Download Presentation

Loading in 3 Seconds

This presentation is the property of its rightful owner.

X

Sponsored Links

- 73 Views
- Uploaded on
- Presentation posted in: General

Math 307 Spring, 2003 Hentzel Time: 1:10-2:00 MWF Room: 1324 Howe Hall Instructor: Irvin Roy Hentzel Office 432 Carver Phone 515-294-8141

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Math 307

Spring, 2003

Hentzel

Time: 1:10-2:00 MWF

Room: 1324 Howe Hall

Instructor: Irvin Roy Hentzel

Office 432 Carver

Phone 515-294-8141

E-mail: [email protected]

http://www.math.iastate.edu/hentzel/class.307.ICN

Text: Linear Algebra With Applications, Second Edition

- Friday, February 14, Section 3.1
- Page 106 Problems 24,26,44
- Main Idea: Review what we have been doing and use new words.
- Key Words: Kernel, Image, Codomain, Span,
< V1, V2, ..., Vr >

- Goal: Learn uses and limitations of the notation for the span of a set of vectors.

- We know how to solve AX = B. We reduce
- it to the Row Canonical Form and read off the
- answers.
- The answer is of the form
- X = X0 + a1 X1 + a2 X2 + ... ar Xr

There are three general possibilities for the outcome.

(1) There are no solutions at all.

This happens when RHS has a stair step one.

(2) There is exactly one solution.

This happens when RHS is the only column

with no stair step one.

(3) There are infinitely many solutions.

This happens when RHS and at least one other

column has no stair step one.

Case (1) implies that there must be some B's such that AX = B is not possible.

We then ask the related question, “For what B's does there exist a solution for AX = B?”

The answer to the question is a set.

We actually know of how to express the solution right now.

From our knowledge of matrix multiplication,

AX = B

B is a linear combination of the

columns of the matrix A.

The SPAN of a set of vectors is the set of all linear combinations of the vectors in the set.

Thus we say:

AX = B

B is in the span of the columns of A.

In matrix theory often we often have to answer

questions where there are an infinite number

of possible answers.

A convenient way of expressing all answers is

to give a few such answers and then say that

the complete set of answers is the span of the

vectors listed.

Notice that when we solved AX = B, we wrote

the answer as

X = X0 + a1 X1 + a2 X2 + ... + ar Xr

We said that the set of all answers were of the

form:

X = X0 + S

where S was some element in <X1, X2, ..., Xr>.

The solutions can be expressed in many, many ways.

We could also express them differently as

X = Y0 + b1 Y1 + b2 Y2 + ... + bs Ys

or

X = Y0 + T

where T is some element in <Y1, Y2, ..., Ys>.

The answers are hard to compare. They can both be

correct, but look quite different.

We can say that:

1. r = s.

(If not, one can be removed without changing the span.)

2. <X1,X2, ..., Xr > = <Y1, Y2, ..., Ys>

The two spans are equal, not the individual vectors.

3. X0 and Y0 can be replaced by any solution.

Because of 2, we are interested in knowing when two

spans are the same.

In less mathematical terms, look at the forest,

not the trees.

The forest is the span. The trees are the

individual vectors.

| 2 | | 2 | |-1 | | 3 |

Example. < | 3 |, |-4 | > = < | 2 |, | -2 | >

| 5 | |-2 | | 1 | | 1 |

These two spans are the same plane.

The normal to the plane is the red line.

If we change the view point we can

see that the spans are the same plane.

We can see that the spans are the same by

| a |

seeing which vectors | b | are in the span.

| c |

| 2 | | 2 | | a | |-1 | | 3 | | a |

x| 3 | + y | -4 | = | b | x | 2 | + y | -2 | = | b |

| 5 | | -2 | | c | | 1 | | 1 | | c |

| 2 2 a | | -1 3 a |

| 3 -4 b | | 2 -2 b |

| 5 -2 c | | 1 1 c |

| 2 2 a | | -1 3 a |

| 3 -4 b | | 2 -2 b |

| 5 -2 c | | 1 1 c |

| 1 1 a/2 | | 1 -3 -a |

| 3 -4 b | | 2 -2 b |

| 5 -2 c | | 1 1 c |

| 1 1 a/2 | | 1 -3 -a |

| 0 -7 -3a/2+b | | 0 4 2a+b |

| 0 -7 -7a/2+c | | 0 4 a+c |

| 1 1 a/2 | | 1 -3 -a |

| 0 -7 -3a/2+b | | 0 4 2a+b |

| 0 0 - a-b+c | | 0 0 -a-b+c |

| 1 0 2 a/7 + b/7 | | 1 0 a/2 + 3b/4 |

| 0 1 +3 a/14 - b/7 | | 0 1 a/2 + b/4 |

| 0 0 -a – b + c | | 0 0 - a - b + c |

| a |

The vector | b | is in the span

| c |

-a-b+c = 0,

Which simply says:

"the third coordinate is the sum of the first two.“

Notice that this is true for all the vectors given.

We were interested in showing that the two

spans were the same and we have

accomplished that.

We can go on and show how to get the

| a |

vector | b | in each case.

|a+b|

| 2 | | 2 | | a |

(2a/7 + b/7) | 3 | + (3a/14 - b/7) | -4 | = | b |

| 5 | | -2 | |a+b|

|-1 | | 3 | | a |

(a/2 + 3b/4)| 2 | +(a/2 + b/4)|-2 | = | b |

| 1 | | 1 | |a+b|

The IMAGE of a matrix A, is the span of the

columns of A.

B is in the image of A AX = B has a solution,

B is not in the image of A AX = B is inconsistent.

Proof: This is self evident because AX is a

linear combination of the columns of A.

There is a distinction between kernel and null space,

but in math 307, we ignore it so, in this course,

Kernel is the same thing as the Null Space.

(It is the null space of a matrix A, but the kernel of the

linear transformation whose matrix is A)

The NULL SPACE of a matrix A is the set of all vectors

X such that AX = 0.

The null space is < X1,X2, ..., Xr > for the

system of linear equations AX = 0.

(* NOTICE X0 is missing *)

Example:

For the matrix | 0 1 |

| 0 0 |.

What is the Range?

What is the Kernel?

Range = | c |

| 0 |

Kernel = | c |

| 0 |

Both the kernel and the range are the X-axis.

For the matrix | 1 2 3 |

| 2 1 0 |

| 3 3 3 |

What is the range?

What is the kernel?

|1 2 3 | | 1 2 3 | | 1 2 3 | | 1 0 -1 |

|2 1 0 | ~ | 0 -3 -6 | ~ | 0 1 2 | ~ | 0 1 2 |

|3 3 3 | | 0 -3 -6 | | 0 0 0 | | 0 0 0 |

| 1 |

The kernel is the span of |-2 |.

| 1 |

| 1 | | 2 |

The range is the span of | 2 | | 1 |

| 3 | | 3 |

The third column is unnecessary because it can be generated from the first two. How?

The elements of the null space are dependence

relations.

Since | 1 2 3 | | 1 | | 0 |

| 2 1 0 | |-2 | = | 0 |

| 3 3 3 | | 1 | | 0 |

We can view this using the columns as:

C1 - 2 C2 + C3 = 0

so:

C3 = -C1 + 2 C2.

Thus C3 is not needed since C3 is in the span of the first

two.

Which columns should you use in your

spanning set?

A useful rule is: Use the columns which

generate stair step ones.

The stair step ones determine the columns

for the spanning set of the column space.

The stair step ones determine the columns for the spanning set of the column space.

The non-stair step ones determine the columns for the spanning set of the null space.

Pictorially speaking.

X-------------------->AX

-------

Rm Rn

0 range

kernel

Question. What is the Column space of

| 4 7 2 4 |

A = | 3 5 1 2 |

| 1 2 1 2 |

It will be a subspace of R^3.

| 4 7 2 4 x| | 1 2 1 2 z| | 1 2 1 2 z| |1 0-3-6 2y-5z|

| 3 5 1 2 y|~| 3 5 1 2 y|~| 0-1-2-4 y-3z|~|0 1 2 4 -y+3z|

| 1 2 1 2 z| | 4 7 2 4 x| | 0-1-2-4 x-4z| |0 0 0 0 x-y-z|

| x | | y+z|

| y | is in the range if and only if it is of the form | y |.

| z | | z |

- Another way to specify the range is to use the columns of the original matrix indicated by the stair step ones.
| 4| | 7 | | 4s+7t |

The range is the span of | 3| | 5 | or | 3s+5t |

| 1| | 2 | | s+2t |

But with this designation, it is harder to tell if something is in the range or not.

- Grading Scale
- 93-100 A
- 90-92 A-
- 87-89 B+
- 83-86 B
- 80-82 B-
- 75-79 C+
- 70-74 C
- 65-69 C-
- 64 D+
- 61-63 D
- 60 D-