4.1 Vector Spaces and Subspaces
Definition: A vector space V is a non-empty
set of objects called vectors on which are
defined two operations called addition and
multiplication by scalars, subject to the 10
axioms below.
An axiom is a fact that we accept without
proof. The axioms must hold for all u, v, and
w in V and all scalars c and d.
1. u + v is in V (closure)
2. u + v = v + u
3. (u + v) + w = u + (v + w)
4. There is a vector (called the zero vector) 0
in V such that:
u+0=u
5. For each u in V there is a vector –u in V
satisfying u + (–u) = 0.
6. cu is in V
7. c(u + v) = cu + cv
8. (c + d)u = cu + du
9. (cd)u = c(du)
10. 1u = u
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Vector Space Examples
Example: Let
 a b 

M 22  
: a, b, c, d are real

 c d 

Note: the zero vector is
0 0
0 0


VERIFY THAT THE AXIOMS HOLD:
Example: Let n ≥ 0 be an integer and let
Pn be the set of all polynomials of degree at
most n ≥ 0
Members of Pn have the form
pt   a0  a1t    ant n
where a0 , a1 ,...an are real numbers, and t is a
real variable.
VERIFY AXIOMS 1, 4, AND 6
2
n


p
t

a

a
t



a
t
0
1
n
1. Closure: Let
and
qt   b0  b1t    bnt n and c a scalar.
The polynomial p + q is defined to be
p  qt   pt   qt 
a0  b0   a1  b1 t    an  bn t n
which is also a polynomial of degree at most
n. So p + q is in Pn.
4. The zero vector here is 0t   0
We need to verify that p  0t   pt 
Let pt  be as in #1.
p  0t   a0  a1t    ant n  0
 a0  a1t    ant n
 pt 
6. Let pt  be as in #1.
cpt   c a0  a1t    ant n


 ca0  ca1t    cant n
This is still a polynomial of degree at most n
since the coefficients cai are still real
numbers.
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Subspaces
Definition: A subspace of a vector space V is
a subset H of V that has the following
properties:
a. The zero vector of V is in H.
b. For each u and v in H, u + v is in H. (H is
closed under vector addition.)
c. For each u in H, and each scalar c, cu is in
H. (H is closed under scalar multiplication.)
If H itself satisfies these properties, then H is
itself a vector space.
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Example: Let
a 

 

H   0  : a, b are real
 b 

 

Claim: H is a subspace of R3.
a. The zero vector of R3 is in H if we let
a = b = 0.
b. Adding two vectors in H always produces
another vector whose second entry is 0,
therefore the sum of two vectors in H is also
in H.
c. Multiplying a vector in H by a scalar
produces another vector in H since for any
scalar c, ca and cb are real and c0 = 0.
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Example: Let
 x 

H  
: x are real

 x  1

Is H a subspace of R2?
Does H satisfy properties a, b, and c?
►All three must be satisfied for H to be a
subspace, so if H fails even one it is not a
subspace
We can see that property a fails because the
zero vector in R2 is not in H, since it would be
impossible to have x  0 and x  1 0 at the
same time.
Thus, H is not a subspace.
►Another way to show that H is not a
subspace is to choose u and v in H and add
them.
0 
u 
Let
1
1 
v 
 2
1
u v   
3 which is not in H.
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A shortcut for determining subspaces.
Theorem 1: If v1, v 2 ,, v p  are in V, then
Span v1 , v 2 ,, v p  is a subspace of V.
Proof:
a. 0  0 v1  0 v 2    0 v p


b. To show that Span v1 , v 2 ,, v p is closed
under vector addition, we choose two
arbitrary vectors in Span v1 , v 2 ,, v p  .
u  a1v1  a2 v 2    a p v p
v  b1v1  b2 v 2    bp v p
u  v  a1v1  a2 v 2    a p v p 
 b1v1  b2 v 2    bp v p 
 a1v1  b1v1    b p v p  a p v p
 a1  b1 v1    b p  a p v p


So, u + v is in Span v1 , v 2 ,, v p .
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

c. To show that Span v1 , v 2 ,, v p is closed
under scalar multiplication, we choose an
arbitrary scalar c and an arbitrary vector in


Span v1 , v 2 ,, v p .
v  b1v1  b2 v 2    b p v p
cv  cb1v1    bp v p 
 cb1v1    cbp v p


So, cv is in Span v1 , v 2 ,, v p .
To sum up:
1. To show that H is a subspace of a vector
space use Theorem 1 (If v1, v 2 ,, v p  are in V,
then Span v1 , v 2 ,, v p  is a subspace of V.).
2. To show that a set is not a subspace of a
vector space, provide a specific example
showing that at least one of the axioms a, b, c
from the definition is violated.
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Example: Let
V  a  2b,2a  3b; a, b are real
Is V a subspace of R2?
Why or why not?
It looks like it could be, so we write the
vectors in V in column form for a closer look.
 a  2b   a   2b 
2a  3b  2a    3b

   

1   2 
 a    b 
2  3
So, any vector in V can be written as a linear
1
2
v 
and v 2   
combination of 1 2
 3 ,
so V  Spanv1,v 2 . Thus, V is a subspace of
R2 by Th. 1.
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Example: Let
a  2b 




H   a  1  : a, b are real
 a 




Is H a subspace of R3?
This looks like the previous, but note the
second entry. There is a constant, so the
zero vector is not in H, therefore H cannot be
a subspace of R3.
Example: Is the set H of all matrices of the
form
b
 2a
3a  b 3b a subspace of


M 22 ,
the set of 2 x
2 matrices?
The zero matrix is in this set for a = b = 0.
If we can write this matrix as a linear
combination of two other matrices A and B,
then we could write the set as Span{A, B}
which is a subspace of M 22 .
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b
 2a
3a  b 3b 


 2 a 0  0 b 




3
a
0
b
3
b

 

 2 0   0 1
 a
 b


3
0
1
3

 

Therefore
2 0 0 1 
H  Span
,



3 0 1 3  so, H is a subspace
of M 22 .
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