Is a subset of linearly independent vectors indep?

[CookieSalesman]

Homework Statement

True or false.

If v1... v4 are linearly independent vectors in 4D space, then {v1,v2,v3} is also linearly independent.
There's a hint: Think about (c for different constant) cv1+cv2+cv3+0*v4=0

I know that linear independence means there's only the trivial solution... which means that the only constants or "weights" on vectors v1 through v4 are all zero.

But if you don't regard v4, can't the situation change?

Or also... what does this "hint" suggest" I thought that this could mean that the problem could be false...?Please just explain the reasoning...

I guess this all has to be linearly independent... because as is already said, the only weights possible are zero, right? But I'm confused. If you're saying v1 through v4 are linearly indep., the fact that they only have the trivial solution is only because v4 is there as well. If you just solve with v1 v2 v3, doesn't that change the problem...?I'm confused.

(No, I didn't use the template, but then again it didn't really apply, and so I tried to use it in spirit)

 

[Samy_A]

Homework Statement

True or false.

If v1... v4 are linearly independent vectors in 4D space, then {v1,v2,v3} is also linearly independent.
There's a hint: Think about (c for different constant) cv1+cv2+cv3+0*v4=0

I know that linear independence means there's only the trivial solution... which means that the only constants or "weights" on vectors v1 through v4 are all zero.

But if you don't regard v4, can't the situation change?

Or also... what does this "hint" suggest" I thought that this could mean that the problem could be false...?Please just explain the reasoning...

I guess this all has to be linearly independent... because as is already said, the only weights possible are zero, right? But I'm confused. If you're saying v1 through v4 are linearly indep., the fact that they only have the trivial solution is only because v4 is there as well. If you just solve with v1 v2 v3, doesn't that change the problem...?I'm confused.

(No, I didn't use the template, but then again it didn't really apply, and so I tried to use it in spirit)

You gave the correct reason: "I know that linear independence means there's only the trivial solution... which means that the only constants or "weights" on vectors v1 through v4 are all zero."
I just bolded the all.

If $v_1,v_2,v_3$ were linearly dependent, you can find three scalars $c_1, c_2, c_3$, not all 0, such that $c_1v_1+c_2v_2+c_3v_3=0$.
But then you also have $c_1v_1+c_2v_2+c_3v_3+0v_4=0$. What conclusion can you draw from this last equation?

 

[CookieSalesman]

Oh...

Umm I'm honestly not sure what conclusion to draw.

That v4 is zero...? Or that... they're all linearly dependent? I seriously don't get it... Sorry but I feel you basically restated the hint, and I tried to figure how the hint was relevant earlier but I really didn't see how.

 

[PeroK]

Oh...

Umm I'm honestly not sure what conclusion to draw.

That v4 is zero...? Or that... they're all linearly dependent? I seriously don't get it... Sorry but I feel you basically restated the hint, and I tried to figure how the hint was relevant earlier but I really didn't see how.

It seems to me that you are being confused by the mathematical side of this. The problem itself is actually quite easy, but it's the mathematical terminology that is confounding you. Let me rephrase the question non-mathematically:

If you were given four things and tried to do something with them, but found you couldn't. Then, you took one away and tried to do the same task with the three that remain, it would be incredible if you achieve with three what you couldn't achieve with all four. After all, if you could do it with three, you could do it with all four, just by leaving the one you don't need on the side.

I don't know if this will help, but think of linear dependence as the task of putting a linear combination of vectors together to make 0. And, linear independence as the impossibility of achieving that task.

 

[HallsofIvy]

If a subset is independent, that means that the only linear combination that is equal to 0 is with all coefficients equal to 0. If that were NOT true, there would be such a linear combination that has some coefficients non-zero. Add the other vectors, in the original set, to that linear combination with 0 coefficients. Do you see how that gives a contradiction?

 

[CookieSalesman]

Oh, I think I understand. So it seems trivial, although I didn't completely understand your last sentence.
But basically that if the original set of four were linearly independent, since v4 already had a 0 with it there's absolutely no different removing it.

Any subset of lin indep vectors are basically lin indep? Or at least all have 0 weights.

 

[Samy_A]

Any subset of lin indep vectors are basically lin indep?

Yes.

Informally, you can say that the more vectors you have, the easier it becomes to find scalars (what you call the weights) that will give a linear combination of these vectors equal to 0.
Conversely, if you can't find such scalars for 4 given vectors, surely you won't be able to find them if you have only 3 of these vectors.

 

[CookieSalesman]

Thanks everyone