Linear algebra/Linearly independent vectors

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


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

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

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

  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.


Let a, b be two numbers such that:

  • a(1,1)+b(-3,2)=(0,0)


  • (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)
  • ...
  • En=(0,0,0, ,1)

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


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

  • a1E1+a2E2+...+anEn=0


  • a1En+a2E2+....+anEn(a1,a2, ,an)

then ai=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 et and e2t in V are linearly independent.


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

  • a*et+b*e2t=0

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

  • (1) a*et+b*e2t=0

Then differentiate this relation, giving:

  • (2) a*et+2b*e2t=0

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

  • b*e2t=0, and thus b=0.

From the first relation we get:

  • a*et=0 and hence a=0.