Math 296 (Winter 2012). Honors Mathematics II
Solutions for Problem Set 1
1.1. Observe that (a + bi)(a − bi) = a2 − (bi2 ) = a2 + b2 . If one or both of a, b is nonzero,
then a2 + b2 6= 0. So if we put c = a/(a2 + b2 ) and d = −b/(a2 + b2 ), we will have
(a + bi)(c + di) = 1, i.e., 1/(a + bi) = c + di.
1.2. Omitted.
1.3. Let w = −v. Then by definition, v + w = w + v = 0, which means that v satisfies
the defining property of −w. So v = −w = −(−v).
1.4. Suppose that a 6= 0. Then v = 1 · v = (a−1 · a) · v = a−1 · (a · v) = a−1 · 0 = 0, where
in the last step we used Proposition 1.5 in the book. So if a 6= 0, then v = 0, as required.
1.5. (a) This is a subspace of F3 by a special case of example (4) on page 3 of the lecture
notes about vector spaces.
(b) This is not a subspace because it does not contain the zero vector.
(c) This is not a subspace because it is not closed under addition. Indeed, the vectors
v = (1, 1, 0) and w = (0, 0, 1) belong to the given subset but v + w does not.
(d) This is a subspace of F3 for the same reason as in (a).
1.6. Take U = Q2 to be the subset of R2 consisting of all vectors with rational coordinates. Then U is clearly closed under addition and
√ under taking additive inverses.
However, U is not a subspace because (1, 0) ∈ U and 2 · (1, 0) 6∈ U .
1.7. Take U ⊂ R2 to be the subset consisting of all vectors (x1 , x2 ) such that either
x1 , x2 ≥ 0 or x1 , x2 ≤ 0. It is clear that U is closed under multiplication by scalars.
However, U is not a subspace: (1, 0) ∈ U and (0, −1) ∈ U , but (1, 0) + (0, −1) 6∈ U .
1.8. Consider
any collection {Ui }i∈I of subspaces of V (where I is some index set) and
T
let U = i∈I Ui denote their intersection. We have 0 ∈ U because 0 ∈ Ui for every i ∈ I.
Next let u, v ∈ U . Then u, v ∈ Ui for each i, so u + v ∈ Ui for each i, so u + v ∈ U .
Similarly, if λ ∈ F, then λ · u ∈ Ui for each i because u ∈ Ui , which implies that λ · u ∈ U .
So we checked all three requirements for U to be a subspace of V .
1.9. Let U, W ⊂ V be two subspaces. Suppose first that U ∪ W is a subspace of V
and assume, to obtain a contradiction, that neither of U and W is contained in the other
one. Then we can find u ∈ U and w ∈ W such that u 6∈ W and w 6∈ U . On the other
hand, we have u + w ∈ U ∪ W because U ∪ W is assumed to be a subspace of V . So
either u + w ∈ U or u + w ∈ W . If u + w ∈ U , then w = (u + w) − u ∈ U , which is a
contradiction. Similarly, if u + w ∈ W , then u ∈ W , which is also a contradiction.
Conversely, suppose that one of U and W is contained in the other one. Then U ∪ W
is equal to one of them, whence it is a subspace of V .
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1.10. We claim that U + U = U . Indeed, we have u1 + u2 ∈ U whenever u1 , u2 ∈ U
(because U is a subspace), so U + U ⊂ U . Conversely, if u ∈ U , then u = u + 0 ∈ U + U ,
whence U ⊂ U + U .
1.11. Yes, this operation is both associative and commutative. Suppose U1 , U2 , U3 ⊂ V
are subspaces. Then U1 + U2 consists of all elements of the form u1 + u2 , where u1 ∈ U1
and u2 ∈ U2 . But u1 + u2 = u2 + u1 , so we see that U1 + U2 = U2 + U1 .
Similarly, (U1 + U2 ) + U3 consists of all elements of the form (u1 + u2 ) + u3 and U1 +
(U2 + U3 ) consists of all elements of the form u1 + (u2 + u3 ), where uj ∈ Uj for j = 1, 2, 3.
But we always have (u1 +u2 )+u3 = u1 +(u2 +u3 ), whence (U1 +U2 )+U3 = U1 +(U2 +U3 ).
1.12. It is clear that {0} is the additive identity for the operation of addition of subspaces
of V . The only subspace that has an additive inverse is {0}. Indeed, we have {0} + {0} =
{0}, so {0} is its own additive inverse. On the other hand, if U, W ⊂ V are arbitrary
subspaces such that U 6= {0}, then U ⊂ U + W , which shows that U + W 6= {0}, whence
U cannot have an additive inverse.
1.13. This assertion is false because the assertion of problem 15 is false, and the counterexample for problem 15 (see below) is also a counterexample for this problem.
1.14. Let W ⊂ P(F ) consist of all polynomials p(z) for which the coefficients of z 2
and of z 5 are equal to 0. It is clear that W is a subspace of P(F ). Moreover, we have
P(F ) = U ⊕ W . Indeed, this is the same as saying that every polynomial p(z) can be
written uniquely as p(z) = az 2 + bz 5 + f (z), where a, b ∈ F and f (z) ∈ W . However, this
is essentially a tautology.
1.15. The given assertion is false. Take V = R2 , let U1 ⊂ V consist of all vectors of the
form (a, 0), let U2 ⊂ V consist of all vectors of the form (0, a), and let W ⊂ V consist of
all vectors of the form (a, a), where a ∈ R in each case.
It is clear that U1 , U2 , W are subspaces of R2 and U1 6= U2 . However, we claim that
R = U1 ⊕ W and R2 = U2 ⊕ W . First note that U1 ∩ W = {0} = U2 ∩ W . Next, given any
(x1 , x2 ) ∈ R2 , we have (x1 , x2 ) = (x1 −x2 , 0)+(x2 , x2 ) and (x1 , x2 ) = (0, x2 −x1 )+(x1 , x1 ),
which shows that R2 = U1 + W and R2 = U2 + W . Using Proposition 1.9 in the book, we
conclude that R2 = U1 ⊕ W and R2 = U2 ⊕ W , completing the solution.
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