MATHEMATICS 201-NYC-05
Vectors and Matrices
Martin Huard
Fall 2015
XV - Vector Spaces and Subspaces
1. Describe the zero vector (the additive identity) for the following vector spaces.
a) 4
b) C  ,  
c) M 2,3
d) P3
e) V   x, y  : x, y  , x  0 with the following operations :
 x1 , y1    x2 , y2    x1 x2 , y1  y2 
k  x1 , y1    x1k , ky1 
2. Describe the additive inverse of a vector for the following vector spaces.
a) 4
b) C  ,  
c) M 2,3
d) P3
e) V   x, y  : x, y  , x  0 with the following operations :
 x1 , y1    x2 , y2    x1 x2 , y1  y2 
k  x1 , y1    x1k , ky1 
3. Determine whether the given set, together with the indicated operations, is a vector space. If
it is, prove that each axiom is satisfied, if it is not, identify the axioms that fail.
a) M 2,3 with standard operation
3
b)
with standard operation
c) P3 with the standard operation
d) The set  x, y  : x  0, y  0 with standard operations
a 1
e) The set of all 2  2 matrices of the form 
 with standard operations
1 b
f) The set ax5 : a   .
g)
2
with the following operations :  x1 , y1    x2 , y2    x1  x2 , y1  y2 
 x1 , y1    kx1 , y1 
with the following operations :  x1 , y1    x2 , y2    x1 , 0 
k  x1 , y1    kx1 , ky1 
with the following operations :  x1 , y1    x2 , y2    x1 x2 , y1 y2 
k  x1 , y1    kx1 , ky1 
with the following operations :  x1 , y1    x2 , y2    x1  x2 , y1  y2 
k  x1 , y1    k 2 x1 , k 2 y1 
k
h)
2
i)
2
j)
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XV – Vector Spaces and Subspaces
Math NYC
4. Consider the set V whose only element is moon, that is, V  moon . Is this set a vector
space under the following operations?
moon  moon = moon
k (moon) = moon
for every real number k
5. Determine whether the subset W of 3 , with the standard operations, is a vector space.
Justify your answer. (Hint: Show that W is a subspace of 3 ).
a) W   a, b, 0  : a, b  
b) W   a,1,1 : a  
c) W   a, b, a  b  : a, b 
e) W   a, b  a, b  : a, b 


d) W   a, b, ab  : a, b 
g) W   x, y , z  : 2 x  y  z  3  0
h) W   x, y, z  : x  2t , y  t , z  5t , t 
i) W  u 
3
: u  w   2, 1,5
6. Determine whether the subset W
Justify your answer.
a b 
a) W   
 : a , b, c , d 
c
d



b) W
c) W
d) W
e) W

f) W   x, y , z  : x  2 y  z  0

of M 2,2 with the standard operations is a vector space.



is the set of 2  2 matrices A such that det( A)  0
is the set of 2  2 symmetric matrices A
is the set of diagonal 2  2 matrices.
a b

 
: a , b, c  

0 c 

a b 

: a  b  c  d  0
f) W   

 c d 

7. Determine whether the subset W of C  ,   is a subspace of C  ,   . Justify your
answer.
a) The set of nonnegative functions: f  x   0
b) The set of all even functions: f   x   f  x 
c) The set of all odd functions: f   x    f  x 
d) The set of all constant functions: f  x   c , c 
.
e) The set of all functions such that f  0   0
f) The set of all functions such that f  0   1
Fall 2015
Martin Huard
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XV – Vector Spaces and Subspaces
Math NYC
ANSWERS
1. a)  0, 0, 0, 0 
0 0 0 
c) 

0 0 0 
b) f ( x)  0
d) p( x)  0  0 x  0 x2  0 x3
e) (1,0)
b)   f   x    f  x 
2. a) u   u1 , u2 , u3 , u4 
-u   u1 , u2 , u3 , u4 
d) p  x   a0  a1 x  a2 x 2  a3 x3
 p  x   a0  a1 x  a2 x 2  a3 x3
 a11
3. a) Let A  
 a21
e) u   x1 , y1 
-u 

1
x1
,  y1
a
a13 
a
c) A   11 12

 a21 a22 a23 
a12 a13 
 a
- A   11

 a21 a22 a23 

a13 
b
b 
c
c 
b
c
, B   11 12 13  and C   11 12 13 

a22 a23 
b21 b22 b23 
c21 c22 c23 
 a11  b11 a12  b12 a13  b13 
1. A  B  
 is a 2  3 matrix
 a21  b21 a22  b22 a23  b23 
 a11  b11 a12  b12 a13  b13   b11  a11 b12  a12 b13  a13 
2. A  B  
  b  a
  B A
a

b
a

b
a

b
b

a
b

a
21
21
22
22
23
23
21
21
22
22
23
23

 

a12 a13   b11  c11 b12  c12 b13  c13 
a
3. A   B  C    11


 a21 a22 a23  b21  c21 b22  c22 b23  c23 
a12
 a11   b11  c11  a12   b12  c12  a13   b13  c13  


 a21   b21  c21  a22   b22  c22  a23   b23  c23  
  a11  b11   c11  a12  b12   c12  a13  b13   c13 
=

 a21  b21   c21  a22  b22   c22  a23  b23   c23 
a12  b12 a13  b13   c11 c12 c13 
a  b
=  11 11


 a21  b21 a22  b22 a23  b23  c21 c22 c23 
=  A  B  C
 a11 a12 a13  0 0 0 
4. A  023  


 a21 a22 a23  0 0 0 
 a  0 a12  0 a13  0   a11
  11

 a21  0 a22  0 a23  0   a21
Fall 2015
Martin Huard
a12
a22
a13 
A
a23 
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XV – Vector Spaces and Subspaces
Math NYC
 a11 a12 a13   a11 a12 a13 
5. A    A   


 a21 a22 a23   a21 a22 a23 
a12  a12 a13  a13  0 0 0 
a  a
  11 11


 a21  a21 a22  a22 a23  a23  0 0 0 
 023
 ka
6. kA   11
 ka21
ka12
ka22
ka13 
is a 2  3 matrix
ka23 
 k  a11  b11  k  a12  b12  k  a13  b13  
7. k  A  B   

 k  a21  b21  k  a22  b22  k  a23  b23  
 ka  kb11 ka12  kb12 ka13  kb13 
  11

 ka21  kb21 ka22  ka22 ka23  ka23 
 kA  kB
  k  l  a11  k  l  a12  k  l  a13 
8.  k  l  A  

 k  l  a21  k  l  a22  k  l  a23 
 ka  la11 ka12  la12 ka13  la13 
  11

 ka21  la21 ka22  la22 ka23  la23 
 kA  lA
 k  la11  k  la12  k  la13  
9. k  lA   

 k  la21  k  la22  k  la23  
  kl  a11

 kl  a21
  kl  A
 kl  a12  kl  a13 
 kl  a22  kl  a23 
1a11 1a12 1a13   a11
10. 1A  

1a21 1a22 1a23   a21
a13 
=A
a23 
a12
a22
b) Let u   u1 , u2 , u3  , v   v1 , v2 , v3  and w   w1 , w2 , w3 
1. u  v   u1  v1 , u2  v2 , u3  v3   3
2. u  v   u1  v1 , u2  v2 , u3  v3    v1  u1 , v2  u2 , v3  u3   v  u
3. u   v  w    u1   v1  v1  , u2   v2  w2  , u3   v3  w3  
   u1  v1   w1 ,  u2  v2   w2 ,  u3  v3   w3 
= u  v   w
4. u  0   u , u , u    0,0,0    u  0, u  0, u  0    u , u , u   u
1
2
3
1
2
3
1
2
3
5. u +  -u    u , u , u    u , u , u   u  u , u  u , u  u    0,0,0   0
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XV – Vector Spaces and Subspaces
Math NYC
6. ku   ku1 , ku2 , ku3  
3
7. k  u  v   k  u1  v1 , u2  v2 , u3  v3 
  k  u1  v1  , k  u2  v2  , k  u3  v3  
  ku1  kv1 , ku2  kv2 , ku3  kv3 
 ku  kv
8.  k  l  u    k  l  u1 ,  k  l  u2 ,  k  l  u3 
  ku1  lu1 , ku2  lu2 , ku3  lu3 
 ku  lu
9. k  lu   k  lu1 , lu2 , lu3     kl  u1 ,  kl  u2 ,  kl  u3    kl  u
10. 1u  1u1 ,1u2 ,1u3    u1 , u2 , u3  = u
c) Let p( x)  a3 x3  a2 x 2  a1 x  a0 q( x)  b3 x3  b2 x 2  b1 x  b0 and
r ( x)  c3 x3  c2 x 2  c1 x  c0
1. p( x)  q( x)   a3  b3  x 3   a2  b2  x 2   a1  b1  x   a0  b0  is a 3rd degree
polynomial
2. p( x)  q( x)   a3  b3  x3   a2  b2  x 2   a1  b1  x   a0  b0 
  b3  a3  x3   b2  a2  x 2   b1  a1  x   b0  a0 
 q( x)  p( x)
3. p( x)   q( x)  r ( x)   a3 x 3  a2 x 2  a1 x  a0   b3  c3  x 3   b2  c2  x 2   b1  c1  x   b0  c0 
  a3   b3  c3   x 3   a2   b2  c2   x 2   a1   b1  c1   x   a0   b0  c0  
   a3  b3   c3  x3    a2  b2   c2  x 2    a1  b1   c1  x    a0  b0   c0 
=  a3  b3  x3   a2  b2  x 2   a1  b1  x   a0  b0   c3 x 3  c2 x 2  c1 x  c0
=  p( x)  q( x)   r( x)
4. p( x)  0   a3  0  x3   a2  0  x 2   a1  0  x   a0  0 
 a3 x3  a2 x 2  a1 x  a0
 p( x )
5. p( x) +  -p( x)    a3  a3  x3   a2  a2  x 2   a1  a1  x   a0  a0 
 0 x3  0 x 2  0 x  0
0
6. kp( x)  ka3 x  ka2 x 2  ka1 x  ka0 is a 3rd degree polynomial
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Fall 2015
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XV – Vector Spaces and Subspaces
Math NYC
7. k  p( x)  q( x)   k   a  b  x 3   a  b  x 2   a  b  x   a  b  
3
3
2
2
1
1
0
0
  ka3  kb3  x 3   ka2  kb2  x 2   ka1  kb1  x   ka0  kb0 
 ka3 x 3  ka2 x 2  ka1 x  ka0  kb3 x 3  kb2 x 2  kb1 x  kb0
 kp( x)  kq( x)
8.  k  l  p( x)   k  l  a3 x3   k  l  a2 x 2   k  l  a1 x   k  l  a0
  ka3 x3  ka2 x 2  ka1 x  ka0    la3 x3  la2 x 2  la1 x  la0 
 kp( x)  lp( x)
9. k  lp( x)   k  la3 x3  la2 x 2  la1 x  la0   kla3 x 3  kla2 x 2  kla1x  kla0   kl  p( x )
10. 1p( x)  1a3 x3  1a2 x 2  1a1 x  1a0  a3 x 3  a2 x 2  a1x  a0 = p( x)
d) Axiom 5 is not satisfied, there is no -u in the set such that u +  -u   0 because
-u   1 u   u1 , u2  is not in the set.
Axiom 6 is not satisfied because if k  0 then ku   ku1 , ku2  is not in the set, ku1 and ku2
being <0.
 a11 1  b11 1   a11  b11
e) Axiom 1 is not satisfied since A  B  


 1 a22   1 b22   2
in the set.
0 0 
Axiom 4 is not satisfied since 
 is not in the set.
0 0 

is not
a22  b22 
2
  a11 1 
Axiom 5 is not satisfied since - A  
 is not in the set.
 1  a22 
k 
 ka11
Axiom 6 in not satisfied since kA  
is not in the set if k  1
ka22 
 k
f) Let p( x)  ax5 , q( x)  bx5 and r( x)  cx5 .
1. p( x)  q( x)  ax 5  bx 5   a  b  x 5 is in the set
2. p( x)  q( x)   a  b  x 5   b  a  x 5  q( x)  p( x)
3. p( x)   q( x)  r ( x)   ax 5   b  c  x 5   a   b  c   x 5    a  b   c  x 5
  a  b  x 5  cx 5 =  p( x)  q( x)   r ( x)
4. p( x)  0   a  0  x 5  ax 5  p( x)
5. p( x) +  -p( x)    a  a  x 5  0 x 5  0
6. kp( x)  kax5 is in the set
7. k  p( x)  q( x)   k   a  b  x5    ka  kb  x5  kax5  kbx5  kp( x)  kq( x)
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8.
9.
 k  l  p( x)   k  l  ax5  kax5  lax 5  kp( x)  lp( x)
k  lp( x)   k  lax5   klax5   kl  p( x)
10. 1p( x)  1ax5  ax5 = p( x)
g) The set is not a vector space since axiom 8 fails. For example, let k  1, l  2 and
u  1,1 .
 k  l  u  1  2 1,1   1  21,1  3,1
 k u    l u   1 1,1    2 1,1   1,1   2,1  3, 2 
Thus  k  l  u   k u    l u  .
Axioms 4 and 5 also fail.
h) The set is not a vector space because axiom 2 fails. For example, let u  1, 2  and
v   2,1 .
u  v  1, 2    2,1  1, 0 
v  u   2,1  1, 2    2, 0 
Thus u  v  v  u .
Axioms 4, 5 and 8 also fail.
i) Axiom 4 fails since u  0   u1  0, u2  0    0,0   u if u  0
Axiom 5 and 7 also fail.
j) Axiom 8 fails
k  l 

u   k  l  u1 ,  k  l  u2 ,  k  l  u3
2
2
2

  k 2u1  2klu1  l 2u1 , k 2u2  2klu2  l 2u2 , k 2u3  2klu3  l 2u3 
 k
u

2kl

u  l
u
  k u   l u 
Axiom 5 also fails.
4. Yes. It is similar to the vector space V  0 .

5. a) Yes 1. u  v   u1 , u2 , 0    v1 , v2 , 0    u1  v1 , u2  v2 , 0   W
2. ku  k  u1 , u2 , 0    ku1 , ku2 , 0   W
Thus W is a subspace of 3
b) No 1. u  v   u1 ,1,1   v1 ,1,1   u1  v1 , 2, 2   W
c) Yes 1. u  v   u1 , u2 , u1  u2    v1 , v2 , v1  v2    u1  v1 , u2  v2 ,  u1  v1    u2  v2   W
2. ku  k  u1 , u2 , u1  u2    ku1 , ku2 , ku1  ku2   W
Thus W is a subspace of 3
d) No 2. ku  k  u1 , u2 , u1u2    ku1 , ku2 , ku1u2   W since  ku1  ku2   k 2u1u2  ku1u2 if
k  1.
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e) Yes 1. u  v   u1 , u2  u1 , u2    v1 , v2  v1 , v2    u1  v1 ,  u2  v2    u1  v1  , u2  v2   W
2. ku  k  u1 , u2  u1 , u2    ku1 , ku2  ku1 , u2   W
Thus W is a subspace of 3
f) Yes If u W then u1  2u2  u3  0 and if v W then v1  2v2  v3  0 .
1. u  v   u1 , u2 , u3    v1 , v2 , v3    u1  v1 , u2  v2 , u3  vv 
Since  u1  v1   2  u2  v2    u3  v3    u1  2u2  u3    v1  2v2  v3   0  0  0
then u  v W .
2. ku  k  u1 , u2 , u3    ku1 , ku2 , ku3 
Since ku1  2ku2  ku3  k  u1  2u2  u3   k 0  0
then ku W .
Thus W is a subspace of 3
g) No If u W then 2u1  u2  u3  3 and if v W then 2v1  v2  v3  3 .
1. u  v   u1 , u2 , u3    v1 , v2 , v3    u1  v1 , u2  v2 , u3  vv 
Since 2  u1  v1    u2  v2    u3  v3    2u1  u2  u3    2v1  v2  v3   3  3  6
then u  v W .
h) Yes If u W then u   2t , t ,5t  and if v W then v   2s,  s,5s  .
1. u  v   2t , t ,5t    2s, s,5s 
  2t  2s, t  s,5t  5s    2  t  s  ,   t  s  ,5  t  s   W
2. ku  k  2t , t ,5t    2kt , kt ,5kt    2  kt  ,   kt  ,5  kt    W
Thus W is a subspace of 3
i) Yes If u , v  W , then u w  0 and v w  0
1. u  v W since  u  v  w  u w  v w  0  0  0
2. ku W since  ku  w  k  u w   k 0  0
Thus W is a subspace of
3
6. a) No. We do not always have closure under scalar multiplication.
1 1  12
1 1
1
1
For a example if k  2 and A= 
 W , then 2 1 1   1

 2
1 1

W
1
2
1
2
1 0 
0 0
b) No. Let A= 
and B  

 . Then det  A   det  B   0 so A, B  W .
0 0 
0 1 
1 0
Since det  A  B  
 1 , then A  B W , so we do not have closure under
0 1
addition.
c) Yes. Let A and B be symmetric matrices, AT  A and BT  B .
T
1.  A  B   AT  BT  A  B , hence A  B W
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XV – Vector Spaces and Subspaces
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2.  kA  kAT  kA , hence kA W
T
Hence W is a subspace of M 2,2 .
a11 0 
b11 0 
d) Yes. Let A= 
and B  

 be in W.
 0 a 22 
 0 b22 
0 
 a11  b11
W
1. A  B  
a22  b22 
 0
0 
 ka11
2. kA  
 W
 0 ka22 
Hence W is a subspace of M 2,2 .
 a11 a12  b11
e) Yes 1. A  B  

 0 a22   0
 a11 a12   ka11
2. kA  k 

 0 a22   0
Hence W is a subspace of M 2,2 .
b12   a11  b11

b22   0
ka12 
W
ka22 
a12  b12 
W
a22  b22 
 a11 a12   b11 b12   a11  b11 a12  b12 
f) Yes 1. A  B  


  W since
 a21 a22  b21 b22   a21  b21 a22  b22 
a11  b11  a12  b12  a21  b21  a22  b22   a11  a12  a21  a22    b11  b12  b21  b22   0
 a11 a12   ka11 ka12 
2. kA  k 

  W since
 a21 a22   ka22 ka22 
ka11  ka12  ka21  ka22  k  a11  a12  a21  a22   0
Hence W is a subspace of M 2,2 .
7. a) No. We do not always have closure under scalar multiplication.
For a example if k  1 , then  f ( x)  W since  f ( x)  0
 f  g  x  W since  f  g   x   f ( x)  g( x)  f ( x)  g( x)   f  g  x 
2.  kf  ( x )  W since  kf  ( x)  kf ( x)  kf  x    kf  x 
c) Yes. 1.  f  g  x   W since  f  g   x   f ( x)  g ( x)  f ( x)  g ( x)    f  g  x 
2.  kf  ( x )  W since  kf  ( x)  kf ( x)   kf  x     kf  x 
d) Yes. 1.  f  g  x   W since  f  g  x   f ( x)  g ( x)  c  d is a constant.
2.  kf  ( x )  W since  kf  ( x )  kf ( x )  kc is a constant.
e) Yes. 1.  f  g  x   W since  f  g  0   f (0)  g (0)  0  0  0 .
2.  kf  ( x )  W since  kf  (0)  kf (0)  k 0  0 .
f) No . 1.  f  g  x   W since  f  g  0   f (0)  g (0)  1  1  2 .
2.  kf  ( x )  W since  kf  (0)  k1  k  1 if k  1 .
b) Yes. 1.
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