Page 213 - 35Linear Algebra
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                                                              Basis and Dimension





                   In chapter 10, the notions of a linearly independent set of vectors in a vector
                   space V , and of a set of vectors that span V were established; any set of
                   vectors that span V can be reduced to some minimal collection of linearly
                   independent vectors; such a minimal set is called a basis of the subspace V .

                   Definition Let V be a vector space. Then a set S is a basis for V if S is
                   linearly independent and V = span S.
                      If S is a basis of V and S has only finitely many elements, then we say
                   that V is finite-dimensional. The number of vectors in S is the dimension
                   of V .

                      Suppose V is a finite-dimensional vector space, and S and T are two dif-
                   ferent bases for V . One might worry that S and T have a different number of
                   vectors; then we would have to talk about the dimension of V in terms of the
                   basis S or in terms of the basis T. Luckily this isn’t what happens. Later in
                   this chapter, we will show that S and T must have the same number of vec-
                   tors. This means that the dimension of a vector space is basis-independent.
                   In fact, dimension is a very important characteristic of a vector space.
                                                                                              n
                   Example 121 P n (t) (polynomials in t of degree n or less) has a basis {1, t, . . . , t },
                   since every vector in this space is a sum

                                          0
                                                           n n
                                                                     i
                                                1
                                        a 1 + a t + · · · + a t ,   a ∈ R ,
                                             n
                   so P n (t) = span{1, t, . . . , t }. This set of vectors is linearly independent; If the
                                                                   0
                                                                        1
                                                                                  n
                                                     n n
                                          1
                                     0
                   polynomial p(t) = c 1 + c t + · · · + c t = 0, then c = c = · · · = c = 0, so p(t) is
                   the zero polynomial. Thus P n (t) is finite dimensional, and dim P n (t) = n + 1.
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