## brinethery 3 years ago Let v1, v2, ..., Vn be a list of nonzero vectors in a vector space V such that no vector in the list is a linear combination of its predecessors. Show that the vectors in the list form an independent set.

1. rulnick

It's almost self-evident from the definition of independence, but here goes. A = no vector is a linear comb of its predecessors B = vectors are indep We will show A => B by showing not B => not A. Assume the vectors are not indep. Then there exists at least one vector that can be expressed as a linear combination of others. Call it v*. If all of the others in the linear combination are predecessors of v* then we are done, so assume not: assume at least one vector in the linear combination follows v*. Then this vector is a linear combination of the others together with v*, which contradicts A. Hence not B => not A, so A => B.

2. brinethery

I just read this twice, and I'll have to really read these definitions and proofs carefully. Thank you for your detailed answer, I appreciate it.

3. rulnick

No problem. It may help to think of this in the case of just three vectors. For example ...

4. rulnick

If you have v1, v2, and v3, and they are dependent, then a v1 + b v2 + c v3 = 0. Which means -a v1 + -c v3 = b v2 ( v2 is a lin comb, but not of its predecessors) but then -a v1 + -b v2 = c v3 (so v3 *is* a lin comb of its predecessors).