Lecture 8 nul col bases dim & rank - section 4-2, 4-3, 4-5 & 4-6

© 2012 Pearson Education, Inc.
Math 337-102
Lecture 8
Null and Column Spaces
Bases
Dimension
Rank

Determinants - Review
 How are |A| and |A-1| related?
 If |A| = 2 and A is 4x4, what is |3A|?
Slide 2.2- 2© 2012 Pearson Education, Inc.

2- 3© 2012 Pearson Education, Inc.
NULL SPACE OF A MATRIX - Definition
Definition: The null space of an mxn matrix A, written
as Nul A, is the set of all solutions of the homogeneous
equation Ax = 0. In set notation,
NUL A = {x : x is in Rn & Ax = 0}
In terms of transformations – set of all x in Rn mapped
to 0 via the linear transformation x  Ax

Null Space - Example


Theorem 2 – All Null Spaces are Subspaces
Theorem 2: The null space of an mxn matrix A
is a subspace of Rn.

Null Space - Example
H is set of all vectors in R4 whose coordinates
(a,b,c,d) satisify:
a – 2b + 5c = d
c – a = b
Is H a subspace of R4?

Finding Spanning Set for Nul A


Finding Spanning Set of Nul Space - Notes
1. The spanning set produced by the
method in Example (1) is automatically
linearly independent because the free
variables are the weights on the
spanning vectors.
2. # vectors in spanning set of Nul(A) = #
free vars in Ax=0.(except if Nul(A)={0})

COLUMN SPACE OF A MATRIX - Definition
 Definition: The column space of an matrix A,
written as Col A, is the set of all linear combinations
of the columns of A.
 Theorem 3: The column space of an mxn matrix A is
a subspace of Rm.
 A typical vector in Col A can be written as Ax for
some x because the notation Ax stands for a linear
combination of the columns of A. That is,
Col A = {b : b = Ax for some x in Rn}
.
m n
1
Col Span{a ,...,a }n
A 

Column Space - Example


COLUMN SPACE OF A MATRIX
 The notation Ax for vectors in Col A also shows that Col
A is the range of the linear transformation .
 The column space of an matrix A is all of Rm iff
the equation Ax=b has a solution for each b in Rm.
x xA
m n

Nul(A) versus Col(A) - Example
 Given
(a) Col(A) is a subspace of?
(b) Nul(A) is a subspace of?
(c) find a non-zero vector in Col(A).
2 4 2 1
2 5 7 3
3 7 8 6
A
 
   
 
  
3
2
u
1
0
 
 
 
 
 
 
3
v 1
3
 
  
 
  

Nul(A) versus Col(A) - Example
(d) Find a non-zero vector in Nul(A)
(e) Is u in Nul(A)?
(f) Is u in Col(A)?
(g) Is v in Col(A)?
(h) Is v in Nul(A)?

KERNEL & RANGE OF A LINEAR TRANSFORMATION
 Subspaces of vector spaces other than Rn are often
described in terms of a linear transformation T:x Ax
instead of a matrix.
 The kernel all u such that T(u)=0 (or null space)
 The range of T is all linear combinations of A  Col(A).

CONTRAST BETWEEN NUL A AND COL A FOR AN
MATRIX Am n
Nul A Col A
1. Nul A is a subspace of
Rn
1. Col A is a subspace of
Rm
2. Nul A is implicitly
defined; i.e., you are
given only a condition
(Ax=0) that vectors in
Nul A must satisfy.
2. Col A is explicitly
defined; i.e., you are
told how to build
vectors in Col A.

MATRIX Am n
3. Hard to find vectors in
Nul A. Row operations
on [A 0]are required.
3. Easy to find vectors in
Col A. The columns of
A are displayed; others
are formed from them.
4. There is no obvious
relation between Nul A
and the entries in A.
4. There is an obvious
relation between Col A
and the entries in A,
since each column of A
is in Col A.

MATRIX Am n
5. A typical vector v in Nul
A has the property that
Av=0.
5. A typical vector v in Col
A has the property that
the equation Ax=v is
consistent.
6. Given a specific vector v,
it is easy to tell if v is in
Nul A. Av = 0?.
6. Given a specific vector
v, it may take time to tell
if v is in Col A. Row
operations on are
required.
 vA

MATRIX Am n
7. Nul A = {0} iff the
equation Ax=0 has only
the trivial solution.
7. Col A = Rm iff the
equation Ax=b has a
solution for every b in
Rm.
8. Nul A = {0} iff the
linear transformation
T:x Ax is one-to-one.
8. Col A = Rm iff the
linear transformation
T:x Ax maps Rn onto
Rm.

Vector Spaces
LINEARLY INDEPENDENT
SETS; BASES

LINEAR INDEPENDENT SETS - Review
 An indexed set of vectors {v1, …, vp} in V is said to
be linearly independent if the vector equation
c1v1 + c2v2 + …+ cpvp = 0 (1)
has only the trivial solution, .
 The set {v1, …, vp} is said to be linearly
dependent if (1) has a nontrivial solution, i.e., if
there are some weights, c1, …, cp, not all zero, such
that (1) holds.
 In such a case, (1) is called a linear dependence
relation among v1, …, vp.

LINEAR INDEPENDENT SETS
Theorem 4: An indexed set {v1, …, vp} of two or more
vectors, with v1 ≠ 0, is linearly dependent iff some vj
(with j > 1) is a linear combination of the preceding
vectors, v1, …, vj-1.

Linearly Independent - Examples
1. p1(t)=1, p2(t)=t, p3(t)=4 – t. Is the set
{p1,p2,p3} linearly independent?
2. Is the set {sin(t), cos(t)} linearly independent?
3. Is the set {sin(t)cos(t),sin(2t) linearly
independent?

Basis - Definition
Definition: Let H be a subspace of a vector space V. An
indexed set of vectors B = {b1, …, bp} in V is a basis for
H if
(i) B is a linearly independent set, and
(ii) The subspace spanned by B coincides with H; i.e.,
H = Span {b1, …, bp}
Thus a basis of V:
 linearly independent
 spans V.

STANDARD BASIS - Rn
 Let e1, …, en be the columns of the matrix, In.
 That is,
 The set {e1, …, en} is called the standard basis for Rn .
See the following figure.
n n
1 2
1 0 0
0 1
e ,e ,...,e
0
0 0 1
n
     
     
       
     
     
     

STANDARD BASIS - Pn
 Standard Basis for Pn: S={1,t,t2,…,tn}

Basis - Example
Show that if A is invertible, then the columns of
A: {a1,…,an} form a basis of Rn.

Basis - Example
𝐯1 =
3
0
−6
, 𝐯2 =
−4
1
7
, 𝐯3 =
−2
1
5
Is set {v1,v2,v3} a basis for R3?

Basis - Example
𝐯1 =
0
2
−1
, 𝐯2 =
2
2
0
, 𝐯3 =
6
16
−5
,
H = Span{v1,v2,v3}
Is set {v1,v2,v3} a basis for H?

THE SPANNING SET THEOREM
Theorem 5: Let S = {v1,v2,…,vp} be a set in V, and let
H = Span{v1,v2,…,vp} .
a. If one of the vectors in S—say, vk—is a linear
combination of the remaining vectors in S, then the
set formed from S by removing vk still spans H.
b. If H ≠{0}, some subset of S is a basis for H.

BASIS FOR COL B
 Example 2: Find a basis for Col B, where
1 2 5
1 4 0 2 0
0 0 1 1 0
b b b
0 0 0 0 1
0 0 0 0 0
B
 
 
      
 
 

BASES FOR COL A
 Theorem 6: The pivot columns of a matrix A form a
basis for Col A.
 Use pivot columns of A, not REF(A)!!!
 Row operations do not change linear dependence
relationship
 BUT …
 Row operations do change Column Space.

BASIS for NUL A
 Solving Ax=0 provides a basis
 Span Nul(A) by definition
 Linearly Independent by arrangement of pivots

Basis - Summary
 Large enough to span
 Small enough to be Linearly Independent

Exercises
𝐯1 =
1
−2
3
, 𝐯2 =
−2
7
−9
Is the set {v1,v2} a basis for R3?
Is the set {v1,v2} a basis for R2?

Exercises
𝐯1 =
1
−3
4
, 𝐯2 =
6
2
−1
, 𝐯3 =
2
−2
3
, 𝐯4 =
−4
−8
9
H = Span{v1,v2,v3,v4}
Find a basis for H.

Vector Spaces
THE DIMENSION OF A
VECTOR SPACE

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?

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
   
   
 
   

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
(CONTINUED)
 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)?

Lecture 8 nul col bases dim & rank - section 4-2, 4-3, 4-5 & 4-6

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