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

1Col Span{a ,...,a }

nA

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

2u

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 1e ,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 =30−6

, 𝐯2 =−417

, 𝐯3 =−215

Is set {v1,v2,v3} a basis for R3?


Basis - Example

𝐯1 =02−1

, 𝐯2 =220

, 𝐯3 =616−5

,

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 0b 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 =1−23

, 𝐯2 =−27−9

Is the set {v1,v2} a basis for R3?

Is the set {v1,v2} a basis for R2?


Exercises

𝐯1 =1−34

, 𝐯2 =62−1

, 𝐯3 =2−23

, 𝐯4 =−4−89

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 =362, 𝐯2 =

−101

, 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 dH 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 81 −2 −4 3 −2−7 8 10 3 −104 −5 −7 0 4

~

1 −2 −4 3 60 3 9 −12 120 0 0 0 00 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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Transcript of Lecture 8 nul col bases dim & rank - section 4-2, 4-3, 4-5 & 4-6