Linear algebra/Linearly independent vectors

Definition: Let V be a vector space over a field K, and let v₁,v₂,..,v_n be elements of V. We say that v₁,v₂, ,v_n are linearly dependent over K if there exist elements a₁,a₂, ,a_n in K not all equal to zero such that:

                           a1*v1+a2*v2+...+an*vn=0.

If there do not exist such elements, then we say that v₁,v₂,...,v_n are linearly independent.

To focus the definition on linear independence, we can say, the vectors v₁,v₂, ,v_n are linearly independent, if and only if the following condition is satisfied:

   Whenever a1,a2,...,an are numbers such that:

                       a1*v1+a2*v2+...+an*vn=0

   then ai=0 for all i=1,2,...,n.

Example I:
Show that the vectors (1,1) and (-3,2) in R² are linearly independent.

Proof:

Let a, b be two numbers such that:

• a(1,1)+b(-3,2)=(0,0)
  

Then:

• (a-3b,a+2b)=(0,0) and
• a-3b=0 and a+2b=0.
  

Solving for a and b, we find that a=0 and b=0.

Example II: Let V=Rⁿ and consider the vectors in Rⁿ:

• E₁=(1,0,0, ,0)
• E₂=(0,1,0, ,0)
• ...
• E_n=(0,0,0, ,1)
  

Then E₁,E₂,...,E_n are linearly independent.

Proof:

Suppose that a₁, a₂, ,a_n are elements of Rⁿ such that

• a₁E₁+a₂E₂+...+a_nE_n=0
  

Since

• a₁E_n+a₂E₂+....+a_nE_n(a₁,a₂, ,a_n)
  

then a_i=0 for all I+{0,1, ,n}.

Example III: (Calculus required) Let V be the vector space of all functions of a real variable t. Then the functions e^t and e^2t in V are linearly independent.

Proof:

Then we must prove that whenever there are 2 numbers a and b, such that:

• a*e^t+b*e^2t=0
  

then a, b=0, for all values of t.
Then suppose that that there are numbers a and b such that:

• (1) a*e^t+b*e^2t=0
  

Then differentiate this relation, giving:

• (2) a*e^t+2b*e^2t=0
  

Subtracting the first relation from the second relation. We obtain:

• b*e^2t=0, and thus b=0.
  

From the first relation we get:

• a*e^t=0 and hence a=0.