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Linear Transformations in Linear Algebra: Lecture Slides by Dr. Arjun Kapoor, Slides of Linear Algebra

Dr. Arjun Kapoor delivered this lecture at Institute of Mathematics and Applications for Linear Algebra course to cover following points: Linear, Transformations, Real, Valued, Functions, Transformation, Matrix, Augmented, Image, Shear

Typology: Slides

2011/2012

Uploaded on 07/03/2012

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Download Linear Transformations in Linear Algebra: Lecture Slides by Dr. Arjun Kapoor and more Slides Linear Algebra in PDF only on Docsity!

Linear Algebra

Linear Transformations

2

Real Valued Functions

Formula Example Description

Function from R to R Function from to R Function from to R

Function from to R

f   x

f  x , y 

f  x , y , z 

f ^ x 1 , x 2 ,, xn 

f   x^  x^2

f  x , y   x^2  y^2

  2 2 2

f x , y , z  x  y  z

  2

2 2

2 1 , 2 , , 1

n

n

x

f x x x x x

 

  

 

2 R 3 R n R

4

Introduction

 Another way to view :

Matrix A is an object acting on x by multiplication to produce a new vector Ax or b.

 Example

Ax = b

1

4 3 1 3 1 5

2 0 5 1 1 8

1

              (^)            

Ax = b

5

Introduction

 Solving Ax = b amounts to finding all vectors x in Rn

that are transformed into the vector b in Rm^ under the action of multiplication by Am×n. The correspondence from x to b is a function from one set of vectors to another. This concept is generalization of f:R - > R.

x (^) b

Multiplication by A

Am×n x = b :

n m

=> T   

7

Introduction

8

Matrix Transformation

 For each x in Rn^ , Tx is computed as Ax , where A is

an m×n matrix. We sometimes denote this Matrix Transformation by x - > Ax. Domain of T lies in Rn when A has n columns and range of T lies in Rm^ when each column of A has m entries.

10

a. Find T(u) , the image of u under the transformation T b. Find an x in R^2 whose image under T is b. c. Is there more than one x whose image under T is b. d. Determine if c is in the range of the transformation T.

Matrix Transformation – Example 1

11

 Solution a.

b. Solve Tx=b for x. That is, solve Ax=b

1 3 5 2 3 5 1 1 1 7 9

       ^     (^)      (^)       (^)    (^)  

T(u) = Au

1

2

1 3 3

3 5 2

1 7 5

x

x

       ^              (^)    (^)  

Matrix Transformation – Example 1

13

c. Any x whose image under T is b must satisfy

We have seen in (b) that there is exactly one x whose

image is b.

1

2

1 3 3

3 5 2

1 7 5

x

x

       ^              (^)    (^)  

Matrix Transformation – Example 1

14

d. The vector c is in the range of T if c is the image of some

x in R^2 , that is, if c=T(x) for some x. This is just another way of asking if the system Ax=c is consistent. To find the answer, row reduce the augmented matrix

The third equation 0 =- 35 shows that the system is

inconsistent. So c is not in the range of T.

1 3 3 1 3 3 1 3 3 1 3 3 3 5 2 0 14 7 0 1 2 0 1 2 1 7 5 0 4 8 0 14 7 0 0 35

 ^   ^   ^        (^)                (^)      (^)    (^)  

Matrix Transformation – Example 1

16

Matrix Transformation

 Let

 Then the transformation T:R^2 - > R^2 defined by Ax is

called the Shear Transformation.

 For example the image of

1 3 0 1

A

      

2 1 3 2 8 is 2 0 1 2 2

                        

17

Matrix Transformation

Shear Transformation

19

Linear Transformations

   

   

1 1 1 2 2 2 3 3 3

1 1 1 1 2 2 2 2 3 3 3 3

1 1 2 2 3 3 1 1 2 2 3 3

1 1

1 2 3 2 1 2 3 2

3 3

a ( u v ) a ( u v ) a ( u v )

a u a v a u a v a u a v

a u a u a u a v a v a v

u v

a a a u a a a v

u v

Au Av

     

     

     

                 

 

Similarly (b) can also be proved

20

Linear Transformations

A transformation T is linear, if

1. T(u+v) = T(u) + T(v) **(for all u,v in the domain of T)

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

 Property ( 1 ) says that the result T(u+v) i.e. first adding

u+v and then applying T is same as applying T first to u and v and then adding T(u) and T(v).