Linear Algebra question:
let [v_1, v_2, v_3] be a set of vectors in m-space. If the transpose of v_i * v_j where i =\= j is zero, prove that the set of vectors is linearly dependent

- anonymous

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- anonymous

okk .. solve this like
a * v1*v2 + b*v1*v3 = 0 (given)
take v1 as common then you get (a*v2+b*v3)*v1 = 0
Assuming that v1 is non zero we get a*v2 + b*v3 = 0 (linearly dependent)

- anonymous

Yeah, forgot to mention that all vectors in the set are assumed to be non-zero

- anonymous

I'm not following that first line of logic. It's given? We know that v1 transpose times any other vector will equal 0 but ...

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- anonymous

but...??

- anonymous

Could you explain how you you get that first statement?

- anonymous

its simple for any i = / j we have v_i*v_j as zero ...
so I would have v1*v2 as zero. I now multilply this with a constant a and we have a*v1*v2 = 0. Similiar is the case for b*v1*v3.
The concept of transpose here is that you cannot multiply two vectors directly. You have to take the transpose of one vector (think .. matrix multiplication). I hope I am clear..

- anonymous

\[V _{i}^{T} * V _{j}\] = 0
means that \[V _{i}*V _{j}\] = 0?

- anonymous

no dude .. its like say you have one column as (1,2,3) and the other as (2,4,6) ... and so if you have to mention v1*v2 in matrix form .. can you directly multiply? they are 3*1 matrices which cannot be multiplied .. you need to take transpose of the first to multiply them ... its more of a convention thing ..

- anonymous

Does the fact that the transpose of one vector multiplied by another giving you zero mean anything special as far as their individual entries go? Because here's the hint they provide in the problem:

- anonymous

"set a1v1 + a2v2 + a3v3 = zero vector and consider (zerovector)^t * zerovector

- anonymous

consider the transpose of the zero vector multiplied by the zero vector? What is that even supposed to mean?

- anonymous

it means that the result is 0 ... imagine a (3*1) 0 matrix and a (1*3) 0 matrix ... you can multiply both right???

- anonymous

yup

- anonymous

are you getting what I am trying to convey here .. ?
forget about matrices ... say you have a vector ai+ bj+ck and other vector as a1i+ b1j+c1k
multiplication would give you aa1 + bb1+ cc1 ...
use this to prove ..

- anonymous

Great, I just realized I have a frickin typo

- anonymous

they want me to prove linear 'independence' not dependence
@#$!

- anonymous

this is a linearly dependent system .. if you have i not equal to j and v_i*v_j = 0

- anonymous

Apparently not, because the question says 'show that the set is linearly independent' (and then everything else applies the same, including the hint and rest of the question)

- anonymous

well ..then you may have another typo in the question .... is it v_i*v_j = 0 for all i not equal to j or is it zero for summation ... if it is for all i not equal to j then this CANNOT be linearly independent...

- anonymous

v_i*v_j for all i not equal to j

- anonymous

thanks for the help anyways man, I appreciate it

- anonymous

Linear Algebra question:
let [v_1, v_2, v_3] be a set of vectors in m-space. If the transpose of v_i * v_j where i =\= j is zero, prove that the set of vectors is linearly in
dependent. All the vectors in the set are non-zero.

- anonymous

What is the definition of linear dependence?

- anonymous

Or independence, either one.

- anonymous

Independence: if solution to the vector equation a1v1 + ... a_n v_n = zero vector is the trivial solution

- anonymous

Okay, so let's assume that they are linearly dependent and then derive a contradiction. That means that
\[a_1 \vec{v_1} + a_2 \vec{v_2} + a_3 \vec{v_3} = 0\]
where the a's are not all zero.

- anonymous

okay

- anonymous

Recall that we take for granted that none of the vectors are the zero vector. What do we get when we take the transpose of a_1 times the whole equation?

- anonymous

I'm sorry, v_1, not a_1

- anonymous

hm...by the parameters of the problem, the only thing that should be left is a_1 v_1 * v_1 ^t
which...does that just equal v_1?

- anonymous

We should get
\[a_1 (v_1^T v_1) = 0\]

- anonymous

okay, yup

- anonymous

furthermore,
\[v_1^Tv_1 \]
is just the length of the vector, is that correct? If so, what do we know this definitely CANNOT be, based on the assumptions we made?

- anonymous

length squared.... :)

- anonymous

Well, it can't be zero, right?

- anonymous

Indeed. So, what must that mean?

- anonymous

a = 0

- anonymous

Perfect. But there's nothing special about v_1, so I could do the same thing with v_2 and v_3, and what conclusion would I be forced to draw about the a's?

- anonymous

All the a's are zero, therefore the trivial solution is the only solution

- anonymous

There ya go :)

- anonymous

Dude, thanks a million

- anonymous

No problem at all, excellent question.

- anonymous

Well, whatever, you know, we're on a math forum, so let's just say my thanks increases without bound

- anonymous

Hahaha good stuff. Good luck with the rest of your work, we'll help you out if you need it.

- anonymous

Thanks again!

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