Proving Linear Independence

Strategy

1. Assume linear dependence for contradiction
2. Derive a contradiction

Example

1. Consider the Standard Basis at $n=3$. That is: $S=\{(1,0,0),(0,1,0),(0,0,1)\}$
2. Suppose for sake of contradiction that there are non-zero coefficients that allow a sum of 0. $(x)(1,0,0)+(y)(0,1,0)+(z)(0,0,1)=\overrightarrow{0}=(0,0,0)$
3. $⟹(x,0,0)+(0,y,0)+(0,0,z)=(0,0,0)$
4. $⟹(x,y,z)=(0,0,0)$
5. Therefore: $x=y=z=0$
6. This contradicts that a Linear Dependent set has non-zero coefficients for the sum of 0
7. Therefore, there are no Non-trivial vectors that are dependent