Announcements

Ï Quiz 2 on Wednesday Jan 27 on sections 1.4, 1.5, 1.7 and 1.8

Ï If you have any grading issues with quiz 1, please discuss with

me asap.

Ï Solution to quiz 1 will be posted on the website by Monday.

Last Class...

A transformation (or function or mapping) T from Rn to Rm is a

rule that assigns to each vector x in Rn a vector T (x) in Rm.

Ï The set Rn is called Domain of T .

Ï The set Rm is called Co-Domain of T .

Ï The notation T :Rn→Rm means the domain is Rn and the

co-domain is Rm.

Ï For x in Rn, the vector T (x) is called the image of x.

Ï Set of all images T (x) is called the Range of T .

Last Class...

A transformation (or function or mapping) T from Rn to Rm is a

rule that assigns to each vector x in Rn a vector T (x) in Rm.

Ï The set Rn is called Domain of T .

Ï The set Rm is called Co-Domain of T .

Ï The notation T :Rn→Rm means the domain is Rn and the

co-domain is Rm.

Ï For x in Rn, the vector T (x) is called the image of x.

Ï Set of all images T (x) is called the Range of T .

Last Class...

A transformation (or function or mapping) T from Rn to Rm is a

rule that assigns to each vector x in Rn a vector T (x) in Rm.

Ï The set Rn is called Domain of T .

Ï The set Rm is called Co-Domain of T .

Ï The notation T :Rn→Rm means the domain is Rn and the

co-domain is Rm.

Ï For x in Rn, the vector T (x) is called the image of x.

Ï Set of all images T (x) is called the Range of T .

Last Class...

A transformation (or function or mapping) T from Rn to Rm is a

rule that assigns to each vector x in Rn a vector T (x) in Rm.

Ï The set Rn is called Domain of T .

Ï The set Rm is called Co-Domain of T .

Ï The notation T :Rn→Rm means the domain is Rn and the

co-domain is Rm.

Ï For x in Rn, the vector T (x) is called the image of x.

Ï Set of all images T (x) is called the Range of T .

Last Class...

A transformation (or function or mapping) T from Rn to Rm is a

rule that assigns to each vector x in Rn a vector T (x) in Rm.

Ï The set Rn is called Domain of T .

Ï The set Rm is called Co-Domain of T .

Ï The notation T :Rn→Rm means the domain is Rn and the

co-domain is Rm.

Ï For x in Rn, the vector T (x) is called the image of x.

Ï Set of all images T (x) is called the Range of T .

Last Class...

A transformation (or function or mapping) T from Rn to Rm is a

rule that assigns to each vector x in Rn a vector T (x) in Rm.

Ï The set Rn is called Domain of T .

Ï The set Rm is called Co-Domain of T .

Ï The notation T :Rn→Rm means the domain is Rn and the

co-domain is Rm.

Ï For x in Rn, the vector T (x) is called the image of x.

Ï Set of all images T (x) is called the Range of T .

Linear Transformation

A transformation (or function or mapping) is Linear if

Ï T (u+v)=T (u)+T (v) for all u and v in the domain of T .

Ï T (cu)= cT (u) for all u and all scalars c .

Linear Transformation

A transformation (or function or mapping) is Linear if

Ï T (u+v)=T (u)+T (v) for all u and v in the domain of T .

Ï T (cu)= cT (u) for all u and all scalars c .

Linear Transformation

A transformation (or function or mapping) is Linear if

Ï T (u+v)=T (u)+T (v) for all u and v in the domain of T .

Ï T (cu)= cT (u) for all u and all scalars c .

Important

If T is a linear transformation

Ï T (0)= (0).

Ï T (cu+dv)= cT (u)+dT (v) for all u and v in the domain of

T .

Important

If T is a linear transformation

Ï T (0)= (0).

Ï T (cu+dv)= cT (u)+dT (v) for all u and v in the domain of

T .

Important

If T is a linear transformation

Ï T (0)= (0).

Ï T (cu+dv)= cT (u)+dT (v) for all u and v in the domain of

T .

Interesting Linear Transformations

Let A=[0 −11 0

]u=

[3

2

],v=

[1

3

]Let T :R2 →R2 a linear transformation de�ned by T (x)=Ax. Find

the images under T of u, v and u+v.

Solution: Image under T of u and v is nothing but

T (u)=[0 −11 0

][3

2

]=

[0.3+ (−1).21.3+0.2

]=

[ −23

]T (v)=

[0 −11 0

][1

3

]=

[0.1+ (−1).31.1+0.3

]=

[ −31

]

Interesting Linear Transformations

Since u+v=[3

2

]+

[1

3

]=

[4

5

],

The image under T of u+v is nothing but

T (u+v)=[0 −11 0

][4

5

]=

[0.4+ (−1).51.4+0.5

]=

[ −54

]The next picture shows what happened here.

Rotation Transformation

Here T rotates u, v and u+v

counterclockwise about the origin through 900.y

x0

u

T (u) v

T (v)

u+v

T (u+v)T

Rotation Transformation

Here T rotates u, v and u+v

counterclockwise about the origin through 900.y

x0

u

T (u) v

T (v)

u+v

T (u+v)T

Rotation Transformation

Here T rotates u, v and u+v

counterclockwise about the origin through 900.y

x0

u

T (u)

v

T (v)

u+v

T (u+v)T

Rotation Transformation

Here T rotates u, v and u+v

counterclockwise about the origin through 900.y

x0

u

T (u) v

T (v)

u+v

T (u+v)T

Rotation Transformation

Here T rotates u, v and u+v

counterclockwise about the origin through 900.y

x0

u

T (u) v

T (v)

u+v

T (u+v)T

Rotation Transformation

Here T rotates u, v and u+v

counterclockwise about the origin through 900.y

x0

u

T (u) v

T (v)

u+v

T (u+v)T

Rotation Transformation

Here T rotates u, v and u+v

counterclockwise about the origin through 900.y

x0

u

T (u) v

T (v)

u+v

T (u+v)

T

Rotation Transformation

Here T rotates u, v and u+v

counterclockwise about the origin through 900.y

x0

u

T (u) v

T (v)

u+v

T (u+v)

T

Rotation Transformation

Here T rotates u, v and u+v

counterclockwise about the origin through 900.y

x0

u

T (u) v

T (v)

u+v

T (u+v)T

Interesting Linear Transformations

Let A=[0 1

1 0

]u=

[3

2

],v=

[1

3

]Let T :R2 →R2 a linear transformation de�ned by T (x)=Ax. Find

the images under T of u and v

Solution: Image under T of u and v is nothing but

T (u)=[0 1

1 0

][3

2

]=

[0.3+ (1).21.3+0.2

]=

[2

3

]T (v)=

[0 1

1 0

][1

3

]=

[0.1+ (1).31.1+0.3

]=

[3

1

]

Re�ection Transformation

Here T re�ects u and v about the line x = y .y

x0

u

Tuv

Tv

Re�ection Transformation

Here T re�ects u and v about the line x = y .y

x0

u

Tuv

Tv

Re�ection Transformation

Here T re�ects u and v about the line x = y .y

x0

u

Tu

v

Tv

Re�ection Transformation

Here T re�ects u and v about the line x = y .y

x0

u

Tu

v

Tv

Re�ection Transformation

Here T re�ects u and v about the line x = y .y

x0

u

Tuv

Tv

Re�ection Transformation

Here T re�ects u and v about the line x = y .y

x0

u

Tuv

Tv

Re�ection Transformation

Here T re�ects u and v about the line x = y .y

x0

u

Tuv

Tv

Re�ection Transformation

Here T re�ects u and v about the line x = y .y

x0

u

Tuv

Tv

Example 6, Section 1.8

Let A=

1 −2 1

3 −4 5

0 1 1

−3 5 −4

, b=

1

9

3

6

Let T be de�ned by by T (x)=Ax. Find a vector x whose image

under T is b and determine whether x is unique.

Solution The problem is asking you to solve Ax= b. In other words,

write the augmented matrix and solve.

Example 6, Section 1.8

Let A=

1 −2 1

3 −4 5

0 1 1

−3 5 −4

, b=

1

9

3

6

Let T be de�ned by by T (x)=Ax. Find a vector x whose image

under T is b and determine whether x is unique.

Solution The problem is asking you to solve Ax= b. In other words,

write the augmented matrix and solve.

1 −2 1 1

3 −4 5 9

0 1 1 3

−3 5 −4 −6

R2-3R1

R4+3R1

=⇒

1 −2 1 1

0 2 2 6

0 1 1 3

0 −1 −1 −3

Divide row 2 by 2

=⇒

1 −2 1 1

0 1 1 3

0 1 1 3

0 −1 −1 −3

1 −2 1 1

0 1 1 3

0 1 1 3

0 −1 −1 −3

R3-R2

R4+R2

1 −2 1 1

0 1 1 3

0 0 0 0

0 0 0 0

Since column 3 doesnot have a pivot, x3 is a free variable. We can

solve for x1 and x2 in terms of x3.{x1 − 2x2 + x3 = 1

x2 + x3 = 3

We have x2 = 3−x3 and

x1 = 1+2x2−x3 = 1+2(3−x3)−x3 = 7−3x3.

1 −2 1 1

0 1 1 3

0 0 0 0

0 0 0 0

Since column 3 doesnot have a pivot, x3 is a free variable. We can

solve for x1 and x2 in terms of x3.{x1 − 2x2 + x3 = 1

x2 + x3 = 3

We have x2 = 3−x3 and

x1 = 1+2x2−x3 = 1+2(3−x3)−x3 = 7−3x3.

The solution is thus

x= x1x2x3

= 7−3x3

3−x3x3

Since we can choose any value for x3, the solution is NOT unique.

Example 10, Section 1.8

Let A=

1 3 9 2

1 0 3 −40 1 2 3

−2 3 0 5

Find all x in R4 that are mapped into

the zero vector by the transformation x 7→Ax for the given matrix

A.

Solution The problem is asking you to solve Ax= 0. In other words,

write the augmented matrix for the homogeneous system and solve.

Example 10, Section 1.8

Let A=

1 3 9 2

1 0 3 −40 1 2 3

−2 3 0 5

Find all x in R4 that are mapped into

the zero vector by the transformation x 7→Ax for the given matrix

A.

Solution The problem is asking you to solve Ax= 0. In other words,

write the augmented matrix for the homogeneous system and solve.

1 3 9 2 0

1 0 3 −4 0

0 1 2 3 0

−2 3 0 5 0

R2-R1

R4+2R1

=⇒

1 3 9 2 0

0 −3 −6 −6 0

0 1 2 3 0

0 9 18 9 0

Divide row 2 by -3 and row 4 by 9

=⇒

1 3 9 2 0

0 1 2 2 0

0 1 2 3 0

0 1 2 1 0

1 3 9 2 0

0 1 2 2 0

0 1 2 3 0

0 1 2 1 0

R3-R2

R4-R2

1 3 9 2 0

0 1 2 2 0

0 0 0 1 0

0 0 0 −1 0

R4+R3

=⇒

1 3 9 2 0

0 1 2 2 0

0 0 0 1 0

0 0 0 0 0

Ï How many pivot columns?

3. Columns 1,2 and 4.

Ï Which is the free variable? x3.

Ï Write the system of equations so that we can express the basic

variables in terms of the free variables.x1 + 3x2 + 9x3 + 2x4 = 0

x2 + 2x3 + 2x4 = 0

x4 = 0

Thus, x2 =−2x3 and

x1 =−3x2−9x3 =−3(−2x3)−9x3 =−3x3. Our solution is thus

x=

x1x2x3x4

=

−3x3−2x3x30

= x3

−3−21

0

Ï How many pivot columns? 3. Columns 1,2 and 4.

Ï Which is the free variable?

x3.

Ï Write the system of equations so that we can express the basic

variables in terms of the free variables.x1 + 3x2 + 9x3 + 2x4 = 0

x2 + 2x3 + 2x4 = 0

x4 = 0

Thus, x2 =−2x3 and

x1 =−3x2−9x3 =−3(−2x3)−9x3 =−3x3. Our solution is thus

x=

x1x2x3x4

=

−3x3−2x3x30

= x3

−3−21

0

Ï How many pivot columns? 3. Columns 1,2 and 4.

Ï Which is the free variable? x3.

Ï Write the system of equations so that we can express the basic

variables in terms of the free variables.x1 + 3x2 + 9x3 + 2x4 = 0

x2 + 2x3 + 2x4 = 0

x4 = 0

Thus, x2 =−2x3 and

x1 =−3x2−9x3 =−3(−2x3)−9x3 =−3x3. Our solution is thus

x=

x1x2x3x4

=

−3x3−2x3x30

= x3

−3−21

0

Chapter 2 Matrix Algebra

De�nitionDiagonal Matrix: A square matrix (same number of rows and

columns) with all non-diagonal entries 0.

Example 1 0 0 0

0 7 0 0

0 0 4 0

0 0 0 3

,

9 0 0

0 0 0

0 0 1

Chapter 2 Matrix Algebra

De�nitionZero Matrix: A matrix of any size with all entries 0.

Example 0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

,

0 0

0 0

0 0

Matrix Addition

Two matrices are equal if

Ï they have the same size

Ï the corresponding entries are all equal

If A and B are m×n matrices, the sum A+B is also an m×n matrix

The columns of A+B is the sum of the corresponding columns of A

and B .

A+B is de�ned only if A and B are of the same size.

Matrix Addition

Two matrices are equal if

Ï they have the same size

Ï the corresponding entries are all equal

If A and B are m×n matrices, the sum A+B is also an m×n matrix

The columns of A+B is the sum of the corresponding columns of A

and B .

A+B is de�ned only if A and B are of the same size.

Matrix Addition

Two matrices are equal if

Ï they have the same size

Ï the corresponding entries are all equal

If A and B are m×n matrices, the sum A+B is also an m×n matrix

The columns of A+B is the sum of the corresponding columns of A

and B .

A+B is de�ned only if A and B are of the same size.

Matrix Addition

Two matrices are equal if

Ï they have the same size

Ï the corresponding entries are all equal

If A and B are m×n matrices, the sum A+B is also an m×n matrix

The columns of A+B is the sum of the corresponding columns of A

and B .

A+B is de�ned only if A and B are of the same size.

Matrix Addition

Let

A= 1 2 3

2 3 4

3 4 5

,B = 0 1 3

2 0 4

0 0 5

,C = 0 1

2 0

0 0

Find A+B , A+C and B +CSolution

A+B = 1+0 2+1 3+3

2+2 3+0 4+4

3+0 4+0 5+5

= 1 3 6

4 3 8

3 4 10

Both A+C and B +C are not de�ned since they are of di�erent

sizes.

Matrix Addition

Let

A= 1 2 3

2 3 4

3 4 5

,B = 0 1 3

2 0 4

0 0 5

,C = 0 1

2 0

0 0

Find A+B , A+C and B +CSolution

A+B = 1+0 2+1 3+3

2+2 3+0 4+4

3+0 4+0 5+5

= 1 3 6

4 3 8

3 4 10

Both A+C and B +C are not de�ned since they are of di�erent

sizes.

Scalar Multiplication

If r is a scalar (number) then the scalar multiple rA is the matrix

whose columns are r times the columns in A. Let

A= 1 2 3

2 3 4

3 4 5

,C = 0 1

2 0

0 0

Find 4A and −2CSolution

4A= 4 8 12

8 12 16

12 16 20

−2C = 0 −2

−4 0

0 0

Basic Algebraic Properties

For all matrices A, B and C of the same size and all scalars r and s

Ï A+B =B +AÏ (A+B)+C =A+ (B +C )

Ï A+0=A

Ï r(A+B)= rA+ rBÏ (r + s)A= rA+ sAÏ r(sA)= (rs)A