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October 5th, 2019, 02:56 AM   #1
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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?
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October 5th, 2019, 06:01 AM   #2
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In your example, z = 3u + v.
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