first classprofessor william haboush 305 altgeld 333-6498 [email protected] 1:30 to 2:30 tuesday...

Professor William Haboush305 [email protected]

1:30 to 2:30 Tuesday1 to 2 WednesdayOften around for much of afternoonWill find in office or lounge upstairs (likely)

Office Hours (may change)

Course is first six chapters of the book

2 midterms + 1 final

Midterm 1 100 points

Midterm 2 100 points

Final 200 points

HW 80 points

Total 480 points

Grading

Just puts them in order and decides A+, A…

Median B-Fair number of A'sLots of homework

First classSunday, August 22, 2010

20:14

Notes

The rigid geometry of n-spaceLeaves straight lines straight and 0 point alone

Google search uses eigenvaluesOften related to computers

Solves systems of linear equations

Gaussian elimination

2x1-x2+x3-x4-x5=1X1+x2+x3 +x4+x5=-2x1-x3-x4-x5=1

Example

2 systems of equations are equivalent if all the solutions of one and all the solutions of the other are the same

2 -1 1 -1 -1 1

1 1 1 1 1 -2

1 0 -1 -1 1 0

1 1 1 1 1 -2

0 -3 -1 -3 -3 5

0 1 2 2 0 -2

1 1 1 1 1 -2

2 -1 1 -1 -1 1

1 0 -1 -1 1 0

Switch rows one and 2

Subtract 2 row one from row 2Subtract row one from row 3, multiply by -1

1 1 1 1 1 -2

0 1 2 2 0 -2

0 -3 -1 -3 -3 5

Switch rows 2 and 3

1 1 1 1 1 -2

0 1 2 2 0 -2

0 0 5 3 -3 1

Row 3 plus 3 row 2

Leave out the x's

What linear algebra isTuesday, August 24, 2010

09:41

Notes

1 0 -1 -1 1 0

0 1 2 2 0 -2

0 0 5 3 -3 1

R1-R2

1 0 -1 -1 1 0

0 1 2 2 0 -2

0 0 1 3/5 -3/5 1/5

R3/5

1 0 -1 -1 1 0

0 1 0 4/5 6/5 -12/5

0 0 1 3/5 -3/5 1/5

R2-2R3

Just put x1,x2,x3 in terms of x4,x5

Replace continuously by simplified systems of equations which are equivalent

Switch equations aroundMultiply an equation by a non-zero constantAdd a multiple of one equation to another

What can one do?

Gaussian elimination always works with any system of equation

Can choose the free variables, but bound variables are then determined

Rank: number of bound variables

Also the dimension of the solution spaceNullity: number of free variables

Notes

Due next ThursdayAssignments always due the following Thursday (7 or 9 days)

Assignment: pg 10: 3cd, 4(c,d), 5bd, 6 bdh: pg 23: 1d -h, 2abc, 5 bcj

If no solution, then inconsistentOtherwise one solution

n equations and n unknowns

will probably have no solutionIf more equations then unknowns, then overdetermined system

If fewer equations then unknowns, then underdetermined system

Terminology:

Always has at least one solution (all x i=0)If more unknowns then equations, then at least one other solution

Homogeneous system (all 0's)

If use reduced Gauss-Jordan form then don't really need to (already has essentially done it)

Back substitution: stressed in the book. If system is in Gauss-Jordan form (not reduced), then put each back into the previous to get answers

Systems of EquationsThursday, August 26, 2010

09:30

Notes

On pg 8

4-x2-x3+x4=0x1+x2+x3+x4=62x1+4x2+x3-2x4=-13x1+x2-2x3+2x4=3

0 -1 1- 1 -4

1 1 1 1 6

2 4 1 -2 -1

3 1 -2 2 3

1 1 1 1 6

0 -1 1- 1 -4

2 4 1 -2 -1

3 1 -2 2 3

1 1 1 1 6

0 1 1 1- 4

0 2 -1 -4 -13

0 -2 -5 -1 -15

1 1 1 1 6

0 1 1 1- 4

0 0 -3 -2 -21

0 0 -3 -3 -7

1 1 1 1 6

0 1 1 1- 4

0 0 -3 -2 -21

0 0 0 -1 14

1 0 0 0 30

0 1 0 0 -79/3

0 0 1 0 49/3

0 0 0 1 -14

Whole thing is augmented matrix

First 4 are matrix of coefficients

In book they forget to put the -4 in

The thing you use to remove everything underneath is called the pivot

Switch R1 and R2

R3-2R1R4-3R1-1*R2

R3-2R2R4+2R2

R4-R3

Reduce it

Example

1 1 1 1 1 1

-1 -1 0 0 1 -1

-2 -2 0 0 3 1

0 0 1 1 3 3

1 1 2 2 4 4

1 1 1 1 1 1

0 0 1 1 2 0

0 0 2 2 5 3

0 0 1 1 3 3

0 0 1 1 3 3

1 1 1 1 1 1

0 0 1 1 2 0

0 0 0 0 1 3

0 0 0 0 1 3

0 0 0 0 1 3

1 1 1 1 1 1

0 0 1 1 2 0

0 0 0 0 1 3

0 0 0 0 0 0

0 0 0 0 0 0

1 1 1 1 0 -2

0 0 1 1 0 -6

0 0 0 0 1 3

0 0 0 0 0 0

System is very underdetermined

ExamplesThursday, August 26, 2010

09:47

Notes

0 0 0 0 0 0

1 1 0 0 0 4

0 0 1 1 0 -6

0 0 0 0 1 3

0 0 0 0 0 0

0 0 0 0 0 0

x1+x2=4x3+x4=-6x5=3

x1=4-x2

x3=-6-x4

x5=3

Solution

Each row starts with a 1Everything below a 1 is a 0Above any INITIAL 1 there is nothing but zeros

Bound variables:x1,x3,x5

Free variables: x2,x4

This is a singular system

Notes

'r' rows'n' columns

An r x n matrix

A rectangle of numbers with a bracket on each end

Use a capital letter to represent a matrix

Can add, subtract, multiply a constant

Matrices and Matrix arithmetic (algebra)Thursday, August 26, 2010

10:20

Notes

Will have a web page eventually

AB != BAAB = 0 does not imply that either A or B is zero

1 -1 2

1 1 -1

x1

x2

x3=

x1-x2+2x3x1+x2-x3

AX=B

If AC=0Then A(M+C)=AM+AC=AM

Cannot simplify further, as it is not commutativeBinomial theorem does not apply

A(AA)=(AA)AAnAm=AmAn=Am+n

Powers only valid with square matrices

However A commutes with all powers of itself

(A+B)(A+B)=AA+AB+BA+BB

Im= Special matrices

1 0 … 0

0 1 … 0

… … 1 …

0 0 … 1

Identity Matrix - plays role of 11 if i=j0 otherwise

Im(ij) =

a11x1+ … +a1mxm=b1

… an1x1+ … +anmxm=bn

Same as AX=B(A-1A)X=A-1BIX=A-1BX=A-1B

Inverse of a matrixSquare matrices only

B is called a right inverse for A if AB=In

B is called a left inverse for A if BA =In

B is an inverse for A if AB=BA=In

-1 2

3 -2

Find the inverse:

-2/-4 -2/-4

-3/-4 -1/-4

-1 2

3 -2

1/2 1/2

3/4 1/4

1 0

0 1=

1/2 1/2

3/4 1/4-1 2

3 -2=

1 0

0 1

Thus it is both a right and left inverse

1 4 3

-1 -2 0

2 2 3

a b

c d Inverse is:

d/(ad-bc) -b/(ad-bc)

-c/(ad-bc) a/(ad-bc)

1/(ad-bc) !=0 and this is the invere

inverse-1/2 -1/2 1/2

1/4 -1/4 -1/4

1/6 1/2 1/6

A right inverse for A exists1.Every system of equations AX=B has a unique solution2.A left inverse for A exists3.Reduced Gauss-Jordan form of A is the identity4.2 implies 1 here and 1 implies 2, thus 1 iff 23 implies 24 implies 2

CA=In

AB=In

These 4 are equivalentThese conditions are the definition of when the square matrix A is non-singular

AA*=A*A=I

B-1A-1(AB)=B-1(A-1A)B=B-1IB=B-1B=ICan extend for more than 2 also(A1A2…Aq)-1=Aq

-1…A1-1

(AB)-1=B-1A-1 as

Matrix algebraTuesday, August 31, 201009:31

Notes

Switch 2 rows1.Add a multiple of row i to row j2.Multiply row j by a non-zero constant3.

Elementary row operations

a11 … a1n

… …

ar1 … arn

Ir

Do E to A where E is an elementary row operation1.Do E to Ir and get E, consider EA2.

2 things to do

1 and 2 produce the same resultTheorem: (Main theorem of linear algebra)

1 -1 2

2 1 1

2 -3 2

R2=R2-2R1

1 -1 2

0 3 -3

2 -3 2

Do E to A

Do E to Ir 1 0 0

0 1 0

0 0 1

1 0 0

-2 1 0

0 0 1

EA 1 0 0

-2 1 0

0 0 1

1 -1 2

2 1 1

2 -3 2

1 -1 2

0 3 -3

2 -3 2

*

=

Proof of theorem is just done by hand sticking them together (like in further maths group theory)

Calculating the inverse

1 4 3 1 0 0

-1 -2 0 0 1 0

2 2 3 0 0 1

1 4 3 1 0 0

0 2 3 1 1 0

0 -6 -3 -2 0 1

1 4 3 1 0 0

0 2 3 1 1 0

0 0 6 1 3 1

1 4 0 1/2 -3/2 -1/2

0 2 0 1/2 -1/2 -1/2

0 0 6 1 3 1

1 0 0 -1/2 -1/2 1/2

0 1 0 1/4 -1/4 -1/4

0 0 1 1/6 1/2 1/6

Put I in after, do row operations to flip other side to be the identity then have inverse on side which originally had identity

Works because of theorem from earlier

A I

A1=E1A E1

A2=E2E1A E2E1

EjEj-1…E1A=Aj EjEj-1…E1

Eventually, Aj=I, thus will be inverse

Thus if reduced gauss-jordan form of A is identity then there is a left inverse

Notes

Homework:Pg 42-43: 1abcde, 2abcd, 4ab, 13Pg 56: 8, 24ab, 25

The result of doing one of the elementary row operations to the identity

1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1

Switch 1st and 3rd and get an elementary matrix

0 0 1 0

0 1 0 0

1 0 0 0

0 0 0 1

Or do any of the row operationsIf there are 2 operations done, it is not elementaryAdding a multiple

1 0 0 0

0 1 0 3

0 0 1 0

0 0 0 1

Multiplying

1 0 0 0

0 4 0 0

0 0 1 0

0 0 0 1

Elementary matrix

Take the r by m matrix A and do the elementary row operation E to it1.Do E to the IR, get E, and then multiply EA2.

2 procedures

1 and 2 produce the same resultTheorem:

Find inverse by the writing adjacent method - proved last week

A I

A1 E1

… …

Aq-1 N

Aq EqN

EqNA=EqAq-1=A by theorem

E1A=A1

Show by induction

A right inverse for A exists1.A left inverse for A exists2.Every system of equations AX=B has a unique solution3.Reduced GJ form of A is I r4.Det(A) != 05.

Equivalent statements

Not every matrix has oneThe LU factorization

Where L is lower triangularU is upper triangular

A=LU

THM: If A can be reduced to an upper triangular matrix using no switches then A=LU for some L and U such that L is lower triangular with 1's on the diagonal and U is upper triangular

Upper triangular matrix uUpper triangular if everything below diagonal is zeroLower triangular if everything above diagonal is zero

Product of two upper triangular matrices is upper triangularProduct of two lower triangular matrices is lower triangular

GLN is the set of non-singular n by n matrices

LaA=Aq, A=L1-1Aq

AqEq…E1=L1

Put -m instead of mPut 1/c instead of cSwitch same two rows (is its own inverse)

Any elementary matrix has an inverse, so this is true

If M1 up through Ml all have inverses (M1-1…Ml

-1) then the inverse of (M1…Ml)-1=Ml-1…M1-1

Inverse of a product is product of inverses in reverse order

Every non-singular matrix can be written as a multiple of L( )U where what is in the brackets has no switches

More MatricesThursday, September 02, 2010

09:37

Notes

1 -1 3 2 1 0

4 -2 1 1 2 1

3 1 1 0 -1 2

4 1 2 1 3 1

Broken up into

A11 A12 A13

A21 A22 A23

A31 A32 A33

Where each is what they are in the aboveA11=[[1,-1][4,-2]]A12=[1,0]

If the sizes match up can multiply partition matrices just as normal numerical matrices

Not used too much, but is occasionally

a11 … a1m

… …

an1 anm

Ai is the ith columnRj is the jth row

Can be also then

A1 … An

OR

B1

…

Bm

BA=[BA1,…,Ban]

If have two matrices, A, B

Row matrix by column matrix is a 1 by 1 matrix or unit matrix - a dot product

b1A1+…bnAn is a linear combination

If AX=B then this system is consistent if and only if B can be written as a linear combination of the columns of A.

This is kind of silly, as it is obvious

Consistency theorem:

Partitioned MatricesThursday, September 02, 2010

10:27

Notes

Website: math.illinois.edu/~haboush/math415-FA10.html

Determinants based on geometry

Can do same thing in 3 dimensions - parallelepiped

Volume

DeterminantsTuesday, September 07, 2010

09:29

Notes

Can do same thing for n-spaceLemma: if two rows are the same then the determinant is zero (as it is collapsed in on itself). OR switch them, it must be the same, but must be negative too, thus is zero.

If reduced GJ form is not equal to I then the determinant is zero

Notes

Will have n! monomialsTotal number of computations is n.n!With Gauss Jordan can do in n3

So 4 by 4 is better by GJ3 by 3 and smaller do formula

n n3 n.n!

1 1 1

2 8 4

3 27 18

4 64 96

5 125 600

6 216 4320

7 343 35280

Can partition matrix to find determinantCould do everything in terms of columns instead of rows

Turn rows into columns and columns into rowsTranspose of matrix

Find the determinant

1 -2 1 4

2 1 1 2

-3 0 1 2

1 1 2 1

Notes

1 -2 1 4

0 5 -1 -6

0 6 4 14

0 3 1 -3

1 -2 1 4

0 3 1 -3

0 -6 4 14

0 5 -1 -6

Switch, so multiply by -1

1 -2 1 4

0 3 1 -3

0 0 6 8

0 0 -8/3 -1

1 -2 1 4

0 3 1 -3

0 0 6 8

0 0 0 23/9

-1*1*3*6*23/9=-46 Thus determinant is

Notes

Polynomials of degree n x n. Question: does the determinant exist? Answer: yes

1 2 3 4 … nr1 r2 … rn

Integeres in wrong order

1 2 … n

r1 r2 … rn

Denoted by a greek letter

1 2 3 4

3 2 1 4

3 4 1 2

3 4 1 2

Number of switches not unique, but whether it is odd or even is(-1)σ=1 if even, -1 if odd

Take n determinantsProof

Can get from one to another switching only insome number of switches

Permutations

Each row has only one '1' and each column has only one '1', the rest are columns

0 0 1 0

0 0 0 1

1 0 0 0

0 1 0 0

Determinant:

1 2 3 4

3 4 1 2

This is σ, and the matrix Sσ'1' is in the 3rd place of the first row'1' is in the 4th place of the second row

More determinantsThursday, September 09, 2010

09:35

Notes

Iterative statement: the determinant of the whole thing. Take an arbitrary row, take ( -1)i+1x1iM1i+… +(-1)1+mMim=det(X)

Notes

Homework due NEXT ThursdayQuestion 3 acg, not 2 DO that

Notes

HMWK 3 due 23rd SeptemberExam on not next Thursday

1 2 3 4 5

4 5 2 1 3

Switch, get negative1.Row by constant, total by constant2.Sum two rows, sum determinant3.

1 2 3 4 5

4 3 5 1 2

Can do column operations as well as row operations

Iterative fact about determinants

Even more determinantsTuesday, September 14, 2010

09:32

Notes

Replace entries by corresponding minorsTransposeSign changesDeterminant is ith row by ith columnDivide by determinant for X-1.

Closed formula for the determinant

Cramer's rule

Notes

Determinant is ith row by ith columnDivide by determinant for X-1.

Closed formula for the determinant

Cramer's rule

Closed formula, but computationally intensive

Say have 30, only want to know x17. Works well if you only want to know one or two of the values in a large system of equations

18th century ruleCramer: interested in Riemann curves, needed to solve large systems of linear equations

Notes

det(AB)=det(A).det(B)

Determinant CalculationsThursday, September 16, 2010

09:33

Notes

A function from V x V to Vi.Binary operation taking 2 elements of V and combining them to obtain a third1.

An operation of F on V2.

A vector space over F is a set V with two kinds of operations (F is a field)

(a+b)+c=a+(b+c)a+b=b+aa+0=0+a=a for all aa+a'=a'+a=0

For +

a*(b*c)=(a*b)*ca*b=b*aa*1=1*a=aAll nonzero a, aa1=a1a

For *

These are a group (commutative)Examples: rationals, real numbers, complex numbers,

Field: a set F with 2 operations + and *

Fields: subsets of complex numbers that are closed under addition and multiplication and follow group definition

We do linear algebra over reals and complexes

u+(v+w)=(u+v)+wu+v=v+uExists 0 such that u+0=0+u=uFor all u, u' exists such that u'+u=u+u'a(u+v)=au+av(a+a')=au+a'ua(uv)=(au)v1v=v

Subset of C, closed under + and *, if x not equal to 0 in F then 1/x is also in F.Usually use reals, but can also do rationals or other stuff

F: field of coefficients

Set V with two operations, one a binary operation VVector space:

Vector SpacesThursday, September 16, 2010

10:27

Notes

Finite dimensional vector space

What is an infinite dimensional vector space?

Notes

Definition: suppose V is a vector space over F. A subspace of V, U is a subset of V so that for each u,u' in U, u+u' is also in U and if u is in U and lambda is in F, then lambda.u is in U and if U is a vector space with these operations

A fortiori: if associativity… holds in V it also holds in U presuming the answer, u, is in U

Exercises - prove that something is or isn't a vector subspace

Notation: S(v1,…vq) is the span of v1 … vq

Lemma: If V is a vector space over F and v1…vq are vectors in V then S(v1…vq) is a vector subspace

Notes

Span is a plane

Notes

Exam on 7 October (2 thursdays)

Pg 116: 3 show complex numbers a+bi is a vector space over R, 6,13, 15Pg 125: 1abe, 2ac, 3adef, 5ac, 6ad,8Pg 137: 2ace

HMWK:

A vector space over F is a set V, with two operations:

V/F

Linear independenceThursday, September 23, 2010

09:34

Notes

1 -1 2 0 1 1 2

2 1 1 2 0 3 0

1 -1 2 2 1 1 1

1 -1 2 0 1 1 2

0 3 -3 2 -2 1 -4

0 0 0 2 0 0 -1

Thus linearly independent

OR do column operations - but determinant is much easier

A bunch of linearly independent vectors that satisfy certain conditionsBasis

Real numbers between zero and one are more numerous than integers

S(v1…vq)=Vi.v1 … vq are linearly independentii.

Def: V is a vector space over F. We say {v1…vq} is said to be a basis of V if

Number of elements in a basis is dimension of vector space

v1 … vq is a basis for V1.v1 … vq is a minimal spanning set for V2.v1 … vq is a maximal linearly independent set3.

Proposition TFAE (the following are equivalent)

Proof:

Dimension is the number in the basis

Notes

Suppose V is a finite dimensional vector space over F. Then any two bases have the same number of elements

Theorem:

Proof

A vector space is finite dimensional if it admits a finite spanning space

Notes

Maximum linear independent set1.Minimal spanning set2.

A basis is a Add that which is not in the span THM: For any set v1…vl which is linearly independent, it can be extended to be a basis

THM: Every spanning set can be diminished to a basisTHM: all bases for same system have same number of elements (proved on Thursday)PROP: if v1…vr are linearly independent and u 1…un span, then r<= n.

L=an(x)dn/dxn+an-1(x)dn-1/dxn-1+…+a1(x)d/dx+a0(x)VL= {f: L(f) = 0}If f and g are members of V L then f+g is in VL and af is in VL - this(VL) is a vector subspace of the continuous functions.

BasisTuesday, September 28, 2010

09:33

Notes

Theorem:

Notes

Jordan-canonical formCn to Cn find a basis such that T =

Matrix TransformationsThursday, September 30, 2010

09:36

Notes

1's below the diagonal, zeros everywhere else

0 0 0 0

1 0 0 0

0 1 0 0

0 0 1 0

EX:

Notes

1's down the diagonal, except the first which is a -1EX:

-1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1

Can find a basis so it can be written this way always for a reflection

Basis for a reflection

On exam will have a problem: here's a matrix, here's a basis write the matrix in that basis

Notes

A: linear transformation from standard basis to standard basisBut want to go to a new coordinate system - M is the equivalent of A in the new coordinate systemS-1AS=M

Notes

First question: list of matrices of different sizesWhat is on the exam?

Then will say:

AB, CA, DAB, DTA… Will be worth 6-8 points

Find the following if they make sense

Next question:

EX: elementary row operationsFind the inverse of some matrix S using some method (will give you the method)

Or might just say to find the inverseMight ask to find adjoint

Typical exam questionsTuesday, October 05, 2010

10:09

Notes

Will be a section saying - which of the following sets is linearly independent? A basis of a given space?

Can give a series of linear equations with, say, 5 unknowns but only 3 equations and asks for the solution of those equations (reduce to gaussian form, write the solution)

1 1 1 1- 1 2

1 2 0 1 -2 3

1 2 1 -2 1 0

1 1 1 1- 1 2

0 1 -1 2 -3 1

0 1 0 -5 6 4

Continue on

1 0 0 7 -8 9

0 1 0 -3 3 -3

0 0 1 -5 6 4 Solution vector

We're covering: chapter 2 (find determinants)Put a big matrix with lots of zeros - find it's determinant by elementary operations to a diagnol and multiply the diagnols Can have cramer's rule - just one of many

Could ask for det(adj(X))

Could ask for det(A)=3, det(B)=2 with A,B 4x4 - find the determinant of 2BA-1BAANS: det(2B)det(A-1)det(B)det(A)=42(2)(1/3)(2)(3)=64

Give a proof: prove that the solution set of a homogeneous system of equations is a vector subset of the space

Show it is a subspaceDo so by showing Av+Av' in K and so is cv

Could have K is the set of v such that Av is equal to zero (presuming all is in V)

Associative, Commutative, 0, inverse, distributive, distributive, associative multipliation, 1.v=v

Know the 8 axioms

Could ask to prove something is a vector space or a subspace of a vector space

Will go through section 3.5 on test

Can ask questions about dimensions - are 4 vectors in 3-space linearly independent - of course not!If there are 3, make matrix and take determinant - non-zero then good

EX:

1 0 1 0 1

1 -1 2 1 1

3 -1 4 1 3 Could ask for definition of basis or if something is a basis

Notes

3 -1 4 1 3

1 0 1 0 1

0 0 1 0 0

0 0 0 0 0

Thus not linearly independent

Could ask for definition of basis or if something is a basisAsk to prove something is a basis -show span and lin. independent

Notes

Homework is online - due next ThursdayLinear transformationsTuesday, October 12, 201009:38

Notes

g

Notes

Dimensions of row space are equal to dimensions of column space

What is im(A)

A times something is a linear combination of the columns of A

This is called the column space of A and is the image of AAX=S(A1…An)

The dimension of the kernel plus the dimension of the image equals the full width of the matrix

Can define the rank and nullity

ExampleTuesday, October 12, 2010

09:49

Notes

Span after elementary row operations is same as span before

Definition: row space

Notes

dim(rowspace(A))=dim(columnspace(A))

THEOREM: Thus dimension of column space is number of initial ones, is the dimensions of row space

Rank(A)+Nullity(A)=width of A (the number of columns)

Rank doesn't change when transposed, nullity mayRank: size of the largest minor with a non-vanishing determinant

Nullity is number of columns minus the rank

a11 a1n

am1 amn

Same as

R1

Rn

When L=A, a matrix

Notes

Linear TransformationsThursday, October 14, 2010

09:54

Notes

Inner products - chapter 5Tuesday, October 19, 201009:34

Notes

Pg 212: 1&2ac, 3,4,5,8,9,11Pg 221: 1ac, 2, 4, 12, 13

(u+v,w)=(u,w)+(v,w)1.(u,v+w)=(u,v)+(u,w)2.c(u,v)=(cu,v)=(u,cv)3.(u,u)>=04.If (u,u)=0 then u = 05.

Def: V is a real vector space. Then an inner product on V is a pairing ( , ): VxV =>R such that for all u,v,w in V, c in the reals

Consider the set of f on R continuous on a set S, such that on any closed interval [a,b], (R\S) intersection [a,b] has a finite number of discontinuities

Definition: f is periodic if f(x+P)=f(x) for all x

Can do Fourier series

Suppose V is finite dimensional with basis v1…vn

Suppose ( , ) is a positive definite inner product (redundant, all inner products are positive definite)

By (3), aij=aji

a11 a1n

an1 ann

A =

Definition: let aij=(vi,vj)

Inner ProductsThursday, October 21, 2010

09:34

Notes

Shortest distance between two points is a straight line

Note:

Notes

Least squares: resolve a vector into components parallel to and perpendicular to a given vector space

Orthogonality and least squaresTuesday, October 26, 2010

09:37

Notes

Truth: If B is rank r and A is non-singular, then BA is rank r

Notes

1ac, 3&4ab, 5Page 239: 1, 3Page 257: 1

New homework: Least squaresThursday, October 28, 201009:35

Notes

Table

y X1 X2 X3

1 0.5 2 1

2 .8 1 0

1.2 .6 1 1

1 .7 1 2

Notes

EX on 229: find the quadratic best squares fit

x 0 1 2 3

y 3 2 4 4

y=Ax2+Bx+C

y X2 X 1

3 0 0 1

2 1 1 1

4 4 2 1

4 9 3 1

C = 3A+B+C=24A+2B+C=49A+3B+C=4

0 0 1

1 1 1

4 2 1

9 3 1

0 1 4 9

0 1 2 3

1 1 1 1

0 1 4 9

0 1 2 3

1 1 1 1

3244

=

ABC

98 36 14

36 14 6

14 6 4

ABC

= 542213

98 36 14 54

36 14 6 22

14 6 4 13

-5 -3 -2 -6

4 1 -1 -2

2 3 7 20

49-3*14=7 49-3*18=-5

-1 -2 -3 -8

0 -5 -15 -42

0 1 1 4

1 2 3 8

0 1 1 4

0 5 15 42

1 2 3 8

0 1 1 4

0 0 10 22

1 2 3 8

0 1 1 4

0 0 1 2.2

1 2 3 8

0 1 0 1.8

0 0 1 2.2

1 0 0 -2.2

0 1 0 1.8

0 0 1 2.2

Example from book: Thursday, October 28, 2010

10:09

Notes

Given 3 vectors(0,-1,2,2)(1,-2,0,2)(3,0,4,5)And a 4th vectorWrite the projection of the 4th on the first 3

Subspace = S(v1…vr), have v not in spanFind proj(v) in that space

This is an orthonormal setSuppose: (vi,vj)=(1 if i=j, 0 otherwise) If one leaves out length one, that is

(vi,vj)=(ci if i=j, 0 otherwise) this is called an orthogonal set.

Orthogonal setsThursday, October 28, 2010

10:19

Notes

Pg 268: 1,5,6,7Pg 280: 1,4,7, 10

HMWK due 11 November:

The columns of Q form an orthonormal basisi.The rows of Q form an orthonormal basisii.For any vectors u,v: (Qu,Qv)=(u,v)iii.For any vector v ||Qv||=||v||iv.QTQ=Iv.QT=Q-1vi.

Theorem: Let Q be an m x n matrix TFAE (the following are equivalent)

RHS: 1/2((u+v,u+v)-(u,u)-(v,v))=1/2((u,u)+(v,v)+2(u,v)-(u,u)-(v,v))=(u,v)=LHSThus ||Qu||=||u|| and ||Q(u+v)||=||u+v|| and ||Qu+Qv||=||u+v||Thus(Qu,Qv)=||Qu+Qv||2-||Qu||2-||Qv||2

And (u,v)=||u+v||2-||u||2-||v||2

Thus iv implies iii and iii implies iv ||Qu||=sqrt((Qu,Qu))=||u||Suppose iii is trueei=(0…i…0)T Thus Qei = ith column of Q, and Qe i

T is the ith rowThus Qe1…Qen are an orthonormal basis and iii <=> IAnd this is true iff columns of Q are an orthonormal basisThe rows form an orthonormal basis by the same argumentShow i implies vCi=Qei= ith columnQTQ=[aij]such that aij=Ci

TCj and CiTCj = 1 if i=j, and 0 if i!=j, thus QTQ=I

(Qu,Qv)=(Qv)TQu=vTQTQu=vTIu=vTu=(u,v), thus implies iii(u,v)=vTu=(Qu,Qv)=(Qv)TQu=vTuThus ei

Tej=(Qej,Qei) or columns of Q, thus (Ci,Cj)If QT = Q-1 (v and vi imply one another, by definition)If Q1=QT then the columns of Q1 are orthonormal, which are the rows of Q and hence 1<=>v,vi

Proof: (u,v)=1/2(||u+v||2-||u||2-||v||2)

Definition: Q is orthogonal if it satisfies any or all of the conditions of the theorem

V=V++V-

V+={u:Su=u} and V-={u:Su=-u}If u is in both V+ and V- then u = 0, thus a direct sumThus V=V+(+)V-

Thus dim(V+)=dim(V)-1V+=ker(S-I)Thus V+ is a hyperplane and V- is a line perpendicular to V+

Show (Sv,Su)=0 where u is in the hyperplane and v is in line

-1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1

Thus not a rotation as det = -1

Matrix is

Example: Let S be a reflection

Thus (S-I)(S+I)=0=S2-S+S-I2=S2-I2=0Let u = (u+Su)/2 + (u-Su)/2S(u+Su)/2=(Su+S2u)/2=(Su+u)/2=(u+Su)/2Every u = u1+u2 where Su1=u1 and Su2=-Su2

Definition: S2=I and S-I is of rank 1

Final Exam: 16 December from 19:00-21:00

0 0 1 0

1 0 0 0

0 0 0 1

0 1 0 0

It just switches order of normal bases

Example:

Definition: A permutation matrix has exactly one 1 in every row and column, and is orthogonal but not necessarily a rotation

We have u1…un and want to convert to an orthonormal basisStart: v1=u1/||u||S(v1…vi)=S(u1…ui)Now: wi+1=ui+1-sum[j=1 to i](ui+1,vj)vj

This is just subtracting the projectionvi+1=wi+1/||wi+1||

Strict Gram Schmidt ProcessModified Gram Schmidt is not making them unit vectors

A = (u1…ur)Now get a new basis, v1…vr which are an orthonormal setWe also know that vi=a1iu1+a2iu2+…+aiiui where aii != 0 due to the fact that it is not in the span of the first i-1 vectors by the definition of basis

a11 a12 …

0 a22 a23

0 0 …

….

...

Thus U=(u1…ur) will be

U is upper triangular r x r matrix with non-zero diagonal matrixLet AU=Q where Q =(v1…vr)A=QU-1=QRThis is called the QR factorization of A

Start with Q = (u1…ur) and A = (v1…vr) which is an orthonormal setAlso A = QR

R=QTA as just multiply by QT

Must calculate R, which is simple

R=QTA

u1Tv1 u1

Tv2 … u1Tvr

u2Tv1

…

(u1T…ur

T)T(v1…vr)=

This is an r x r upper triangular matrixQTQ= r x r identity

AX=bCould try orthonormalizing A with Gramm Schmidt, get QQTAX=QTbNow have an upper triangular system times X equals a vector - this allows for back substitution

If have a least squares problem

( , ) is an abstract inner productv1… vr

u1=v1/||v1||Have u1…ul

PS(u1…ul) = (vl+1,u1)u1+(vl+1,u2)+…+(vl+1,ul)

Gram Schmidt is OK on abstract inner product spaces (but won't ask about it)

Orthonormal Matrixes Tuesday, November 02, 2010

09:37

Notes

Good for orthogonal polynomialsPn: polynomials of degree n or lessPn={a0+a1x+…+anxn}Choose an interval [a,b] with w(x) >0Find an inner product<f,g>W=fnint(f(x)g(x)w(x)dx,x,a,b)Do gram schmidt to the standard polynomials 1,x,x2…xn

Do this and will get Hermite polynomials, H0…Hn - use gram schmidt to relate each of them to the first ones (get iterative relation, don't compute directly)

Say w(x) = (1-x2)(1+x2) on [-1,1] (always will be positive)

On page 272 on the bottom

Let's say w(x)=1-x2

Notes

That is Acv=qv where q,c are scalarsFind vectors such that Av=cv where c is a scalar

Definition: an axis is a line L such that AL=L

An eigenvector for A is a vector v such that Av=cv for some c (usually c is in the field, but occasionally will allow it to be imaginary)

1.

An eigenvalue for A is a value c in the field such that there exists a vector v in V such that Av=cv

2.

det(A-cIn)= ΧA(c) =0i.The characteristic polynomial of A is the polynomial3.

A is an n x n matrix

Eigenvectors and eigenvaluesThursday, November 04, 2010

10:10

Notes

-3*3*3+4*3*3-5*3+6=0

There exists a matrix A. Suppose there exist n different vectors (a basis) so that each is an eigenvectorAvi=civi with vectors v i…vn

S=(v1…vn)Take S-1ASFirst take AS = (Av1,…,Avn)=(c1v1…cnvn)So could multiply by a diagonal matrix and get it

Notes

If there exists a vector v != 0 and a value c such that Av=cv, then c is called an eigenvalue of A and v is a called an eigenvector of A.

I.

If ΧA(x)=det(A-xIn), then ΧA(c) is called the characteristic polynomial of A.II.

A is an n x n matrix.

Example:

-4 -4 2

3 4 -1

-3 -2 3

A =

A-xI3=-4-x -4 2

3 4-x -1

-3 -2 3-x

-4-x -4 2

3 4-x -1

0 2-x 2-x

2-x

-4-x -4 2

3 4-x -1

0 1 1

(2-x)-4-x 2

3 -1

-4-x -4

3 4-x+-1 | | ||[ ]

(2-x)(-4-x+6+-16+x^2+12)(2-x)(x2-x-2)=(2-x)(x-2)(x+1)ΧA(x)=-(x-2)(x-2)(x+1)=(x-2)2(x+1)Thus the eigenvalues are 2,2,-1

Start with 2

-4-2 -4 2

3 4-2 -1

-3 -2 3-2

-6 -4 2

3 2 -1

-3 -2 1

3 2 -1

0 0 0

0 0 0

3x=-2y+zx=-2/3*y+1/3*z

Thus v=(-2y/3+z/3,y,z)T

Those are a basis for the eigenspace that belongs to 2v=y(-2/3,1,0)T+z(1/3,0,1)T

-2/310

1/301

Eigenvectors for 2

For -1-3 -4 2

-3 5 -1

-3 -2 4

-3 -4 2

-3 5 -1

0 3 3

-3 -4 2

0 1 1

0 1 1

-3 0 6

0 1 1

0 0 0

y+z=0-3x+6z=0

y= -zx=2z

2-11

eigenvector

The conjugate of x by a is axa-1.

-230

103

2-11

S =

-2 1 2

3 0 -1

0 3 1

AS=

-2 1 2

3 0 -1

0 3 1

-4 -4 2

3 4 -1

-3 -2 3

=

-4 2 2

6 0 1

0 6 -1

AS=S

2 0 0

0 2 0

0 0 -1

S-1AS=

2 0 0

0 2 0

0 0 -1

Eigenvalues and eigenvectorsTuesday, November 09, 2010

09:34

Notes

A is a matrixv1…vr are eigenvectorsFor λ1….λn where no two c's are identicalTheorem: the vi's are linearly independent

Suppose they're not linearly independent - then there is some c1v1+…+crvr=0

ci1vi1+…+cilvil=0Divide thruA(vi1+ci2vi2+…+cilvil)=Avi1+ci2Avi2+…+cilAvil

Avij=λijvij

vi1+ci2vi2+…+cilvil=0

None are zero - thus this is a one shorter combination, but the one above already was, thus this is not possible

Ci2(λi2-λi1)vi1+ci3(λi3-λi2)vi2+…+cil(λil-λi1)vil=0

QED

There is thus some shortest such combination

Proof:

If ΧA(x)=+/-(x-λ1)r1(x-λ2)r2…(x-λq)rq

And if the eigenspace of λi is of dimension ri. Then there exists a basis of eigenvectors

λ1 v11…v1r1

λ2 v21…v2r2

…

λq vq1…vqrq

That is

And S is all the v'sAnd S-1AS= λ1

λ2

...

But not necessarily possible - take A = 1 1

0 1

1-x 1

0 1-x| |=(1-x)2

Subtract: 0 1

0 0

Solution space is merely (1 0)T and is only eigenvector, thus cannot be basis

When working over the complexes can find a matrix such that it is all the λ's in order (but can be more than one of each) and some 1's in the row immediately above, but everywhere else are zeros

Jordan Canonical form

Example:

λ1 1 0 0 0

0 λ1 0 0 0

0 0 λ1 0 0

0 0 0 λ2 1

0 0 0 0 λ2

Theorem: Let A be an n x n matrix. Suppose A has n eigenvalues with multiplicities. That is ΧA(x)=(x-λ1)r1…(x-λq)rq where none are equal and r1+…rq=n. and suppose the eigenspace of λi

is the of dimension ri. Then there exists S such that S-1AS=[matrix with diagonal of λi's

Corollary: suppose ΧA(x) has n distinct eigenvalues. Then there exists a basis of eigenvalues and A can be diagonalized. That is A is regular and semisimple.

A subspace of 1 dimension less, in general, thus takes up little of whole space

Most of the time the matrices are nice like this - only not if the determinant (in whatever dimension) happens to be zero

Notes

Say we have a vector space over C, need for eigenvalues of certain matricesSay z is complex, z=a+biz*=a-biNorm of z: N(z)=zz*z*w*=(zw)*z*+w*=(z+w)*N(zw)=N(z)N(w)N(z)=a2+b2.|z|=sqrt(N(z))|zw|=|z||w|z=|z|(cos(θ)+isin(θ))=|z|cis(θ)eiθ=cis(θ)

Now take the vector space Cn, where C is the complex numbers

0 1

-1 0

x2+1=0λ=+/-iFind eigenvectors for i and -i

-i 1

-1 -i

1 i

0 0

ab =0

x=-iy-iyy=y*(-i,1)T

Eigenvector for i is (-i,1)T

Eigenvector for -i is (i,1)T

z-1=z*/N(z)

i 1

-1 i

-1 i

0 0

x=yi=yiy

-i i

1 1

S=

S-1AS=-i 0

0 -i

Dot product in normal way won't work with complexes

Hermite Hermitian

(λu,v)=λ(u,v)1.(u,λv)=λ*(u,v)2.

(u,v+v')=(u,v)+(u,v')i.(u+u',v)=(u,v)+(u',v)3.

(u,v)=(v,u)*4.(u,u) is in the reals, and (u,u)>=0, and if (u,u)=0 then u = 05.

Let V be a C vector space. A Hermitian inner product is a pairing u,v goes to (u,v) such that

Say we have Cn

Complex numbersThursday, November 11, 201009:41

Notes

A=

1 0 0

-2 1 3

1 1 -1

y=(y1,y2,…,yn)

Example of linear systems of differential equations with constant coefficientsThursday, November 11, 201010:32

Notes

1+i 2-2i 5-3i

3+i 1-i i

3+7i -i 4-i

Find the gauss jordan form of (1+i)-1=(1-i)/2

1 2-2i 1-4i

0 -7+3i

Very tedious with complex matrices

(5-3i)(1-i)/2=1-4i

(1-i)-(2-2i)(3+i)=(1-i)-(8-4i)=(-7+3i)

i 1

2+i -i

Make

To an orthonormal basis

Complex matricesTuesday, November 16, 2010

09:31

Notes

Definition: M is Hermitian iff MH=M

Notes

Definition: U is unitary if UH=U-1.That is UHU=In.Corollary: U is unitary iff the columns (or the rows) form an orthonormal basis. As with the above (the dot product matrix) - by definition it will only be and identity matrix if the rows or columns are an orthornormal basis as then (C i,Cj)=1 iff i=j, and if i!= j, (C i,Cj)=0

a b

c d

Ha* c*

b* d*= =

a b

c d

a*=a, thus a is reald*=d, thus d is realb*=c, thus b and c are complex conjugatesc*=b

Any real number Any complex number

Bar of the complex number in top right Any real number

a11 a12 … a1n

a12*

…

a1n* ann

This means there is a unitary change of basisProve it by inductionTrue for 1 x 1 (n=1), triviallyAssume it is true for n = k, that is k x k matricesSuppose m is a k+1 x k+1 matrix

That is Mw1=λw1

M has an eigenvector, w1.

Presume w1 is length 1 (can be done by dividing by its length if it is not already)Make a change of basisTake w1 and (Cw1)perp, W is a subspace, looking at the set of v such that (w,v)=0, that is v is perpendicular to all of W. Thus if w1,w2 are in Wperp.V=Cn.

Let M be any matrix. Then there exists a unitary matrix U so that UHMU is upper triangular. U is not necessarily unique.

Notes

V=Cn.(Cw1)perp=W is a subspace of VTake (w1,u2…un) and do gram schmidt to it, this is a full basis of Cn.Will now get (w1…wn) as a basis , BIn this basis Mw1=λw1

λ Stuff

0

…

0

[M]B=

Thus UHMU is a k x k matrixBy inductive hypothesis can get another one such that has first column of 0's below top and stuff above it. Continue getting these down to a 1x1 and put together

M'

1 0

0 VH

λ R

0 M' =

1 RV

0 VHM'V

The product of 2 unitary matrices is unitaryUV(UV)H(UV)=VHUHUV=VHV=I

This is on the final, but not this midterm

Notes

first classprofessor william haboush 305 altgeld 333-6498 [email protected] 1:30 to 2:30 tuesday...

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