HW 4: If a set of vectors is linearly dependent then at least one of the vectors is a scalar multiple of another one.
FALSE. Consider the set {(1, 0), (0, 1), (2, 3)}. This set is linearly dependent because the last vector is a linear combination of the first two, but neither vector is a scalar multiple of another one.
To prove this statement, we need to first establish the definition of linear dependence. Then, we can use this definition to show that if a set of vectors is linearly dependent, then at least one of the vectors is a scalar multiple of another one.
Definition of Linear Dependence:
A set of vectors {v_1, v_2, …, v_n} is linearly dependent if there exist scalars c_1, c_2, …, c_n, not all zero, such that:
c_1*v_1 + c_2*v_2 + … + c_n*v_n = 0
In other words, a set of vectors is linearly dependent if there is a non-trivial linear combination of the vectors that equals the zero vector.
Proof:
Suppose we have a set of linearly dependent vectors: {v_1, v_2, …, v_n}. Then, by the definition of linear dependence, there exist scalars c_1, c_2, …, c_n, not all zero, such that:
c_1*v_1 + c_2*v_2 + … + c_n*v_n = 0
Without loss of generality, we can assume that c_1 is not equal to zero (otherwise, we can just index the vectors differently). Then, we can solve for v_1:
v_1 = (-c_2/c_1)*v_2 – … – (c_n/c_1)*v_n
This shows that v_1 can be expressed as a linear combination of the remaining vectors, with coefficients that are scalars. Therefore, v_1 is a scalar multiple of a linear combination of the other vectors.
Thus, we have shown that if a set of vectors is linearly dependent, then at least one of the vectors is a scalar multiple of another one.
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