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.

Question

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.

rulnick · Answer

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.

anonymous · Answer

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.

rulnick · Answer

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

rulnick · Answer

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).