Linearly Independent Sets and Bases

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SUMMARY

The discussion focuses on determining a basis for the vector space spanned by the column vectors {(1,0,0,1), (-2,1,-1,1), (6,-1,2,-1), (5,-3,3,-4), (0,3,-1,1)}. The key method involves checking for linear combinations among the vectors to identify and remove any that are not linearly independent. The process continues until the smallest set of vectors is found, where the only solution to the equation α₁V₁ + α₂V₂ + ... + αₙVₙ = 0 is the trivial solution (all α's equal to zero). This smallest set defines the basis and the dimension of the vector space.

PREREQUISITES
  • Understanding of linear independence in vector spaces
  • Familiarity with vector notation and operations
  • Knowledge of basis and dimension concepts in linear algebra
  • Ability to solve linear equations involving vectors
NEXT STEPS
  • Study the concept of linear combinations in vector spaces
  • Learn about the Gaussian elimination method for determining linear independence
  • Explore the definition and properties of vector space dimensions
  • Investigate the Gram-Schmidt process for orthogonalizing a set of vectors
USEFUL FOR

Students of linear algebra, educators teaching vector space concepts, and anyone interested in understanding the foundations of linear independence and basis in mathematical contexts.

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.
 
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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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