Theorem 1.8. Let V be a vector space and u1, U2, ..., Un be distinct vectors in V. Then B = {ul, u2, ..., Un} is a basis for V if and only if each v E V can be uniquely expressed as a linear combination of vectors of B, that is, can be expressed in the form v = alu1 + azu2 + . . . + anun for unique scalars a1, 2, . . ., an. Proof. Let B be a basis for V. If v E V, then v E span(B) because span(3) = V. Thus v is a linear combination of the vectors of B. Suppose that v = alu1 + azu2 + . . . + anun and v = bju1 + bzu2 + . .. + bnun are two such representations of v. Subtracting the second equation from the first gives 0 = (a1 - b1)u1 + (a2 - b2)u2 + . . . + (an -bn)un. Since B is linearly independent, it follows that al - b1 = a2 - b2 = .. . = an - bn = 0. Hence a1 = b1, a2 = b2, . .., an = bn, and so v is uniquely expressible as a linear combination of the vectors of B. The proof of the converse is an exercise. Theorem 1.8 shows that if the vectors u1, u2, . . ., Un form a basis for a vector space V, then every vector in V can be uniquely expressed in the form v = alu1 + azu2 + . . . + anunfor appropriately chosen scalars a1, a2, ..., an. Thus v determines a unique n-tuple of scalars (a1, a2, . .., an) and, conversely, each n-tuple of scalars de- termines a unique vector v E V by using the entries of the n-tuple as the coefficients of a linear combination of u1, u2, ..., Un. This fact suggests that V is like the vector space F", where n is the number of vectors in the basis for V. We see in Section 2.4 that this is indeed the case. In this book, we are primarily interested in vector spaces having finite bases. Theorem 1.9 identifies a large class of vector spaces of this type.19. Complete the proof of Theorem 1.8