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Winter 2017 Ma 1b Analytical Problem Set 2 Solutions
1. (5 pts) From Ch. 1.10 in Apostol: Problems 1,3,5,7,9. Also, when appropriate exhibit a basis
for S.
Solution. (1.10.1) Yes, S is a subspace of V3 with basis {(0, 0, 1), (0, 1, 0)} and dimension 2.
Proof. First of all (0, 0, 0) ∈ S so S is nonempty. For any (x1 , y1 , z1 ), (x2 , y2 , z2 ) ∈ S we have
x1 = 0 and x2 = 0 so x1 + x2 = 0 and (x1 , y1 , z1 ) + (x2 , y2 , z2 ) = (x1 + x2 , y1 + y2 , z1 + z2 ) ∈
S, and S is closed under vector addition. For any a ∈ R and (x1 , y1 , z1 ) ∈ S we have
x1 = 0 so ax1 = 0 and a(x1 , y1 , z1 ) = (ax1 , ay1 , az1 ) ∈ S and S is closed under scalar
multiplication. Thus S is a subspace of V3 . {(0, 0, 1), (0, 1, 0)} is independent in S because
for any a, b ∈ R, a(0, 0, 1) + b(0, 1, 0) = (0, 0, 0) gives us (0, b, a) = (0, 0, 0) and a = b = 0.
L({(0, 0, 1), (0, 1, 0)}) = S because for any (x, y, z) ∈ S, x = 0 so (x, y, z) = (0, y, z) =
y(0, 1, 0) + z(0, 0, 1). Thus {(0, 0, 1), (0, 1, 0)} is a basis of S, and S has dimension 2.
(1.10.3) Yes, S is a subspace of V3 with basis {(−1, 0, 1), (−1, 1, 0)} and dimension 2.
Proof. First of all (0, 0, 0) ∈ S so S is nonempty. For any (x1 , y1 , z1 ), (x2 , y2 , z2 ) ∈ S we have
x1 + y1 + z1 = 0 and x2 + y2 + z2 = 0 so (x1 + x2 ) + (y1 + y2 ) + (z1 + z2 ) = (x1 + y1 + z1 ) +
(x2 + y2 + z2 ) = 0 and (x1 , y1 , z1 ) + (x2 , y2 , z2 ) = (x1 + x2 , y1 + y2 , z1 + z2 ) ∈ S, and S is
closed under vector addition. For any a ∈ R and (x1 , y1 , z1 ) ∈ S we have x1 + y1 + z1 = 0
so ax1 + ay1 + az1 = 0 and a(x1 , y1 , z1 ) = (ax1 , ay1 , az1 ) ∈ S and S is closed under scalar
multiplication. Thus S is a subspace of V3 . {(−1, 0, 1), (−1, 1, 0)} is independent in S because
for any a, b ∈ R, a(−1, 0, 1) + b(−1, 1, 0) = (0, 0, 0) gives us (−a − b, b, a) = (0, 0, 0) and
a = b = 0. L({(−1, 0, 1), (−1, 1, 0)}) = S because for any (x, y, z) ∈ S, x + y + z = 0 so
(x, y, z) = (−y − z, y, z) = y(−1, 1, 0) + z(−1, 0, 1). Thus {(−1, 0, 1), (−1, 1, 0)} is a basis of
S, and S has dimension 2.
(1.10.5) Yes, S is a subspace of V3 with basis {(1, 1, 1)} and dimension 1.
Proof. First of all (0, 0, 0) ∈ S so S is nonempty. For any (x1 , y1 , z1 ), (x2 , y2 , z2 ) ∈ S we have
x1 = y1 = z1 and x2 = y2 = z2 so x1 + x2 = y1 + y2 = z1 + z2 and (x1 , y1 , z1 ) + (x2 , y2 , z2 ) =
(x1 + x2 , y1 + y2 , z1 + z2 ) ∈ S, and S is closed under vector addition. For any a ∈ R and
(x1 , y1 , z1 ) ∈ S we have x1 = y1 = z1 so ax1 = ay1 = az1 and a(x1 , y1 , z1 ) = (ax1 , ay1 , az1 ) ∈
S and S is closed under scalar multiplication. Thus S is a subspace of V3 . {(1, 1, 1)} is
independent in S because for any a ∈ R, a(1, 1, 1) = (0, 0, 0) gives us (a, a, a) = (0, 0, 0) and
a = 0. L({(1, 1, 1)}) = S because for any (x, y, z) ∈ S, x = y = z so (x, y, z) = (x, x, x) =
x(1, 1, 1). Thus {(1, 1, 1)} is a basis of S, and S has dimension 1.
(1.10.7) No, S is not a subspace of V3 since (1, 1, 0), (1, −1, 0) ∈ S but (1, 1, 0) + (1, −1, 0) =
(2, 0, 0) 6∈ S.
(1.10.9) Yes, S is a subspace of V3 with basis {(1, 2, 3)} and dimension 1.
Proof. First of all (0, 0, 0) ∈ S so S is nonempty. For any (x1 , y1 , z1 ), (x2 , y2 , z2 ) ∈ S we have
y1 = 2x1 , z1 = 3x1 , y2 = 2x2 , and z2 = 3x2 , so y1 + y2 = 2(x1 + x2 ), z1 + z2 = 3(x1 + x2 ), and
(x1 , y1 , z1 ) + (x2 , y2 , z2 ) = (x1 + x2 , y1 + y2 , z1 + z2 ) ∈ S, and S is closed under vector addition.
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Winter 2017 Ma 1b Analytical Problem Set 2 Solutions
For any a ∈ R and (x1 , y1 , z1 ) ∈ S we have y1 = 2x1 and z1 = 3x1 so ay1 = 2ax1 , az1 = 3ax1 ,
and a(x1 , y1 , z1 ) = (ax1 , ay1 , az1 ) ∈ S and S is closed under scalar multiplication. Thus S is
a subspace of V3 . {(1, 2, 3)} is independent in S because for any a ∈ R, a(1, 2, 3) = (0, 0, 0)
gives us (a, 2a, 3a) = (0, 0, 0) and a = 0. L({(1, 2, 3)}) = S because for any (x, y, z) ∈ S,
y = 2x and z = 3x, so (x, y, z) = (x, 2x, 3x) = x(1, 2, 3). Thus {(1, 2, 3)} is a basis of S, and
S has dimension 1.
2. (8 pts) From Ch. 1.10 in Apostol: Problem 24.
Solution. (a) We adapt the proof of Theorem 1.7(a). Let B = {x1 , . . . , xk } be any independent set of elements in S. If L(B) = S, then B is a basis for S. If not, then there is some
element y in S which is not in L(B). Adjoin this element to B and consider the new set
B 0 = {x1 , . . . , xk , y}. If this set were dependent there would be scalars c1 , . . . , ck+1 not
all zero such that
k
X
ci xi + ck+1 y = O.
i=1
Since the x1 , . . . , xk are independent, we would not have ck+1 = 0 or else c1 = · · · = ck = 0
by the linear independence of x1 , . . . , xk , contrary to our assumption that B is dependent.
Thus we would have ck+1 6= 0 and could solve this equation for y and find that y ∈ L(B),
contradicting our assumption that y is not in L(B). Therefore, the set B 0 is independent
but contains k + 1 elements. If L(B 0 ) = S, then B 0 is a basis of S. If L(B 0 ) 6= S, we
can argue with B 0 as we did with B, getting a new set B 00 which contains k + 2 elements
and is independent. We can repeat the process until we arrive at a basis in a total of
dim V − k steps or less, or else we eventually obtain an independent set with dim V + 1
elements, contradicting Theorem 1.5. This basis will be finite and have at most dim V
elements, so S is finite dimensional and dim S ≤ dim V .
(b) The “if” part holds trivially by substitution. The “only if” part follows from applying
Theorem 1.7(b) to the basis for S constructed in part (a). The basis for S is a basis for
V , so taking the span gives us S = V .
(c) Apply Theorem 1.7(a).
(d) Consider the example V = V2 and S = R × {0} ⊆ V2 . {(1, −1), (1, 1)} is a basis for V
which contains no basis for S because (1, −1), (1, 1) 6∈ S and ∅ is not a basis for S since
S contains nonzero elements.
3. Let V be a vector space and U, W subspaces of V . Prove that if U ∩ W = {0} then
dim L(U, W ) = dim U + dim W . (Do not appeal to a more general result for dim L(U, W )).
Solution. If U or W is infinite dimensional then suppose for the sake of contradiction that
L(U, W ) is finite dimensional. Since U and W are subspaces of L(U, W ) that implies that U
and W are finite dimensional by part (a) of the previous problem, so we get a contradiction.
Thus we get L(U, W ) infinite dimensional. It remains to consider the case where U and W
are both finite dimensional.
Let S = {x1 , . . . , xn } be a basis for U and T = {y1 , . . . , ym } be a basis for W . A finite linear
combination of elements of U or W is a finite linear combination of elements of S ∪ T because
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Winter 2017 Ma 1b Analytical Problem Set 2 Solutions
we can rewrite each element of U as a finite linear combination of elements of S, rewrite each
element of W as a finite linear combination of elements of T , and distribute to get a finite
linear combination of elements of S ∪ T . Thus, L(U, W ) ⊂ L(S, T ). Since S ∪ T ⊂ U ∪ W , it
follows that L(U, W ) ⊃ L(S, T ) and hence L(U, W ) = L(S, T ) (ie. S ∪ T spans L(U, W )).
S ∪ T is linearly independent by the following argument. Suppose
n
X
ai xi +
i=1
Then
m
X
bj yj = 0.
j=1
n
X
ai xi = −
m
X
i=1
bj yj ,
j=1
and since U ∩ W = {0} we have
n
X
ai xi =
i=1
m
X
bj yj = 0.
j=1
Since S is linearly independent and T is linearly independent, we get ai = 0 for all 1 ≤ i ≤ n
and bj = 0 for all 1 ≤ j ≤ m. Therefore, S ∪ T is linearly independent.
S∪T spans L(U, W ) and is linearly independent so S∪T is a basis of L(U, W ) and dim L(U, W ) =
|S ∪ T | = |S| + |T | = dim U + dim W .
4. (4 pts) From Ch. 2.4. in Apostol: Problems 11-14. Determine whether T is linear.
Solution.
(2.4.11) Yes, T is linear.
Proof. In V2 , vector addition and scalar multiplication are defined in terms of rectangular coordinates, so it is necessary to describe what T does to points given in rectangular coordinates.
(r, θ) is (x, y) = (r cos θ, r sin θ) in rectangular coordinates, and (r, θ + ϕ) is
(r cos(θ + ϕ), r sin(θ + ϕ)) = (r(cos θ cos ϕ − sin θ sin ϕ), r(sin θ cos ϕ + cos θ sin ϕ))
= (r cos θ cos ϕ − r sin θ sin ϕ, r sin θ cos ϕ + r cos θ sin ϕ)
= (x cos ϕ − y sin ϕ, y cos ϕ + x sin ϕ),
so T (x, y) = (x cos ϕ − y sin ϕ, y cos ϕ + x sin ϕ). Suppose (x1 , y1 ), (x2 , y2 ) ∈ V2 . Then
T ((x1 , y1 ) + (x2 , y2 )) = T (x1 + x2 , y1 + y2 )
= ((x1 + x2 ) cos ϕ − (y1 + y2 ) sin ϕ, (y1 + y2 ) cos ϕ + (x1 + x2 ) sin ϕ)
= (x1 cos ϕ − y1 sin ϕ, y1 cos ϕ + x1 sin ϕ)
+ (x2 cos ϕ − y2 sin ϕ, y2 cos ϕ + x2 sin ϕ)
= T (x1 , y1 ) + T (x2 , y2 )
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Winter 2017 Ma 1b Analytical Problem Set 2 Solutions
so T preserves vector addition. Suppose a ∈ R and (x1 , y1 ) ∈ V2 . We have
T (a(x1 , y1 )) = T (ax1 , ay1 )
= (ax1 cos ϕ − ay1 sin ϕ, ay1 cos ϕ + ax1 sin ϕ)
= a(x1 cos ϕ − y1 sin ϕ, y1 cos ϕ + x1 sin ϕ)
= aT (x1 , y1 )
so T preserves scalar multiplication. Therefore T is linear.
(2.4.12) Yes, T is linear.
Proof. T maps a point with polar coordinates (r, θ) onto the point with polar coordinates
(r, ϕ − θ), where ϕ is the angle of elevation of the fixed line. In V2 , vector addition and scalar
multiplication are defined in terms of rectangular coordinates, so it is necessary to describe
what T does to points given in rectangular coordinates. (r, θ) is (x, y) = (r cos θ, r sin θ) in
rectangular coordinates, and (r, ϕ − θ) is
(r cos(ϕ − θ), r sin(ϕ − θ)) = (r(cos ϕ cos θ + sin ϕ sin θ), r(sin ϕ cos θ − cos ϕ sin θ))
= (r cos θ cos ϕ + r sin θ sin ϕ, r cos θ sin ϕ − r sin θ cos ϕ)
= (x cos ϕ + y sin ϕ, x sin ϕ − y cos ϕ),
so T (x, y) = (x cos ϕ + y sin ϕ, x sin ϕ − y cos ϕ). Suppose (x1 , y1 ), (x2 , y2 ) ∈ V2 . Then
T ((x1 , y1 ) + (x2 , y2 )) = T (x1 + x2 , y1 + y2 )
= ((x1 + x2 ) cos ϕ + (y1 + y2 ) sin ϕ, (x1 + x2 ) sin ϕ − (y1 + y2 ) cos ϕ)
= (x1 cos ϕ + y1 sin ϕ, x1 sin ϕ − y1 cos ϕ)
+ (x2 cos ϕ + y2 sin ϕ, x2 sin ϕ − y2 cos ϕ)
= T (x1 , y1 ) + T (x2 , y2 )
so T preserves vector addition. Suppose a ∈ R and (x1 , y1 ) ∈ V2 . We have
T (a(x1 , y1 )) = T (ax1 , ay1 )
= (ax1 cos ϕ + ay1 sin ϕ, ax1 sin ϕ − ay1 cos ϕ)
= a(x1 cos ϕ + y1 sin ϕ, x1 sin ϕ − y1 cos ϕ)
= aT (x1 , y1 )
so T preserves scalar multiplication. Therefore T is linear.
(2.4.13) No, T is not linear. T ((0, 1)) + T ((1, 0)) = (1, 1) + (1, 1) = (2, 2) 6= (1, 1) = T (1, 1) =
T ((0, 1) + (1, 0)) so T does not preserve vector addition.
(2.4.14) Yes, T is linear.
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Winter 2017 Ma 1b Analytical Problem Set 2 Solutions
Proof. In V2 , vector addition and scalar multiplication are defined in terms of rectangular coordinates, so it is necessary to describe what T does to points given in rectangular coordinates. (r, θ) is (x, y) = (r cos θ, r sin θ) in rectangular coordinates, and (2r, θ) is
(2r cos θ, 2r sin θ) = (2x, 2y) so T (x, y) = (2x, 2y). Suppose (x1 , y1 ), (x2 , y2 ) ∈ V2 . Then
T ((x1 , y1 ) + (x2 , y2 )) = T (x1 + x2 , y1 + y2 ) = (2(x1 + x2 ), 2(y1 + y2 ))
= (2x1 + 2x2 , 2y1 + 2y2 )
= (2x1 , 2y1 ) + (2x2 , 2y2 )
= T (x1 , y1 ) + T (x2 , y2 )
so T preserves vector addition. Suppose a ∈ R and (x1 , y1 ) ∈ V2 . Then
T (a(x1 , y1 )) = T (ax1 , ay1 ) = (2ax1 , 2ay1 ) = a(2x1 , 2y1 ) = aT (x1 , y1 )
so T preserves scalar multiplication. Therefore T is linear.
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