# Linear Algebra: Span, Linear Independence Proof

1. Sep 30, 2012

### miglo

1. The problem statement, all variables and given/known data
Suppose v_1,v_2,v_3,...v_n are vectors such that v_1 does not equal the zero vector
and v_2 not in span{v_1}, v_3 not in span{v_1,v_2}, v_n not in span{v_1,v_2,...v_(n-1)}
show that v_1,v_2,v_3,....,V_n are linearly independent.

2. Relevant equations
linear independence, span

3. The attempt at a solution
he gave us a hint, which was to use induction
heres what i have so far
for the base case n=1
v_1 does not equal 0
so for cv_1=0, c must equal 0 making v_1 linearly independent
then assume v_n is linearly independent to show v_(n+1) is linearly independent
since v_n is linearly independent, then v_1,v_2,v_3,v_(n-1) are all linearly independent as well, my books states this as a remark to linear independence so i assume i can use it
and v_(n+1) not in span{v_1,...v_n}
therefore c_1v_1+c_2v_2+....+c_nv_n+c_(n+1)v_(n+1)=0 if either
c_(n+1)v_(n+1)=-c_1v_1-c_2v_2-.....-c_nv_n
or c_(n+1)v_(n+1)=0
the former isnt true since its not in the span of all the vectors before it so then the latter must hold true

this is where i started doubting myself because then i would have to show that v_(n+1) is not zero and im unsure on how to do that, also im a beginner with proofs so im not even sure if im doing this correctly using induction

2. Sep 30, 2012

### Dick

You are really close. You can say v_(n+1) is not the zero vector. The zero vector is in the span of any set of vectors. Try and restate your argument knowing that.

Last edited: Sep 30, 2012
3. Sep 30, 2012

### miglo

so can i just say since the zero vector is in the span of any set of vectors and v_(n+1) is not in the span of all the vectors before it then v_(n+1) is not the zero vector??
if thats correct then c_(n+1) must equal zero thus showing that all the vectors are linearly independent

4. Sep 30, 2012

### Dick

Yes, that's pretty much it. If c_(n+1) is nonzero then v_(n+1) is in the span, contradiction. If c_(n+1) is zero then it shows they are linearly independent. Well done. You are better at proofs than you thought.

5. Sep 30, 2012

cool thanks!