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

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0okk .. 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
 4 years ago
Best ResponseYou've already chosen the best response.0Yeah, forgot to mention that all vectors in the set are assumed to be nonzero

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I'm not following that first line of logic. It's given? We know that v1 transpose times any other vector will equal 0 but ...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Could you explain how you you get that first statement?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0its 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
 4 years ago
Best ResponseYou've already chosen the best response.0\[V _{i}^{T} * V _{j}\] = 0 means that \[V _{i}*V _{j}\] = 0?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no 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
 4 years ago
Best ResponseYou've already chosen the best response.0Does 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
 4 years ago
Best ResponseYou've already chosen the best response.0"set a1v1 + a2v2 + a3v3 = zero vector and consider (zerovector)^t * zerovector

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0consider the transpose of the zero vector multiplied by the zero vector? What is that even supposed to mean?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it means that the result is 0 ... imagine a (3*1) 0 matrix and a (1*3) 0 matrix ... you can multiply both right???

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0are 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
 4 years ago
Best ResponseYou've already chosen the best response.0Great, I just realized I have a frickin typo

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0they want me to prove linear 'independence' not dependence @#$!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0this is a linearly dependent system .. if you have i not equal to j and v_i*v_j = 0

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Apparently 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
 4 years ago
Best ResponseYou've already chosen the best response.0well ..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
 4 years ago
Best ResponseYou've already chosen the best response.0v_i*v_j for all i not equal to j

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thanks for the help anyways man, I appreciate it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Linear Algebra question: let [v_1, v_2, v_3] be a set of vectors in mspace. 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 nonzero.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0What is the definition of linear dependence?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Or independence, either one.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Independence: if solution to the vector equation a1v1 + ... a_n v_n = zero vector is the trivial solution

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Okay, 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
 4 years ago
Best ResponseYou've already chosen the best response.0Recall 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
 4 years ago
Best ResponseYou've already chosen the best response.0I'm sorry, v_1, not a_1

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hm...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
 4 years ago
Best ResponseYou've already chosen the best response.0We should get \[a_1 (v_1^T v_1) = 0\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0furthermore, \[v_1^Tv_1 \] 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
 4 years ago
Best ResponseYou've already chosen the best response.0length squared.... :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Well, it can't be zero, right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Indeed. So, what must that mean?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Perfect. 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
 4 years ago
Best ResponseYou've already chosen the best response.0All the a's are zero, therefore the trivial solution is the only solution

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Dude, thanks a million

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0No problem at all, excellent question.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Well, whatever, you know, we're on a math forum, so let's just say my thanks increases without bound

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Hahaha good stuff. Good luck with the rest of your work, we'll help you out if you need it.
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