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October 5th, 2019, 02:56 AM  #1 
Newbie Joined: Oct 2019 From: Hong Kong Posts: 1 Thanks: 0  Span and linear dependence
Dear everyone, l really get confused when reading some textbooks on linear algebra : Suppose there are two vectors in R3, u= (3,1,0) and v = (1,6,0). Firstly, u and v are linearly independent because neither vector is a multiple of the other. IF w is a linear combination of u and v, then {u,v,w} is linearly dependent and w is in span {u,v}. So , it joined to conclusion that any set {u,v,w} in R3 with u, and v linearly independent. The set {u,v,w} will be linearly dependent if and only if w is in the plane spanned by u and v . l wonder if this theorem is true if x[SUB]3[/SUB] is not equal to 0 for the above case. (That is not a x1x2 plane with x3 =0) Besides, are the following statements true ? 1) If x and y are linearly independent, and if {x,y,z} is linearly dependent, then z is in Span {x,y} 2) lf x and y are linearly independent , and if z is in Span {x,y}, then {x,y,z} is linearly dependent. Now for a practice problem: There are four vectors u= (3,2,4) v= (6,1,7) w= (0,5,2) z= (3,7, 5) It shows that each pairs of the above vectors are linearly independent because neither vector is a multiple of the other. But, {u,v,w,z} is linearly dependent, because there are more vectors than entries in them. So my question arised, how could{u,v,w,z} be linearly dependent when w is NOT in span{u,v,z} ? According to the definition of linearly dependent, the vector set need to have at least one vector which is a linear combination of the others. So is the above example means something like u is in span {v,z,w} or v = x1u+ x2z + x3w? 
October 5th, 2019, 06:01 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 21,035 Thanks: 2271 
In your example, z = 3u + v.


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