{v1, v2} linearly independent if and only if
span {v1,v2} is a plane
{v1, v2} are linearly independent if and only if neither of the vectors can be expressed as a linear combination of the other vector.
To prove this statement, we need to show that if {v1, v2} are linearly independent, then neither vector can be expressed as a linear combination of the other vector; and conversely, if neither vector can be expressed as a linear combination of the other vector, then {v1, v2} are linearly independent.
First, assume that {v1, v2} are linearly independent. This means that for any scalars a and b, the vector equation a*v1 + b*v2 = 0 has only the trivial solution a=0, b=0. If v1 can be expressed as a linear combination of v2, then there exist scalars a and b not both zero such that v1 = a*v2 + b*v1. Then, 0 = -a*v2 – b*v1 = (-a)*v2 + (-b)*v1, showing that v1 and v2 are linearly dependent, which contradicts the assumption that they are linearly independent. Similarly, if v2 can be expressed as a linear combination of v1, then v1 and v2 are linearly dependent, which is a contradiction.
Conversely, assume that neither v1 nor v2 can be expressed as a linear combination of the other vector. Suppose that we have scalars a and b such that a*v1 + b*v2 = 0. We need to show that a=0 and b=0. Suppose, for the sake of contradiction, that at least one of a or b is nonzero, say a is nonzero. Then we can write v1 = (-b/a)*v2, which contradicts the assumption that neither v1 nor v2 can be expressed as a linear combination of the other vector. Similarly, we can assume b is nonzero, and write v2 = (-a/b)*v1 to derive another contradiction. Therefore, a=b=0 is the only solution to a*v1 + b*v2 = 0, which implies that {v1,v2} are linearly independent.
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