The document contains announcements and information about an upcoming exam: - A quiz and test are scheduled. Sample exams and review sessions will be provided. - Exam 1 will cover several sections of the textbook and the professor will be available for questions. - Tips are provided for studying including doing homework, examples, and practicing sample exams. - Sections about subspaces and column/null spaces of matrices are summarized, including properties and examples.

1. Announcements
Quiz 2 after lecture.
Test 1 will be on Feb 1, Monday in class on sections 1.1-1.5,
1.7-1.8, 2.1-2.3 and 2.8-2.9
Sample Exam 1 will be on the website by Thursday evening
Review for Exam 1 after tomorrow's lecture
I will be in oce all day friday. Feel free to stop by any time if
you have questions.

2. Tips for Exam
Do your homework problems including T/F questions (Check
whether you have the latest homework set, I did some
trimming)
Do the examples we did in class
Planning to have problems worth 20 points from chapter 1 and
80 points from chapter 2.
Do the sample exam yourself with a 50 min time limit. I will
post the solutions only by Saturday noon so that you will do it
yourself rst.
Not an exam with lots of tedious calculations.

3. Section 2.8 Subspaces of Rn
Consider any set of vectors H in Rn . This set can be called a
subspace of Rn if it satises the following properties.

4. Section 2.8 Subspaces of Rn
Consider any set of vectors H in Rn . This set can be called a
subspace of Rn if it satises the following properties.
1. The zero vector 0 should be in H .

5. Section 2.8 Subspaces of Rn
Consider any set of vectors H in Rn . This set can be called a
subspace of Rn if it satises the following properties.
1. The zero vector 0 should be in H .
2. H must be closed under addition. This means, if u and v are 2
vectors in H , then their sum u + v must be in H .

6. Section 2.8 Subspaces of Rn
Consider any set of vectors H in Rn . This set can be called a
subspace of Rn if it satises the following properties.
1. The zero vector 0 should be in H .
2. H must be closed under addition. This means, if u and v are 2
vectors in H , then their sum u + v must be in H .
3. H must be closed under scalar multiplication. This means, if u
is a vector in H and c is any scalar, the product c u must be in
H.

7. Important Example of Subspace
Consider 2 vectors u and v in Rn . Let H be the set of all linear
combinations (or span) of u and v.
1. The vector 0u + 0v is a linear combination of u and v. So the
zero vector is in H .

8. Important Example of Subspace
Consider 2 vectors u and v in Rn . Let H be the set of all linear
combinations (or span) of u and v.
1. The vector 0u + 0v is a linear combination of u and v. So the
zero vector is in H .
2. Consider 2 dierent linear combinations of u and v.
For example, c1 u + c2 v and c3 u + c4 v.

9. Important Example of Subspace
Consider 2 vectors u and v in Rn . Let H be the set of all linear
combinations (or span) of u and v.
1. The vector 0u + 0v is a linear combination of u and v. So the
zero vector is in H .
2. Consider 2 dierent linear combinations of u and v.
For example, c1 u + c2 v and c3 u + c4 v.
Their sum will be c1 u + c2 v + c3 u + c4 v = (c1 + c3 )u + (c2 + c4 )v.
This is again a linear combination of u and v which means the
sum belongs to H .

10. Important Example of Subspace
Consider 2 vectors u and v in Rn . Let H be the set of all linear
combinations (or span) of u and v.
1. The vector 0u + 0v is a linear combination of u and v. So the
zero vector is in H .
2. Consider 2 dierent linear combinations of u and v.
For example, c1 u + c2 v and c3 u + c4 v.
Their sum will be c1 u + c2 v + c3 u + c4 v = (c1 + c3 )u + (c2 + c4 )v.
This is again a linear combination of u and v which means the
sum belongs to H .
3. Also consider r (c1 u + c2 v) = (rc1 )u + (rc2 )v. The scalar product
of a linear combination is also a linear combination.
Thus H =Span{u, v} is a subspace of Rn .

11. Column Space of a Matrix
The column space of a matrix A is the set of all linear
combinations of the columns of A. It is denoted by Col A.
Checking whether a vector b is in Col A for any matrix A is same
as

12. Column Space of a Matrix
The column space of a matrix A is the set of all linear
combinations of the columns of A. It is denoted by Col A.
Checking whether a vector b is in Col A for any matrix A is same
as
1. Checking whether b is a linear combination of the columns of
A which is same as

13. Column Space of a Matrix
The column space of a matrix A is the set of all linear
combinations of the columns of A. It is denoted by Col A.
Checking whether a vector b is in Col A for any matrix A is same
as
1. Checking whether b is a linear combination of the columns of
A which is same as
2. Checking whether the system Ax = b is consistent (one or
many solutions)

14. Example 8, section 2.8
−3 −2 0 1
       
Letv1 =  0 , v2 =  2 , v3 = −6and p =  14 .
6 3 3 −9
How many vectors are in Col A? Determine if p is in Col A where
A = v1 v2 v3 .

15. Example 8, section 2.8
−3 −2 0 1
       
Letv1 =  0 , v2 =  2 , v3 = −6and p =  14 .
6 3 3 −9
How many vectors are in Col A? Determine if p is in Col A where
A = v1 v2 v3 .
Solution: Remember Col A is the set of all possible linear
combinations of the columns of A. So there are innitely many
vectors in it.

16. Example 8, section 2.8
−3 −2 0 1
       
Let v1 =  0 , v2 =  2 , v3 = −6
and p =  14 .
6 3 3 −9
How many vectors are in Col A? Determine if p is in Col A where
A = v1 v2 v3 .
Solution: Remember Col A is the set of all possible linear
combinations of the columns of A. So there are innitely many
vectors in it.
To determine if p is in Col A, write the augmented matrix and
check the consistency. This also determines whether p is in the
subspace of R3 generated (spanned) by v1 , v2 and v3 . (See
Problem 5 in your homework)

17. Example 8, section 2.8
Solution: Start with the augmented matrix
 
 −3 −2 0 1 
 
 
 0 2 −6 14
 

 
 
6 3 3 −9
 
Divide R2 by 2 andalso 
 −3 −2 0 1 
 
 
 0 2 −6 14 R3+2R1
 

 
 
6 3 3 −9
 

18. Example 8, section 2.8
 
 −3 −2 0 1 
 
 
 0 1 −3 7
 

R3+R2
 
 
0 −1 3 −7
 
−3 −2 0 1
 
 0 1 −3 7 
0 0 0 0

19. Example 8, section 2.8
 
 −3 −2 0 1 
 
 
 0 1 −3 7
 

R3+R2
 
 
0 −1 3 −7
 
−3 −2 0 1
 
 0 1 −3 7 
0 0 0 0
The system is consistent and so p is in Col A.

20. Null Space of a Matrix
The null space of a matrix A is the set of all solutions of the
homogeneous equation Ax = 0. It is denoted by Nul A.
Finding the vectors in Nul A is same as
1. Solving Ax = 0 which means

21. Null Space of a Matrix
The null space of a matrix A is the set of all solutions of the
homogeneous equation Ax = 0. It is denoted by Nul A.
Finding the vectors in Nul A is same as
1. Solving Ax = 0 which means
2. Finding the basic variables and free variables, then

22. Null Space of a Matrix
The null space of a matrix A is the set of all solutions of the
homogeneous equation Ax = 0. It is denoted by Nul A.
Finding the vectors in Nul A is same as
1. Solving Ax = 0 which means
2. Finding the basic variables and free variables, then
3. Expressing the basic in terms of free variables, then

23. Null Space of a Matrix
The null space of a matrix A is the set of all solutions of the
homogeneous equation Ax = 0. It is denoted by Nul A.
Finding the vectors in Nul A is same as
1. Solving Ax = 0 which means
2. Finding the basic variables and free variables, then
3. Expressing the basic in terms of free variables, then
4. Writing the solution in the vector form.

24. Basis
Denition
A basis of any subspace H of Rn is a
1. linearly independent set in H

25. Basis
Denition
A basis of any subspace H of Rn is a
1. linearly independent set in H
2. that spans H

26. Simplest Example
Example
1 0
Consider the vectors e1 = ,e = .
0 2 1
1 0
These vectors are linearly independent because I2 = has 2
0 1
pivot columns (no free variables).
Consider any point in R2 , say (3,2). We can write
3 1 0
=3 +2
2 0 1
We have a linear combination (span) of the given vectors.
This is true for ANY point (vector) in R2 .

27. Simplest Example
1 0
The vectors e1 = and e2 = are called the Standard Basis
0 1
Vectors of R2 In general, the vectors
1 0 0
     
0 1 0
     
e1 = 0, e2 = 0, . . . , en = 0 forms the standard basis for Rn .
     
. . .
. . .
. . .
0 0 1

28. Simplest Example
1 0
The vectors e1 = and e2 = are called the Standard Basis
0 1
Vectors of R2 In general, the vectors
1 0 0
     
0 1 0
     
e1 = 0, e2 = 0, . . . , en = 0 forms the standard basis for Rn .
     
. . .
. . .
. . .
0 0 1
Look Carefully: These vectors are the columns of the respective
identity matrix.

29. Finding Basis for Col A
This is easy!!!
1. Look for the pivot columns in the echelon form of A
2. Pick the corresponding columns from A

30. Finding Basis for Nul A
This is easy but takes more steps!!!
1. Express the basic variables in terms of free variables
2. Write the solution in the vector form
3. The vector multiplying each free variable belongs to Nul A.

31. Example 24, section 2.8
Given A and an echelon form of A. Find a basis for Col A and Nul
A.
−3 9 −2 −7 1 −3 6 9
   
A =  2 −6 4 8  ∼  0 0 4 5 
3 −9 −2 2 0 0 0 0
Here columns and 3 are pivot columns. So a basis for
Solution: 1
 −3 −2 
   
Col A is  2  ,  4 .
3 −2
To nd a basis for Nul A, write the system of equations, express the
basic variables x1 and x3 in terms of the free variables x2 and x4 .

32. Example 24, section 2.8
x1 − 3x2 + 6x3 + 9x4 = 0
4x3 + 5x4 = 0
Thus we have

33. Example 24, section 2.8
x1 − 3x2 + 6x3 + 9x4 = 0
4x3 + 5x4 = 0
Thus we have x3 = − 5 x4 and
4

34. Example 24, section 2.8
x1 − 3x2 + 6x3 + 9x4 = 0
4x3 + 5x4 = 0
Thus we have x3 = − 5 x4 and
4
x1 = 3x2 − 6x3 − 9x4 = 3x2 − 6(− 5 x4 ) − 9x4
4
=3x2 + 30 x4 − 9x4 = 3x2 − 6 x4 .
4 4

35. Example 24, section 2.8
x1 3x2 − 6 x4 3 −1.5
       
4
x   x2  = x 1 + x  0 .
 2 
So,   =  5
x3   − 4 x4  2 
0 4 −1.25
    
  
x4 x4 0 1
 3 −1.5 
   
 1
   0 

   

A basis for Nul A is thus

 ,  .
0 −1.25
0 1

 

 

36. Example 26, section 2.8
Given A and an echelon form of A. Find a basis for Col A and Nul
A.
3 −1 7 3 9 3 −1 7 0 6
   
A =  −2 2 −2 7 4  ∼  0 2 4 0 3 
5 
 
 −5 9 3 3   0 0 0 1 1 
   
−2 6 6 3 7 0 0 0 0 0
Solution:    1, 2  4 are pivot columns. So a basis for
Here columns   and
 3 −1 3 
 −2
   2  7

      
Col A is   ,   ,  .

−5  9  3
 −2

6 3 

To nd a basis for Nul A, write the system of equations, express the
basic variables x1 , x2 and x4 in terms of the free variables x3 and x5 .

37. Example 26, section 2.8
3x1 x2 7x3 6x5 0

− + + =
x2 + 4x3 + 3x5 = 0
x4 x5 0

+ =
Thus we have
x4 = −x5 ,

38. Example 26, section 2.8
3x1 x2 7x3 6x5 0

− + + =
x2 + 4x3 + 3x5 = 0
x4 x5 0

+ =
Thus we have
x4 = −x5 ,
2x2 = −4x3 − 3x5 =⇒ x2 = −2x3 − 1.5x5 and

39. Example 26, section 2.8
3x1 x2 7x3 6x5 0

− + + =
x2 + 4x3 + 3x5 = 0
x4 x5 0

+ =
Thus we have
x4 = −x5 ,
2x2 = −4x3 − 3x5 =⇒ x2 = −2x3 − 1.5x5 and
3x1 = x2 − 7x3 − 6x5 =−2x3 − 1.5x5 −7x3 − 6x5 =−9x3 − 7.5x5 .
x1 =−3x3 − 2.5x5

40. Example 26, section 2.8
x1 −3x3 − 2.5x5 −3 −2.5
       
x  −2x − 1.5x  2 −1.5
 2  3 5
So, x3  =  x3  = x3  1  + x5  0 .
   
    
  
x4   −x5 0  −1 
       

x5 x5 0 1
−3 −2.5
   
 2  −1.5
A basis for Nul A is thus
   
 1  ,  0 .
   
 0   −1 
   
0 1

41. Example 16, section 2.8
−4 2
Does , and form a basis for R2 . Why(not)?
6 −3
Solution: Remember the 2 conditions for a set to be basis?
1. linearly independent
2. span
Here the rst vector is -2 times the second vector. (Or, the
determinant of the matrix formed by these 2 vectors as columns is
0). So it is a linearly dependent set and hence not a basis

42. Example 18, section 2.8
1 −5 7
     
Does ,  1  , −1 and 0form a basis for R3 . Why(not)?
−2 2 5
Solution: Idea: Check whether the matrix formed by these vectors
is invertible. (3 pivot columns and rows)
 
1 −5 7
R2-R1
 
R3+2R1
 
 
1 −1 0
 
 
 
 
−2 2 5
 

43. Example 18, section 2.8
 
 1 −5 7 
 
 
0 4 −7
 
 
R3+2R2
 
 
0 −8 19
 
1 −5 7
 
 0 4 −7 
0 0 5

44. Example 18, section 2.8
 
 1 −5 7 
 
 
0 4 −7
 
 
R3+2R2
 
 
0 −8 19
 
1 −5 7
 
 0 4 −7 
0 0 5
We have 3 pivot positions and hence this matrix is invertible
(columns linearly independent). So the 3 given vectors form a basis.