Prove or Disprove.

Steven G said:
What is the problem here. c1v1 + c2v2 +...+ ckvk = 0 implies that c1=...=cn=0 since v1,...,vn is linear independent set. Now if b=0 as well we have a problem. What is that problem?
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Then it means that {v1...vk, u} is also linearly independent even though we assumed it to be linearly dependent.
 
Yes, you need to show that b is not 0.

Now either b=0 or b is not 0. If you could show that there is a contradiction if b=0, then you must conclude that b =/=0
 
Steven G said:
What is the problem here. c1v1 + c2v2 +...+ ckvk = 0 implies that c1=...=cn=0 since v1,...,vn is linear independent set. Now if b=0 as well we have a problem. What is that problem?
Click to expand...
If c1v1 + c2v2 +...+ ckvk = 0 implies that c1=...=ck=0 since v1,...,vk is a linear independent set, if b=0 as well, doesn't it imply that {v1...vk, u} is a linearly independent set as well since all scalars c1=...=ck=b=0, and this contradicts the assumption that {v1...vk, u} is a linearly dependent set?
 
Steven G said:
You really have to start thinking more about what is going on.
Why must b=0?
What does that imply?
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Or perhaps it is related to the fact that since we supposed {v1...vk, u} is a linearly dependent set, if b=0, then u cannot be written as a linear combination of v1...vk?
Or is this wrong as well, because not all vectors can necessarily be written as linear combination of the other vectors in a linearly dependent set?

So basically the fact that if b=0, then u cannot be written as a linear combination of the other vectors v1..vk does not contradicts the presumption that v1...vk, u is a linearly dependent set?
 
pka said:
Suppose that [imath]\gamma_0 {\bf u}+\gamma_1 V_1++\gamma_2 V_2+\cdots+\gamma_k V_k=\bf 0[/imath]
What would it imply if [imath]\gamma_0\ne 0~\&~(\exists j\ge 1)[\gamma_j\ne 0]~?[/imath]
[imath][/imath][imath][/imath]
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It would imply that the set {v1...vk, u} is linearly dependent, since at least two coefficients in the linear combination are nonzero.
 
If b=0, then c1, c,2,..., ck MUST BE ALL ZERO. Then the set v1,v2,...,vn, u is a L.I. set.
 
TheWrathOfMath said:
It would imply that the set {v1...vk, u} is linearly dependent, since at least two coefficients in the linear combination are nonzero.
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Does that mean that [imath]{\bf{u}}\in\text{span}\left\{v_1,v_2\cdots v_k\right\}~?[/imath]

[imath][/imath][imath][/imath][imath][/imath]
 
Here is your proof.
I will show you this proof so that you know what Linear Algebra is all about. It is about putting ideas together.

Proof:
We will proceed by proving the contrapositive of the theorem.
We are given that v1...vk is a set of Linear Independent vectors
Suppose {v1...vk, u} is a linearly dependent set.
(we need to show that u can be written as a linear combination of {v1,...,vk}

Case 1: u=0.
Then u = 0v1 + ... +0vn.
(So we wrote u as a linear combination of {v1,...,vn})

Case 2: u is not the zero vector.
Since v1...vk is a set of Linear Independent vectors we know that the only linear combination that sums to the zero vector is the trivial one.
(**)Since {v1...vk, u} is a linearly dependent set we know that there exists c1, ..., ck, b where c1,...,ck, b are not all 0, such that
c1v1 + c2v2 +...+ cmvk + bu = 0.

Claim b =/= 0. Suppose b=0.
Then c1v1 + c2v2 +...+ cmvk + bu =c1v1 + c2v2 +...+ cmvk= 0. But c1v1 + c2v2 +...+ cmvk= 0 implies c1=...=ck =0 since {v1,...,vk} is given to be linear independent. So ALL the scalars c1=....=ck=b=0. But by ** this is not true. So b=/=0

We know (from above) that c1v1 + c2v2 +...+ cmvk + bu = 0 with b=/0. Now you can write u as a Linear Combination of v1,...vn.

Combining both cases we see that we can write u as a linear combination of {v1,...vk}.

Is this clear?
 
Steven G said:
Here is your proof.
I will show you this proof so that you know what Linear Algebra is all about. It is about putting ideas together.

Proof:
We will proceed by proving the contrapositive of the theorem.
We are given that v1...vk is a set of Linear Independent vectors
Suppose {v1...vk, u} is a linearly dependent set.
(we need to show that u can be written as a linear combination of {v1,...,vk}

Case 1: u=0.
Then u = 0v1 + ... +0vn.
(So we wrote u as a linear combination of {v1,...,vn})

Case 2: u is not the zero vector.
Since v1...vk is a set of Linear Independent vectors we know that the only linear combination that sums to the zero vector is the trivial one.
(**)Since {v1...vk, u} is a linearly dependent set we know that there exists c1, ..., ck, b where c1,...,ck, b are not all 0, such that
c1v1 + c2v2 +...+ cmvk + bu = 0.

Claim b =/= 0. Suppose b=0.
Then c1v1 + c2v2 +...+ cmvk + bu =c1v1 + c2v2 +...+ cmvk= 0. But c1v1 + c2v2 +...+ cmvk= 0 implies c1=...=ck =0 since {v1,...,vk} is given to be linear independent. So ALL the scalars c1=....=ck=b=0. But by ** this is not true. So b=/=0

We know (from above) that c1v1 + c2v2 +...+ cmvk + bu = 0 with b=/0. Now you can write u as a Linear Combination of v1,...vn.

Combining both cases we see that we can write u as a linear combination of {v1,...vk}.

Is this clear?
Click to expand...
Crystal.

But why is the explanation I provided for why b cannot be equal to zero flawed, then?
Perhaps I needed to be slighty clearer, but it appears to me that the overall idea I had was true.

I wrote: "Then it means that {v1...vk, u} is also linearly independent even though we assumed it to be linearly dependent."

Similarly, I wrote in a different comment that:
"If c1v1 + c2v2 +...+ ckvk = 0 implies that c1=...=ck=0, since v1,...,vk is a linearly independent set, if b=0 as well, doesn't it imply that {v1...vk, u} is a linearly independent set as well since all scalars c1=...=ck=b=0, and this contradicts the assumption that {v1...vk, u} is a linearly dependent set?" (Therefore b must not be equal to zero so that not all scalars are equal to zero).

Why is my overall idea flawed?
 
Just because c1v1 + c2v2 + ... + ckvk + bu =0 you can not conclude that {v1, v2, ..., vk, u} is a Linear Independent set. I hope that is not what you are saying.
Given k+1 vectors, can you think of a linear combination of these vectors thad yield 0.
 
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