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Lecture 16
General Vector Spaces
Basis of Row Space and Column Space
Rank and Nullity
Covered Range : 4.8 +4.9
2021/11/17Lec16_LA21_ Singling Lee 1
Terminologies
Row Space ;
Column Space ;
Null Space ;
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Recap Row(A), Col(A)and Null(A)
Definition 2 : A is an m×nmatrix A=[aij]
Row(A) ⊆ Rn: Row space of A ⇒ span{r1,r2,…,rm}
Col(A) ⊆ Rm: Column space of A ⇒ span{c1,c2,…,cn}
Null(A) ⊆ Rn: Null space of A ⇒ Solution of Ax=0
Note : Ax=0 always has a solution x=0 (Orthogonal to each row vector)
The solution set is a vector subspace of Rn
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Recap Col (A) and the Solution of Ax=b
Theorem4.7.1 Ax = b is consistent if and only if b is in the column space of A.
Means b ∈ Col(A) = span {c1, c2, … ,cn}.
Ax : Linear combinations of column vectors of A .
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Recap Orthogonality of Row(A) and Null(A)
Theorem 3.4.3 : If 𝐴 is an 𝑚 × 𝑛 matrix, then the solution set of the homogeneous linear system 𝐴𝐱 = 𝟎 consists of all vectors in 𝑅𝑛 that are orthogonal to every row vector of 𝐴.
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Recap Relation Between Ax = 0 and Ax = b
Theorem 4.8.2 : If x0 is any solution of a consistent linear system of Ax = b and ifS={v1 ,v2 , … , vk} is a basis for the null pace of A, then every solution of Ax = b can be expressed in the form
x = x0 + c1v1 + c2v2 + … + ckvk
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Recap Subspace of Ax = 0 by Spanning (II)
Example 6 :
x2 = r, x4 = s, x5 = t
Solution : 𝐱 = (−3𝑟 − 4𝑠 − 2𝑡, 𝑟, −2𝑠, 𝑠, 𝑡, 0)
parametric equation : (x1, x2, x3, x4, x5 , x6) =
r (-3, 1, 0, 0, 0, 0) + s (-4, 0, -2, 1, 0, 0) + t (-2, 0, 0, 0, 1, 0)
⇒ 𝐯𝟏 = −3,1,0,0,0,0 , 𝐯2 = −4,0, −2,1,0,0 , 𝐯3 = (−2,0,0,0,1,0)
linearly ind. span the solution space and dimension = 3
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1 3 0 4 2 0 00 0 1 2 0 0 00 0 0 0 0 1 00 0 0 0 0 0 0
Recap Row Operations : Row(A) ,Null(A)
Theorem 4.8.3 :
a) Row equivalent matrices have same row space
b) Row equivalent matrices have same null space
Elementary row operations do not change the row space and null space of a matrix
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Recap Bases for Row(A) and Col(A) on ref Matrix
Theorem 4.8.4 :
If a matrix R is in row echelon form, then the rowvectors with the leading 1's (the nonzero rowvectors) form a basis for the row space of R, and thecolumn vectors with the leading 1's of the rowvectors from a basis for the column space of R.
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Recap Bases for Row(A) and Col(A) on ref Matrix
Example 3 pp 268 : (Row-Echelon Matrix)
𝑅 =
1 −2 5 0 30 1 3 0 00 0 0 1 00 0 0 0 0 Theorem 4.8.4
𝐫𝟏 = 1 −2 5 0 3𝐫𝟐 = 0 1 3 0 0𝐫𝟑 = 0 0 0 1 0
𝐜𝟏 =
1000
, 𝐜𝟐 =
−2100
, 𝐜𝟒 =
0010
⇒
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Recap Find the Basis of Row(A)
Find the basis of Row(A):
1. Use Gaussian Elimination to construct R : R : ref matrix of A,
2. Basis of Row Space of A are formed by “rows with leading 1's” in R.
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Recap Basis for Column Space Theorem 4.8.5 : If A and B are row equivalent matrices,
1) A given set of column vectors of A is linearly independent iff the corresponding column vectors of Bis linearly independent
2) A given set of column vectors of A forms a basis of the column space of A iff the corresponding column vectors of B forms a basis for the column space of B
How to use this theorem?
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Recap Examples for Basis of Col(A) Example 5 pp.270 Basis for Col(A)
𝐴 =
1 −3 4 −2 5 42 −6 9 −1 8 22 −6 9 −1 9 7
−1 3 −4 2 −5 −4
⇒
R =
1 −3 4 −2 5 40 0 1 3 −2 −60 0 0 0 1 50 0 0 0 0 0
The correspond vectors in A :
c1, c2, c5 is a basis of Col(A)
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Reduce A to a row-echelon matrix (Gaussian Elimination) R
𝐜1′ =
1000
𝐜3′ =
4100
𝐜5′ =
5−210
⇒
c1’, c3’, c5’,
basis of Col(R)
𝐜𝟏 =
122
−1
𝐜𝟑 =
499
−4
𝐜5 =
589
−5
Lecture 16
General Vector Spaces
Basis of Row Space and Column Space
Rank and Nullity
Covered Range : 4.8 +4.9
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Basis from the Rows of AExample 6 pp. 270: Find a basis for row(A) spanned by r1, r2, r3, r4 of A
Consisting entirely of row vectors from A {使用原來的vectors r1, r2, r3, r4 }
Use AT:
→ column 1, 2, 4 for AT
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1 2 0 0 3
2 5 3 2 6
0 5 15 10 0
2 16 18 8 6
A
1 2 0 2 1 2 0 2
2 5 5 6 0 1 5 10
0 3 15 18 0 0 0 1
0 2 10 8 0 0 0 0
3 6 0 6 0 0 0 0
TA
1 2 31, 2,0,0,3 2, 5, 3, 2,6 2,6,18,8,6 r r r
→ consists entirely of row vectors from A
Find Basis Vectors and Linear Comb. of Nonbasis Vectors
Example 8 pp. 272 :
Find a subset of vectors v1=(1, -2, 0, 3) v2=(2, -5,-3, 6) v3=(0, 1, 3, 0) v4=(2, -1, 4, -7) v5=(5, -8, 1, 2) that forms a basis
1) Use original vector as column vectors Column Space
2) Express each vector by the basis
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1 2 0 2 5−2 −5 1 −1 −80 −3 3 4 13 6 0 −7 2
⇒
1 0 2 0 10 1 −1 0 10 0 0 1 10 0 0 0 0
𝐯𝟏 𝐯𝟐 𝐯𝟑 𝐯𝟒 𝐯𝟓 𝐰𝟏 𝐰𝟐 𝐰𝟑 𝐰𝟒 𝐰 𝟓
{𝐯𝟏 , 𝐯𝟐 , 𝐯𝟒 } ⇐ {𝐰𝟏 , 𝐰𝟐 , 𝐰𝟒 }
Find Basis Vectors and Linear Comb. of Nonbasis Vectors
Example 8 pp. 272 :
2) Express each vector by the basis
(b) w3 = 2w1 - w2 v3 = 2v1 - v2
w5 = w1 + w2 + w4 v5 = v1 + v2 + v4
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0130
= 2
1−203
−
2−5−36
𝐯3 𝐯1 𝐯2
5−812
=
1−230
+
2−5−36
+
2−147
𝐯5 𝐯1 𝐯2 𝐯4
1 2 0 2 5−2 −5 1 −1 −80 −3 3 4 13 6 0 −7 2
⇒
1 0 2 0 10 1 −1 0 10 0 0 1 10 0 0 0 0
𝐯𝟏 𝐯𝟐 𝐯𝟑 𝐯𝟒 𝐯𝟓 𝐰𝟏 𝐰𝟐 𝐰𝟑 𝐰𝟒 𝐰 𝟓
{𝐯𝟏 , 𝐯𝟐 , 𝐯𝟒 } ⇐ {𝐰𝟏 , 𝐰𝟐 , 𝐰𝟒 }
Since 𝐸𝐯𝒋 = 𝑐1𝐸𝐯𝟏 + ⋯ + 𝑐𝑘𝐸𝐯𝒌 ⇒ 𝐯𝒋 = 𝑐𝟏𝐯𝟏 + ⋯ + 𝑐𝒌𝐯𝒌!
Steps to Find a Basis of Vectors and Linear Comb. Of Nonbasis Vectors
Given S = {v1, v2, …, vk } ⊆ Rn
Find a basis ⊆ S for span(S)
1) Matrix A has v1, v2, …, vk as it column vectors
2) Reduce A to reduced row-echelon form R.
3) Columns that contain leading 1’s in R⇒ The corresponding vectors of A are the basis
4) Express each nonbasis column vector of R as a linear combination of the leading 1 columns in R!
5) Replace the linear combination of nonbasis column vectors in R by the corresponding column vectors in A
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Recap General Questions
Question 1 : What are the relationships between Ax = b and Row Space, Column Space and Null Space of the coefficient matrix A ?
Question 2: What are the relationships among Row(A), Col(A) and Null(A) of a given matrix A ?
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Conclusion
Ax = b is consistent if and only if b ∈ Col(A),
Use row echelon form (ref) to find basis of Row(A) and Col(A).
Use reduced row echelon form (rref) to find basis of Null(A).
General solutions of Ax = b : Special solution + Span (Basis of Null(A))
Orthogonality of Row(A) and Null(A).
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Lecture 16
General Vector Spaces
Basis of Row Space and Column Space
Rank and Nullity
Covered Range : 4.8 +4.9
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dim(Row(A)) = dim(Col(A))Theorem 4.9.1 : The row space and the column space
have the same dimension !
Basis of Row(A) : The leading one rows in R (row-echelon matrix)!
Basis of Col(A) : Column vectors in A corresponding to the leading one columns in R (row-echelon matrix)!
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Rank and Nullity
Definition 1 : Given a matrix A : m×n
Rank(A) = dim(row space of A) = dim(column space of A)
Nullity(A) = dim(null space of A)
Rank(A) = Rank(AT)
Rank(A) ≦min(m , n)
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Examples for Rank and Nullity (I) Example 1 pp 277 : Rank and Nullity
The solution space of Ax=0
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A=
−1 2 0 4 5 −332
−7−5
22
04
16
41
4 −9 2 −4 −4 7
->
1 0 −4 −28 −37 1300
10
−20
−120
−160
50
0 0 0 0 0 0
From Row1: 𝑥1−4𝑥3 −28 𝑥4 −37 𝑥5 + 13𝑥6 = 0
Row2: 𝑥2 −2𝑥3 − 12𝑥4 −16 𝑥4 + 5𝑥6 = 0
𝑥1 =4𝑥3 + 28𝑥4 + 37𝑥5 − 13𝑥6 ; 𝑥2 = 2𝑥3 +12𝑥4 + 16𝑥5 − 5𝑥6
𝑥3= r , 𝑥4 = 𝑠, 𝑥5 = 𝑡, 𝑥6 = 𝑢𝑥1= 4𝑟 + 28𝑠 + 37𝑡 − 13𝑢𝑥2= 2𝑟 + 12𝑠 + 16𝑡 − 5𝑢
Examples for Rank and Nullity (II)
The Solution of Ax= 0 with 4 parameters
Rank =2 , dim(Null(A)) = 4; Nullity = 4
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𝑥3= r , 𝑥4 = 𝑠, 𝑥5 = 𝑡, 𝑥6 = 𝑢
A=
−1 2 0 4 5 −332
−7−5
22
04
16
41
4 −9 2 −4 −4 7
->
1 0 −4 −28 −37 1300
10
−20
−120
−160
50
0 0 0 0 0 0
𝑥1= 4𝑟 + 28𝑠 + 37𝑡 − 13𝑢𝑥2= 2𝑟 + 12𝑠 + 16𝑡 − 5𝑢
𝑥1
𝑥2
𝑥3
𝑥4
𝑥5
𝑥6
= r
421000
+ s
28120100
+ t
37160010
+ u
−13−50001
Find the basis of Null(A) (I)
Find the basis of Null(A):
1. Use Gaussian Elimination to construct RR : RR : Reduced row-echelon matrix of A,
2. Basis of Null(A) (Rn) are formed by :
Finding Free variables (parameters) xk : without leading 1's columns in R
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Find the basis of Null(A) (II)
Find the basis of Null(A) (⊆Rn) :
Finding Free variables (parameters) xk : without leading 1's columns in R
Basis of Null (A) in Rn :
(a) having 1 at the position k for the corresponding vector,
(b) reverse the sign in column k of R at the appropriate positions in the vector,
(c) zeros at the positions for the other free variables!
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Basis of Null(A) (II)
The Solution of Ax= 0 with 4 parameters
Rank =2 , dim(Null(A)) = 4; Nullity = 4
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𝑥3= r , 𝑥4 = 𝑠, 𝑥5 = 𝑡, 𝑥6 = 𝑢
A=
−1 2 0 4 5 −332
−7−5
22
04
16
41
4 −9 2 −4 −4 7
->
1 0 −4 −28 −37 1300
10
−20
−120
−160
50
0 0 0 0 0 0
𝑥1= 4𝑟 + 28𝑠 + 37𝑡 − 13𝑢𝑥2= 2𝑟 + 12𝑠 + 16𝑡 − 5𝑢
𝑥1
𝑥2
𝑥3
𝑥4
𝑥5
𝑥6
= r
421000
+ s
28120100
+ t
37160010
+ u
−13−50001
reduced row echelon matrix
← Linearly Ind.
Dimension Theorem for MatrixTheorem 4.9.2: If A is a m×n matrix then Rank(A) + Nullity(A) = n
From Reduced Row-Echelon form.
Rank(A) + = n
Rank(A) + Nullity(A) = n
Similarly : Rank(AT) + Nullity(AT) = m
Rank(A) + Nullity(AT) = m
→ If Rank(A) = r then , Nullity(A) = n-r ; Nullity(AT) = m - r
number of free
variables
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Nullity and Number of Parameters
Theorem 4.9.3: If A is an m×n matrix then
(a) Rank(A) = number of leading one variables in rref of matrix A;
(b) Nullity(A) = number of parameters (free variables)
in general solution of Ax=0
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Rank, Nullity, and Linear Systems
Example 4 pp. 279
a) Find the number of parameters in general solution of Ax = 0 if A is a 5 x 7 matrix of rank 3!
b) Find the rank of a 5 x 7 matrix for which Ax=0 has a two-dimensional solution space.
Sol:
a) Nullity(A) = n- rank(A) = 7-3 = 4
b) Nullity(A) = 2. rank(A) = 7-2 = 5
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Equivalent Statements
Theorem 4.9.4 : If 𝐴𝐱 = 𝐛 is a consistent linear system of 𝑚 equations in n unknown, and if 𝐴 has rank 𝑟, then the general solution of the system contains 𝑛 − 𝑟 parameters.
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Fundamental Spaces of a Matrix
Theorem 4.9.5: If A is any matrix, rank(A)=rank(AT)
If A : m × n matrix
rank(A) + nullity(A) = n
rank(A) + nullity(AT) = m
dim[row(A)] = dim[col(A)]= r = rank(A)
dim[null(A)]=n-r dim[null(AT)]=m-r
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The Consistence Theorem
If Ax = b and A is m×n , then the following are equivalent.
a) Ax = b is consistent.
b) b ∈ Col (A)
c) Rank(A) = Rank( [A | b] )
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Condition for Ax = b has ≦1 Solution
A is m×n, the following are equivalent.
1) Ax = 0 has only the trivial solution.
2) The column vectors of A are linearly independent
3) Ax = b has at most one solution (none or one)for every m×1 vector b
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Orthogonal Complement
Definition 2: W is a subspace of Rn, the set of all vectors in Rn that are orthogonal to every vector in W is called the orthogonal complement(正交補
餘) of W, denote W⊥
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Properties of Orthogonal Complements (I)
Theorem 4.9.6: W: subspace of Rn
a) W⊥ is a subspace of Rn
b) The only vector common to W and W⊥ is 0
c) The orthogonal complement of W⊥ is W
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Examples for Orthogonal Complements
1. In R2 , what is the orthogonal complement W⊥ of a line W through the origin?
2. In R3 , what is the orthogonal complement W⊥ of a plane W through the origin ?
3. In R3 , what is the orthogonal complement of zerovector ?
4. In R3 , what is the orthogonal complement W⊥ of subspace W spanned by vectors (1,5,1) and (2,2,2)?
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Orthogonal Comp. for row(A) and null(A)
Theorem 4.9.7: A: m × n matrix
a) Null(A) and Row(A) are orthogonal comp. in Rn
from Ax=0 : each element in row space is orthogonal to any solution x;
b) Null(AT) and Col(A) are orthogonal comp. in Rm
from ATy=0
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Fundamental Theorem of Linear Algebra
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Solving Ax = b for an 𝑚 × 𝑛 matrix A with rank(A) = r
Equivalent StatementsTheorem 4.9.8: If A is an n×n matrix, no duplicate
rows and columns, then the following are equivalent.
a) A is invertible.
b) Ax=0 has only the trivial solution.
c) The reduced row-echelon form of A is In.
d) A is expressible as a product of elementary matrices.
e) Ax=b is consistent for every n×1 matrix b.
f) Ax=b has exactly one solution for every n×1 matrix b.
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Equivalent Statements
g) det(A) ≠0.
h) The column vectors of A are linearly ind.
i) The row vectors of A are linearly ind.
j) The column vectors of A span Rn. (col(A) = Rn )
k) The row vectors of A span Rn. (row(A) = Rn )
l) The column vectors of A form a basis for Rn.
m) The row vectors of A form a basis for Rn.
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Equivalent Statements [cont.]
n) A has rank n.
o) A has nullity 0
p) The orthogonal complement of the null space of A is Rn.
q) The orthogonal complement of the row space of A is {0}.
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