Does Scalar Multiplication Prove Zero Vector or Zero Scalar?

[Saladsamurai]

Man. I have always taken a statement like this for granted, so I have no idea how to prove it.

Homework Statement

$$\text{Prove that if }a\in\mathbf{F}\text{ and }v\in V\text{ and }av=0\text{ then a=0 or v=0}$$

Seriously? What about saying

$$\text{Let }v=(x_1,...,x_n) \text{ be a vector space over \textbf{F} and }a\in\mathbf{F}$$

Suppose that $$av=0$$

then

$$a(x_1,...,x_n)=\mathbf{0}\Rightarrow (ax_1,...,ax_n)=(0,...,0)$$

therefore a=0 or v=0.Seems like I just said a whole lot...but I am not sure if I actually proved anything. Is using the definition of scalar mult over V enough? Or am I missing something else?

 

[Office_Shredder]

v is an arbitrary vector in an arbitrary vector space, it doesn't need to be in Fⁿ

Start like this: If a=0, then av=0 (prove this). Then, assume a is NOT zero, and av=0. Can you prove v is the zero vector?

 

[snipez90]

First of all, you have a slight typo in your proof. In the second to last line, you have av = (x_1, ..., x_n), which is probably not what you intended.

While your steps can be easily turned into a proof, your last line (beginning with "therefore") does not justify the claim. One way to complete the proof is by simple case examination. If a = 0, then you are done, since you end up with (0*x_1, ... , 0*x_n) = (0, ..., 0). Now what if a =/= 0? Remember, you need to use the properties of a field to justify your conclusions, or else reaching the last equality before your last line is kind of pointless.

 

[Saladsamurai]

v is an arbitrary vector in an arbitrary vector space, it doesn't need to be in Fⁿ

Oh..I was thinking that Fⁿ is how we represent an arbitrary vector space. How do we represent an arbitrary vector space?

First of all, you have a slight typo in your proof. In the second to last line, you have av = (x_1, ..., x_n), which is probably not what you intended.

Righto! Typo fixed. Once I figure out the proper way to represent an arbitrary vector space, I will continue.

 

[Saladsamurai]

Okay. I know that it may seem like I am getting hung up on little things, but how do you represent the elements of an arbitrary vector space if not by (x1,...,xn)?

I would just say that V is an arbitrary vector space, but than when I want to actually apply some operation on V, I need to use the elements of V. Know what i mean?

So, if I say that V is an arbitrary vector space then v=(...)<--- what goes in here?

 

[Saladsamurai]

Okay. I know that it may seem like I am getting hung up on little things, but how do you represent the elements of an arbitrary vector space if not by (x1,...,xn)?

I would just say that V is an arbitrary vector space, but than when I want to actually apply some operation on V, I need to use the elements of V. Know what i mean?

So, if I say that V is an arbitrary vector space then v=(...)<--- what goes in here?

Anyone help me out on this notational issue?

 

[HallsofIvy]

You could choose some basis, say ${v_1, v_2, \cdot\cdot\cdot, v_n}$ and then write $v= x_1v_1+ x_2v_2+ \cdot\cdot\cdot+ x_nv_n$. But you don't "need to use the elements of V". That is almost never a good way to prove general statements about vector spaces. You need to use the definitions for a general vector space. The definition of "scalar product" requires (a+ b)v= av+ bv and a(u+ v)= au+ av as well as a(bv)= (ab)v. Suppose a is NOT 0. Then, since a is a non-zero number, it has a reciprocal (1/a). (1/a)(av)= ((1/a)a)v= v. And therefore?

 

[Saladsamurai]

so Halls, it's better to just deal with the symbolic representation of the a vector in the vector space rather than with its elements?

Okay then. Now with regard to your approach to the proof: Suppose that a is non-zero. Therefore (1/a) exists.
(1/a)(av)=((1/a)a)v=v
I really don't see where we are going with this v=v ... how does that help?

 

[Office_Shredder]

If av=0_V (the zero vector in V)

then

1/a(av) = (1/a)0_v

Hence

v=?

 

[Saladsamurai]

Okay. Halls. Is this what you mean?

$$\text{Assume that a and v }\ne 0\text{ therefore }1/a \text{ exists. Also assume that av=0}$$

Multiplying both sides by 1/a we have:

(1/a)av=0
((1/a)a)v=0
v=0 which is a contradiction.

Does this do it? I really have no idea how to use a contradiction.

 

[jgens]

I don't think that's what he's looking for. Clearly, if av = 0 and a = 0 the proof is trivial. Now suppose that a != 0. If av = 0 then (a^-1)(a)(v) = 0(a^-1) . . .

 

[Saladsamurai]

I don't think that's what he's looking for. Clearly, if av = 0 and a = 0 the proof is trivial. Now suppose that a != 0. If av = 0 then (a^-1)(a)(v) = 0(a^-1) . . .

I don't see how this is any different from what I wrote?

Okay. Halls. Is this what you mean?

$$\text{Assume that a and v }\ne 0\text{ therefore }1/a \text{ exists. Also assume that av=0}$$

Multiplying both sides by 1/a we have:

(1/a)av=0
((1/a)a)v=0
v=0 which is a contradiction.

Does this do it? I really have no idea how to use a contradiction.

 

[jgens]

It's not really that much different, it's just clearer and more concise if you don't do this with a contradiction. You don't need to assume that v != 0, it should follow from the proof.

 

[Saladsamurai]

You don't need to assume that v != 0, it should follow from the proof.

AH! Okay! I am with you now. And then i assume I would just do the same thing we v, yes?

 

[jgens]

Yup! That should work out.

 

[Saladsamurai]

Thanks everyone. Get ready for a lot more of these!

 

[HallsofIvy]

I don't think that's what he's looking for. Clearly, if av = 0 and a = 0 the proof is trivial. Now suppose that a != 0. If av = 0 then (a^-1)(a)(v) = 0(a^-1) . . .

I don't see how this is any different from what I wrote?

What's different is that he did NOT say "Assume that a and V $\ne 0$" as you did before. Assume that $a\ne 0$ so that 1/a exists. (1/a)(av)= ((1/a)a)v= v= 0. And, since av= 0, that is the same as (1/a)(0)= 0 so v= 0. That's all. Either a= 0 or $a\ne 0$ and then v= 0.