Redirected from Linearly independent
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or, more concisely:
(Note that the zero on the right is the zero element in V, not the zero element in K.)
If there do not exist such field elements, then we say that v1,v2,...,vn are linearly independent. An infinite subset of V is said to linearly independent if all its finite subsets are.
To focus the definition on linear independence, we can say that the vectors v1,v2,..,vn are linearly independent, if and only if the following condition is satisfied:
Proof:
Let a, b be two real numbers such that:
Then:
Solving for a and b, we find that a = 0 and b = 0.
Then e1,e2,...,en are linearly independent.
Proof:
Suppose that a1, a2, ,an are elements of Rn such that
Since
then ai = 0 for all i in {1,..,n}.
Proof:
Suppose a and b are two real numbers such that
for all values of t. We need to show that a=0 and b=0. In order to do this,
we differentiate equation (1) to get
which also holds for all values of t.
Subtracting the first relation from the second relation, we obtain:
and, by plugging in t = 0, we get b = 0.
From the first relation we then get:
and again for t = 0 we find a = 0.
A linear dependence among vectors v1,...,vn is a vector (a1,...,an) with n scalar components, not all zero, such that a1v1+...+anvn=0. If such a linear dependence exists, then the n vectors are linearly dependent. It makes sense to identify two linear dependences if one arises as a non-zero multiple of the other, because in this case the two describe the same linear relationship among the vectors. Under this identification, the set of all linear dependences among v1, ...., vn is a projective space.
See also:
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