Download Math 3013 Problem Set 4: Determining Subspaces and Solving Linear Systems - Prof. Birne Bi and more Assignments Linear Algebra in PDF only on Docsity!
Math 3013
Problem Set 4
Problems from §1.6 (pgs. 99-101 of text): 1,3,5,7,9,11,17,19,21,35,37,
- (Problems 1,3,4,7,9 in text). Determine whether the indicated subset is a subspace of the given R
n .
(a) W = {[r, −r] | r ∈ R} in R 2
- It suffices to show that if v 1 and v 2 are in W then so is any linear combination of v 1 and v 2. Set
v 1 = [r 1 , −r 1 ] , v 2 = [r 2 , −r 2 ]
Then
c 1 v 1 + c 2 v 2 = c 1 [r 1 , −r 1 ] + c 2 [r 2 , −r 2 ]
= [c 1 r 1 + c 2 r 2 , −c 1 r 1 − c 2 r 2 ]
= [(c 1 r 1 + c 2 r 2 ) , − (c 1 r 1 + c 2 r 2 )] ∈ W
(b) W = {[n, m] | n and m are integers} in R
2
- This subset is not closed under scalar multiplication for
[1, 1] ∈ W but
2 [1, 1] =
[
]
∈ / W
Since this subset is not closed under scalar multiplication it cannot be a subspace.
(c) W = {[x, y, z] | x, y, z ∈ R and z = 3x + 2} in R 3
- Consider two arbitrary vectors in W
v 1 = [x 1 , y 1 , 3 x 1 + 2] , v 2 = [x 2 , y 2 , 3 x 2 + 2]
we have
v 1 − v 2 = [x 1 − x 2 , y 1 − y 2 , 3(x 1 − x 2 ) + 0] ∈/ W
Since the difference of two vectors in W does not lie in W , W is not a subspace.
(d) W = {[x, y, z] | x, y, z ∈ R and z = 1, y = 2x} in R 3
- Consider two arbitrary vectors in W
v 1 = [x 1 , 2 x 1 , 1] , v 2 = [x 2 , 2 x 2 , 1]
we have
v 1 − v 2 = [x 1 − x 2 , 2(x 1 − x 2 ), 0] ∈/ W
(e) W = {[2x 1 , 3 x 2 , 4 x 3 , 5 x 4 ] | xi ∈ R} in R
4
1
- Consider two arbitrary vectors in R
4
x = [x 1 , x 2 , x 3 , x 4 ] , x
′ = [x
′ 1 , x
′ 2 , x
′ 3 , x
′ 4 ]
Then the vectors
v 1 = [2x 1 , 3 x 2 , 4 x 3 , 5 x 4 ] , v 2 = [2x
′ 1 , 3 x
′ 2 , 4 x
′ 3 , 5 x
′ 4
]
will be in W. We have
c 1 v 1 + c 2 v 2 = [2c 1 x 1 , 3 c 1 x 2 , 4 c 1 x 3 , 5 c 1 x 4 ] + [2c 1 x
′ 1 ,^3 c^2 x
′ 2 ,^4 c^2 x
′ 3 ,^5 c^2 x
′ 4 ]
= [2 (c 1 x 1 + c 2 x
′ 1 )^ ,^ 3 (c^1 x^2 +^ c^2 x
′ 2 )^ ,^ 4 (c^1 x^3 +^ c^2 x
′ 3 )^ ,^ 5 (c^4 x^1 +^ c^2 x
′ 4 )]
= [2x
′′ 1 ,^3 x
′′ 2 ,^4 x
′′ 3 ,^5 x
′′ 4 ]
This vector belongs to W since
x
′′ = [c 1 x 1 + c 2 x
′ 1 , c 1 x 2 + c 2 x
′ 2 , c 1 x 3 + c 2 x
′ 3 , c 4 x 1 + c 2 x
′ 4
] ∈ R
4
Since an arbitrary linear combinations of two vectors in W also lies in W , W is a subspace.
- (Problem 11 in text). Prove that the line y = mx is a subspace of R
2
. (Hint: write the line as
W = {[x, mx] | x ∈ R}.)
- It suffices to show that an arbitrary linear combinations of two vectors in W also lies in W. Set
v 1 = [x 1 , mx 1 ] , v 2 = [x 2 , mx 2 ]
Then
c 1 v 1 + c 2 v 2 = [c 1 x 1 + c 2 x 2 , c 1 mx 1 + c 2 mx 2 ]
= [(c 1 x 1 + c 2 x 2 ) , m (c 1 x 1 + c 2 x 2 )] ∈ W
Hence, W is a subspace.
- (Problems 17,19 and 21 in text). Find a basis for the solution set of the following homogeneous linear
systems.
(a)
3 x 1 + x 2 + x 3 = 0
6 x 1 + 2x 2 + 2x 3 = 0
− 9 x 1 − 3 x 2 − 3 x 3 = 0
- This linear system corresponds to the following augmented matrices
R 3 → R 2 − 2 R 1
R 3 → R 3 + 3R 1
The latter augmented matrix corresponds to
3 x 1 + x 2 + x 3 = 0
Which is, effectively, one equation for three unknowns. Solving for x 1 in terms of x 2 and x 3 we
obtain
x 1 = −
(x 2 + x 3 )
So any vector of the form [
x 2 −
x 3 , x 2 , x 3
]
= x 2
[
]
[
]
which is equivalent to
x 1 = −
(7x 3 + 6x 4 )
x 2 =
(x 3 − x 4 )
Hence, any vector of the form
[
(7x 3 + 6x 4 ),
(x 3 − x 4 ), x 3 , x 4
]
= x 3
[
]
[
]
will be a solution. Thus, the vectors
e 1 =
[
]
, e 2 =
[
]
provide a basis for the solution space of the original set of equations.
(c)
x 1 − x 2 + 6x 3 + x 4 − x 5 = 0
3 x 1 + 2x 2 − 3 x 3 + 2x 4 + 5x 5 = 0
4 x 1 + 2x 2 − x 3 + 3x 4 − x 5 = 0
3 x 1 − 2 x 2 + 14x 3 + x 4 − 8 x 5 = 0
2 x 1 − x 2 + 8x 3 + 2x 4 − 7 x 5 = 0
- First we’ll row-reduce the corresponding augmented matrix until it is in row-echelon form.
1 − 1 6 1 − 1
3 2 − 3 2 5
4 2 − 1 3 − 1
3 − 2 14 1 − 8
2 − 1 8 2 − 7
R 2 → R 2 − 3 R 1
R 3 → R 3 − 4 R 1
R 4 → R 4 − 3 R 1
R 5 → R 5 − 2 R 1
R 2 → R 2 − 5 R 5
R 3 → R 3 − 6 R 5
R 4 → R 4 − R 5
R 2 ←→ R 5
R 3 → R 3 − 6 R 5
R 2 ←→ R 5
R 3 → −R 3
R 4 → −
1 2
R 4
R 5 → (1/6)(R 5 − R 3 )
R 1 → R 1 + R 2
R 3 → −R 3
R 4 → −
1 2
R 4
R 5 → (1/4)(R 5 − R 3 )
This last augmented matrix corresponds to the following linear system:
x 5 = 0
x 4 = 0
x 3 = −x 4 + 29x 5 = 0
x 2 = 4 x 3 + 5x 5 = 0
x 1 = x 2 − 6 x 3 − x 4 + x 5 = 0
Thus, the only solution vector is 0 = [0, 0 , 0 , 0 , 0]. One can think of the zero vector as the basis
vector for the solution subspace, however, it’s probably better to think of the solution space as the
zero vector.
- (Problems 35 and 37 in text). Solve the following linear systems and express the solution set in a form
that illustrates Theorem 1.18.
(a)
(1) 2 x 1 − x 2 + 3x 3 = − 3
4 x 1 + 4x 2 − x 4 = 1
To solve the linear system we row reduce
[ 2 − 1 3 0
4 4 0 − 1
]
R 2 → (1/6)) (R 2 − 2 R 1 ) ⇒
[
1 6
7 6
]
R 1 →
R 1 +
R 2 ⇒
[
1 12 0 1 − 1 −
1 6
25 12 7 6
]
From which we can infer
x 1 + x 3 −
x 4 = −
x 2 − x 3 −
x 4 =
or
x 1 = −x 3 +
x 4 −
x 2 = x 3 +
x 4 +
So a solution vector will have the form [
−x 3 +
x 4 −
, x 3 +
x 4 +
, x 3 , x 4
]
[
]
- x 3 [− 1 , 1 , 1 , 0] + x 4
[
]
Setting
p =
[
]
and
h = r [− 1 , 1 , 1 , 0] + s
[
]
[
−r +
s
, r +
s
, r, s
]
- (Problem 38 in text). Mark each of the following statements True or False.
a. A linear system with fewer equations than unknowns has an infinite number of solutions.
- False. If the system is inconsistent, it won’t have any solutions.
b. A consistent linear system with fewer equations than unknowns has an infinite number of solutions.
c. If a square linear system Ax = b has a solution for every choice of column vector b, then the solution is
unique for each choice of b.
- True. This follows from Theorems 1.12 and 1.16.
d. If a square system Ax = 0 has only the trivial solution x = 0 , then Ax = b has a unique solution for
every column vector b with the appropriate number of components.
- True. This follows from Corollary 2 and Theorem 1.16
e. If a linear system Ax = 0 has only the trivial solution x = 0 , then Ax = b has a unique solution for
every column vector b with the appropriate number of components.
- False. Consider what happens when the matrix A is of the form
A =
Then the corresponding system of equations
x 1 = 0
x 2 = 0
has a unique solution. Now consider a non-homogeneous linear system with the same matrix A;
say
Ax = b =
b 1
b 2
b 3
The corresponding augmented matrix will be
b 1
b 2
b 3
and so we’ll have an inconsistent set of equations whenever b 3 6 = 0.
f. The sum of two solution vectors of any linear system is also a solution vector of the system.
- False. Consider the following linear system:
x 1 + x 2 = 2
x 1 + x 3 = 2
and the following two solution vectors
v 1 = [1, 1 , 0] , v 2 = [1, 0 , 1]
If we set
v 3 = v 1 + v 2 = [2, 1 , 1]
then, v 3 will not be a solution vector.
g. The sum of two solution vectors of any homogeneous linear system is also a solution vector of the system.
h. A scalar multiple of a solution vector of any homogeneous linear system is also a solution vector of the
system.
- True. The solution space of a homogeneous linear system is a subspace: therefore it’s closed under
scalar multiplication.
i. Every line in R
2 is a subspace of R
2 generated by a single vector.
x ∈ R
2 | x = (1, 1) + (1, −1)t ; t ∈ R
is not a subspace.
j. Every line in R
2 through the origin in R
2 is a subspace of R
2 generated by a single vector.
