4.5. BASIS AND DIMENSION OF A VECTOR SPACE 135 4.5 Basis and Dimension of a Vector Space In the section on spanning sets and linear independence, we were trying to understand what the elements of a vector space looked like by studying how they could be generated. We learned that some subsets of a vector space could generate the entire vector space. Such subsets were called spanning sets. Other subsets did not generate the entire space, but their span was still a subspace of the underlying vector space. In some cases, the number of vectors in such a set was redundant in the sense that one or more of the vectors could be removed, without changing the span of the set. In other cases, there was not a unique way to generate some vectors in the space. In this section, we want to make this process of generating all the elements of a vector space more reliable, more e¢ cient. 4.5.1 Basis of a Vector Space De…nition 297 Let V denote a vector space and S = u1; u2; :::; un a subset of V . S is called a basis for V if the following is true: f g 1. S spans V . 2. S is linearly independent. This de…nition tells us that a basis has to contain enough vectors to generate the entire vector space. But it does not contain too many. In other words, if we removed one of the vectors, it would no longer generate the space. A basis is the vector space generalization of a coordinate system in R2 or R3. Example 298 We have already seen that the set S = e1; e2 where e1 = (1; 0) 2 f g and e2 = (0; 1) was a spanning set of R . It is also linearly independent for the only solution of the vector equation c1e1 + c2e2 = 0 is the trivial solution. Therefore, S is a basis for R2. It is called the standard basis for R2. These vectors also have a special name. (1; 0) is i and (0; 1) is j. 3 Example 299 Similarly, the standard basis for R is the set e1; e2; e3 where f g e1 = (1; 0; 0), e2 = (0; 1; 0) and e3 = (0; 0; 1). These vectors also have a special name. They are i; j and k respectively. 2 Example 300 Prove that S = 1; x; x is a basis for P2, the set of polynomials of degree less than or equal to 2. We need to prove that S spans P2 and is linearly independent. S spans P2. We already did this in the section on spanning sets. A typical polynomial of degree less than or equal to 2 is ax2 + bx + c. S is linearly independent. Here, we need to show that the only solution to a (1) + bx + cx2 = 0 (where 0 is the zero polynomial) is a = b = c = 0. 136 CHAPTER 4. VECTOR SPACES From algebra, we remember that two polynomials are equal if and only if their corresponding coe¢ cients are equal. The zero polynomial has all its coe¢ cients equal to zero. So, a (1) + bx + cx2 = 0 if and only if a = 0, b = 0, c = 0. Which proves that S is linearly independent. We will see more examples shortly. The next theorem outlines an important di¤erence between a basis and a spanning set. Any vector in a vector space can be represented in a unique way as a linear combination of the vectors of a basis.. Theorem 301 Let V denote a vector space and S = u1; u2; :::; un a basis of V . Every vector in V can be written in a unique wayf as a linear combinationg of vectors in S. Proof. Since S is a basis, we know that it spans V . If v V , then there 2 exists scalars c1; c2; :::; cn such that v =c1u1 + c2u2 + ::: + cnun. Suppose there is another way to write v. That is, there exist scalars d1; d2; :::; dn such that v =d1u1 + d2u2 + ::: + dnun. Then, c1u1 + c2u2 + ::: + cnun = d1u1 + d2u2 + ::: + dnun. In other words, (c1 d1) u1 + (c2 d2) u2 + ::: + (cn dn) un = 0. Since S is a basis, it must be linearly independent. The unique solution to (c1 d1) u1 +(c2 d2) u2 +:::+(cn dn) un = 0 must be the trivial solution. It follows that ci di = 0 for i = 1; 2; :::; n in other words ci = di for i = 1; 2; :::; n. Therefore, the two representations of v are the same. Remark 302 We say that any vector v of V has a unique representation with respect to the basis S. The scalars used in the linear representation are called the coordinates of the vector. For example, the vector (x; y) can be represented in the basis (1; 0) ; (0; 1) by the linear combination (x; y) = x (1; 0) + y (0; 1). Thus, x andfy are the coordinatesg of this vector (we knew that!). De…nition 303 If V is a vector space, and B = u1; u2; :::; un is an ordered basis of V , then we know that every vector v of Vfcan be expressedg as a linear combination of the vectors in S in a unique way. In others words, there exists unique scalars c1; c2; :::; cn such that v = c1u1 + c2u2 + ::: + cnun. These scalars are called the coordinates of v relative to the ordered basis B. Remark 304 The term "ordered basis" simply means that the order in which we list the elements is important. Indeed it is since each coordinate is with respect to one of the vector in the basis. We know that in R2, (2; 3) is not the same as (3; 2). Example 305 What are the coordinates of (1; 2; 3) with respect to the basis (1; 1; 0) ; (0; 1; 1) ; (1; 1; 1) ? f g One can indeed verify that this set is a basis for R3. Finding the coordinates of (1; 2; 3) with respect to this new basis amounts to …nding the numbers (a; b; c) such that a (1; 1; 0) + b (0; 1; 1) + c (1; 1; 1) = (1; 2; 3). This amounts to solving 4.5. BASIS AND DIMENSION OF A VECTOR SPACE 137 1 0 1 a 1 the system 1 1 1 b = 2 . The solution is 2 0 1 1 3 2 c 3 2 3 3 4 5 4 5 4 5 1 a 1 0 1 1 b = 1 1 1 2 2 c 3 2 0 1 1 3 2 3 3 4 5 4 1 5 4 5 = 1 2 2 3 4 5 The next theorem, deals with the number of vectors the basis of a given vector space can have. We will state the theorem without proof. Theorem 306 Let V denote a vector space and S = u1; u2; :::; un a basis of V . f g 1. Any subset of V containing more than n vectors must be dependent. 2. Any subset of V containing less than n vectors cannot span V . Proof. We prove each part separately. 1. Consider W = v1; v2; :::; vr a subset of V where r > n. We must show f g that W in dependent. Since S is a basis, we can writ each vi in term of elements in S. More speci…cally, there exists constants cij with 1 i r and 1 j n such that vi = ci1u1 + ci2u2 + ::: + cinun. Consider the linear combination r r djvj = dj (cj1u1 + cj2u2 + ::: + cjnun) = 0 j=1 j=1 X X d1c11 + d1c12 + ::: + d1c1n = 0 d2c21 + d2c22 + ::: + d2c2n = 0 So, we must solve 8 . where the unknowns > . <> drcr1 + drcr2 + ::: + drcrn = 0 are d1; d2; :::; dr. Since> we have more unknowns than equations, we are guaranteed that this:> homogeneous system will have a nontrivial solution. Thus W is dependent. 2. Consider W = v1; v2; :::; vr a subset of V where r < n. We must show that W doesf not span Vg. We do it by contradiction. We assume it does span V and show this would imply that S is dependent. Suppose that there exists constants cij with 1 i n and 1 j r such that ui = ci1v1 + ci2v2 + ::: + cirvr. Consider the linear combination n r djuj = dj (cj1v1 + cj2v2 + ::: + cjrvr) = 0 j=1 j=1 X X 138 CHAPTER 4. VECTOR SPACES d1c11 + d1c12 + ::: + d1c1r = 0 d2c21 + d2c22 + ::: + d2c2r = 0 So, we must solve 8 . where the unknowns > . <> dncn1 + dncn2 + ::: + dncnr = 0 are d1; d2; :::; dn.> Since we have more unknowns than equations, we are guaranteed that this:> homogeneous system will have a nontrivial solution. Thus S would be dependent. But it can’t be since it is a basis. Corollary 307 Let V denote a vector space. If V has a basis with n elements, then all the bases of V will have n elements. Proof. Assume that S1 is a basis of V with n elements and S2 is another basis with m elements. We need to show that m = n. Since S1 is a basis, S2 being also a basis implies that m n. If we had m > n, by the theorem, S2 would be dependent, hence not a basis. Similarly, since S2 is a basis, S1 being also a basis implies that n m. The only way we can have m n and n m is if m = n. 4.5.2 Dimension of a Vector Space All the bases of a vector space must have the same number of elements. This common number of elements has a name. De…nition 308 Let V denote a vector space.