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anonymous
 5 years ago
Anyone good with vector spaces?
anonymous
 5 years ago
Anyone good with vector spaces?

This Question is Closed

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0a collection of vectors that contains theta must be linearly dependent. Thus theta cannot be contained in a basis. How do you go about proving this? I know that

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0...I know that to be linearly dependent there must be scalars that are not zero in the collection of vectors, Does theta mean zero, or some sort of value?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0...theta is also zero...and the collection of vectors must = zero in the end

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You have to prove some axioms regarding vector spaces. Suppose you had \[\{R^2= <x,y>  x,y \in R\}\] Define Addition as: \[ <x_1,y_1> + <x_2,y_2> = <x_1+x_2,y_1+y_2>\] Then: \[ <x_1,y_1> + <x_2,y_2> = <x_1+x_2,y_1+y_2>=<a,b>\] \[\in R= <x,y>\] Therefore, closed under addition, since the addition of two elements = an element inside \[R^2\]. Rinse and repeat for the other ones.
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