Just take any three which are linearly independent! I can see more than one such triple.
Reduce the set { [ 1 2 3], [1 1 1], [2 0 1], [1 0 0] , [1 0 1] , [-2 0 -1], [0 0 0] [3 3 3] } to a basis of r^3
if I stare at it long enough I can sort of see patterns where some vectors are simply combination of others with different scalars but is there a more precise method? I realize there are 3 independent vectors and 5 dependent, I just can't seem to figure it out.
Can anyone explain this more accuratly?
I know there is such a thing as a rejection rule where you apply a factor to all vectors and and their sum and it's suppose to give 0
0= k[ 1 2 3]+ s[1 1 1]+ t[2 0 1] +c[1 0 0] + b[1 0 1] + a[-2 0 -1]+ v[0 0 0]+ p[3 3 3]
and you make a system
like so
k+s+2t+c+b-2a+0v+3p=0
2k+s+0t+0c+0b+0a+0v+3p=0
3k+s+t+0c+b-a+0v+3p=0
but i can't seem to understand what to do beause there will be no answer to this since unknown's are greater thant the number of equations. sorry if my terms aren't correct, I speak french and I didn't learn the "english" math terms...
thanks again!
An idea together with a rather important though sometimes missed lemma:
A set of vectors is linearly dependent iff there is one vector which is a lin. combination of the ones PRECEEDING it.
The above means that no matter how you order a set of vectors, if this set is lin. dependent then there must be a vector lin. dep. on the previous ones.
Well, in your case now just begin checking from left to right: clearly the 2nd vector is not lin. dep. on the first one (it is not a scalar multiple of it).
Now check the 3rd one is not a lin. comb. of the first two and you're done.
Tonio