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. - PowerPoint PPT PresentationTRANSCRIPT
Math 307Spring, 2003
Hentzel
Time: 1:10-2:00 MWFRoom: 1324 Howe Hall
Instructor: Irvin Roy HentzelOffice 432 Carver
Phone 515-294-8141E-mail: [email protected]
http://www.math.iastate.edu/hentzel/class.307.ICNText: 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 |
| 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.
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