Prove this set of vectors is dependant

**jayshizwiz** · Jul 4th 2010, 07:55 AM

(v1,v2 ,...,vk ,w) are vectors in vector space V

{ v1 − w,v2 − w,...,vk − w } spans V and w∉Sp{v1,v2 ,...,vk}

prove { v1,v2 ,...,vk } is dependant.

Attempt

I think it has something to do with this:

since { v1 − w,v2 − w,...,vk − w } - containing k vectors - spans V

that means (v1,v2 ,...,vk ,w) - containing k+1 vectors - are dependent

and maybe since w∉Sp{v1,v2 ,...,vk}

that means that in (v1,v2 ,...,vk ,w) w is not a linear combination of (v1,...,vk)

which then means that { v1,v2 ,...,vk } is dependent

What do you guys think??

**dwsmith** · Jul 4th 2010, 09:59 AM

Check out the 2nd post in the sticky. It may help you.

**HallsofIvy** · Jul 4th 2010, 12:23 PM

Originally Posted by jayshizwiz

(v1,v2 ,...,vk ,w) are vectors in vector space V

{ v1 − w,v2 − w,...,vk − w } spans V and w∉Sp{v1,v2 ,...,vk}

prove { v1,v2 ,...,vk } is dependant.

Attempt

I think it has something to do with this:

since { v1 − w,v2 − w,...,vk − w } - containing k vectors - spans V

that means (v1,v2 ,...,vk ,w) - containing k+1 vectors - are dependent

and maybe since w∉Sp{v1,v2 ,...,vk}

that means that in (v1,v2 ,...,vk ,w) w is not a linear combination of (v1,...,vk)

which then means that { v1,v2 ,...,vk } is dependent

What do you guys think??

That does NOT follow. The fact that w in not a linear combination of {v1, v2, ..., vk} does not prove that {v1, v2, ..., vk} is dependent. $\vec{k}$ is not a linear combination of $\{\vec{i}, \vec{j}\}$ but they are not dependent. What that means is that {v1, v2, ..., vk, w} are dependent- but you already knew that.

All of this is correct:-

Since {v1-w, v2-w, v3-w, ..., vk- w} spans V, then V is of dimension at most k. Since {v1, v2, v3, ..., vk, w} contains k+1 vectors, it is dependent.

Since w is NOT in sp(v1, v2, ..., vk), w is not a linear combination of (v1, ..., vk).

But you still have to prove your last statement- you cannot just say "that means"!

Why does it mean that {v1, v2, ..., vk} is dependent? It does mean that (v1, v2, ..., vk) does not span V.

A "basis" for an n-dimensional vector space, V, has three properties:
1) it spans V.
2) it is independent.
3) it contains exactly n vectors.

And, most importantly for this problem, if any two of those are true, so is the third!

We know that {v1, v2, ..., vk} contains k vectors and that V has dimension k. So if that set were also independent, it would satify (2) and (3) and so ...

**jayshizwiz** · Jul 5th 2010, 12:38 AM

I am trying to put all the information together. Maybe one of you guys can complete this for me:

{ v1 − w,v2 − w,...,vk − w } spans V

* $\lambda_1(v1-w) + \lambda_2(v2-w) +...+ \lambda_k(vk -w) = w$ since w $\in$ V

We also know that

w∉Sp{v1,v2 ,...,vk}

** $\lambda_1v1 + \lambda_2v2 +...+ \lambda_kvk + \lambda_{k+1}w = 0$

which gives us $\lambda_1 =...\lambda_{k+1} = 0$

but on the other hand since

{ v1 − w,v2 − w,...,vk − w } spans V

the dimension of V is k.

so ** must be linear dependent since there are k+1 vectors.

So shouldn't this amount of info already be enough to tell me that (v1,...,vk) is dependent. since there are k+1 vectors we know that ** is dependent. and since we know that w isn't in sp(v1,...vk) that must mean that v1,...,vk is dependent and has infinite solutions to $\lambda_1v1 + \lambda_2v2 +...+ \lambda_kvk = 0$

There's also this extra biti of info that must be important somehow, I'm just not sure why:

when I open the parethesis in * and switch the order around I get

$\lambda_1v1 + \lambda_2v2 +...+ \lambda_kvk - (\lambda_1w + \lambda_2w +...+ \lambda_kw) = w$

$\lambda_1v1 + \lambda_2v2 +...+ \lambda_kvk = w + \lambda_1w + \lambda_2w +...+ \lambda_kw$

So if someone can complete my ideas that would be soo great! Thanks!

(I'm just a bit nervous since I have a final on Thursday and I still have so much to study)

**HallsofIvy** · Jul 5th 2010, 05:24 AM

Originally Posted by jayshizwiz

I am trying to put all the information together. Maybe one of you guys can complete this for me:

* $\lambda_1(v1-w) + \lambda_2(v2-w) +...+ \lambda_k(vk -w) = w$ since w $\in$ V

We also know that ** $\lambda_1v1 + \lambda_2v2 +...+ \lambda_kvk + \lambda_{k+1}w = 0$

which gives us $\lambda_1 =...\lambda_{k+1} = 0$

but on the other hand since the dimension of V is k.

so ** must be linear dependent since there are k+1 vectors.

So shouldn't this amount of info already be enough to tell me that (v1,...,vk) is dependent. since there are k+1 vectors we know that ** is dependent. and since we know that w isn't in sp(v1,...vk) that must mean that v1,...,vk is dependent and has infinite solutions to $\lambda_1v1 + \lambda_2v2 +...+ \lambda_kvk = 0$

Again, you simply say "that must mean...". WHY must that mean the set is dependent? That is whole point of theproblem.

There's also this extra biti of info that must be important somehow, I'm just not sure why:

when I open the parethesis in * and switch the order around I get

$\lambda_1v1 + \lambda_2v2 +...+ \lambda_kvk - (\lambda_1w + \lambda_2w +...+ \lambda_kw) = w$

$\lambda_1v1 + \lambda_2v2 +...+ \lambda_kvk = w + \lambda_1w + \lambda_2w +...+ \lambda_kw$

So if someone can complete my ideas that would be soo great! Thanks!

(I'm just a bit nervous since I have a final on Thursday and I still have so much to study)

The simplest way is what I said before:

A "basis" for an n-dimensional vector space, V, has three properties:
1) it spans V.
2) it is independent.
3) it contains exactly n vectors.

And, most importantly for this problem, if any two of those are true, so is the third!

We know that {v1, v2, ..., vk} contains k vectors and that V has dimension k. So if that set were also independent, it would satify (2) and (3) and so ...

**jayshizwiz** · Jul 5th 2010, 06:48 AM

The simplest way is what I said before:

A "basis" for an n-dimensional vector space, V, has three properties:
1) it spans V.
2) it is independent.
3) it contains exactly n vectors.

And, most importantly for this problem, if any two of those are true, so is the third!

We know that {v1, v2, ..., vk} contains k vectors and that V has dimension k. So if that set were also independent, it would satify (2) and (3) and so ...

Ok, i finally realized something...since w isn't in sp(v1,...,vk) that must mean (v1,...,vk) does not span V - since w is in V.

O, and since we know k is the dimension of V, if we had a set of k vectors that were linear independent, it would be a basis for V. But since we know (v1,...vk) does not span V, that tell us that they are dependent...

Come on, I think I finally got it...

Agree???

**HallsofIvy** · Jul 6th 2010, 07:39 AM

Originally Posted by jayshizwiz

Ok, i finally realized something...since w isn't in sp(v1,...,vk) that must mean (v1,...,vk) does not span V - since w is in V.

O, and since we know k is the dimension of V

I said earlier "V has dimension k" but now I don't where I got that! We know that a set of k vectors, { v1 − w,v2 − w,...,vk − w }, spans V so the dimension of V is at most v (a basis is the smallest set of vectors that spans V).

Now we have two possibilities. The dimension of V is k, in which case your argument is correct, or the dimension of k is less than k and then it follows that any set of k vectors is dependent (a basis is the largest set of independent vectors).

, if we had a set of k vectors that were linear independent, it would be a basis for V. But since we know (v1,...vk) does not span V, that tell us that they are dependent .

Come on, I think I finally got it...

Agree???

**jayshizwiz** · Jul 6th 2010, 08:16 AM

I said earlier "V has dimension k" but now I don't where I got that! We know that a set of k vectors, { v1 − w,v2 − w,...,vk − w }, spans V so the dimension of V is at most v (a basis is the smallest set of vectors that spans V).

Now we have two possibilities. The dimension of V is k, in which case your argument is correct, or the dimension of k is less than k and then it follows that any set of k vectors is dependent (a basis is the largest set of independent vectors).

So it doesn't matter if the basis is K or k-1... We know the basis is at most K in which case my argument is correct.

But if the basis is less than k, then (v1,...vk) is dependent since there are more vectors than the basis...

So either way, it's dependent...

So we don't need to know if { v1 − w,v2 − w,...,vk − w } is a basis of V. We get enough info just by the fact that it spans V...

Now agree???

**HallsofIvy** · Jul 6th 2010, 04:45 PM

Yes, that is correct.

**jayshizwiz** · Jul 6th 2010, 09:57 PM

Ya!! I'm gonna rock this test!

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