How Can We Prove Linear Dependence in Vector Spaces?

[autre]

I have to prove:

Let u_{1} and u_{2} be nonzero vectors in vector space U. Show that {u_{1},u_{2}} is linearly dependent iff u_{1} is a scalar multiple of u_{2} or vice-versa.

My attempt at a proof:

(\rightarrow) Let {u_{1},u_{2}} be linearly dependent. Then, \alpha_{1}u_{1}+ \alpha_{2}u_{2}=0 where \alpha_{1} \not= \alpha_{2}...I'm stuck here in this direction

(\leftarrow) Fairly trivial. Let and u_{1} = -u_{2}. Then \alpha_{1}u_{1}+ \alpha_{2}u_{2}=0 but \alpha_{1} \not= \alpha_{2}.

Any ideas?

 

[spamiam]

Then, \alpha_{1}u_{1}+ \alpha_{2}u_{2}=0 where \alpha_{1} \not= \alpha_{2}...I'm stuck here in this direction

Look at the definition of linear dependence again. That's not what linear dependence tells you about the scalars. It tells you that \alpha_1 and \alpha_2 are not both...? Fixing this definition will also help finish the proof.

(\leftarrow) Fairly trivial. Let and u_{1} = -u_{2}. Then \alpha_{1}u_{1}+ \alpha_{2}u_{2}=0 but \alpha_{1} \not= \alpha_{2}.

Maybe I'm missing something, but you can't just assume that u_1 = -u_2 to prove the reverse direction. You're only given that one is a scalar multiple of the other, so you only know u_1 = c u_2 for some scalar c.

 

[gordonj005]

"\rightarrow"

\alpha_{1}u_{1}+ \alpha_{2}u_{2}=0

looking at the definition, what is the condition on \alpha_{1} and \alpha_{2} for {u_{1}, u_{2}} to be linearly dependant?

"\leftarrow"

in this part you have to assume u_{1} = c u_{2}, perhaps the negative you put in your original will give you a hint as to what to do for the first part.

 

[autre]

Thanks for the input guys.

Look at the definition of linear dependence again.

(\rightarrow) Let {u_{1},u_{2}} be linearly dependent. Then, \alpha_{1}u_{1}+ \alpha_{2}u_{2}=0 where \alpha_{1}, \alpha_{2} are not both 0. Therefore, if \alpha_{1}u_{1}+ \alpha_{2}u_{2}=0, \alpha_{1}u_{1} = -\alpha_{2}u_{2}. Is that good?

For part 2:

(\leftarrow) Let and u_{1} = cu_{2}. Why does that mean that the coefficients aren't both 0?

 

[gordonj005]

now you're getting somewhere for the "\rightarrow" part.
so if \alpha_{1}u_{1} = -\alpha_{2}u{2} where either \alpha_{1} or \alpha_{2} is non zero (or maybe both are non-zero), what can you do now that you couldn't before?

for the second part, you have to somehow relate u_{1} = c u_{2} to your findings from part one.

 

[autre]

If \alpha_{1}u_{1} = -\alpha_{2}u_{2}, then one is a scalar multiple of another as required by the direction, right? What more do I need to do?

 

[gordonj005]

I know it seems obvious, but you have to explicitly state:
assume one of \alpha_{1}, \alpha_{2} is non zero (by the definition of linear dependance). for the sake of argument we take \alpha_{1} to be the non zero coefficiant, and since it is non-zero we can divide both sides by that coefficiant.
which leads us to : u_{1} = \frac{-\alpha_{2}u_{2}}{\alpha_{1}}. therefore if the set {u_{1}, u_{2}} is linearly dependant, one must be a scalar multiple of the other as desired.

the "\leftarrow" is just a reversal of "\rightarrow"
its a lot more powerful to prove a set of n elements than one of just 2, if you're looking for good practice i'd suggest trying that.

 

[spamiam]

For part 2:

(\leftarrow) Let and u_{1} = cu_{2}. Why does that mean that the coefficients aren't both 0?

So now you need to find scalars \alpha_1, \alpha_2 not both zero such that \alpha_1 u_1 + \alpha_2 u_2 =0. Can you see a way to use the information u_1 = c u_2 to choose scalars so this is true? Try rearranging the equation in your post.

 

[Deveno]

as gordonj005 pointed out, the proof of (→) breaks down into 2 cases.

you can avoid this difficulty by noting that, in point of fact:

\alpha_1u_1 = -\alpha_2u_2 \implies \alpha_1,\alpha_2 \neq 0 since, for example:

\alpha_1 = 0 \implies -\alpha_2u_2 = 0 \implies \alpha_2 = 0 since u_2 \neq 0.

so you are free to divide by α₁ or α₂.

you almost had the (←) in your first go-round. your mistake was this: assuming α₁ = 1. just use "c" where c is the multiple of u₁ that u₂ is.

why do you know that c ≠ 0 (because u₁ is _______)?

 

[autre]

why do you know that c ≠ 0 (because u1 is _______)?

a non-zero vector! Just curious, how would I go about this part of the proof if it weren't specified that u1, u2 were nonzero vectors?

 

[Deveno]

suppose u1 = 0. then {u1,u2} is linearly dependent no matter what u2 is:

au1 + 0u2 = 0, for any non-zero value of a.

the same goes if u2 = 0.

so the statement:{u1,u2} is linearly dependent iff u1 is a scalar multiple of u2 (and vice-versa), is no longer true.

however, in actual practice, no one ever tries to decide if the 0-vector is part of a basis, because including it automatically makes a set linearly dependent. so one just wants to decide if a set of non-zero vectors is linearly independent or not.