subspace, col space, basis

Announcements

Ï Quiz 2 after lecture.

Ï 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

Ï Sample Exam 1 will be on the website by Thursday evening

Ï Review for Exam 1 after tomorrow's lecture

Ï I will be in o�ce all day friday. Feel free to stop by any time ifyou have questions.

Tips for Exam

Ï Do your homework problems including T/F questions (Checkwhether you have the latest homework set, I did sometrimming)

Ï Do the examples we did in class

Ï Planning to have problems worth 20 points from chapter 1 and80 points from chapter 2.

Ï Do the sample exam yourself with a 50 min time limit. I willpost the solutions only by Saturday noon so that you will do ityourself �rst.

Ï Not an exam with lots of tedious calculations.

Section 2.8 Subspaces of Rn

Consider any set of vectors H in Rn. This set can be called asubspace of Rn if it satis�es 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 2vectors in H, then their sum u+v must be in H.

3. H must be closed under scalar multiplication. This means, if uis a vector in H and c is any scalar, the product cu must be inH.

Important Example of Subspace

Consider 2 vectors u and v in Rn. Let H be the set of all linearcombinations (or span) of u and v.

1. The vector 0u+0v is a linear combination of u and v. So thezero vector is in H.

2. Consider 2 di�erent linear combinations of u and v.For example, c1u+c2v and c3u+c4v.Their sum will be c1u+c2v+c3u+c4v= (c1+c3)u+ (c2+c4)v.This is again a linear combination of u and v which means thesum belongs to H.

3. Also consider r(c1u+c2v)= (rc1)u+(rc2)v. The scalar productof a linear combination is also a linear combination.

Thus H=Span{u,v} is a subspace of Rn.

2. Consider 2 di�erent linear combinations of u and v.For example, c1u+c2v and c3u+c4v.

Their sum will be c1u+c2v+c3u+c4v= (c1+c3)u+ (c2+c4)v.This is again a linear combination of u and v which means thesum belongs to H.

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 sameas

1. Checking whether b is a linear combination of the columns ofA which is same as

2. Checking whether the system Ax= b is consistent (one ormany solutions)

Example 8, section 2.8

Let v1 =−306

, v2 =−223

, v3 = 0−63

and p= 114−9

.How many vectors are in Col A? Determine if p is in Col A whereA= [

v1 v2 v3].

Solution: Remember Col A is the set of all possible linearcombinations of the columns of A. So there are in�nitely manyvectors in it.

To determine if p is in Col A, write the augmented matrix andcheck the consistency. This also determines whether p is in thesubspace of R3 generated (spanned) by v1, v2 and v3. (SeeProblem 5 in your homework)

Let v1 =−306

, v2 =−223

, v3 = 0−63

and p= 114−9

v1 v2 v3].

Let v1 =−306

, v2 =−223

, v3 = 0−63

and p= 114−9

v1 v2 v3].

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

6 3 3 −9

R3+2R1

−3 −2 0 1

0 1 −3 7

0 −1 3 −7

−3 −2 0 10 1 −3 70 0 0 0

The system is consistent and so p is in Col A.

−3 −2 0 1

0 1 −3 7

0 −1 3 −7

−3 −2 0 10 1 −3 70 0 0 0

The system is consistent and so p is in Col A.

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.

De�nitionA basis of any subspace H of Rn is a

1. linearly independent set in H

2. that spans H

De�nitionA basis of any subspace H of Rn is a

1. linearly independent set in H

2. that spans H

Simplest Example

Example

Consider the vectors e1 =[10

], e2 =

These vectors are linearly independent because I2 =[1 00 1

]has 2

pivot columns (no free variables).

Consider any point in R2, say (3,2). We can write[32

]We have a linear combination (span) of the given vectors.This is true for ANY point (vector) in R2.

Simplest Example

The vectors e1 =[10

]and e2 =

]are called the Standard Basis

Vectors of R2 In general, the vectors

100...0

, e2 =010...0

, . . . , en =

000...1

forms the standard basis for Rn.

Look Carefully: These vectors are the columns of the respectiveidentity matrix.

Simplest Example

The vectors e1 =[10

]and e2 =

]are called the Standard Basis

Vectors of R2 In general, the vectors

100...0

, e2 =010...0

, . . . , en =

000...1

forms the standard basis for Rn.

Look Carefully: These vectors are the columns of the respectiveidentity matrix.

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

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.

Given A and an echelon form of A. Find a basis for Col A and NulA.

A= −3 9 −2 −7

2 −6 4 83 −9 −2 2

∼ 1 −3 6 9

0 0 4 50 0 0 0

Solution: Here columns 1 and 3 are pivot columns. So a basis for

Col A is

−323

−24−2

To �nd a basis for Nul A, write the system of equations, express thebasic variables x1 and x3 in terms of the free variables x2 and x4.

{x1 − 3x2 + 6x3 + 9x4 = 0

4x3 + 5x4 = 0

Thus we have

x3 =−5

4x4 and

x1 = 3x2−6x3−9x4 = 3x2−6(−5

4x4)−9x4

=3x2+ 30

4x4−9x4 = 3x2− 6

{x1 − 3x2 + 6x3 + 9x4 = 0

4x3 + 5x4 = 0

Thus we have x3 =−5

4x4 and

x1 = 3x2−6x3−9x4 = 3x2−6(−5

4x4)−9x4

=3x2+ 30

4x4−9x4 = 3x2− 6

{x1 − 3x2 + 6x3 + 9x4 = 0

4x3 + 5x4 = 0

Thus we have x3 =−5

4x4 and

x1 = 3x2−6x3−9x4 = 3x2−6(−5

4x4)−9x4

=3x2+ 30

4x4−9x4 = 3x2− 6

3x2− 6

−1.50

−1.251

A basis for Nul A is thus

−1.50

−1.251

Given A and an echelon form of A. Find a basis for Col A and NulA.

3 −1 7 3 9−2 2 −2 7 5−5 9 3 3 4−2 6 6 3 7

3 −1 7 0 60 2 4 0 30 0 0 1 10 0 0 0 0

Solution: Here columns 1, 2 and 4 are pivot columns. So a basis for

Col A is

3−2−5−2

−1296

To �nd a basis for Nul A, write the system of equations, express thebasic variables x1, x2 and x4 in terms of the free variables x3 and x5.

3x1 − x2 + 7x3 + 6x5 = 0

x2 + 4x3 + 3x5 = 0x4 + x5 = 0

Thus we havex4 =−x5,

2x2 =−4x3−3x5 =⇒ x2 =−2x3−1.5x5 and3x1 = x2−7x3−6x5=−2x3−1.5x5−7x3−6x5=−9x3−7.5x5.x1=−3x3−2.5x5

3x1 − x2 + 7x3 + 6x5 = 0

x2 + 4x3 + 3x5 = 0x4 + x5 = 0

Thus we havex4 =−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

3x1 − x2 + 7x3 + 6x5 = 0

x2 + 4x3 + 3x5 = 0x4 + x5 = 0

Thus we havex4 =−x5,2x2 =−4x3−3x5 =⇒ x2 =−2x3−1.5x5 and3x1 = x2−7x3−6x5=−2x3−1.5x5−7x3−6x5=−9x3−7.5x5.x1=−3x3−2.5x5

−3x3−2.5x5−2x3−1.5x5

−x5x5

−32100

−2.5−1.50−11

A basis for Nul A is thus

−32100

−2.5−1.50−11

Does ,

[−46

[2−3

]form a basis for R2. Why(not)?

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, thedeterminant of the matrix formed by these 2 vectors as columns is0). So it is a linearly dependent set and hence not a basis

Does ,

11−2

−5−12

form a basis for R3. Why(not)?

Solution: Idea: Check whether the matrix formed by these vectorsis invertible. (3 pivot columns and rows)

1 −5 7

1 −1 0

−2 2 5

R3+2R1

1 −5 7

0 4 −7

0 −8 19

R3+2R2

1 −5 70 4 −70 0 5

We have 3 pivot positions and hence this matrix is invertible(columns linearly independent). So the 3 given vectors form a basis.

1 −5 7

0 4 −7

0 −8 19

R3+2R2

1 −5 70 4 −70 0 5

We have 3 pivot positions and hence this matrix is invertible(columns linearly independent). So the 3 given vectors form a basis.

subspace, col space, basis

Education

vectors u

span of u

linear combination of

set of vectors h

r c1 u

product c u

vector b

subspaces of rn