Lecture 9 dim & rank - 4-5 & 4-6

© 2012 Pearson Education, Inc.
Vector Spaces
THE DIMENSION OF A
VECTOR SPACE

5- 2© 2012 Pearson Education, Inc.
DIMENSION OF A VECTOR SPACE
Theorem 9: If a vector space V has a basis
B = {b1, … ,bn} then any set in V containing more
than n vectors must be linearly dependent.

Theorem 10: If a vector space V has a basis of n
vectors, then every basis of V must consist of
exactly n vectors.

Definition: If V is spanned by a finite set, then V is
said to be finite-dimensional, and the dimension
of V, written as dim V, is the number of vectors in a
basis for V. The dimension of the zero vector
space {0} is defined to be zero. If V is not spanned
by a finite set, then V is said to be infinite-
dimensional.

Dimension - Examples
 Dim Rn?
 Dim Pn?
 𝐯1 =
3
6
2
, 𝐯2 =
−1
0
1
, H=Span{v1,v2}, Dim
H?
Slide 2.2- 5© 2012 Pearson Education, Inc.

Dimension - Examples
Find the dimension of the subspace
3 6
5 4
: , , , in
2
5
a b c
a d
H a b c d
b c d
d
   
      
   
    

Geometric Interpretation of Subspaces of R3
 0-dimensional  origin
 1-dimensional  line through the origin
 2-dimensional  plane through the origin
 3-dimensional  all of R3

SUBSPACES OF A FINITE-DIMENSIONAL
SPACE
Theorem 11: Let H be a subspace of a finite-
dimensional vector space V. Any linearly
independent set in H can be expanded, if
necessary, to a basis for H. Also, H is finite-
dimensional and dim H ≤ dim V

THE BASIS THEOREM
Theorem 12: Let V be a p-dimensional vector
space,
p ≥ 1. Any linearly independent set of exactly p
elements in V is automatically a basis for V. Any
set of exactly p elements that spans V is
automatically a basis for V.

THE DIMENSIONS OF NUL A AND COL A
 dim Nul(A) = # free vars in Ax=0
 dim Col(A) = # pivot columns in A

DIMENSIONS OF NUL A AND COL A
 Example 2: Find the dimensions of the null
space and the column space of
3 6 1 1 7
1 2 2 3 1
2 4 5 8 4
A
   
   
 
   

© 2012 Pearson Education, Inc.
Vector Spaces
RANK

THE ROW SPACE
 If A is an matrix, each row of A has n
entries and thus can be identified with a vector in
Rn.
 The set of all linear combinations of the row
vectors is called the row space of A and is
denoted by Row A.
 Since the rows of A are identified with the
columns of AT, we could also write Col AT in
place of Row A.
m n

THE ROW SPACE
Theorem 13: If two matrices A and B are row
equivalent, then their row spaces are the same. If
B is in echelon form, the nonzero rows of B form a
basis for the row space of A as well as for that of
B.
Important:
 Use pivot cols of A for basis of Col(A)
 Use pivot rows of REF(A) for basis of Row(A)

THE ROW SPACE - Example
Example 1: Find bases for the row space, the
column space, and the null space of the matrix
2 5 8 0 17
1 3 5 1 5
3 11 19 7 1
1 7 13 5 3
   
 
 
 
   

RANK- Definition
 Definition: rank of A is the dim Col(A).
 Since Row A is the same as Col AT, the
dimension of the row space of A is the rank of
AT.
 The dimension of the null space is sometimes
called the nullity of A.

THE RANK THEOREM
Theorem 14: The dimensions of the column
space and the row space of an mxn matrix A
are equal. This common dimension, the rank
of A, also equals the number of pivot
positions in A and satisfies the equation:
rank A + dim Nul(A) = n

THE RANK THEOREM
Proof:
dim Col(A)  # pivot columns in REF(A) = #pivots
dim Row(A)  # nonzero rows in REF(A) = #pivots
dim Nul(A)  #free vars in Ax=0 = #nonpivot cols

THE RANK THEOREM - Example
Example 2:
a. If A is a 7x9 matrix with a two-dimensional null
space, what is the rank of A?
b. Could a 6x9 matrix have a two-dimensional null
space?

THE INVERTIBLE MATRIX THEOREM
(CONTD)
 Theorem: Let A be an nxn matrix. Then the
following statements are each equivalent to the
statement that A is an invertible matrix.
m. The columns of A form a basis of Rn.
n. Col(A) = Rn
o. dim Col(A) = n
p. rank A = n
q. Nul(A) = {0}
r. dim Nul(A) = 0

Practice Problem
𝐴 =
2 −1 1 −6 8
1 −2 −4 3 −2
−7 8 10 3 −10
4 −5 −7 0 4
~
1 −2 −4 3 6
0 3 9 −12 12
0 0 0 0 0
0 0 0 0 0
1. Find rank A
2. Find dim Nul(A)
3. Find basis for Col(A)
4. Find basis for Row(A)
5. What is the next step to find basis for Nul(A)?
6. How many pivots in REF(AT)?

Markov Chains
State 1 State 2
Probability going
from 1 to 2
Prob
staying
in state
1
Probability going
from 2 to 1
Prob
staying
in state
2

Stochastic Matrix
 Square matrix
 Columns are probability vectors
 aij is probability of going from state i to state j

Markov Chain
 Sequence of state vectors, xk, such that
xk+1 = Pxk
 Next state depends only on current state

Markov Chains - Example
City Suburbs
.4
.6
.3
.7

Stochastic Matrix - Example
 Find the stochastic matrix for the previous
example.
 Use the stochastic matrix to predict next
year’s populations if this year’s
populations are:
 City: 600,000
 Suburbs: 400,000

Steady State Vector
 Pq= q
 Every stochastic matrix has a steady state
vector
 Theorem 18: If P is an nxn regular stochastic
matrix, then P has a unique steady-state
vector q. Further, if x0 is any initial state, the
Markov chain converges to q.
 Regular: some power of P contains only
strictly positive entries

Regular Matrix - Example
 Is P =
1 .3
0 .7
a regular stochastic matrix?

Steady State Vector - Example
 Find the steady state vector for the
city/suburb example.

Lecture 9 dim & rank - 4-5 & 4-6

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Lecture 9 dim & rank - 4-5 & 4-6