Question 1. If v1, v2, v3, v4 are in R^4 and v3 = 2v1 + v2, then {v1, v2, v3, v4} must be linearly dependent.
Answer: True, since 2v1 + 1v2 -1v3 + 0 v4 = 0.
Question 2. If v1, v2, v3, v4 are in R^4 and v3 = 0, then {v1, v2, v3, v4} must be linearly dependent.
Answer: True, since 0v1 + 0v2 + 1v3 + 0 v4 = 0.
Question 3. If v1, v2, v3, v4 are in R^4 and v3 is not a linear combination of v1, v2, v4, then {v1, v2, v3, v4} must be linearly independent.
Answer: False. For example, if v4 = v1 + v2, then 1v1 + 1 v2 + 0 v3 - 1 v4 = 0.
Question 4. If v1, v2, v3, v4 are in R^4 and {v1, v2, v3}is linearly dependent, then {v1, v2, v3, v4} must also be linearly dependent.
Answer: True, if x1v1 + x2v2 + x3v3 = 0 with x1, x2, x3 not all zero, then x1v1 + x2v2 + x3v3 + 0 v4 = 0 and still the coefficients are not ALL zero.
Question 5. If v1, v2, v3, v4 are linearly independent vectors in R^4, then {v1, v2, v3} must also be linearly independent.
Answer: True. If {v1, v2, v3}were dependent, then there would be an equation if x1v1 + x2v2 + x3v3 = 0 with x1, x2, x3 not all zero. But then x1v1 + x2v2 + x3v3 + 0 v4 = 0 and still the coefficients are not ALL zero so this contradicts the fact that {v1, v2, v3, v4} is linearly independent. Note: Questions 4 and 5 are logically equivalent.
Question 6. The columns of A are linearly independent if the equation Ax = 0 has the trivial solution.
Answer: False. For any matrix, Ax = 0 has the trivial solution. It's the nontrivial solutions that make the difference.
Question 7. If S is a linearly dependent set, then each vector in S is a linear combination of the others.
Answer: False. For example, v1 = (1,0), v2 = (2,0) and v3 = (1,1). Then v2 = 2v1 but v3 is not a linear combination of v1 and v2, since it is not a multiple of v1. But 2v1 - 1v2 + 0 v3 = 0.
Question 8. . The columns of any 4x5 matrix A are linearly dependent.
Answer: True. There is at least one free variable in the general solution of Ax = 0 (since there are 5 variables and at most 4 pivots). Thus there are an infinite number of solutions, not just the trivial solution.