1. Announcements
Quiz 2 on Wednesday Jan 27 on sections 1.4, 1.5, 1.7 and 1.8
If you have any grading issues with quiz 1, please discuss with
me asap.
Solution to quiz 1 will be posted on the website by Monday.
2. 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 .
3. 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 .
The set Rn is called Domain of T .
4. 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 .
The set Rn is called Domain of T .
The set Rm is called Co-Domain of T .
5. 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 .
The set Rn is called Domain of T .
The set Rm is called Co-Domain of T .
The notation T : Rn → Rm means the domain is Rn and the
co-domain is Rm .
6. 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 .
The set Rn is called Domain of T .
The set Rm is called Co-Domain of T .
The notation T : Rn → Rm means the domain is Rn and the
co-domain is Rm .
For x in Rn , the vector T (x) is called the image of x.
7. 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 .
The set Rn is called Domain of T .
The set Rm is called Co-Domain of T .
The notation T : Rn → Rm means the domain is Rn and the
co-domain is Rm .
For x in Rn , the vector T (x) is called the image of x.
Set of all images T (x) is called the Range of T .
9. Linear Transformation
A transformation (or function or mapping) is Linear if
T (u + v) = T (u) + T (v) for all u and v in the domain of T .
10. Linear Transformation
A transformation (or function or mapping) is Linear if
T (u + v) = T (u) + T (v) for all u and v in the domain of T .
T (c u) = cT (u) for all u and all scalars c .
12. Important
If T is a linear transformation
T (0) = (0).
13. Important
If T is a linear transformation
T (0) = (0).
T (c u + d v) = cT (u) + dT (v) for all u and v in the domain of
T.
14. Interesting Linear Transformations
0 −1 3 1
Let A =
1 0
u= 2
,v =
3
Let T : R2 → R2 a linear transformation dened 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 −1 3 = 0.1.+ +−12.2 = −2
1 0 2
3 ( )
3 0. 3
0 −1 1 0.1 + (−1).3 −3
T (v) = 1 0 3 = 1.1 + 0.3 = 1
15. Interesting Linear Transformations
3 1 4
Since u+v = + = ,
2 3 5
The image under T of u+v is nothing but
T (u+v) = 0 −1
1 0
4
5
=
0.4 + (−1).5
1. 4 + 0. 5
=
−5
4
The next picture shows what happened here.
16. Rotation Transformation
Here T rotates u, v and u+v
counterclockwise about the origin through 900 .
y
x
0
17. Rotation Transformation
Here T rotates u, v and u+v
counterclockwise about the origin through 900 .
y
u
x
0
18. Rotation Transformation
Here T rotates u, v and u+v
counterclockwise about the origin through 900 .
y
T (u)
u
x
0
19. Rotation Transformation
Here T rotates u, v and u+v
counterclockwise about the origin through 900 .
y
T (u) v
u
x
0
20. Rotation Transformation
Here T rotates u, v and u+v
counterclockwise about the origin through 900 .
y
T (u) v
u
T (v)
x
0
21. Rotation Transformation
Here T rotates u, v and u+v
counterclockwise about the origin through 900 .
y
u+v
T (u) v
u
T (v)
x
0
22. Rotation Transformation
Here T rotates u, v and u+v
counterclockwise about the origin through 900 .
y
u+v
T (u+v)
T (u) v
u
T (v)
x
0
23. Rotation Transformation
Here T rotates u, v and u+v
counterclockwise about the origin through 900 .
y
u+v
T (u+v)
T (u) v
u
T (v)
x
0
24. Rotation Transformation
Here T rotates u, v and u+v
counterclockwise about the origin through 900 .
y
u+v
T
T (u+v)
T (u) v
u
T (v)
x
0
25. Interesting Linear Transformations
0 1 3 1
Let A =
1 0
u= 2
,v =
3
Let T : R2 → R2 a linear transformation dened 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 3 = 0133 + 1)22 = 2
1 0 2
. +( .
. 0. 3
0 1 1 0.1 + (1).3 3
T (v) = 1 0 3 = 1.1 + 0.3 = 1
34. Example 6, Section 1.8
1 −2 1 1
3 −4 5 9
Let A = , b =
0 1 1 3
−3 5 −4 6
Let T be dened by by T (x) = Ax. Find a vector x whose image
under T is b and determine whether x is unique.
35. Example 6, Section 1.8
1 −2 1 1
3 −4 5 9
Let A = , b =
0 1 1 3
−3 5 −4 6
Let T be dened 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.
39. 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 .
40. The solution is thus
x1
7 − 3x3
x= x2 = 3 − x3
x3 x3
Since we can choose any value for x3 , the solution is NOT unique.
41. Example 10, Section 1.8
1 3 9 2
1 0 3 −4
Let A = Find all x in R4 that are mapped into
0 1 2 3
−2 3 0 5
the zero vector by the transformation x → Ax for the given matrix
A.
42. Example 10, Section 1.8
1 3 9 2
1 0 3 −4
Let A = Find all x in R4 that are mapped into
0 1 2 3
−2 3 0 5
the zero vector by the transformation x → 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.
47. How many pivot columns? 3. Columns 1,2 and 4.
Which is the free variable?
48. How many pivot columns? 3. Columns 1,2 and 4.
Which is the free variable? x3 .
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
x1 −3x3 −3
x=
x2 −2x
=
−2
= x3
3
x3 x3
1
x4 0 0
49. Chapter 2 Matrix Algebra
Denition
Diagonal Matrix: A square matrix (same number of rows and
columns) with all non-diagonal entries 0.
Example
1 0 0 0
9 0 0
0 7 0 0
,
0 0 0
0 0 4 0
0 0 1
0 0 0 3
50. Chapter 2 Matrix Algebra
Denition
Zero 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
51. Matrix Addition
Two matrices are equal if
they have the same size
the corresponding entries are all equal
52. Matrix Addition
Two matrices are equal if
they have the same size
the corresponding entries are all equal
If A and B are m × n matrices, the sum A + B is also an m × n matrix
53. Matrix Addition
Two matrices are equal if
they have the same size
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 .
54. Matrix Addition
Two matrices are equal if
they have the same size
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 dened only if A and B are of the same size.
56. Matrix Addition
Let
1 2 3 0 1 3 0 1
A= 2 3 4 ,B = 2 0 4 ,C = 2 0
3 4 5 0 0 5 0 0
Find A + B , A + C and B + C
Solution
1+0 2+1 3+3 1 3 6
A+B = 2+2 3+0 4+4 = 4 3 8
3+0 4+0 5+5 3 4 10
Both A + C and B + C are not dened since they are of dierent
sizes.
57. 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
1 2 3 0 1
A= 2 3 4 ,C = 2 0
3 4 5 0 0
Find 4A and −2C
Solution
4 8 12
4A = 8 12 16
12 16 20
0 −2
−2C = −4 0
0 0
58. Basic Algebraic Properties
For all matrices A, B and C of the same size and all scalars r and s
A+B = B +A
(A + B ) + C = A + (B + C )
A+0 = A
r (A + B ) = rA + rB
(r + s )A = rA + sA
r (sA) = (rs )A
