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Vector space and subspace
Math 112, week 8
Goals:
• Vector space, subspace, span.
• Null space, column space.
• Linearly independent, bases.
Suggested Textbook Readings: Sections §4.1, 4.2, 4.3
Week 8: Vector space and subspace
Review of R2 :
 
 
u1
v1
Addition: ~u =  , ~v =  , ~u + ~v =
u2
v2
 
u1
Multiplication by scalars: ~u =  , c is a scalar, c~u =
u2
Show that the following properties are satisfied:
1. The sum ~u + ~v is in R2 .
2. ~u + ~v = ~v + ~u.
3. (~u + ~v ) + w
~ = ~u + (~v + w).
~
4. The zero vector satisfies: ~u + ~0 = ~u.
5. The vector −~u is also in R2 , and ~u + (−~u) = ~0.
6. The scalar multiple c~u is in R2 .
7. c(~u + ~v ) = c~u + c~v .
8. (c + d)~u = c~u + d~u.
9. c(d~u) = (cd)~u.
10. 1~u = ~u.
These ten properties are known as the Axioms of vector space.
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Definition (Vector space):
A vector space is a non-empty set V of objects, called vectors, on
which are defined two operations, called addition and multiplication
by scalars (real numbers), subject to the ten axioms listed below:
1. The sum of ~u, ~v , denoted by ~u + ~v , is in V .
2. ~u + ~v = ~v + ~u.
3. (~u + ~v ) + w
~ = ~u + (~v + w).
~
4. There is a zero vector ~0 in V such that ~u + ~0 = ~u.
5. For each ~u in V , there is a vector −~u in V such that ~u + (−~u) = ~0.
6. The scalar multiple c~u is in V .
7. c(~u + ~v ) = c~u + c~v .
8. (c + d)~u = c~u + d~u.
9. c(d~u) = (cd)~u.
10. 1~u = ~u.
The spaces Rn are premier examples of vector spaces.
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Examples of vector spaces.
1. Let S be the space of all infinite sequence of numbers:
{yk } = (y0 , y1 , y2 , · · · )
If {zk } is another element of S, define
The sum {yk } + {zk } =
The scalar multiplication c{yk } =
2. The set Pn consists of all polynomials of degree at most n:
p~(t) = a0 + a1 t + a2 t2 + · · · + an tn
If ~q(t) = b0 + b1 t + b2 t2 + · · · + bn tn is another element in Pn , define
The sum p~ + ~q:
The scalar multiplication c~p:
3. Let P be the set of all polynomials with real coefficients, with sum
and scalar multiplication as above. Then it is a vector space.
Week 8: Vector space and subspace
Definition (Subspace)
A subspace of a vector space V is a subset H of V such that:
a. The zero vector of V is in H.
b. H is closed under vector addition. That is,
c. H is closed under multiplication by scalars. That is,
Note: Properties of a. b. c. guarantee that H is itself a vector space.
Examples:
1. Zero subspace: The set H = {~0} is a subspace of V .
2. The set
 
s
 
H = { t  : s and t are real}
 
0
is a subspace of R3 .
3. R2 is not a subspace of R3 .
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4. A plane in R3 which is not through the origin is not a subspace of
R3 .
5. A line in R2 which is not through the origin is not a subspace of R2 .
Example 1: Given v~1 and ~v2 in a vector space V , let H = Span{~v1 , ~v2 }.
Show that H is a subspace of V .
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Theorem: If ~v1 , · · · , ~vk are in a vector space V , then Span{~v1 , · · · , ~vk }
is a subspace of V .


a − 3b


b−a
,
Example 2: Let H be the set of all vectors of the form 


 a 


b
where a, b are arbitrary real numbers. Show that H is a subspace of R4 .
Example 3: Let W be the union of the first and third quadrants
x
in the xy-plane. That is, W = {  : xy ≥ 0}. Is W a subspace of R2 ?
y
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Definition (Null space). The null space of an m × n matrix A, written
as NulA, is the set of all solutions of A~x = ~0. In set notation,
NulA = {~x : ~x ∈ Rn and A~x = ~0}
 

3
1 −2 −2

, and let ~u = 
Example 4: Let A = 
1,
 
−5 11 3
1

is ~u in the null space of A?
Theorem: The null space of an m × n matrix A is a subspace of Rn .
Equivalently, the solution set of the linear system A~x = ~0 with m equations and n variables is a subspace of Rn .
Proof.
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An explicit description of Nul A
Example 5: Find a spanning set for the null space of the matrix


−3 6 −1 1 −7


A =  1 −2 2 3 −1


2 −4 5 8 −4
Note:
1. The spanning set produced by the method in Example 5 is automatically linearly independent.
2. The number of vectors in the spanning set for Nul A equals the
number of free variables in the equation A~x = ~0.
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Definition (Column space).
The column space of an m × n matrix A, written as Col A, is the set of
all linear combinations of the columns of A. If A = [~a1 · · · ~an ], then
ColA = Span{~a1 , · · · , ~an }
Theorem: The column space of an m × n matrix is a subspace of Rm .
Example 6: Find a matrix A such that W = ColA.


6a − b


W = { a + b  : a, b ∈ R}


−7a
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The contrast between Null A and Col A.


 2 4 −2 1


Example 7: A = −2 −5 7 3


3 7 −8 6
a. If Col A is a subspace of Rk , what is k?
b. If Nul A is a subspace of Rk , what is k?


 2 4 −2 1


Example 8: A = −2 −5 7 3


3 7 −8 6
a. Find a nonzero vector in Col A.
b. Find a nonzero vector in Nul A.
Week 8: Vector space and subspace
Example

2

A = −2

3
12
9: Let


−2 1

−5 7 3,

7 −8 6
4

 
3
 
3
−2
 

~u = 
−1
  and ~v = 
 
−1
 
3
0
a. Determine if ~u is in Nul A. Could ~u be in Col A?
b. Determine if ~v is in Col A. Could ~v be in Nul A?
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Contrast between Nul A and Col A for an m × n matrix A.
Nul A
1. Nul A is a subspace of Rn
Col A
1. Col A is a subspace of Rm
2. Nul A is implicitly defined; 2. Col A is explicitly defined.
(A~x = ~0)
3. Any vector ~v in Nul A has
3. Any vector ~v in Col A has
the property that A~v = ~0.
the property that A~x = ~v
is consistent.
4. Given a specific vector ~v ,
4. Given a specific vector ~v ,
it is easy to tell if ~v is
it may take time to tell if ~v is
in Nul A.
in Col A.
5. Nul A = {~0} if and only if
A~x = ~0 has only the trivial
5. Col A = Rm if and only if
A~x = ~b has a solution for every ~b
solution.
every ~b in Rm .
Week 8: Vector space and subspace
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Definition (Linearly independent):
A set of vectors {~v1 , · · · , ~vk } in vector space V is said to be linearly
independent if the vector equation
c1~v1 + c2 v~2 + · · · + ck~vk = ~0
has only the trivial solution: c1 = 0, · · · , ck = 0.
(Linearly dependent): The set {~v1 , · · · , ~vk } is linearly dependent if the
vector equation has a non-trivial solution.
Theorem: A set of vectors {~v1 , · · · , ~vk } is linearly dependent if and only
if at least one of the vectors is a linear combination of the others.
Example 10: Determine if the following sets of vectors are linearly independent.
   
4 6
   
1. {−2 , −3}
   
9
6
       
−1
1
2
5
2. {  ,   ,   ,  }
4
3
1
8

    
 3  0 −6
     
3. { 5  , 0 ,  5 }
     
−1
0
4
     
1 2 3
     
4. {2 , 3 , 5}
     
3
4
7
Week 8: Vector space and subspace
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Definition (Basis): Let H be a subspace of a vector space V .
The vectors B = {~b1 , · · · , ~bk } in V is a basis for H if
a. B is a linearly independent set, and
b. the subspace spanned by B coincides with H; that is,
H = Span { ~b1 , · · · , ~bk }
Example 11: Let A be an n × n invertible matrix, then the columns of
A form a basis for Rn .
Example 12: Determine if {~e1 , · · · , ~en } is a basis for Rn :
 
1
 
0

~e1 = 
 ..  ,
.
 
0
 
0
 
1

~e2 = 
 ..  ,
.
 
0
··· ,
 
0
 
0

~en = 
 .. 
.
 
1
Week 8: Vector space and subspace
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 
 
−2
−4
3
 
 
 
Example 13: Let ~v1 =  0  , ~v2 =  1  , ~v3 =  1 
 
 
 
5
7
−6


Determine if {~v1 , ~v2 , ~v3 } is a basis for R3 .


 
 
0
2
6
 
 
 
Example 14: ~v1 =  2  , ~v2 = 2 , ~v3 =  16 , and H = Span{~v1 , ~v2 , ~v3 }.
 
 
 
−1
0
−5
Note that ~v3 = 5~v1 + 3~v2 , and show that Span{~v1 , ~v2 , ~v3 } = Span{~v1 , ~v2 }.
Then find a basis for H.
Week 8: Vector space and subspace
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The Spanning Set Theorem:
Let S = {~v1 , · · · , ~vk } be a set in V , and let H = Span{~v1 , · · · , ~vk },
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 6= {~0}, some subset of S is a basis for H.
Proof.
Week 8: Vector space and subspace
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Two views of a basis.
• A basis is a spanning set that is as small as possible.
• A basis is also a linearly independent set that is as large as possible.
Bases for Nul A. See the method in Example 5, page 9.
Bases for Col A.
Example 15: Find a basis for ColB, where

1

0
B = [~b1 · · · ~b5 ] = 

0

0
4 0
2

0

0 1 −1 0


0 0 0 1

0 0 0 0
Week 8: Vector space and subspace
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What about a matrix A that is not in RREF? How can we find a basis
for Col A?
Example 16: Find a basis for ColA, where

1

3
A = [~a1 · · · ~a5 ] = 

2

5

4 0 2 −1

12 1 5 5 


8 1 3 2

20 2 8 8