1. ## Reducing the set

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.

2. Just take any three which are linearly independent! I can see more than one such triple.

3. could you give a little more detailed answer? I guess the numbers just sort of blur together for me

4. ## explanation?

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!

5. As Bruno J. suggested, there are many correct answers to this. Do you know what "independent" means? If so, Bruno J.'s answer should be enough.

6. Thanks!

7. Originally Posted by PolyMECABAC
Thanks!

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