# Reducing the set

• Sep 30th 2009, 06:50 PM
zodiacbrave
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.
• Sep 30th 2009, 07:28 PM
Bruno J.
Just take any three which are linearly independent! I can see more than one such triple.
• Sep 30th 2009, 07:42 PM
zodiacbrave
could you give a little more detailed answer? I guess the numbers just sort of blur together for me
• Oct 1st 2009, 05:53 PM
PolyMECABAC
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!
• Oct 2nd 2009, 02:54 PM
HallsofIvy
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.
• Oct 7th 2009, 10:30 AM
PolyMECABAC
Thanks!
• Oct 7th 2009, 11:02 AM
tonio
Quote:

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