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December 11th, 2014, 12:17 PM  #1 
Newbie Joined: Nov 2014 From: earth Posts: 4 Thanks: 0  why if the determinant is zero then not span V
If we want to see if a set of vectors spans a vector space V, then lets say the set A spans a vector space V only If every linear combination of A produces V, then Span(A)=V if we forme the coeficient matrix of the system formed by c1s1+..+cnsn=u where si are the vectors in set A and u is any vector in V , if the determinant is zero, then there is at least one choice of u for which this system will not have a solution and hence can not be written as a linear combination of these vectors?why is this? Note: i know that if determinant is nonzero then it will have exactly on solution for each u.i know that if the determinant is zero then it can have infinitily many solutions or no solution. Supose that the coefficient matrix is an n*n matrix In other words, if the determinant is a nonzero each u will have exactly one solution if it is zero what happens?and why? Note : i tried to edit the other post but i couldnt and i also dont know how to delete the other thread 

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span(A∪B) = span(A) span(B),span and determinant,how do you know when to use a determinent to solve for a span,what does it mean when the determinant is 0 and span,if determinent is zero then vector is,Determinant of a span set is zero?
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