IMT, col space again

Announcements

Ï Please bring any grade related questions regarding quiz 2without delay.

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 today evening

Ï Review for Exam 1 after this lecture

Ï I will be in o�ce all day tomorrow. Feel free to stop by anytime if you 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.

Last Class..

The column space of a matrix A is the set of all linearcombinations of the columns of A. It is denoted by Col A.

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.

Last Class...

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

1. linearly independent set in H

2. that spans H

Last Class...

To �nd basis for Col A

1. Look for the pivot columns in the echelon form of A

2. Pick the corresponding columns from A

Last Class...

Finding Basis for Nul A

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.

Section 2.9 Dimension and Rank

De�nitionLet H be a non-zero subspace. The number of vectors in any

basis for H is called the dimension of H. It is denoted by dim H.

The dimension of the zero subspace {0} is zero.


Example

1. R3 has dimension 3. Every basis for R3 has 3 vectors.

2. A straight line through 0 in R3 is one dimensional.


De�nitionThe Rank of a matrix A is the dimension of the column space of A.

Finding Rank of A

To �nd the rank of a matrix A,

1. Count the number of vectors in Col A.

2. Again, this is exactly the number of pivot columns.

3. Again, this is exactly the number of basic variables.

Finding dim Nul A

1. Count the number of vectors in Nul A.

2. Again, this is exactly the number of non-pivot columns.

3. Again, this is exactly the number of free variables.

Rank + dim Nul A=??

The Rank Theorem

TheoremIf a matrix has n columns, rank A + dim Nul A =n.

Useful

The Basis Theorem

TheoremLet H be any p dimensional subspace of Rn.

1. Any linearly independent set of exactly p elements in H is a

basis for H.

2. Any set of p elements of H that spans H is a basis for H.

Revisit the Invertible Matrix Theorem (IMT)

Let A be an invertible n×n matrix.

1. dim Col A =n (Remember it MUST have n pivot columns?)

2. rank A = n.

3. Nul A={0} (Only trivial solution, no free variables or non-pivotcolumns)

4. dim Nul A = 0

5. Col A = Rn (all columns are pivot columns)

Example 10, section 2.9

Given A and an echelon form of A. Find a basis for Col A and NulA. Find dim Col A and dim Nul A. What is the rank of A?

A=

1 −2 9 5 41 −1 6 5 −3−2 0 −6 1 −24 1 9 1 −9

∼

1 −2 9 5 40 1 −3 0 −70 0 0 1 −20 0 0 0 0

Solution: Here columns 1, 2 and 4 are pivot columns.

So a basis for

Col A is

11−24

,

−2−101

,

5511

.


Given A and an echelon form of A. Find a basis for Col A and NulA. Find dim Col A and dim Nul A. What is the rank of A?

A=

1 −2 9 5 41 −1 6 5 −3−2 0 −6 1 −24 1 9 1 −9

∼

1 −2 9 5 40 1 −3 0 −70 0 0 1 −20 0 0 0 0

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

Col A is

11−24

,

−2−101

,

5511

.


What is dim Col A?

3, since there are 3 vectors.

What is rank A? 3 (same as dim Col A)


What is dim Col A? 3, since there are 3 vectors.

What is rank A? 3 (same as dim Col A)

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.

x1 − 2x2 + 9x3 + 5x4 + 4x5 = 0

x2 − 3x3 − 7x5 = 0x4 − 2x5 = 0

Thus we havex4 = 2x5,x2 = 3x3+7x5 andx1 = 2x2−9x3−5x4−4x5=2(3x3+7x5)−9x3−5(2x5)−4x5=−3x3.


x1 − 2x2 + 9x3 + 5x4 + 4x5 = 0x2 − 3x3 − 7x5 = 0

x4 − 2x5 = 0

Thus we havex4 = 2x5,

x2 = 3x3+7x5 andx1 = 2x2−9x3−5x4−4x5=2(3x3+7x5)−9x3−5(2x5)−4x5=−3x3.


x1 − 2x2 + 9x3 + 5x4 + 4x5 = 0x2 − 3x3 − 7x5 = 0

x4 − 2x5 = 0

Thus we havex4 = 2x5,x2 = 3x3+7x5 and

x1 = 2x2−9x3−5x4−4x5=2(3x3+7x5)−9x3−5(2x5)−4x5=−3x3.


x1 − 2x2 + 9x3 + 5x4 + 4x5 = 0x2 − 3x3 − 7x5 = 0

x4 − 2x5 = 0

Thus we havex4 = 2x5,x2 = 3x3+7x5 andx1 = 2x2−9x3−5x4−4x5=2(3x3+7x5)−9x3−5(2x5)−4x5=−3x3.


So,

x1x2x3x4x5

=

−3x3

3x3+7x5x32x5x5

= x3

−33100

+x5

07021

.

A basis for Nul A is thus

−33100

,

07021

.


What is dim Nul A?

2, since there are 2 vectors.

Are the columns of this matrix linearly independent or linearlydependent?Linearly Dependent, since there are free variables.


What is dim Nul A? 2, since there are 2 vectors.




Are the columns of this matrix linearly independent or linearlydependent?

Linearly Dependent, since there are free variables.

Review

You must know

1. All the above stu�. Given a matrix A, (and some times itsechelon form as well, otherwise you have to do it), �ndingbasis for Col A, �nding Nul A, their dimensions, rank etc.

2. Inverse of a 3×3 matrix using the algorithm we learned inclass.

3. Problems where you have to check linear combination, lineardependence (this automatically tests your knowledge aboutconsistency, homogeneous equations, pivots, basic and freevariables, span etc etc)

4. The invertible matrix theorem with all bells and whistles. Bothas a "�ll in the blank" question and given a matrix to decidewhether it is invertible.

5. Finding matrix products, inverses of 2×2 square matrices,using the inverse to solve a simple system of 2 equations, 2variables.

Re-visit Linear combination/Span

Example 26, section 1.3 Let A= 2 0 6

−1 8 51 −2 1

and b= 10

33

.Let W be the set of all linear combinations of the columns of A. Isb in W ?(In other words is b in the span of the columns of A?)

What are we supposed to check here?

Consistency!!! (Thisproblem has nothing to do with free variables, basic variables etcetc. Be very precise with your conclusion).

Re-visit Linear combination/Span

Example 26, section 1.3 Let A= 2 0 6

−1 8 51 −2 1

and b= 10

33

.Let W be the set of all linear combinations of the columns of A. Isb in W ?(In other words is b in the span of the columns of A?)

What are we supposed to check here? Consistency!!! (Thisproblem has nothing to do with free variables, basic variables etcetc. Be very precise with your conclusion).

Start with sugmented matrix, divide the �rst row by 31 0 3 5

−1 8 5 3

1 −2 1 3

R1+R2

R3-R1

1 0 3 50 8 8 80 −2 −2 −2

Divide R2 by 8 and R3 by -2

1 0 3 50 1 1 10 1 1 1

1 0 3 5

0 1 1 1

0 1 1 1

R3-R2

1 0 3 50 1 1 10 0 0 0

The system is consistent and so b is in W .

Re-visit Linear (in)dependence

Example 8, section 1.7 Determine if the columns of the givenmatrix form a linearly independent set. Justify your answer.

A= 1 −3 3 −2

−3 7 −1 20 1 −4 3

What are we supposed to check here? You did this on yesterday'squiz.

Absence of free variables (only trivial or zero solution) for thehomogeneous system. If there are no free variables (all columns arepivot columns) then we have linear independence.

Re-visit Linear (in)dependence

Example 8, section 1.7 Determine if the columns of the givenmatrix form a linearly independent set. Justify your answer.

A= 1 −3 3 −2

−3 7 −1 20 1 −4 3

What are we supposed to check here? You did this on yesterday'squiz. Absence of free variables (only trivial or zero solution) for thehomogeneous system. If there are no free variables (all columns arepivot columns) then we have linear independence.

1 −3 3 −2 0

−3 7 −1 2 0

0 1 −4 3 0

R2+3R1

1 −3 3 −2 00 −2 8 −4 00 1 −4 3 0

Divide R2 by -2 1 −3 3 −2 0

0 1 −4 2 00 1 −4 3 0

1 −3 3 −2 0

0 1 −4 2 0

0 1 −4 3 0

R3-R2

1 −3 3 −2 00 1 −4 2 00 0 0 1 0

Third column is non-pivot which means x3 is free (non-trivialsolution). So the columns are linearly dependent. (No need to writeout the solution here)

Re-visit Matrix Product, Inverse

Given A=[3 27 4

]and B =

[3 −47 8

]. Find AB and A−1.

Use the above A−1 to solve{3x1 + 2x2 = 27x1 + 4x2 = 4

AB =[3 27 4

][3 −47 8

]=

[3.3+2.7 3.(−4)+2.87.3+4.7 7.(−4)+4.8

]=

[23 449 4

]

Re-visit Matrix Product, Inverse

Given A=[3 27 4

], det A= (3)(4)− (7)(2)=−2 6= 0.

Flip the diagonal elements, change signs of the o�-diagonalelements. [

4 −2−7 3

]Divide each element by det A=−2 and we get

A−1 =[ −2 17/2 −3/2

]

To solve {3x1 + 2x2 = 27x1 + 4x2 = 4

,

�nd the product A−1b where b=[24

]

x=[x1x2

]=

[ −2 17/2 −3/2

]︸︷︷︸

A−1

[24

]︸︷︷︸

b

=[01

]

Inverse of 3×3 matrix

Please go through example(s) for the inverse of 3×3 matrices wedid in class

Method:

1. Augment the given matrix A with the 3×3 identity matrix.

2. Do row reductions on both A and the augmented identitymatrix till A becomes the 3×3 identity matrix.

3. The new matrix formed from the 3×3 identity matrix is ourA−1.

Don't forget I have open o�ce hours all day tomorrow

Feel free to email me if you cannot make it to my o�ce

Good Luck!!!!!!!

IMT, col space again

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Transcript of IMT, col space again