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Math 307 Spring, 2003 Hentzel Time: 1:10-2:00 MWF Room: 1324 Howe Hall

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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 by Otto Bretscher

Wednesday, Feb 19 Chapter 3.3

Page 130 Problems 22, 30, 52

Main Idea: We label the vectors using an

index system.

Key Words: Dimension, basis.

Goal: Learn the basic results about linearly

independent spanning sets.

Previous Assignment

Page 118 Problem 16

Use paper and pencil to decide whether the given vectors

are linearly independent.

| 1 | | 4 | | 7 | | 0 |

a| 2 |+b| 5 |+c| 8 | = | 0 |

| 3 | | 6 | | 9 | | 0 |

a b c RHS

| 1 4 7 0 |

| 2 5 8 0 |

| 3 6 9 0 |

| 1 4 7 0 |

| 0 -3 -6 0 |

| 0 -6 -12 0 |

| 1 4 7 0 |

| 0 1 2 0 |

| 0 -6 -12 0 |

| 1 0 -1 0 |

| 0 1 2 0 |

| 0 0 0 0 |

| a | | 1 |

| b | = |-2 |

| c | | 1 |

They are not linearly independent. They satisfy this

dependence relation:

| 1 | | 4 | | 7 | | 0 |

| 2 | -2 | 5 |+ | 8 | = | 0 |

| 3 | | 6 | | 9 | | 0 |

Page 118 Problem 28.

| 1 1 1 |

Find a basis of the image of the matrix | 1 2 5 |.

| 1 3 7 |

| 1 1 1 a |

| 1 2 5 b |

| 1 3 7 c |

| 1 1 1 a |

| 0 1 4 -a+b |

| 0 2 6 -a+c |

| 1 0 -3 2a - b |

| 0 1 4 -a + b |

| 0 0 -2 a -2b+c |

| 1 0 -3 2a - b |

| 0 1 4 -a + b |

| 0 0 1 -a/2 +b-c/2 |

| 1 0 0 a/2 +2b -3c/2 |

| 0 1 0 a - 3b +2c |

| 0 0 1 -a/2+ b -c/2 |

This says that you can get any vector you want.

|1| |1| |1| |a|

(a/2+2b-3c/2)|1|+(a-3b+2c)|2|+(-a/2+b-c/2)|5|=|b|

|1| |3| |7| |c|

| 1 | | 1 | | 1 |

One basis for the image is | 1 | , | 2 | , | 5 |

| 1 | | 3 | | 7 |

| 1 | | 0 | | 0 |

Another basis for the image is | 0 |, | 1 |, | 0 |.

| 0 | | 0 | | 1 |

Page 118 Problem 34

Consider the 5x4 matrix A = [V1 V2 V3 V4].

| 1 |

Suppose that | 2 | is in the kernel of A.

| 3 |

| 4 |

Write V4 as a linear combination of V1 V2 V3.

We are given that V1 + 2 V2 + 3 V3 + 4 V4 = 0.

Thus V4 = -1/4 V1 -2/4 V2 -3/4 V3.

There are many many bases for a particular vector

space. There is nothing unique about bases

except for one thing.

Two bases of the same space will always have

the same number of vectors.

We call this number the dimension of the space.

So our first task is to show that it is really

true that two bases of the same vector

space always have the same number of

elements.

We actually show something stronger.

An independent set is always smaller than a

spanning set.

Theorem. Consider vectors V1, V2, ..., Vi and

W1, W2, ..., Ws in a subspace of Rn. If the V's

are linearly independent and the W are a

spanning set, then i <= s.

Proof:

[V1 V2 ... Vi ] = [W1 W2 ... Ws] [X1 X2 ... Xi]

nxi nxs sxi

Independent Spanning set Coefficients

If i > s the matrix [x1 x2 ... xi ] has this shape.

.

Then there is a vector X =/= 0 such that

[X1 X2 ... Xi] X = 0.

But then :

[V1 V2 ... Vi] X = 0

as well and this contradicts the fact that the Vi's are linearly

independent. Thus we have to agree that i <= s. That means,

the number of vectors in any spanning set must be at least as

large as the number of vectors in any linearly independent set.

Theorem: All Bases of a subspace V of Rn have

the same number of vectors.

Proof: Given bases V1, V2, ..., Vp and

W1, W2, ..., Wq, then

p <= q and q <= p. Thus p = q.

Definition: The number of elements in any basis of a vector

subspace V is called the dimension of V.

Theorem: Consider a subspace V of Rn with dim(V) = m.

(a) We can find at most m linearly independent vectors in V.

(b) We need at least m vectors to span V.

(c) If m vectors in V are linearly independent, then they form a

basis of V.

(d) If m vectors span V, then they form a basis of V.

Find a basis of the kernel of the matrix.

A = | 1 2 0 3 0 |

| 2 4 1 9 5 |

a b c

x y z w u

| 1 2 0 3 0 |

| 0 0 1 3 5 |

| x | |-2 | | -3 | | 0 |

| y | | 1 | | 0 | | 0 |

| z | = a | 0 | + b | -3 | + c |-5 |

| w | | 0 | | 1 | | 0 |

| u | | 0 | | 0 | | 1 |

<---These are a basis-- >

of the Null Space

RCF(A) = | 1 2 0 3 0 |

| 0 0 1 3 5 |

x x

Use these columns of A for a basis of the Range

of A. The Range of A is also the Column Space

of A.

A basis of Range (A) is | 1 | | 0 |

| 2 |, | 1 |

Notice that the elements of the null space

provide the dependence relations for the

columns which do not contain stair step

ones.

When you do Row Canonical form, you are finding a basis of the row space.

When you multiply on the left by an invertible

matrix P, the row space is preserved.

Proof: If P is any matrix, then the rows of PA are

linear combinations of the rows of A.

Thus RS(PA) c RS(A).

When P is invertible, then we get the reverse

inclusion since

RS(A) = RS(P-1 P A) c RS(PA) c RS(A).

Thus RS(PA) = RS(A) when P is invertible.

In particular, RS(A) = RS(RCF(A)) and the non

zero rows of RCF(A) are linearly independent due

to the stair step ones.

Thus the dimension of the RS(A) is the number of

non zero rows in the RCF(A).

The Dimension of the Row Space of A is called

the row rank of A. The Dimension of the column

space is called the column rank of A.

Theorem: The Dimension of the Row Space

of a Matrix is always the same as the

Dimension of the column space.

Row Rank = Column Rank

Proof that Row Rank = Column Rank.

Given a matrix A, choose columns C1 C2 ... Cr to be a basis of the column space.

| x11 ... x1n |

| |

A = [ C1 C2 ... Cr ] | |

| |

| xr1 ... xr n |

rxn

So A has a spanning set of the rows with r

vectors. Thus

Dimension of the Column Space is >= the

Dimension of the Row Space.

We can reverse the argument and prove the

reverse inequality. Thus the dimension of the

Row Space and Column Space are equal.