Download MAT 1341E: DGD 4 1. Show that W = {f ∈ F [0,3] | 2f(0)f(3) = 0} is not

MAT 1341E: DGD 4
SANGHOON BAEK
1. Show that W = {f ∈ F [0, 3] | 2f (0)f (3) = 0} is not a subspace of F [0, 3].
{
{
1 if x = 0
0 if x = 0
Solution: Let f (x) =
and g(x) =
for 0 ≤ x ≤ 3. Then
0 if x ̸= 0
1 if x ̸= 0
f, g ∈ W . But 2(f + g)(0)(f + g)(3) = 2(f (0) + g(0))(f (3) + g(3)) = 2 ̸= 0. Hence,
f +g ∈
/ W , i.e., W is not a subspace.
2. Show that W = {f ∈ F [0, 1] | f (0) = f (1)} is a subspace of F [0, 1].
Solution: If f, g ∈ W , then (f + g)(0) = f (0) + g(0) = f (1) + g(1) = (f + g)(1). For
any k ∈ R and any f ∈ W , kf (0) = kf (1). Hence, W is a subspace.
3. Show that {(1, 2), (1, 1)} is linearly independent. Show that {(0, 0), (1, 1)} is not
linearly independent.
Solution: If a(1, 2) + b(1, 1) = (0, 0) for some a, b ∈ R, then a = b = 0. Therefore,
{(1, 2), (1, 1)} is linearly independent. Any set containing the zero vector is linearly
dependent.
4. Let W = {(x, y, z) ∈ R3 |x + y − z = 0} = {(x, y, z) ∈ R3 |(1, 1, −1) · (x, y, z) = 0}.
(a) Find a basis of W . (b) Give a geometric description of W . (c) Extend your basis
in (a) to a basis of R3 .
Solution: (a) : Let (x, y, z) ∈ W . Then (x, y, z) = (x, y, x + y) = x(1, 0, 1) +
y(0, 1, 1). Therefore, span{(1, 0, 1), (0, 1, 1)} = W . If a(1, 0, 1) + b(0, 1, 1) = 0 for
some a, b ∈ R, then a = b = 0. Hence, they are linearly independent. Therefore, {(1, 0, 1), (0, 1, 1)} is a basis of W . (b) : W is the plane through the origin
with normal vector (1, 1, −1). (c) : As (1, 0, 0) ∈
/ span{(1, 0, 1), (0, 1, 1)} (because
if (1, 0, 0) = a(1, 0, 1) + b(0, 1, 1) for some a, b ∈ R, then a = 1, b = 0, and
a + b = 0 but this is impossible), {(1, 0, 1), (0, 1, 1), (1, 0, 0)} is linearly independent.
As dim(R3 ) = 3, {(1, 0, 1), (0, 1, 1), (1, 0, 0)} is a basis of R3 .
5. Show that sin x2 ∈
/ span{sin2 x, cos2 x}.
Solution: Assume that sin x2 = a sin2 x + b cos2 x for some a, b ∈ R. If x = π, then b =
/ span{sin2 x, cos2 x}.
1. If x = 2π, then b = 0. This is a contradiction. Hence, sin x2 ∈
6. Let W = {“ax + by = c” | a, b, c ∈ R} be the set of linear equations with real
coefficients in the variables x and y. We define the operations: addition of equations,
denoted here by “⊕” and multiplication by scalars, denoted here by “~”, as follows:
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“ax + by = c” ⊕ “dx + ey = f ” := “(a + d)x + (b + e)y = c + f ”
and
∀k ∈ R, k ~ “ax + by = c” := “ka x + kb y = k c”.
Find a basis of W .
Solution: Note that a ~ “x = 0” := “ax = 0”. Similarly, b ~ “y = 0” := “by = 0”
and c ~ “0 = 1” := “0 = c”. Hence,
“ax + by = c” = (a ~ (“x = 0”)) ⊕ (b ~ (“y = 0”)) ⊕ (c ~ (“0 = 1”)).
Therefore, {“x = 0”, “y = 0”, “0 = 1”} is a spanning set. Suppose that (a ~ (“x =
0”)) ⊕ (b ~ (“y = 0”)) ⊕ (c ~ (“0 = 1”)) = “0 = 0”. Then a = b = c = 0. Hence,
{“x = 0”, “y = 0”, “0 = 1”} is linearly independent.
7. Suppose {u, v, w} spans a vector space V . Which of the following statements is always
true?
(a) Any 4 vectors in V are linearly independent.
(b) Any 3 vectors in V are linearly independent.
(c) Any 4 vectors in V are linearly dependent.
(d) Any 3 vectors in V are linearly dependent.
(e) dim(span{u, v, w}) = 3.
Solution: Answer: (c).
(a) False: As the number of elements of a spanning set is always bigger or equal to
the number of elements of a linearly independent set, any 4 or more vectors in
V should be linearly dependent
(b) False: Take u = 0. Then {0, v, w} is linearly dependent.
(c) True
(d) False: Take V = R3 , u = e1 , v = e2 , w = e3 . Then {u, v, w} spans V = R3 and
is linearly independent.
(e) False: Take V = R and u = v = w be any nonzero vector in R . Then {u, v, w}
spans V = R and dim(span{u, v, w}) = 1.
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