Linearly dependent set of vectors?

[sutupidmath]

suppose that $$\{ v_1,v_2,v_3\}$$ is a set of dependent vectors. Prove that $$\{v_1,(v_1+v_2),v_3\}$$ is also a set of dependent vectors?

Ok, I have two ways of going about it. My concern is about the first one, it looks like it is not correct the way i did it, i think so, because i managed to find some kind of counter example, but i cannot see why it doesn't hold,because it looks logical to me.
Method 1.

Since $$\{ v_1,v_2,v_3\}$$ lin. dependent set, we know that the vector equation:

$$av_1+bv_2+cv_3=0$$ has a non trivial solution, that is at least one of a,b,c is a non-zero scalar. Now, let's take a linear combination of the vectors $$\{v_1,(v_1+v_2),v_3\}$$

$$av_1+b(v_1+v_2)+cv_3=0$$ my goal is to show that this eq. has a nontrivial sol. that is that one of, a,b,c must be nonzero. So:

$$av_1+b(v_1+v_2)+cv_3=0=>av_1+bv_1+bv_2+cv_3=0=>(av_1+bv_2+cv_3)+bv_1=0=>0+bv_1=0$$ SO, i concluded then that $$v_1=0, or, b=0$$ Then if $$v_1=0$$ we know that every set that contains the zero vector is dependent, so i concluded that the set $$\{v_1,(v_1+v_2),v_3\}$$ is also dependent. On the other hand, if b=0, then either a, or c must be different from zero, so again this set is lin. dependent.

I am not convinced that this works because of this:

Let$$v_1=[1,2,3]^T,v_2=[2,-1,4]^T,v_3=[0,5,2]^T$$ then none of these vectors is zero, and they are also lin. dependent. so if i try to follow the same logic as above, in here to show that the set $$\{v_1,(v_1+v_2),v_3\}$$ is lin. dependent i end up with a contradiction:

$$bv_2=0$$ but, b is not necessarly zero. SO, where have i gone wrong? I mean why the above doesn't work?

Here is the second method that i used to prove that, this is way shorter, I again took a lin. comb on the vectors in the set:$$\{v_1,(v_1+v_2),v_3\}$$

THat is:$$av_1+b(v_1+v_2)+cv_3=0$$ then after a little manipulation we have:

$$(a+b)v_1+bv_2+cv_3=0$$ now since $$v_1,v_2,v_3$$ are lin.dependent then one of a+b, b, c must be different from zero. SO, we can automatically conclude that

:$$av_1+b(v_1+v_2)+cv_3=0$$ has nontrivial sol, hence it it dependent set.



Please enlighten me, i am kind of confused?

Thnx in advance!

 

[morphism]

suppose that $$\{ v_1,v_2,v_3\}$$ is a set of dependent vectors. Prove that $$\{v_1,(v_1+v_2),v_3\}$$ is also a set of dependent vectors?

Ok, I have two ways of going about it. My concern is about the first one, it looks like it is not correct the way i did it, i think so, because i managed to find some kind of counter example, but i cannot see why it doesn't hold,because it looks logical to me.
Method 1.

Since $$\{ v_1,v_2,v_3\}$$ lin. dependent set, we know that the vector equation:

$$av_1+bv_2+cv_3=0$$ has a non trivial solution, that is at least one of a,b,c is a non-zero scalar. Now, let's take a linear combination of the vectors $$\{v_1,(v_1+v_2),v_3\}$$

$$av_1+b(v_1+v_2)+cv_3=0$$ my goal is to show that this eq. has a nontrivial sol. that is that one of, a,b,c must be nonzero.

It looks like you're using a,b,c for two different things.

Here is the second method that i used to prove that, this is way shorter, I again took a lin. comb on the vectors in the set:$$\{v_1,(v_1+v_2),v_3\}$$

THat is:$$av_1+b(v_1+v_2)+cv_3=0$$ then after a little manipulation we have:

$$(a+b)v_1+bv_2+cv_3=0$$ now since $$v_1,v_2,v_3$$ are lin.dependent then one of a+b, b, c must be different from zero.

Why?

 

[sutupidmath]

It looks like you're using a,b,c for two different things.

Well, yeah, that is confusing, but i didn't mean for a,b,c to be the same things in both places. This means that we can use say r,m,n or any other constant in the second part. But, if we do so, then would what i did there hold?

I'm going to come back, after i have tried this way, i think things will change a little bit, if we don't use the same a,b,c.

 

[sutupidmath]

Why?

Because since v_1,v_2,V_3, are lin. dependent it means that the vect. eq.

(a+b)v_1+bv_2+cv_3=0, has nontrivial solutions.
RIght?

 

[morphism]

Because since v_1,v_2,V_3, are lin. dependent it means that the vect. eq.

(a+b)v_1+bv_2+cv_3=0, has nontrivial solutions.
RIght?

Yes, but that doesn't mean it only has nontrivial solutions!

 

[sutupidmath]

Yes, but that doesn't mean it only has nontrivial solutions!

Well, yeah that is true,but since ther are non-trivial solutions we know that those vectors are lin. dependent, that was my point.

 

[Defennder]

I think mathman's point was that you didn't show how this:

$$(a+b)v_1+bv_2+cv_3=0$$ now since $$v_1,v_2,v_3$$ are lin.dependent then one of a+b, b, c must be different from zero.

implies this:

SO, we can automatically conclude that

:$$av_1+b(v_1+v_2)+cv_3=0$$ has nontrivial sol, hence it it dependent set.

I'll think of it this way:

We know that either or all of a+b, b, c must be non-zero. Suppose $$c \neq 0$$. Then this means that in the original equation $$av_1 + b(v_1+v_2) + cv_3 = 0$$ has non-trivial solutions. We already know that c is nonzero, so it doesn't matter whether a,b are themselves non zero. Hence the equation has non-zero solutions, namely c. The same reasoning applies for b. Supposing b is non-zero, then it doesn't matter what a,c are because it means in the vector equation that there is a non zero solution, namely b.

Based on this reasoning, we see that in order for there to be only the trivial solution, b and c must be zero, and a must also be zero. But a need not be zero. We can see from the modified equation (the one where you grouped coffefficients of vectors together) that if neither b,c are non-zero, then (a+b) has to be non-zero. And that means that $$a \neq -b$$. Since we know that both b,c are necessarily 0, (because otherwise we would have already proven by above that the original equation has non-trivial solutions), then we can conclude that $$a \neq 0$$ after substituting $$b=0$$. And this implies that $$\{v_1,(v_1+v_2),v_3\}$$ are also linearly dependent.