m211f2012 Slides Sec2.2 2.3handout

7/28/2019 m211f2012 Slides Sec2.2 2.3handout

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MATH 211 Winter 2013

Lecture Notes(Adapted by permission of K. Seyffarth)

Sections 2.2 & 2.3

Sections 2.2 & 2.3 Page 1/1


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Example

The linear system

x1 + x2 x3 + 3x4 = 2x1 + 4x2 + 5x3 2x4 = 1x1 + 6x2 + 3x3 + 4x4 = 1

has coefficient matrix A and constant matrix B, where

A =

1 1 1 31 4 5 2

1 6 3 4

and B =

211

.

Using (matrix) addition and scalar multiplication, we can rewrite this

system as

x1

11

1

+ x2

146

+ x3

15

2

+ x4

32

4

=

21

1

Sections 2.2 & 2.3 An Alternate Form for a Linear System Page 2/1


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This example illustrates the fact that solving a system of linear equationsis equivalent to finding the coefficients of a linear combination of thecolumns of the coefficient matrix A so that the result is equal to theconstant matrix B.

Sections 2.2 & 2.3 An Alternate Form for a Linear System Page 3/1


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Notation and Terminology

R: the set of real numbers.

Rn: set of columns (with entries from R) having n rows.

1

103

R4

,

6

5

R2,

23

7

R3

.

The columns ofRn are also called vectors or n-vectors.

To save space, a vector is sometimes written as the transpose of arow matrix.

1 1 0 3T

R4

Sections 2.2 & 2.3 Notation and Terminology Page 4/1


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Definition (The Matrix-Vector Product)

Let A =

A1 A2 An

be an m n matrix with columns

A1,

A2, . . . ,

An, and X = x

1x

2. . . x

nT

any n-vector.The product AX is defined as the m-vector given by

x1A1 + x2A2 + xnAn,

i.e., AX is a linear combination of the columns of A (and the coefficientsare the entries of X, in order).

This means that if a system of m linear equations in n variables has them n matrix A as its coefficient matrix, the n-vector B as its constantmatrix, and the n-vector X as the matrix of variables, then the system canbe written as the matrix equation

AX = B.

Sections 2.2 & 2.3 The Matrix-Vector Product Page 5/1


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Theorem (2.2 Theorem 1)

Every system of linear equations has the form AX = B where A is thecoefficient matrix, X is the matrix of variables, and B is theconstantmatrix.

AX = B is consistent if and only if B is a linear combination of the

columns of A.If A =

A1 A2 . . . An

, then X =

x1 x2 . . . xn

Tis a

solution to AX = B if and only if x1, x2, . . . , xn are a solution to thevector equation

x1A1 + x2A2 + xnAn = B.



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Example

Let

A =

1 0 2 12 1 0 1

3 1 3 1

and Y =

21

14

1 Compute AY.

2 Can B =

11

1

be expressed as a linear combination of the columns

of A? If so, find a linear combination that does so.



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Example (continued)

1 AY = 2

12

3

+ (1)

0

1

1

+ 1

20

3

+ 4

1

1

1

=

09

12

2 Solve the system AX = B for X =

x1 x2 x3 x4

T. To do this,

put the augmented matrix

A B

in reduced row-echelon form.

1 0 2 1 12 1 0 1 13 1 3 1 1

1 0 0 1 1

70 1 0 1 570 0 1 1 37

Since there are infinitely many solutions, simply choose a value for x4.Taking x4 = 0 gives us

11

1

= 1

7

12

3

5

7

01

1

+ 3

7

20

3

.



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Example (Example 5, p. 44.)

If A is the m n matrix of all zeros, then AX = 0 for any n-vector X.

If X is the n-vector of zeros, then AX = 0 for any m n matrix A.



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Properties of Matrix-Vector Multiplication


Let A and B be m n matrices, X, Y Rn be n-vectors, and k R be ascalar.

1 A(X + Y) = AX + AY

2 A(kX) = k(AX) = (kA)X

3 (A + B)X = AX + BX



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Definition

Given a linear system AX = B, the system AX = 0 is called the associatedhomogeneous system.


Suppose that X1 is a particular solution to the system of linear equationsAX = B. Then every solution to AX = B has the form X2 = X1 + X0 forsome solution X0 to the associated homogeneous system AX = 0.

How do we use this result?

Sections 2.2 & 2.3 The Associated Homogeneous System Page 11/1


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Example

The system of linear equations AX = B, with

A =

2 1 3 30 1 1 11 1 3 0

and B =

42

1

has solution

X =

1 2s t2 + s tst

=

1200

+ s

2110

+ t

1101

, s, t R.



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Example (continued)

Furthermore, X1 =

1

200

is a particular solution to AX = B (obtained bysetting s = t = 0), while

X0 = s

2

110

+ t11

01

, s, t R

is the general solution, in parametric form, to the associated homogeneoussystem AX = 0.

Example 7 (p. 46) is similar.



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The Dot Product

Definition

The dot product of two n-tuples (a1, a2, . . . , an) and (b1, b2, . . . , bn) is the

number (scalar)a1b1 + a2b2 + + anbn.


Suppose that A is an m n matrix and that X is an n-vector. Then theith entry of AX is the dot product of the ith row of A with X.

Example

Compute the product

1 0 2 12 1 0 1

3 1 3 1

21

14

Sections 2.2 & 2.3 The Dot Product Page 14/1


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Definition

The n n identity matrix, denoted In is the matrix having ones on its maindiagonal and zeros elsewhere, and is defined for all n 2.

Example 11 (p. 49) shows that for any n-vector X, InX = X.

Definition

Let n 2. For each j, 1 j n, we denote by Ej the jth column of In.


Let A and B be m n matrices. If AX = BX for every X Rn, thenA = B.

Why?



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Problem

Find examples of matrices A and B, and a vector X = 0, so that

AX = BX but A = B.



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Matrix Transformations

Examples

In R2, reflection in the x-axis transforms a

b

to ab

.

In R2, reflection in the y-axis transforms

ab

to

a

b

.

These are examples of transformations ofR2

.

Definition

A transformation is a function T : Rn Rm, sometimes written

R

n T

Rm

,

and is called a transformation from Rn to Rm.If m = n, then we say T is a transformation ofRn.

What do we mean by function?Sections 2.2 & 2.3 Matrix Transformations Page 17/1


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Example (Defining a transformation by specifying its action.)T : R3 R4 defined by

T a

bc

=

a + bb+ ca cc b

is a transformation.

Sections 2.2 & 2.3 Matrix Transformations Page 18/1


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Definition (Equality of Transformations)

Suppose S : Rn Rm and T : Rn Rm are transformations. Then

S = T if and only if S(X) = T(X) for every X Rn

.



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Definition (Matrix Transformation)

Let A be an m n matrix. The transformation T : Rn Rm defined by

T(X) = AX for each X Rn

is called the matrix transformation induced by A.

Example

In R2, reflection in the x-axis, which transforms

ab

to

a

b

, is a

matrix transformation because

ab

= 1 0

0 1 a

b

.

What about reflection in the y-axis?



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Example

The transformation T : R3 R4 defined by

T

ab

c

=

a + bb+ ca cc b

is a matrix transformation. Why?Because T is induced by the matrix

A =

1 1 0

0 1 11 0 10 1 1

Check!



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Example (Example 14, p. 51.)

R2

: R2 R2

denotes counterclockwise rotation about the origin through an angle of 2radians. Then

R2 a

b = b

a .

Furthermore, R2

is a matrix transformation, as can be seen by the factthat

ba

=

0 11 0

ab

.



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Definition

R : R2 R

2

denotes counterclockwise rotation about the origin through and angle of .

Problem

Suppose that

ab

R2. Then R

ab

=?

ProblemIs R a matrix transformation?


Some other matrix transformations


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Some other matrix transformations

1 If A is the m n matrix of all zeros, then the transformation induced

by A, namely T(X) = AX for all X Rn,

is called the zero transformation from Rn to Rm, and is written T=0.

2 If A is the n n identity matrix, then the transformation induced by

A is called the identity transformation onRn

, and is written 1Rn .3 Example 15, p. 52: x-expansion, x-compression, y-expansion, and

y-compression ofR2.

4 Example 16, p. 52: an x-shear ofR2.From this you should be able to define a y-shear ofR2, and find thematrix that induces it.

5 Exercise 11(b)(c), p. 54: reflection in the line y = x and reflectionin the line y = x.



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Not all transformations ofR2

are matrix transformations.Example

Let T : R2 R2 be defined by

T(X) = X + 11

for all X R2.

Why is T not a matrix transformation?



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2.3 Matrix Multiplication

Sections 2.2 & 2.3 Matrix Multiplication Page 26/1

Matrix Multiplication


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Matrix Multiplication

Definition (Product of two matrices.)

Let A be an m n matrix and B =

B1 B2 Bk

an n k matrix,whose columns are B1, B2, . . . , Bk. The product of A and B is the matrix

AB = A

B1 B2 Bk

=

AB1 AB2 ABk

,

i.e., the first column of AB is AB1, the second column of AB is AB2, etc.


Example


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Example

Let A and B be matrices,

A =1 0 32 1 1

and B =

1 1 20 2 41 0 0

Then AB has columns

AB1 =1 0 3

2 1 1

101

, AB2 =1 0 3

2 1 1

120

,

and AB3 = 1 0 3

2 1 1 2

40

Thus, AB =

4 1 2

1 4 0

.



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Let A be an m n matrix, and B an n k matrix. Then

A(BX) = (AB)X for all k-vectors X Rk.

What does this mean? Why is this important?


Another way to think about matrix multiplication


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Another way to think about matrix multiplication


Let A be an m n matrix and B and n k matrix. Then the (i,j)-entry

of AB is the dot product of row i of A with column j of B.

Example

Use the above theorem to compute

1 0 3

2 1 1

1 1 20 2 41 0 0



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Definition (Compatibility for Matrix Multiplication)Let A and B be matrices. In order for the product AB to exist, thenumber of rows in B must be equal to the number of columns in A.

Assuming that A is an m n matrix, the product AB is defined if and

only if B is an n k matrix for some k. If the product is defined,then A and B are said to be compatible for (matrix) multiplication.

Given that A is m n and B is n k, the product AB is an m kmatrix.

Sections 2.2 & 2.3 Compatibility for Matrix Multiplication Page 31/1


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Example (revisited)

As we saw earlier,

231 0 3

2 1 1

33 1 1 20 2 4

1 0 0

=

234 1 2

1 4 0

Note that the product

33

1 1 2

0 2 4

1 0 0

23

1 0 32 1 1

does not exist.


Example


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Example

Let

A =

1 23 0

1 4

and B = 1 1 2 0

3 2 1 3

Does AB exist? If so, compute it.

Does BA exist? If so, compute it.

AB =

7 5 4 63 3 6 011 7 2 12

BA does not exist



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Example

Let

G = 1

1

and H =

1 0

Does GH exist? If so, compute it.

Does HG exist? If so, compute it.

GH =

1 01 0

HG = 1 In this example, GH and HG both exist, but they are not equal. Theyarent even the same size!


E l


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Example

Let

P = 1 02 1

and Q = 1 10 3

Does PQ exist? If so, compute it.

Does QP exist? If so, compute it.

PQ =

1 12 1

QP = 1 16 3

In this example, PQ and QP both exist and are the same size, butPQ= QP.



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Fact

The four previous examples illustrate an important property of matrixmultiplication.

In general, matrix multiplication is not commutative, i.e., theorder of the matrices in the product is important.


E l


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Example

Let

U = 2 00 2

and V = 1 23 4

Does UV exist? If so, compute it.

Does VU exist? If so, compute it.

UV =

2 46 8

VU = 2 4

6 8

In this particular example, the matrices commute, i.e., UV = VU.



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Let A, B, and C be matrices of appropriate sizes, and let k R be ascalar.

1 IA = A and AI = A where I is an identity matrix.

2 A(BC) = (AB)C (associative property)

3 A(B + C) = AB + AC (distributive property)

4 (B + C)A = BA + CA (distributive property)

5 k(AB) = (kA)B = A(kB)

6 (AB)T = BTAT

Work through Example 7 on p. 61: simplifying algebraic expressionsinvolving matrix multiplication.

Sections 2.2 & 2.3 Properties of Matrix Multiplication Page 38/1

Elementary Proofs


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Example

Let A and B be m

n matrices, and let C be an n

k matrix. Prove thatif A and B commute with C, then A + B commutes with C.

Proof.

We are given that AC = CA and BC = CB. Consider (A + B)C.

(A + B)C = AC + BC

= CA + CB

= C(A + B)

Since (A + B)C = C(A + B), A + B commutes with C.

Examples 8 and 9 on p. 61 are similar types of elementary proofs.Omit Example 10 on pp. 6162.

Sections 2.2 & 2.3 Properties of Matrix Multiplication Page 39/1

Block Multiplication


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Example

Let A be an m n matrix. Let B be an n k matrix with columnsB1, B2, . . . , Bk, i.e., B =

B1 B2 Bk

. This represents a

partition of B into blocks in this example, the blocks are the columns ofB. We can now write

AB = A

B1 B2 Bk

=

AB1 AB2 ABk

Here, the columns of AB, namely AB1, AB2, . . . , ABk, can be thought ofas blocks of AB.

If A is an m n matrix and B is an n k matrix, and if A and B arepartitioned compatibly into blocks in some way, then the computation ofthe product AB may be simplified.

Sections 2.2 & 2.3 Block Multiplication Page 40/1


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Example

A =

2 1 3 11 0 1 2

0 0 1 00 0 0 1

B =

1 2 0

1 0 0

0 5 11 1 0



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Example

A =

2 1 3 11 0 1 2

0 0 1 00 0 0 1

B =

1 2 0

1 0 0

0 5 11 1 0



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Example

A =

2 1 3 11 0 1 2

0 0 1 00 0 0 1

B =

1 2 0

1 0 0

0 5 11 1 0



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Example

A =

2 1 3 11 0 1 2

0 0 1 00 0 0 1

B =

1 2 0

1 0 0

0 5 11 1 0


Example (continued)


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Example (continued)

Let

A =

2 1 3 1

1 0 1 20 0 1 00 0 0 1

=

A1 A20 I2

,

and let

B =

1 2 01 0 0

0 5 11 1 0

=

B1 0B2 B3

.

Then

AB =

A1 A20 I2

B1 0B2 B3

.


Example (continued)


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p ( )

AB = A1 A20 I2

B1 0B2 B3

=

A1B1 + A2B2 A1(0) + A2B3(0)B1 + I2B2 (0)0 + I3B3

= A1B1 + A2B2 A2B3B2 B3

Now recall that

A =

2 1 3 11 0 1 2

0 0 1 00 0 0 1

= A1 A2

0 I2

, B =

1 2 01 0 0

0 5 11 1 0

= B1 0

B2 B3

.

Now compute A1B1, A2B2 and A2B3.


Example (continued)


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p ( )

A1B1 =

2 11 0

1 2

1 0

=

3 41 2

A2B2 =

3 11 2

0 51 1

=

1 142 3

A2

B3

= 3 11 2

10 = 3

1

Now,

AB = A1B1 + A2B2 A2B3

B2 B3

=

4 18 33 5 1

0 5 11 1 0

=

4 18 33 5 10 5 11 1 0


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