Download 1 SUBSPACE TEST Strategy: We want to see if H is a

SUBSPACE TEST
Strategy: We want to see if H is a subspace of V.
1.) Is the zero vector of V also in H? If no, then H is not a subspace of V. If yes, then move on
to step 2.
2.) Identify c, u , v , and list any “facts”.
3.) Is u + v in H? If yes, then move on to step 4. If no, then give a specific example to show
that u and v are in H but u + v is not in H. This shows that H is not a subspace of V.
4.) Is cu in H? If yes, then H is a subspace of V. If no, then give a specific example to show
that u is in H but cu is not in H. This shows that H is not a subspace of V.
Note: If H is defined with the variables u, v, or c, feel free to use different letters. For example,
let your scalar be d instead of c or let your two vectors be g and h instead of u and v. There is
nothing special about using c, u, and v. Just be sure to be consistent. Don't start using d as your
scalar and end up using c as your scalar.
1
Example 1
 a + b + c 




Let H =  2a + 3b  a, b, c ∈ R . Is H a subspace of R 3 ?
 4c 




1.) Is the zero vector of R 3 also in H?
a + b + c  0

  
We need to see if the equation  2a + 3b  =  0  has a solution. A solution to this equation is
 4c   0 

  
a = b = c = 0 . So 0 is in H.
2.) Identify d, u , v , and list any “facts”. (I’m going to use d as my scalar because c has already
been used.)
 u1 + u 2 + u 3 
 v1 + v 2 + v 3 




d ∈ R , u =  2u1 + 3u 2  , and v =  2v 1 + 3v 2 




4u 3
4v 3




3.) Is u + v in H?
 u1 + u 2 + u 3 + v 1 + v 2 + v 3   (u1 + v 1 ) + (u 2 + v 2 ) + (u 3 + v 3 )

 

u + v =  2u1 + 3u 2 + 2v 1 + 3v 2  = 
2(u1 + v 1 ) + 3(u 2 + v 2 )





4u 3 + 4v 3
4(u 3 + v 3 )

 

Here a = u1 + v 1 , b = u 2 + v 2 , and c = u 3 + v 3 . So u + v is in H.
4.) Is du in H?
 d(u1 + u 2 + u 3 )  du1 + du 2 + du 3 

 

du =  d(2u1 + 3u 2 )  =  2du1 + 3du 2 
 


d(4u 3 )
4du 3

 

Here a = du1 , b = du 2 , and c = du 3 . So du is in H. This means that H is a subspace of R 3 .
2
Example 2
 a 

 

 b 

Let W =   : c = a + 2b & d = a − 3b . Is W a subspace of R 4 ?
 c 

 d 



1.)
Is the zero vector of R 4 also in W?
a  a  0
  
  
b  b  0
has a solution.
We need to see if the equation   = 
=
c
a + 2b   0 
  
  
 d   a − 3b   0 
  
  
A solution to this equation is a = b = 0 . So 0 is in W.
2.)
Identify k, u , v , and list any “facts”.
 u1 
 v1 
 
 
u2 
v 
k ∈R , u =   , v =  2 
u3
v
 
 3
u 
v 
 4
 4
Our “facts” are u 3 = u1 + 2u 2 , u 4 = u1 − 3u 2 , v 3 = v 1 + 2v 2 , and v 4 = v 1 − 3 v 2 .
3.)
Is u + v in W?
 u1 + v 1 


 u2 + v 2 
u+v =
u + v3 
 3

u + v 
4
 4
Here a = u1 + v 1 , b = u 2 + v 2 , c = u 3 + v 3 , and d = u 4 + v 4 . We need to know if c = a + 2b and d = a − 3b .
c = u 3 + v 3 = (u1 + 2u 2 ) + (v 1 + 2v 2 ) = (u1 + v 1 ) + 2(u 2 + v 2 ) = a + 2b
d = u 4 + v 4 = (u1 − 3u 2 ) + (v 1 − 3v 2 ) = (u1 + v 1 ) − 3(u 2 + v 2 ) = a − 3b
So u + v is in W.
4.)
Is ku in W?
 ku 1 


 ku 2 
ku = 
ku 
 3
 ku 
 4
Here a = ku1 , b = ku 2 , c = ku 3 , and d = ku 4 . We need to know if c = a + 2b and d = a − 3b .
c = ku 3 = k (u1 + 2u 2 ) = ku1 + 2ku 2 = a + 2b and d = ku 4 = k (u1 − 3u 2 ) = ku1 − 3ku 2 = a − 3b
So ku is in W. This means that W is a subspace of R 4 .
3
Example 3
 2a + b 




Let H =  2b + 4c − 1 a, b, c ∈ R  . Is H a subspace of R
 a + b + c 




1.) Is the zero vector of R
3
3
?
also in H?
 2a + b   0 

  
Does the equation  2b + 4c − 1 =  0  have a solution? No, this equation can be written as the
 a + b + c  0

  
2a + b = 0
system 2b + 4c = 1 . Looking at the augmented matrix
a+b+c =0
2 1 0 0  1 0 − 10

 

 0 2 4 1  ~  0 1 2 0  we can see
 1 1 1 0   0 0 0 1

 

that this system has no solution. This means that 0 is not in H. So H is not a subspace of R
3
.
4
Example 4
 abc 




Let H =  2a + 3b  a, b, c ∈ R  . Is H a subspace of R 3 ?
 4c 




1.) Is the zero vector of R 3 also in H?
 abc   0 

  
We need to see if the equation  2a + 3b  =  0  has a solution. A solution to this equation is
 4c   0 

  
a = b = c = 0 . So 0 is in H.
2.) Identify d, u , v , and list any “facts”. (I’m going to use d as my scalar because c has already
been used.)
 u1 * u 2 * u 3 
 v1 * v 2 * v 3 




d ∈ R , u =  2u1 + 3u 2  , and v =  2v 1 + 3v 2 




4u 3
4v 3




3.) Is u + v in H?
 u1 * u 2 * u 3 + v 1 * v 2 * v 3   u 1 * u 2 * u 3 + v 1 * v 2 * v 3 

 

u + v =  2u1 + 3u 2 + 2v 1 + 3v 2  =  2(u1 + v 1 ) + 3(u 2 + v 2 ) 

 

4u 3 + 4v 3
4(u 3 + v 3 )

 

Looking at the last two entries we see that a = u1 + v 1 , b = u 2 + v 2 , and c = u 3 + v 3 .
However, abc = (u1 + v 1 )(u 2 + v 2 )(u 3 + v 3 ) ≠ u1 * u 2 * u 3 + v 1 * v 2 * v 3 , in general.
8
 27 
 


For example, let u1 = u 2 = u 3 = 2 , v 2 = 3 , and v 1 = v 3 = −3 . So u = 10  , v =  3  , and
8
 − 12 
 


 35   abc 
 


13 − 3b
 13 − 3b 
u + v =  13  =  2a + 3b  . Now we have that c = −1 , a =
, and 
 * b * (− 1) = 35
2
 2 
 − 4   4c 

 

or 3b 2 − 13b − 70 = 0 . Using the quadratic formula, we get b =
13 ± 169 − 840
∉ R . So, in
6
general, u + v is not in H. This means that H is not a subspace of R 3 .
5
EXERCISES
n
n
For problems 1-5, determine if H is a subspace of R . If so, write it as the span of a set of vectors in R and the
column space of a matrix.
1.)

 a + b + 2c 

H = 
 ab + c 



a, b, c ∈ R


2.)

 a + 2b 

H =  a + b + c 


0


3.)

 x + 2y − z 

H =  4y + 6z 
 7 x − 3 y + 14z 


4.)

 2x + 3 y + 4z + 6 w 

y−z

H = 

1


 x + y − z − w 

5.)

a − b + 7c



 3a − 4b + 8c − 9d 
H = 

d−c




6a − 17d





a, b, c ∈ R 





x, y, z ∈ R





x, y, z, w ∈ R 






a, b, c, d ∈ R



4 6
 1 2


3
6.) Let A =  − 1 5 − 3 2  . Is the solution set of A x = 0 as subspace of R ? Is the solution set of A x = 0 as
 1 2
3 4 

4
subspace of R ?
7.)
4 6
 1 2
 1


 
3
Let A =  − 1 5 − 3 2  and b =  2  . Is the solution set of A x = b as subspace of R ? Is the solution set
 1 2
4
3 4 

 
4
of A x = b as subspace of R ?
n
For problems 8-14, determine if W is a subspace of R .
8.)

 a 

W =  b 
 c 
 



abc = 0


9.)

 x 

W =  y 
 z 
 



x + y + z = 0



 x 
10.) W =  
 y 

sin x =

π

2


 r 

11.) W =  s 
 t 
 



r + s ≤ t



 x 
 y 
12.) W =  
 z 
 w 




2x + 3 y = z + w 




 a 
 b 
13.) W =  
 c 
 d 




a + b = 3c and b + c = 4d



14.) Let T : R n → R m be a linear transformation.

Determine if W = x

n
subspace of R .
()

x ∈ R n and T x = 0  is a

6
ANSWERS TO THE ODD EXERCISES
2
1.)
H is not a subspace of R .
3.)
 1 
 
H is a subspace of R where H = span  0 ,
 7 
 
 2


 4 ,
− 3


2 − 1
1
 − 1


 
A
=
0
4
6 .

 6  = ColA with
 14 
 7 − 3 14 
 


 1 
 
 3 
4
5.) H is a subspace of R where H = span ,
 0 
 6 

 − 1


 − 4
 0 ,


 0


 7
 
 8
 − 1,
 
 0
 
3
3
0
 0 
1 −1 7




8 − 9
 − 9 
3 − 4
=
ColA
with
.
A
=


0
1
0 −1
1




 − 17 
6
0
0 − 17 



4
7.)
The solution set of A x = b is not a subspace of R or R .
9.)
W is a subspace of R .
3
3
11.) W is not a subspace of R .
4
13.) W is a subspace of R .
7