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 dependent?

"$\leftarrow$"

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 dependent, 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.