# D if v 1 v 2 and v 3 is a basis for v and w 1 w 2 and

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(d) If v 1 , v 2 , and v 3 is a basis for V , and w 1 , w 2 , and w 3 is a basis for W , then these six vectors cannot be independent and some linear combination (with some coefficients 6 = 0) is zero: c i v i + d i w i = 0, or c i v i = - d i w i is a vector in both subspaces. (e) (i) False: The stated conditions guarantee that if a solution exists then it is unique, but this does not guarantee that a solution will exist. For an A ∈ < MxN matrix, if N < M then the columns do not span all of < M and there will not be a solution for vectors b that are not in the span of the columns of A . (ii) True: In < M there cannot be more than M linearly independent vectors (the same number of vectors that would form a basis for that space). In our case, since N = 7 > M = 5 the claim that all vectors cannot be linearly independent is true. 3