complex number entries.
If r and s are complex numbers in the matrix A, then as n get large if ||r|| < 1 and ||s|| < 1 the powers of A will get close to the zero matrix ,  if r=s=1 the powers of A will always be A,  and otherwise the powers of A will diverge .

Polynomials with complex coefficients.
Because multiplication and addition make sense for complex numbers, we can consider polynomials with coefficients that are complex numbers and use a complex number for the variable, making a complex polynomial a function from the complex numbers to the complex numbers.
This can be visualized using one plane for the domain of the polynomial and a second plane for the co-domain, target, or range of the polynomial.

The Fundamental Theorem of Algebra: If f is a non constant  polynomial with complex number coefficients then there is at least on complex number z*  where f(z*) = 0.

For more on complex numbers see: Dave's Short Course on Complex Numbers,

Fields: The structure used for "solving linear equations," finding inverses for matrices, and doing other parts of matrix algebra.

Definition- Axioms See Fields.
Examples: R, Q, C, Z2, Zp, where p is a prime number, F4, {f where f is a rational function defined on all but a finite number of real numbers}, algebraic numbers = {z: z is a complex number which is the root of a polynomial with integer coefficients}.
• Polynomials with coefficients in a field. [We will study these in some detail later].
• Matrices with entries in a field.

• Discussion of the properties of a field with 4 elements- from the homework assignment.
• A few more properties of Fields:
• Suppose F is a field.
• If  a and b are in F and ab = 0 then either a =0 or b=0.

• Proof: Case 1. If a=0 then we are done.
Case 2. If a is not 0, then a has an inverse... c where ca=1. Then b=1b= (ca)b=  c(ab) = c0 = 0.
Thus either  a =0 or b=0. IRMC   [I rest my case.]
• If  a, b and c are in F a+b = a+c implies b=c.

• Proof: Let k be the element of the field where k + a = 0 Then
b = 0 + b = (k +a)+b =k +(a+b)= k +(a+c) =(k +a)+c = 0 + c = c.
• If  a, b and c are in F with  a not equal to 0 ab = ac implies b=c.

• Proof: Similar to the last result.
• Let  n·1 stand for  1 + 1 + 1 + ... + 1 with n summands. Either (i) for all n,  n·1 is not 0 in which case we say the field has "characteristic 0" , or (ii) for some n, n·1=0. In the case n·1=0, there is a smallest n for which n·1=0, in which case we say the field has "characteristic n". For example: R, Q, and C all have characteristic 0, while Z2, Zp, where p is a prime number, and any finite field such as  F4, all have a non zero characteristic.
• If the characteristic of F is not zero, then it is a prime number.

• Proof: If n is not a prime, n = rs with 1<r,s<n. Let a = 1+1+...+1 r times and b= 1+1+...+1 s times. Then ab=n·1=0, so either a = 0 or b = 0, contradicting the fact the n was supposed to be the SMALLEST natural number for which n·1=0. IRMC.

• Solving equations with fields:
Suppose F is a field and `alpha` and `beta` are elements of F and X is a variable. We say that the equation `alpha` X = `beta` has a solution in F if there is an element of F, `gamma` where `alpha gamma = beta`.
Proposition: If `alpha \ne 0` then  there is an element of F, `gamma`,  where `alpha gamma = beta`.
Preface: We did some work on the side to "solve" the equation.
This analysis did the work of finding the candidate for `gamma` and seeing it was in F.
Proof: Suppose `alpha  in F and alpha  \ne 0.`
Then by the field axioms, there is an element of F, `alpha^{-1}` where `alpha^{-1} alpha = 1`.
Let  `gamma = alpha^{-1} beta`. Then `alpha gamma = alpha (alpha^{-1} beta) = (alpha alpha^{-1}) beta = 1 beta = beta`. IRMC!
• Systems of linear equations:
One linear equation with n unknowns with coefficients in a field F:
`alpha_1,alpha_2,...,alpha_n , and beta in F`
`alpha_1X_1 + alpha_2X_2 + ... + alpha_nX_n = beta `
k linear equationswith n unknowns with coefficients in a field F:
`alpha_{11}X_1 + alpha_{12}X_2 + ... + alpha_{1n}X_n = beta_1 `
`alpha_{21}X_1 + alpha_{22}X_2 + ... + alpha_{2n}X_n = beta _2`
.
.
.
`alpha_{k1}X_1 + alpha_{k2}X_2 + ... + alpha_{kn}X_n = beta _k`
The techniques learned in your previous linear algebra course can still be used for F since those techniques relied solely on the arithmetic that makes sense in any field.
• Motivation Question II

• Coke or Pepsi?
• Initial Sample: out of 1000 people.... 600 liked coke, 400 liked pepsi. ... v0 = (600,400)

• A. First resample: 500 of 600 stayed loyal to Coke, 100 switched to Pepsi.
300 of 400 stayed loyal to Pepsi , 100 switched to Coke.  so  v1 = (600,400)
• B. Alternative Resample: 400 of 600 stayed loyal to Coke, 200 switched to Pepsi.
•                       300 of 400 stayed loyal to Pepsi , 100 switched to Coke.  so  v1 = (500,500)

• Example A.
v0TA = v1 where

TA= (
 5/6 1/6 1/4 3/4

)

Example B.
v0TB = v1 where

TB= (
 2/3 1/3 1/4 3/4

)

Questions: If the pattern in case B continues with the proportions of those switching remaining the same:
what can we say about vn  when n is large?
Does the distribution of vn when n is large depend on v_0?
Is there some initial distribution that would remain unchanged under the pattern of switching in B?
i.e. is there a distribution v*0 = (c*,p*)  where v*0 = v*1 =  ... = v*n.
Solve the equation:  (c*, 1000-c*) TB = (c*, 1000-c*).

How are these questions related to Motivation Question I?

v0Tn = vn.