Answer:
Explanation:
To prove the statement "A Set S with two or more vectors is linearly independent if and only if no vector in S is expressible as a linear combination of the other vectors in S," we must show that both directions of the statement are true. That is, we must show that if a set S is linearly independent, then no vector in S is expressible as a linear combination of the other vectors in S, and conversely, if no vector in S is expressible as a linear combination of the other vectors in S, then the set S is linearly independent.
First, let's assume that the set S is linearly independent. This means that for any vectors v1, v2, ..., vn in S, the equation a1v1 + a2v2 + ... + anvn = 0 has only the trivial solution a1 = a2 = ... = an = 0. We will prove that no vector in S is expressible as a linear combination of the other vectors in S.
Suppose, for the sake of contradiction, that there exists a vector v in S that can be expressed as a linear combination of the other vectors in S. That is, there exist vectors v1, v2, ..., vn-1 in S such that v = b1v1 + b2v2 + ... + bn-1vn-1, where not all of the bi's are zero. Without loss of generality, assume that b1 is nonzero. Then we can write v1 as a linear combination of the other vectors in S:
v1 = (1/b1)v - (b2/b1)v2 - ... - (bn-1/b1)vn-1
Substituting this expression for v1 in the equation a1v1 + a2v2 + ... + anvn = 0, we get:
a1[(1/b1)v - (b2/b1)v2 - ... - (bn-1/b1)vn-1] + a2v2 + ... + anvn = 0
Multiplying both sides by b1 and rearranging, we get:
(a1/b1)v + (-a1b2/b1)v2 + ... + (-a1bn-1/b1)vn-1 + a2v2 + ... + anvn = 0
This is a linear combination of vectors in S that equals the zero vector, and not all of the coefficients are zero, since a1 is nonzero. But this contradicts the assumption that S is linearly independent, so our assumption that there exists a vector in S that can be expressed as a linear combination of the other vectors in S must be false. Therefore, no vector in S is expressible as a linear combination of the other vectors in S.
Now let's assume that no vector in S is expressible as a linear combination of the other vectors in S. We will prove that the set S is linearly independent. Suppose, for the sake of contradiction, that there exist vectors v1, v2, ..., vn in S such that the equation a1v1 + a2v2 + ... + anvn = 0 has a nontrivial solution, where not all of the ai's are zero. Without loss of generality, assume that a1 is nonzero. Then we can write v1 as a linear combination of the other vectors in S:
v1 = (-a2/a1)v2 - ... - (an/a1)vn
Substituting this expression for v1 in the equation a1v1 + a2v2 + ... + anvn = 0, we get:
0 = a1(-a2/a1)v2 + ... + a