Linearly Independent Sets and Bases

  • Thread starter Hashmeer
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  • #1
Hashmeer
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Homework Statement


So I'm trying to find a basis for the space that is spanned by the given vectors.

{(1,0,0,1) (-2,1,-1,1) (6,-1,2,-1) (5,-3,3,-4) (0,3,-1,1)} These are written as column vectors.


Homework Equations


None really (that I know of)


The Attempt at a Solution


So I think I need to check to see if any of these vectors are linear combinations of the others and then remove those vectors. I'm kinda confused by the whole Basis idea maybe if someone can explain that it will help me understand where I need to be headed with this problem.
 

Answers and Replies

  • #2
LAHLH
409
1
You need to start by thinking about [tex] \alpha_1\vec{V_1}+\alpha_2\vec{V_2}+............+\alpha_n\vec{V_n}=0 [/tex], where the V's are your n vectors, and alphas are just some coefficients.

Now if this equation has a solution (other than the trivial one of all the [tex] \alpha[/tex]'s being zero) then it means that is possible to write one or more of your vectors in terms of the others in the set (Think about this. You can just rearrange the equation and solve for a particular vector in terms of some of ther others...)

This means your vectors are not linearly independent. You keep doing this reducing your set of vectors until you get to the smallest set of vectors for which the solution to the above equation can only be the trivial alphas are zero one. The dimension of a vector space is then the maximum number of these LI vectors, and they are said to 'span' the vector space.

So you can start out by writing [tex] \alpha_1 (1,0,0,1)+\alpha_2 (-2,1,-1,1)+.....=0 [/tex]
 

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