(a + bi) + (c + di
= ((a + c) + (b + d)i) (By definition of complex addition)
= ((c + a) + (d + b)i) (Real addition is commutative)
= (c + di) + (a + bi) (By definition of complex addition)
A2: Similar to A1; pick three complex numbers, use the definition of complex addition
as often as necessary, together with the known associativity of real addition,
that complex addition is associative.
A4: To show existence of the additive inverse, choose an
arbitrary complex number (say,
x = a + bi) and construct its additive inverse. This will be made easy by your
knowledge of real additive inverses.
A5: We must prove that scalar multiplication distributes
over complex addition. Let
a + bi; c + di ∈C, and let ∈R. Then
a((a + bi) + (c + di))
= ((a + c) + (b + d)i) (Def'n complex addition)
= (a + c) + (b + d)i (Def'n of scalar mult. in C)
= ( a + c) + ( b + d)i (Distributivity in R)
= ( a + bi) + ( c + di) (Def'n of complex addition)
= (a + bi) + (c + di) (Def'n of scalar mult. in C)
A6: Similar to A5.
A7: Use definition of scalar multiplication in C and associativity of real multiplication.
A8: Use definition of scalar multiplication in C and the
fact that 1 is the multiplicative
identity in R.
Proof: Let V be a vector space. We know that V contains at least one zero
since V satisfies the axioms. We must show, then, that V contains at most one
element. So suppose that v and w are zeros in V . Then
v = v + w (Since w is a zero)
= w + v (Since addition commutes)
= w (Since v is a zero)
Thus uniqueness is proven, and it now makes sense to reserve a special symbol
denote the zero element. 2
3.1.8 Let x, y, and z be vectors in a vector space V .
Prove the additive cancellation law: if
x + y = x + z, then y = z.
Proof: Suppose x + y = x + z. Then
y = 0 + y (A3, A1)
= (-x + x) + y (A4)
= -x + (x + y) (A2)
= -x + (x + z) (Since x + y = x + z)
= (-x + x) + z (A2)
= 0 + z (A4)
= z (A1, A3):
3.1.11 Let V be the set of all ordered pairs of real numbers with addition defined in the usual
fashion by (x1, x2)+(y1, y2) = (x1+y1, x2+y2), but with scalar multiplication defined by
(x1, x2) = ( αx1, x2). Is V a vector space with these operations? Justify your
Solution: No, this is not a vector space. Axiom 6 fails:
unless x2 = 0.
3.1.12 Let R+ denote the set of positive real numbers. Define the ope ration of
plication, denoted ο, by α ο x = x^α for any real
α and x ∈R+. Define addition, denoted
, by x ⊕ y = x y for all x; y ∈R+. (The dot re presents the usual multiplication
reals.) Is R+ a vector space when equipped with these operations? Prove your
Solution: Yes, this is a vector space. To prove this, we must verify that the
hold. Here is a partial proof:
A1: Use the definition of ⊕, together with the commutativity of ordinary real multipli-
A2: Use the definition of ⊕, together with the associativity of ordinary real multiplica-
A3: The zero element is the number 1, since for any x ∈ R+
we have x ⊕1 = x ۰1 = x.
A4: The additive inverse in this oddball space is the
usual multiplicative inverse. That
is, for any x ∈R+, 1/x ∈R+, and x ⊕1/x = x
۰ 1/x = 1. By the preceding
argument, 1 is the zero element.
A5: Let α ∈R and x; y ∈R+. Then
A6: Let α, β,x ∈R. Then
A7: Let α, β ∈R, and x ∈R+. Then
A8: Let x ∈R. Then 1۰ x = x1 = x, where the first equality is by our local definition
of scalar multiplication and the second is by the usual laws of exponents .
3.1.16 We can define a one-to-one correspondence between the elements of Pn and
Show that if p ↔ a and q
↔ b then
1. p ↔ a for every scalar .
2. p + q ↔ a + b.
Proof: For p(x) as given in the introduction to the problem, and for any choice
Now given p(x) as above, and given q(x) = b1 +b2x+
… +bnxn-1 and its
image b = (b1; b2…,bn),