DEFINITION:
A vector space is a nonempty set V of
objects, called vectors, on which are defined two operations, called addition and
multiplication by scalars (real numbers),
subject to the following 10 axioms (or
rules):
1. The sum of ū and v̄, denoted by
ū + v̄, is in V.
2. ū + v̄ = v̄ + ū.
3. (ū + v̄) + w̄ = ū + (v̄ + w̄).
4. There is a zero vector 0̄ in V such
that ū + 0̄ = ū.
5. For each ū in V, there is a vector
−ū in V such that ū + (−ū) = 0̄.
6. The scalar multiple of ū by c, denoted by cū, is in V.
7. c(ū + v̄) = cū + cv̄.
8. (c + d)ū = cū + dū.
9. c(dū) = (cd)ū.
10. 1 · ū = ū.
These axioms must hold for all vectors
ū, v̄, and w̄ in V and all scalars c and d.
EXAMPLE:



x1







x2 
: x1, . . . , xn ∈ R
1. Rn = 
.


.






xn
2. The set Pn of polynomials of degree
at most n:
p̄(t) = antn + . . . + a2t2 + a1t + a0
where the coefficients an, . . . , a0 and the
variable t are real numbers.
3. The set of all real-valued functions
defined on R.
DEFINITION:
A subspace of a vector space V is a subset H of V that has 3 properties:
1. The zero vector of V is in H.
2. H is closed under vector addition.
That is, for each ū and v̄ in H, the sum
ū + v̄ is in H.
3. H is closed under multiplication by
scalars. That is, for each ū in H and
each scalar c, the vector cū is in H.
EXAMPLE:
The set consisting of only the zero vector 0̄ in a vector space V is a subspace of
V, called the zero subspace and written
as {0̄}.
WARNING:
R2 is not a subspace of R3, because R2
is not a subset of R3.
EXAMPLE:
The set
 

 s

H =  t  : s and t are real numbers


0
is a subspace of R3.
THEOREM:
If v̄1, . . . , v̄p are in a vector space V, then
Span {v̄1, . . . , v̄p} is a subspace of V.
EXAMPLE:
Let
 
 
1
1
v̄1 =  2  v̄2 =  1  .
3
2
By the Theorem above
Span{v̄1, v̄2}
is a subspace of R3.
EXAMPLE:
Let H be the set of all vectors of the
form


4a − b
 2b 


 a − 2b 
a−b
where a and b are arbitrary scalars. Show
that H is a subspace of R4.
SOLUTION:
We have


 


4a − b
4
−1
 2b 
0
 2

 = a   +b 

 a − 2b 
1
 −2 
a−b
1
−1
| {z }
| {z }
v̄1
v̄2
We see that
H = Span{v̄1, v̄2}
therefore H is a subspace of R4 by the
Theorem above.
EXAMPLE:
Let H be the set of all vectors of the
form


a−b
b−c


c−a
b
where a, b and c are arbitrary scalars.
Find a set S of vectors that spans H or
show that H is not a vector space.
SOLUTION:
We have








a−b
1
−1
0
b−c
 0
 1
 −1 

 = a
 +b 
 +c 

c−a
 −1 
 0
 1
b
0
1
0
| {z }
| {z }
| {z }
v̄1
v̄2
v̄3
and we see that H is a vector space and
{v̄1, v̄2, v̄3}
spans H.
EXAMPLE:
Let H be the set of all vectors of the
form


3a + b
 4 
a − 5b
where a and b are arbitrary scalars. Show
that H is not a vector space.
SOLUTION:
H is not a vector space, since 0̄ 6∈ H
(the second entry is always nonzero).
EXAMPLE:
Let H be the set of
form:  
a
(a)  b 
0
 
a
(b)  b  , where
c
 
a
(c)  a 
1
 
a
(d)  b  , where
c
 
a
(e)  b  , where
c
all vectors of the
a≥0
a + 2b − c = 0
2a − b + c = 1.
Is H a subspace of R3?
SOLUTION:
(a) We have
 
 
 
1
0
a
 b = a0 + b1,
0
0
0
which is a linear combination and therefore a subspace of R3.
(b) It is not a subspace, since it is not
closed under multiplication by a scalar.
In fact,
 


a
−a
(−1)  b  =  −b  ,
c
−c
and the first entry is negative, which
contradicts the initial assumption.
(c) It is not a subspace, since there is no
a zero vector (the last entry is always
nonzero).
(d) We have
 


 
 
0
1
a
a
 b =  b  = a0 + b1,
2
1
a + 2b
c
which is a linear combination and therefore a subspace of R3.
(e) We have


 
a
a
,
 b = 
b
1 − 2a + b
c
which is not a subspace, since there is
no a zero vector. In fact,
 


0
a

 = 0
b
0
1 − 2a + b
if and only if a = 0 and b = 0. But in
this case the last entry in nonzero, since
1 − 2a + b = 1 − 2 · 0 + 0 = 1.
Contradiction.
DEFINITION:
The null space of an m × n matrix A,
written as Nul A, is the set of all solutions to the homogeneous equation
Ax̄ = 0̄.
DEFINITION0:
The null space of an m × n matrix A is
the set of all x̄ in Rn that are mapped
into the zero vector 0̄ in Rm by the linear transformation
x̄ 7−→ Ax̄.
EXAMPLE:
Let
1 −2 −1
.
2 −3 −4
 
5
Determine if ū =  2  belongs to the
1
null space of A.
A=
SOLUTION:
Since
 
5
1 −2 −1  
0
2 =
Aū =
,
0
2 −3 −4
1
ū is in Nul A.
EXAMPLE:
Let
1 −3 −2
.
−5 9 1


5
Determine if ū =  3  belongs to the
−2
null space of A.
A=
SOLUTION:
Since
Aū =


5
1 −3 −2 
0

3 =
,
0
−5 9 1
−2
ū is in Nul A.
THEOREM:
The null space of an m × n matrix A
is a subspace of Rn. Equivalently, the
set of all solutions to a system Ax̄ = 0̄
of m homogeneous linear equations in n
unknowns is a subspace of Rn.
EXAMPLE:
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
SOLUTION:
We find the general solution of Ax̄ = 0̄:


1 −2 0 −1 3 0
[A 0̄] ∼  0 0 1 2 −2 0  ,
0 0 0 0 0 0
therefore
x1 − 2x2 − x4 + 3x5 = 0
x3 + 2x4 − 2x5 = 0,




x1
2x2 + x4 − 3x5
 x2 


x
2







so 
 x3  =  −2x4 + 2x5 
 x4 


x4
x5
x5




 
1
−3
2
1
 0
 0




 





= x2  0  +x4  −2  +x5  2 
,
0
 1
 0
0
0
1
| {z }
| {z }
| {z }
ū
v̄
so Nul A =Span {ū, v̄, w̄}.
w̄
DEFINITION:
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.
REMARK:
So, if A = [ā1 . . . ān], then
Col A = Span{ā1, . . . , ān}.
THEOREM:
The column space of an m × n matrix is
a subspace of Rm.
EXAMPLE:
Let

2 4 −2
A =  −2 −5 7
3 7 −8
Find a nonzero vector in
nonzero vector in Nul A.

1
3.
6
Col A and a
SOLUTION:
1. Any column of A is anonzero
vector

2
in Col A. For example,  −2  =
3
 






1
4
−2
2
= 1  −2  +0  −5  +0  7  +0  3  .
6
8
7
3
2. To find a nonzero vector in Nul
A, we row reduce the augmented matrix [A 0̄]:


1 0 9 0 0
[A 0̄] ∼  0 1 −5 0 0  ,
0 0 0 1 0
therefore any vector


x1
 x2 
=
x̄ = 
 x3 
x4


−9x3
 5x3 


 x3 
0
is in Nul A. For example, if we put x3 =
1, we get


−9
 5

ū = 
 1
0
is in Nul A.