Download If ax = 0, then either a = 0 or x = 0. Proof

Solution Set for Homework 13
1 Theorem 15.3
(d) Statement: If ax = 0, then either a = 0 or x = 0.
Proof: Suppose a 6= 0 then
ax = 0
1
1
=⇒ (ax) = .0
a
a
1
=⇒ ( .a)x = 0 [Using (b) and associative law for scalar multiplication]
a
=⇒ x = 0.
(e) Statement: If ax = ay and a 6= 0 then x = y.
Proof:
ax = ay
=⇒ a(x − y) = 0 [Distributive law].
Now proceed as in the previous part.
(f) Statement: If ax = bx and x 6= 0 then a = b.
Proof: Follow previous two parts.
(g) Statement: −(x + y) = (−x) + (−y) = −x − y.
Proof: Use distributive law.
(h) Statement: x + x = 2x, x + x + x = 3x, and in general,
Proof: Use distributive law and method of induction.
Pn
i=1
x = nx.
2 Vector addition and scalar multiplication is define in usual way.
(4) V := {f : R → R|2f (0) = f (1)}.
Answer: Yes, V is a vector space.
Axiom 1: Closure under addition:
Let f, g ∈ V then 2f (0) = f (1) and 2g(0) = g(1). Now observe that f + g is also a
real valued function defined on R, also
2[(f + g)(0)] = 2[f (0) + g(0)]
= 2f (0) + 2g(0) [distributive law of real numbers]
= f (1) + g(1)
= (f + g)(1)
and hence f + g ∈ V .
1
2
Axiom 2: Closure under scalar multiplication:
Let α ∈ R and let f ∈ V . Then
2[(α.f )(0)] = 2[αf (0)]
= (2.α)f (0) [associative law for real numbers]
= (α.2)f (0) [commutative law for real numbers]
= α(2f (0)) [associative law for real numbers]
= α.f (1) [as f ∈ V ]
= (α.f )(1).
Thus α.f ∈ V .
Axiom 3: Commutative Law:
Let f, g ∈ V be two functions. Then
(f + g)(x) = f (x) + g(x)
= g(x) + f (x) [commutative law for real numbers]
= (g + f )(x)
and hence f + g = g + f .
Axiom 4: Associative Law:
Let f, g, h ∈ V be functions. Then
[(f + g) + h](x) = (f + g)(x) + h(x)
= [f (x) + g(x)] + h(x)
= f (x) + (g(x) + h(x)) [associative law for real numbers]
= f (x) + (g + h)(x)
= [f + (g + h)](x)
and hence (f + g) + h = f + (g + h).
Axiom 5: Existence of Zero Element:
Consider the zero function 0. Notice that 2.0(0) = 0 = 0(1) and hence 0 ∈ V . Now
let f ∈ V be a function then
(f + 0)(x) = f (x) + 0(x)
= f (x) + 0
= f (x) [0 is the zero element of R]
hence f + 0 = f .
Axiom 6: Existence of Negative:
Let f be a function in V , then 2f (0) = f (1) and hence
− 2f (0) = −f (1)
=⇒ 2(−f (0)) = −f (1) [Associative law of real numbers]
=⇒ 2[(−f )(0)] = (−f )(1)
and hence −f ∈ V . Also f (x) + (−f )(x) = f (x) − f (x) = 0, therefore f + (−f ) = 0.
SOLUTION SET FOR HOMEWORK 13
3
Axiom 7: Associative Law for scalar multiplication:
Let α, β ∈ R and f ∈ V be given.
[α(βf )](x) = α(βf )(x)
= α(βf (x))
= (αβ)f (x) [Associative law of real numbers]
= [(αβ)f ](x)
hence α(βf ) = (αβ)f .
Axiom 8: Distributive Law for Addition in V :
Let f, g ∈ V and α ∈ R then
[α(f + g)](x) = α(f + g)(x)
= α(f (x) + g(x))
= αf (x) + αg(x) [distributive property of R]
= (αf )(x) + (αg)(x)
= [(αf ) + (αg)](x)
and hence α(f + g) = (αf ) + (αg).
Axiom 9: Distributive Law for Addition of Numbers:
Let α, β ∈ R and f ∈ V then
[(α + β)f ](x) = (α + β)f (x)
= αf (x) + βf (x) [distributive property of R]
= (αf )(x) + (βf )(x)
= [(αf )(x) + (βf )](x)
therefore (α + β)f = (αf )(x) + (βf ).
Axiom 10: Existence of Identity:
Let f ∈ V be any function then
(1.f )(x) = 1.f (x)
= f (x) [1 is identity in R]
so 1.f = f .
Hence V is a vector space.
(11) V := {f : R → R|f is a increasing function}.
Answer: V is not a vector space as the function f defined by, f (x) = x for all x ∈ R, is in
V but −f is not in V . Hence V is not closed under scalar multiplication.
3 Vector addition and scalar multiplication is define in usual way.
(13) V := {f : [0, 1] → R|f is integrable function and
Answer: V is a vector space. Verify!
R1
0
f (x)dx = 0}.
R1
(14) V := {f : [0, 1] → R|f is integrable function and 0 f (x)dx ≥ 0}.
Answer: V is not a vector space as constant function 1 belongs to V but constant function
−1 doesn’t belong.
(16) It is a vector space.
4
4 Vector addition and scalar multiplication is define in usual way.
(18) It is a vector space.
(21) It is a vector space.
5 Vector addition and scalar multiplication is define in usual way.
(23) V := {(x, y, z) ∈ R3 |x = 0 or y = 0}.
Answer: V is not a vector space as (1, 0, 0), (0, 1, 0) ∈ V but (1, 0, 0) + (0, 1, 0) = (1, 1, 0) ∈
/
V.
(24) V := {(x, y, z) ∈ R3 |y = 5x}.
Answer: V is a vector space. Verify!
(25) ] V := {(x, y, z)R3 |z = 0, 2x + 4y = 1}.
Answer: NO, V is not a vector space. Observe that (0, 0, 0) ∈
/ V , which is the zero element
in usual addition.
(27) It is a vector space.
6 V := {x ∈ R|x > 0}.
Answer: V is a vector space under given vector addition and scalar multiplication.
Observe that zero element is 1 and inverse of an element x is x1 . Now verify all the axioms.