(a) (b) (c) The list (1,0,..., 0), (0, 1, 0, . . . , 0), ....

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(a) (b) (c) The list (1,0,..., 0), (0, 1, 0, . . . , 0), . . . , (0, . . . , 0, 1) is a basis of F", called the standard basis of Fn. The list (1, 2), (3, 5) is a basis of F2. The list (1,2,-4), (7,-5, 6) is linearly independent in F³ but is not a basis of F3 because it does not span F³. (d) The list (1,2). (3.5). (4, 13) spans F2 but is not a basis of F² because it is not linearly independent. The list (1, 1,0), (0, 0, 1) is a basis of {(x, x, y) € F³ : x, y € F}. The list (1,-1,0), (1, 0, -1) is a basis of (e) (f) {(x, y, z) € F³: x + y + z = 0}. The list 1, 2,..., zm is a basis of Pm (F). (a) (b) (c) The list (1,0,..., 0), (0, 1, 0, . . . , 0), . . . , (0, . . . , 0, 1) is a basis of F", called the standard basis of Fn. The list (1, 2), (3, 5) is a basis of F2. The list (1,2,-4), (7,-5, 6) is linearly independent in F³ but is not a basis of F3 because it does not span F³. (d) The list (1,2). (3.5). (4, 13) spans F2 but is not a basis of F² because it is not linearly independent. The list (1, 1,0), (0, 0, 1) is a basis of {(x, x, y) € F³ : x, y € F}. The list (1,-1,0), (1, 0, -1) is a basis of (e) (f) {(x, y, z) € F³: x + y + z = 0}. The list 1, 2,..., zm is a basis of Pm (F).

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A The list 1 0 0 0 1 0 0 0 0 1 is a basis of F called the standard basis of F This list forms a basis of F because it is linearly independent and spans F Each vector in this list is linearly independe... View the full answer