Vector space and fields question

[ilyas.h]

Homework Statement

Let V be a vector space over the field F. The constant, a, is in F and vectors x, y in V.

(a) Show that a(x - y) = ax - ay in V .

(b) If ax = 0_V show that a = 0_F or x = 0_V .

Homework Equations

axiom 1: pv in V, if v in V and p in F.
axiom 2: v + v' in V if v, v' in V

The Attempt at a Solution

(a) x - y = x + (-y)====> (x+ (-y)) in V (axiom 2)
====> a(x + (-y)) in V (axiom 1)(b)...?

how would I start part (b)?

 

[Orodruin]

How about starting by assuming that both a and x are different from their resoective zeros and trying to find a contradiction?

 

[ilyas.h]

How about starting by assuming that both a and x are different from their resoective zeros and trying to find a contradiction?

if they are both non-zero then they cannot bultiply together to give a zero vector i.e. either one (or both) have to be their zero's.

I feel as though I am missing something, my solution is too simple.

 

[Orodruin]

Yes, but why can they not multiply to zero if they are both non-zero? This is the entire question.

 

[ilyas.h]

Yes, but why can they not multiply to zero if they are both non-zero? This is the entire question.

just looked up all the axioms from a Field and a vector space, from what I can see:

a Field doesn't have the multiplicative axiom: 0a = 0. On the other hand, a vector space does have this multiplicative axiom: 0v = 0.

(b) If ax = 0_V show that a = 0_F or x = 0_V .

ax =/= 0_F because that would assume that an element in the Field contains this specific multiplicative axiom.

ergo, ax = 0_V iff a = 0_F or x = 0_V

 

[Orodruin]

Field doesn't have the multiplicative axiom: 0a = 0

Note that, while not an axiom, it is trivial to show that this is true from the distributive property of multiplication.

ax =/= 0_F because that would assume that an element in the Field contains this specific multiplicative axiom.

I am not really sure what you are trying to argue here.

 

[ilyas.h]

Note that, while not an axiom, it is trivial to show that this is true from the distributive property of multiplication.
I am not really sure what you are trying to argue here.

a in F. v in V. Let's refer to it as an axiom for the sake of simplicity.

since a is in the field, you cannot have 0a = 0, because this multiplicative axiom is not in any Field, given the axioms in a Field.

Therefore, ax =/= 0_F, because if x=0_V, then you have a0 = 0_F, and this is a contradiction (given the axioms in a Field).

Therefore, ax = 0_V, because a vector space can contain the axiom 0v = 0, and it holds when either a = 0_F or v = 0_V.
I could be wrong, or perhaps i didnt explain it properly.

 

[Orodruin]

you cannot have 0a = 0

As I mentioned, the relation 0a = 0 is true for all fields. You have that:
(b + 0)a = (b)a = ba
But at the same time, the distributive property gives
(b + 0)a = ba + 0a
Adding the additive inverse of ba to both these relations gives
ba + (-ba) = 0 = ba + 0a + (-ba) = 0a
And therefore 0a = 0.

 

[ilyas.h]

As I mentioned, the relation 0a = 0 is true for all fields. You have that:
(b + 0)a = (b)a = ba
But at the same time, the distributive property gives
(b + 0)a = ba + 0a
Adding the additive inverse of ba to both these relations gives
ba + (-ba) = 0 = ba + 0a + (-ba) = 0a
And therefore 0a = 0.

i misinterpreted your reply.

If that's the case, could we use the inverse?

assume a and x are non-zero's:

ax = 0_V

ax*a^-1 = 0_V * a^-1
x = 0_V

contradiction. x must equal 0_V.

on the other hand, there is no multiplicative inverse axiom within a vector subspace, therefore if

ax = 0_V

you CANNOT do:

ax*x^-1 = 0_V x^-1

i.e. you cannot eliminate the x on the LHS, ergo:

ax = 0_V (=/= 0_F), iff a = 0_F or x = 0_V

 

[Orodruin]

ax*a^-1 = 0_V * a^-1
x = 0_V

This is sufficient. You do not need to worry about the existence or non-existence of a multiplicative inverse for V. You assumed the anti-hypothesis of what you wanted to prove, i.e., that both were non-zero, and arrived at a contradiction, thus proving that the anti-thesis is false, meaning that the hypothesis is true.