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June 17th

June 17th

# Partial Solution Set

3.1.3 We are to show that the set C of complex numbers , with scalar multiplication defined
by (a + bi) = a + bi and addition defined by (a + bi) + (c + di) = (a + c) + (b + d)i,
satisfies the eight axioms of a vector space. This is only a partial solution.

A1: Let a + bi; c + di C. Then

(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, to show

A3: The zero element is 0 = (0 + 0i).

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

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.

3.1.4 Use the solution to 3.1.3 as a template for your solution. The objects are different
(matrices rather than complex numbers) and the operations are necessarily defined dif-
ferently, but these differences have no effect on the structure - R^m×n is simply another
vector space. The challenge is to avoid committing yourself to concrete values of m
and/or n
.

3.1.7 Show that the element 0 in a vector space is unique.

Note: This is a standard uniqueness argument. We as sume that we have two zero
elements and then discover that they are identical twins. The proof goes like this :

Proof: Let V be a vector space. We know that V contains at least one zero element,
since V satisfies the axioms. We must show, then, that V contains at most one zero
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 (0) to
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
α ס (x1, x2) = ( αx1, x2). Is V a vector space with these operations? Justify your answer.

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 scalar multi-
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 of
reals.) Is R+ a vector space when equipped with these operations? Prove your answer.

Solution: Yes, this is a vector space. To prove this, we must verify that the axioms
hold. Here is a partial proof:

A1: Use the definition of , together with the commutativity of ordinary real multipli-
cation.

A2: Use the definition of , together with the associativity of ordinary real multiplica-
tion.

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 Rn by

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 of ,

Now given p(x) as above, and given q(x) = b1 +b2x+ … +bnxn-1 and its corresponding
image b = (b1; b2…,bn),

and we're done.

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