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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.
Slide 4.2- 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
Theorem 2 – All Null Spaces are Subspaces
Theorem 2: The null space of an mxn matrix A is a subspace of Rn.
Slide 2.2- 5© 2012 Pearson Education, Inc.
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?
Slide 2.2- 6© 2012 Pearson Education, Inc.
Slide 4.2- 8© 2012 Pearson Education, Inc.
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})
Slide 4.2- 9© 2012 Pearson Education, Inc.
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
Slide 4.2- 11© 2012 Pearson Education, Inc.
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).
Slide 2.2- 12© 2012 Pearson Education, Inc.
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)?
Slide 2.2- 13© 2012 Pearson Education, Inc.
Slide 4.2- 14© 2012 Pearson Education, Inc.
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).
Slide 4.2- 15© 2012 Pearson Education, Inc.
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.
Slide 4.2- 16© 2012 Pearson Education, Inc.
CONTRAST BETWEEN NUL A AND COL A FOR AN
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.
Slide 4.2- 17© 2012 Pearson Education, Inc.
CONTRAST BETWEEN NUL A AND COL A FOR AN
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
Slide 4.2- 18© 2012 Pearson Education, Inc.
CONTRAST BETWEEN NUL A AND COL A FOR AN
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.
Slide 4.3- 20© 2012 Pearson Education, Inc.
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.
Slide 4.3- 21© 2012 Pearson Education, Inc.
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?
Slide 2.2- 22© 2012 Pearson Education, Inc.
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.
Slide 2.2- 23© 2012 Pearson Education, Inc.
Slide 4.3- 24© 2012 Pearson Education, Inc.
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}
Slide 2.2- 25© 2012 Pearson Education, Inc.
Basis - Example
Show that if A is invertible, then the columns of
A: {a1,…,an} form a basis of Rn.
Slide 2.2- 26© 2012 Pearson Education, Inc.
Basis - Example
𝐯1 =30−6
, 𝐯2 =−417
, 𝐯3 =−215
Is set {v1,v2,v3} a basis for R3?
Slide 2.2- 27© 2012 Pearson Education, Inc.
Basis - Example
𝐯1 =02−1
, 𝐯2 =220
, 𝐯3 =616−5
,
H = Span{v1,v2,v3}
Is set {v1,v2,v3} a basis for H?
Slide 2.2- 28© 2012 Pearson Education, Inc.
Slide 4.3- 29© 2012 Pearson Education, Inc.
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.
Slide 4.3- 30© 2012 Pearson Education, Inc.
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
Slide 4.3- 31© 2012 Pearson Education, Inc.
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
Slide 2.2- 32© 2012 Pearson Education, Inc.
Slide 4.3- 33© 2012 Pearson Education, Inc.
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?
Slide 2.2- 34© 2012 Pearson Education, Inc.
Exercises
𝐯1 =1−34
, 𝐯2 =62−1
, 𝐯3 =2−23
, 𝐯4 =−4−89
H = Span{v1,v2,v3,v4}
Find a basis for H.
Slide 2.2- 35© 2012 Pearson Education, Inc.
Slide 4.5- 37© 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.
Slide 4.5- 38© 2012 Pearson Education, Inc.
DIMENSION OF A VECTOR SPACE
Theorem 10: If a vector space V has a basis of n vectors,
then every basis of V must consist of exactly n vectors.
Slide 4.5- 39© 2012 Pearson Education, Inc.
DIMENSION OF A VECTOR SPACE
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?
Slide 2.2- 40© 2012 Pearson Education, Inc.
Dimension - Examples
Find the dimension of the subspace
Slide 2.2- 41© 2012 Pearson Education, Inc.
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
Slide 2.2- 42© 2012 Pearson Education, Inc.
Slide 4.5- 43© 2012 Pearson Education, Inc.
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
Slide 4.5- 44© 2012 Pearson Education, Inc.
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.
Slide 4.5- 45© 2012 Pearson Education, Inc.
THE DIMENSIONS OF NUL A AND COL A
dim Nul(A) = # free vars in Ax=0
dim Col(A) = # pivot columns in A
Slide 4.5- 46© 2012 Pearson Education, Inc.
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
Slide 4.6- 48© 2012 Pearson Education, Inc.
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
Slide 4.6- 49© 2012 Pearson Education, Inc.
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)
Slide 4.6- 50© 2012 Pearson Education, Inc.
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
Slide 4.6- 51© 2012 Pearson Education, Inc.
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
Slide 2.2- 52© 2012 Pearson Education, Inc.
Slide 4.6- 53© 2012 Pearson Education, Inc.
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
Slide 4.6- 54© 2012 Pearson Education, Inc.
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?
Slide 4.6- 55© 2012 Pearson Education, Inc.
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)?
Slide 2.2- 56© 2012 Pearson Education, Inc.