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196              Chapter 4                                              Vector Spaces

                                    the columns of B are linearly independent, and hence N (B)= {0} —recall
                                    (4.3.2). Therefore, the supposition that there exists a basis for V containing
                                    fewer than n vectors must be false, and we may conclude that B is indeed a
                                    minimal spanning set.
                                        Proof of (4.4.2) =⇒ (4.4.1). If B is a minimal spanning set, then B must
                                    be a linearly independent spanning set. Otherwise, some b i would be a linear
                                    combination of the other b ’s, and the set


                                                        B = {b 1 ,..., b i−1 , b i+1 ,..., b n }
                                    would still span V, but B would contain fewer vectors than B, which is im-

                                    possible because B is a minimal spanning set.
                                        Proof of (4.4.3) =⇒ (4.4.1). If B is a maximal linearly independent subset
                                    of V, but not a basis for V, then there exists a vector v ∈V such that
                                    v /∈ span (B) . This means that the extension set

                                                          B∪ {v} = {b 1 , b 2 ,..., b n , v}
                                    is linearly independent—recall (4.3.15). But this is impossible because B is a
                                    maximal linearly independent subset of V. Therefore, B is a basis for V.
                                        Proof of (4.4.1) =⇒ (4.4.3). Suppose that B is a basis for V, but not a
                                    maximal linearly independent subset of V, and let
                                                   Y = {y 1 , y 2 ,..., y k }⊆V,  where  k> n

                                    be a maximal linearly independent subset—recall that (4.3.16) insures the ex-
                                    istence of such a set. The previous argument shows that Y must be a basis
                                    for V. But this is impossible because we already know that a basis must be a
                                    minimal spanning set, and B is a spanning set containing fewer vectors than Y.
                                    Therefore, B must be a maximal linearly independent subset of V.
                                        Although a space V can have many different bases, the preceding result
                                    guarantees that all bases for V contain the same number of vectors. If B 1 and
                                    B 2 are each a basis for V, then each is a minimal spanning set, and thus they
                                    must contain the same number of vectors. As we are about to see, this number
                                    is quite important.


                                                                Dimension
                                       The dimension of a vector space V is defined to be


                                          dim V = number of vectors in any basis for V
                                               = number of vectors in any minimal spanning set for V
                                               = number of vectors in any maximal independent subset of V.
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