Suggested Problems Solution – Sections 4.1-4.2, 4.7
Math 81, Applied Analysis
Instructor: Dr. Doreen De Leon
• Section 4.1: 30, 31, 34, 35
• Section 4.2: 5, 6, 7, 10, 14, 15, 20
• Section 4.7: 1, 6, 7, 9, 10, 11
Section 4.1
30. V is the set of all (x, y, z) such that x + y + z = 0. Show that V is a subspace of R3 .
• zero check
0 = (0, 0, 0) ∈ V since 0 + 0 + 0 = 0.
• closure under vector addition
Let u = (u1 , u2 , u3 ) ∈ V =⇒ u1 + u2 + u3 = 0
v = (v1 , v2 , v3 ) ∈ V =⇒ v1 + v2 + v3 = 0
u + v = (u1 + v1 , u2 + v2 , u3 + v3 ).
We need to verify that u + v ∈ V .
u + v ∈ V if (u1 + v1 ) + (u2 + v2 ) + (u3 + v3 ) = 0.
(u1 + v1 ) + (u2 + v2 ) + (u3 + v3 ) = (u1 + u2 + u3 ) + (v1 + v2 + v3 )
=0+0
=0
So, u + v ∈ V . X
• closure under scalar multiplication
Let c be a scalar and let u = (u1 , u2 , u3 ) ∈ V . Then,
cu = (cu1 , cu2 , cu3 ).
We need to verify that cu ∈ V .
=⇒ We need to verify that (cu1 ) + (cu2 ) + (cu3 ) = 0.
(cu1 ) + (cu2 ) + (cu3 ) = c(u1 + u2 + u3 )
= c(0)
= 0.
1
Therefore, cu ∈ V . X
NOTE: Remember, when checking closure under scalar multiplication, never divide by c, because c could be 0.
Since V ⊂ R3 contains the zero vector and is closed under vector addition and scalar
multiplication, V is a subspace of R3 .
Alternate solution:
Since if u = (u1 , u2 , u3 ) ∈ V , then u1 +u2 +u3 = 0, we can solve for one of the variables,
say u3 in terms of the other two variables; i.e., we can write u3 = −u1 − u2 . So, we
can also do the problem in the following way.
• zero check
0 = (0, 0, 0) ∈ V since 0 + 0 + 0 = 0.
• closure under vector addition
Let u = (u1 , u2 , u3 ) ∈ V =⇒ u = (u1 , u2 , −u1 − u2 )
v = (v1 , v2 , v3 ) ∈ V =⇒ u = (v1 , v2 , −v1 − v2 )
u + v = (u1 + v1 , u2 + v2 , −u1 − u2 − v1 − v2 ).
We need to verify that u + v ∈ V .
(u1 + v1 , u2 + v2 , −u1 − u2 − v1 − v2 ) = (u1 + v1 , u2 + v2 , −(u1 + v1 ) − (u2 + v2 )).
Therefore, u + v ∈ V . X
• closure under scalar multiplication
Let c be a scalar and let u ∈ V . Then,
cu = (cu1 , cu2 , c(−u1 − u2 )).
We need to verify that cu ∈ V .
(cu1 , cu2 , c(−u1 − u2 )) = (cu1 , cu2 , −(cu1 ) − (cu2 )).
Therefore, cu ∈ V . X
Since V ⊂ R3 contains the zero vector and is closed under vector addition and scalar
multiplication, V is a subspace of R3 .
31. V is the set of all (x, y, z) such that 2x = 3y. Show that V is a subspace of R3 .
• zero check
0 = (0, 0, 0) ∈ V since 2(0) = 0 = 3(0).
2
• closure under vector addition
Let u = (u1 , u2 , u3 ) ∈ V =⇒ 2u1 = 3u2
v = (v1 , v2 , v3 ) ∈ V =⇒ 2v1 = 3v2
u + v = (u1 + v1 , u2 + v2 , u3 + v3 ).
We need to verify that u + v ∈ V ,
u + v ∈ V if 2(u1 + v1 ) = 3(u2 + v2 ).
2(u1 + v1 ) = 2u1 + 2v1
= 3u2 + 3v2
= 3(u2 + 3v2 ).
Therefore, u + v ∈ V . X
• closure under scalar multiplication
Let c be a scalar and let u = (u1 , u2 , u3 ) ∈ V . Then,
cu = (cu1 , cu2 , cu3 ).
We need to verify that cu ∈ V .
=⇒ We need to verify that 2(cu1 ) = 3(cu2 ).
2(cu1 ) = c(2u1 )
= c(3u2 )
= 3(cu1 ).
Therefore, cu ∈ V . X
Since V ⊂ R3 contains the zero vector and is closed under vector addition and scalar
multiplication, V is a subspace of R3 .
NOTE: This problem has an alternate solution along the same lines as problem 30.
34. V is the set of all (x, y, z) such that x + y + z = 3 Show that V is not a subspace of
R3 .
0 = (0, 0, 0) ∈
/ V because 0 + 0 + 0 = 0 6= 3. Therefore, V is not a subspace of R3 .
35. V is the set of all (x, y, z) such that z ≥ 0. Show that V is not a subspace of R3 .
Since the condition for a vector to be in the set is that the third component of the
vector must be nonnegative and since scalars can take any value, we suspect that this
set will not be closed under scalar multiplication. We need only show that this is true.
So, let u = (0, 0, 1) ∈ V and let c = −1. Then
cu = (0, 0, −1) ∈
/ V.
Therefore, V is not closed under scalar multiplication.
=⇒ V is not a subspace of R3 .
3
Section 4.2
5. W is the set of all vectors in R4 such that x1 + 2x2 + 3x3 + 4x4 = 0. Is W a subspace
of R4 ?
• zero check
0 = (0, 0, 0, 0) ∈ W since 0 + 2(0) + 3(0) + 4(0) = 0.
• closure under vector addition
Let u = (u1 , u2 , u3 , u4 ) ∈ W =⇒ u1 + 2u2 + 3u3 + 4u4 = 0
v = (v1 , v2 , v3 , v4 ) ∈ W =⇒ v1 + 2v2 + 3v3 + 4v4 = 0
u + v = (u1 + v1 , u2 + v2 , u3 + v3 , u4 + v4 ).
We need to verify that u + v ∈ W .
u + v ∈ V if (u1 + v1 ) + 2(u2 + v2 ) + 3(u3 + v3 ) + 4(u4 + v4 ) = 0.
(u1 + v1 ) + 2(u2 + v2 ) + 3(u3 + v3 ) + 4(u4 + v4 ) = (u1 + 2u2 + 3u3 + 4u4 ) + (v1 + 2v2 + 3v3 + 4v
=0+0
=0
So, u + v ∈ W . X
• closure under scalar multiplication
Let c be a scalar and let u = (u1 , u2 , u3 , u4 ) ∈ W . Then,
cu = (cu1 , cu2 , cu3 , cu4 ).
We need to verify that cu ∈ W .
=⇒ We need to verify that (cu1 ) + 2(cu2 ) + 3(cu3 ) + 4(cu4 ) = 0.
(cu1 ) + 2(cu2 ) + 3(cu3 ) + 4(cu4 ) = c(u1 + 2u2 + 3u3 + 4u4 )
= c(0)
= 0.
Therefore, cu ∈ W . X
Since W ⊂ R4 contains the zero vector and is closed under vector addition and scalar
multiplication, W is a subspace of R4 .
Alternate solution:
Since if u = (u1 , u2 , u3 , u4 ) ∈ W , then u1 + 2u2 + 3u3 + 4u4 = 0, we can solve for
one of the variables, say u1 in terms of the other two variables; i.e., we can write
u1 = −2u2 − 3u3 − 4u4 . So, we can also do the problem in the following way.
4
• zero check
0 = (0, 0, 0) ∈ V since 0 + 2(0) + 3(0) + 4(0) = 0.
• closure under vector addition
Let u = (u1 , u2 , u3 , u4 ) ∈ W =⇒ u = (−2u2 − 3u3 − 4u4 , u2 , u3 , u4 )
v = (v1 , v2 , v3 , v4 ) ∈ V =⇒ u = (−2v2 − 3v3 − 4v4 , v2 , v3 , v4 )
u + v = ((−2u2 − 3u3 − 4u4 ) + (−2v2 − 3v3 − 4v4 ), u2 + v2 , u3 + v3 , u4 + v4 ).
We need to verify that u + v ∈ V .
((−2u2 − 3u3 − 4u4 ) + (−2v2 − 3v3 − 4v4 ), u2 + v2 , u3 + v3 , u4 + v4 )
= (−2(u2 + v2 ) − 3(u3 + v3 ) − 4(u4 + v4 ), u2 + v2 , u3 + v3 , u4 + v4 ).
Therefore, u + v ∈ W . X
• closure under scalar multiplication
Let c be a scalar and let u ∈ W . Then,
cu = (c(−2u2 − 3u3 − 4u4 ), cu2 , cu3 , cu4 ).
We need to verify that cu ∈ V .
(c(−2u2 − 3u3 − 4u4 ), cu2 , cu3 , cu4 ) = (−2(cu2 ) − 3(cu3 ) − 4(cu4 ), cu2 , cu3 , cu4 ).
Therefore, cu ∈ W . X
Since W ⊂ R3 contains the zero vector and is closed under vector addition and scalar
multiplication, W is a subspace of R3 .
6. W is the set of all vectors in R4 such that x1 = 3x3 and x2 = 4x4 . Is W a subspace of
R4 ?
• zero check
0 = (0, 0, 0) ∈ V since 0 = 3(0) and 0 = 4(0).
• closure under vector addition
Let u = (3u3 , 4u4 , u3 , u4 ) ∈ W
v = (3v3 , 4v4 , v3 , v4 ) ∈ W
We need to verify that u + v ∈ W . But, we see that
u + v = (3u3 + 3v3 , 4u4 + 4v4 , u3 + v3 , u4 + v4 )
= (3(u3 + v3 ), 4(u4 + v4 ), u3 + v3 , u4 + v4 ).
Therefore, u + v ∈ W . X
5
• closure under scalar multiplication
Let c be a scalar and u = (3u3 , 4u4 , u3 , u4 ) ∈ W .
We need to verify that cu ∈ W . But,
cu = (c(3u3 ), c(4u4 ), cu3 , cu4 )
= (3(cu3 ), 4(cu4 ), cu3 , cu4 ).
Therefore, cu ∈ W . X
Since W ⊂ R3 contains the zero vector and is closed under vector addition and scalar
multiplication, W is a subspace of R3 .
7. W is the set of all vectors in R2 such that |x1 | = |x2 |. Is W a subspace of R2 ?
In this case, the fact that we have absolute values should raise suspicions about W
being a subspace of R2 . But, what condition could fail? A quick check shows that
0 = (0, 0) ∈ W . What about scalar addition? That is a good possibility, since we
know that |a + b| =
6 |a| + |b|. So,
let u = (1, −1) ∈ W since |1| = 1 = | − 1 and let v = (1, 1) ∈ W (since |1| = |1|). But,
u + v = (2, 0) ∈
/ W since |2| = 2 6= |0|.
Therefore, W is not closed under vector addition.
=⇒ W is not a subspace of R2 .
10. W is the set of all vectors in R2 such that |x1 | + |x2 | = 1. Is W a subspace of R2 ?
In this case, 0 = (0, 0) ∈
/ W since |0| + |0| = 0 + 0 = 0 6= 1. Therefore, W is not a
2
subspace of R .
14. W is the set of all vectors in R4 whose components are all nonzero. Is W a subspace
of R4 ?
Clearly, 0 = (0, 0, 0, 0) 6 W . Therefore, W is not a subspace of R4 .
15. Find two solution vectors u and v such that the solution space is the set of linear
combinations of the form su + tv.
x1 − 4x2 + x3 − 4x4 = 0
x1 + 2x2 + x3 + 8x4 = 0
x1 + x2 + x3 + 6x4 = 0
Solve by Gaussian elimination:






1 −4 1 −4 | 0
1 −4 1 −4 | 0 r ↔ 1 r
1 −4 1 −4 | 0
2
r2 →r2 −r1
6 2
1
2 1
8 | 0 −−
6 0 12 | 0 −−−−
1 0
2 | 0
→ 0
−−−−→ 0
r3 →r3 −r1
1
1 1
6 | 0
0
5 0 10 | 0
0
5 0 10 | 0


1 −4 1 −4 | 0
r3 →r3 −5r2

1 0
2 | 0
−−−−−−→ 0
0
0 0
0 | 0
6
Since there is no leading entry for x3 and x4 , x3 and x4 are arbitrary, so let x3 =
s, x4 = t. Then,
x2 + 2x4 = 0 =⇒ x2 = −2x4 = −2t
x1 − 4x2 + x3 − 4x4 = 0 =⇒ x1 = 4x2 − x3 + 4x4 = 4(−2t) − s + 4t = −s − 4t
So,
 
x1
 x2 

x=
 x3 
x4


−s − 4t
 −2t 

=
 s 
t
  

−s
−4t
 0 −2t
 

=
 s +  0
0
t
 
 
−1
−4
 0
−2

 
= s
 1 + t  0 .
0
1
20. Find a single solution vector u
multiples of u.
x1 +
2x1 +
2x1 +
such that the solution space is the set of all scalar
5x2 + x3 − 8x4 = 0
5x2
− 5x4 = 0
7x2 + x3 − 9x4 = 0
Solve by Gaussian elimination:






1
5 1 −8 | 0
1 5 1 −8 | 0
1
5
1 −8 | 0 r ↔ 3 r
2
r2 →r2 −2r1
5 2
2 5 0 −5 | 0 −
→ 0 −5 2 11 | 0
−−−−−→ 0 −5 −1 11 | 0 −−−−
r3 →r3 −2r1
2
0
0 51
| 0
2 7 1 −9 | 0
0 −3 −1
7 | 0
5


1
5 1 −8 | 0
r3 →5r3

−−−−→ 0 −5 2 11 | 0
0
0 1
2 | 0
Since there is no leading entry for x4 , x4 is arbitrary, so let x4 = r. Then,
x3 + 2x4 = 0 =⇒ x3 = −2x4 = −2r
−2x3 + 11x − 4
−2(−2r) + 11r
−5x2 − 2x3 + 11x4 = 0 =⇒ x2 =
=
= 3r
5
5
x1 + 5x2 + x3 − 8x4 = 0 =⇒ x1 = −5x2 − x3 + 8x4 = −5(3r) − (−2r) + 8r = −5r
7
So,


−5r
 3r

x=
−2r
r
 
−5
 3

= r
−2 .
1
Section 4.7
1. W is the set of all diagonal 3 × 3 matrices. Is W a subspace of M33 ?
• zero check


0 0 0
0 = 0 0 0 ∈ W (it is diagonal).
0 0 0
• closure under vector addition

a1

Let A = 0
0

b1
B = 0
0
Then

0 0
a2 0  ∈ W
0 a3

0 0
b2 0  ∈ W
0 b3

a1 + b1
0
0
a2 + b 2
0  ∈ W. X
A+B = 0
0
0
a3 + b 3

• closure under scalar multiplication
Let c be a scalar and A ∈ W . Then,


ca1 0
0
cA =  0 ca2 0  . X
0
0 ca3
Since W ⊂ M33 contains the zero matrix and is closed under vector addition and scalar
multiplication, W is a subspace of M33 .
6. W is the set of all f such that f (x) 6= 0 for all x. Is W a subspace of F?
The zero element for F is the zero function, f (x) = 0. Therefore, the zero function is
not in W , so W is not a subspace of F.
8
7. W is the set of all f such that f (0) = 0 and f (1) = 1. Is W a subspace of F?
The zero element for F is the zero function, f (x) = 0 (so, f (1) = 0 6= 1). Therefore,
the zero function is not in W , so W is not a subspace of F.
9. W is the set of polynomials of the form a0 + a1 x + a2 x2 + a3 x3 such that a3 6= 0. Is W
a subspace of P3 ?
The zero vector for P3 is 0 = 0 + 0x + 0x2 + 0x3 is not in W , since a3 = 0. Therefore,
W is not a subspace of P3 .
10. W is the set of polynomials of the form a0 + a1 x + a2 x2 + a3 x3 such that a0 = a1 = 0.
Is W a subspace of P3 ?
• zero check
0 = 0 + 0x + 0x2 + 0x3 ∈ W since a0 = a1 = 0.
• closure under vector addition
Let p ∈ W =⇒
=⇒
Let q ∈ W =⇒
=⇒
p(x) = a0 + a1 x + a2 x2 + a3 x3 , where a0 = a1 = 0
p(x) = a2 x2 + a3 x3 .
q(x) = b0 + b1 x + b2 x2 + b3 x3 , where b0 = b1 = 0
p(x) = b2 x2 + b3 x3 .
Then
(p + q)(x) = p(x) + q(x)
= (a2 + b2 )x2 + (a3 + b3 )x3 .
Therefore, p + q ∈ W . X
• closure under scalar multiplication
Let c be a scalar and let p ∈ W =⇒ p(x) = a2 x2 + a3 x3 . Then,
(cp)(x) = cp(x) = (ca2 )x2 + (ca3 )x3 .
Therefore, cp ∈ W . X
Since W ⊂ P3 contains the zero polynomial and is closed under vector addition and
scalar multiplication, W is a subspace of P3 .
11. W is the set of all polynomails of the form a0 + a1 x + a2 x2 + a3 x3 such that a0 + a1 +
a2 + a3 = 0. Is W a subspace of P3 ?
• zero check
0 = 0 + 0x + 0x2 + 0x3 ∈ W since 0 + 0 + 0 + 0 = 0.
9
• closure under vector addition
Let p ∈ W =⇒ p(x) = a0 + a1 x + a2 x2 + a3 x3 , where a0 + a1 + a2 + a3 = 0.
Let q ∈ W =⇒ q(x) = b0 + b1 x + b2 x2 + b3 x3 , where b0 + b1 + b2 + b3 = 0.
Then
(p + q)(x) = p(x) + q(x)
= (a0 + b0 ) + (a1 + b1 )x + (a2 + b2 )x2 + (a3 + b3 )x2
We need to verify that p + q ∈ W . We know that p + q ∈ W if
(a0 + b0 ) + (a1 + b1 ) + (a2 + b2 ) + (a3 + b3 ) = 0.
But,
(a0 + b0 ) + (a1 + b1 ) + (a2 + b2 ) + (a3 + b3 ) = (a0 + a1 + a2 + a3 ) + (b0 + b1 + b2 + b3 )
=0+0
= 0.
Therefore, p + q ∈ W . X
• closure under scalar multiplication
Let c be a scalar and let p ∈ W . Then,
(cp)(x) = cp(x) = (ca0 ) + (ca1 )x + (ca2 )x2 + (ca3 )x3 .
We need to verify that cp ∈ W . We know that cp ∈ W if
(ca0 ) + (ca1 ) + (ca2 ) + (ca3 ) = 0.
But,
(ca0 ) + (ca1 ) + (ca2 ) + (ca3 ) = c(a0 + a1 + a2 + a3 )
= c(0)
= 0.
Therefore, cp ∈ W . X
Since W ⊂ P3 contains the zero polynomial and is closed under vector addition and
scalar multiplication, W is a subspace of P3 .
Alternate Solution: An alternate solution may be obtained by solving for one of the
coefficients, say a4 and proceeding as above.
10