Linear Algebra. Paul Yiu. Department of Mathematics Florida Atlantic University. Fall 2011

Transcription

1 Linear Algebra Paul Yiu Department of Mathematics Florida Atlantic University Fall 2011

2 Linear Algebra Paul Yiu Department of Mathematics Florida Atlantic University Fall A: Vector spaces

3 Fields Definition A field F is a nonempty set with two operations, an addition + and a multiplication such that (i) F is an abelian group under addition with a zero element 0, (ii) F := F \ {0} is an abelian group under multiplication with a unit element 1, (iii) the multiplication distributes over the addition, i.e., λ(µ + ν) =λµ + λν. 1

4 Fields Definition A field F is a nonempty set with two operations, an addition + and a multiplication such that (i) F is an abelian group under addition with a zero element 0, (ii) F := F \ {0} is an abelian group under multiplication with a unit element 1, (iii) the multiplication distributes over the addition, i.e., λ(µ + ν) =λµ + λν. Examples (1) The field R of real numbers. (2) The field C of complex numbers. (3) The finite field F p for a prime p. 2

5 Vector spaces Definition Let F be a field. A vector space over F consists of a nonempty set V with two operations: (1) a vector addition V V V giving V the structure of an abelian group, (2) a scalar multiplication F V V such that 1u = u for u V, (λµ)u = λ(µu), for λ, µ F, (λ + µ)u = λu + µu, for λ, µ F and u V, λ(u + v) = λu + λv, for λ F and u, v V. 3

6 Vector spaces: Example 1 Let F be a field, and V = F n := F F (n copies). Define (i) addition of vectors in V componentwise: (a 1,..., a n ) + (b 1,..., b n ) = (a 1 + b 1,..., a n + b n ); (ii) scalar multiplication also componentwise by λ(a 1,..., a n ) = (λa 1,..., λa n ). This is the standard n-dimensional vector space over F. 4

7 Vector spaces: Example 2 Let V := C(R, R) be the set of continuous functions f : R R. (1) For f, g, V, define f + g by (f + g)(x) = f(x) + g(x) for every x R. (2) For f V and λ R, define λf by (λf)(x) = λf(x). C(R, R) is a vector space over R. 5

8 Algebras A vector space which is also a ring is called an algebra. V = C(R, R) is an algebra since there is a multiplication in V given by (fg)(x) = f(x)g(x) for every x R. 6

9 Example: Quaternion algebra H The quaternion algebra is a 4-dimensional vector space over R, with basis 1, i, j, k, and multiplication defined by 1 i j k 1 1 i j k i i 1 k j j j k 1 i k k j i 1 This is an associative but noncommutative algebra ij = ji. 7

10 Vector subspaces Let V be a vector space over F, and S V. S is a (vector) subspace of V if S is a vector space over F. Examples: (1) For m n, F m is a subspace of F n. (2) Let V = C(R, R), and W = the subset of (everywhere) differentiable functions f : R R. W is a subspace of V. 8

11 Vector subspaces Let V be a vector space over F, and S V. S is a (vector) subspace of V if S is a vector space over F. Lemma. Let V be a vector space over F, and W V. W is a vector subspace of V if and only if (i) v, w W = v + w W, (ii) v W and λ F = λv W. 9

12 Vector subspaces Let V be a vector space over F, and S V. S is a (vector) subspace of V if S is a vector space over F. Lemma. Let V be a vector space over F and W V. W is a vector subspace of V if and only if (i) v, w W = v + w W, (ii) v W and λ F = λv W. Examples: Let V be a vector space over F. (3) V is a subspace of itself. (4) {0} is a subspace of V. (5) If W 1 and W 2 are subspaces of V, then so is W 1 W 2. A proper subspace of V is one which is neither {0} nor V itself. Exercise. subspaces. Show that a vector space cannot be the union of two proper 10

13 Linear combinations Let V be a vector space over F, and A V. A linear combination of vectors in A is a finite sum a A λ aa V (in which only finitely many of the coefficients λ a, a A, are nonzero). Denote by Span(A) the set of all linear combinations of vectors in A, and call it the span of A. Exercise. Span(A) is a subspace of V. A subset A V is a spanning set of V if every vector of V is a linear combination of vectors in A, i.e., Span(A) = V. Exercise. Show that two subsets A and B of a vector space V generate the same subspace if and only if each vector in A is a linear combination of vectors in B and vice versa. 11

14 Example 1 Consider the standard n-dimensional vector space F n. The set of n vectors e 1 = (1, 0, 0..., 0), e 2 = (0, 1, 0,..., 0),. e n = (0, 0, 0,..., 1), is a spanning set since for every vector a = (a 1, a 2,..., a n ) F n, a = a 1 e 1 + a 2 e a n e n. 12

15 Example 2 Let F be a field. Consider the ring of polynomials F[x] as a vector space over F. The family {1, x, x 2,..., x n,... } is a spanning set of F[x]. No spanning set of F[x] can be finite. 13

16 Example 3 Let A be a nonempty set, and F a field. Consider the vector space of scalar valued functions f : A F. The family {f a : a A} defined by { 1, if x = a, f a (x) = 0, otherwise, is a spanning set if and only if A is a finite set. If A is an infinite set, the constant function c : A F defined by c(a) = 1 for every a A is not a linear combination of the family {f a : a A}. 14

17 Linear Algebra Paul Yiu Department of Mathematics Florida Atlantic University Fall B: Linear independence

18 Linear independence A subset A V is linearly dependent if the zero vector in V is a nontrivial linear combination of A, i.e., there exist scalars λ a, a A, not all zero, such that λ a a = 0. a A Otherwise, A is said to be linearly independent. In other words, A is linear independent if and only if λ a a = 0 = λ a = 0 for each a A. a A 1

19 Theorem Let V be a vector space over F. For A V, the following statements are equivalent. (A) A is a minimal spanning set of V. (B) A is a maximal linearly independent set in V. (C) A is a linearly independent spanning set of V. Proof. (A) = (C). Let A be a minimal spanning set. Suppose, for a contradiction, that it is a linearly dependent set. There is a nontrivial linear combination a A λ aa = 0 in which we may assume λ b 0 for a specific b A. For this b, we have b = 1 λ a a. λ b a b Since A is a spanning set, every vector in V is a linear combination of A. As such, it is also a linear combination of A \ {b}. This contradicts the minimality of A as a spanning set. Therefore, A must be linearly independent. 2

20 Theorem Let V be a vector space over F. For A V, the following statements are equivalent. (A) A is a minimal spanning set of V. (B) A is a maximal linearly independent set in V. (C) A is a linearly independent spanning set of V. (C) = (A). Let A be a linearly independent spanning set of V. For b A, if A = A \ {b} is a (smaller) spanning set, then b = a b λ aa, and ( 1)b + a b λ aa = 0. This contradicts the linear independence of A. Therefore, A is a minimal spanning set. 3

21 Theorem Let V be a vector space over F. For A V, the following statements are equivalent. (A) A is a minimal spanning set of V. (B) A is a maximal linearly independent set in V. (C) A is a linearly independent spanning set of V. (B) = (C). Let A be a maximal linearly independent set in V. For each b V \ A, A = A {b} cannot be linearly independent, and we have λ b b + a A λ aa = 0, in which the coefficients are not all zero. Note that λ b 0, for otherwise, A would have been linearly dependent. Thus, b is a linear combination of A. This shows that A is a spanning set. 4

22 Theorem Let V be a vector space over F. For A V, the following statements are equivalent. (A) A is a minimal spanning set of V. (B) A is a maximal linearly independent set in V. (C) A is a linearly independent spanning set of V. (C) = (B). Let A be a linearly independent spanning set of V. For an arbitrary b V \ A, write b = a A λ aa. From this we have ( 1)b + a A λ aa = 0, a nontrivial linear combination showing that A {b} is linearly dependent. Thus, A is maximal as a linearly independent set. 5

23 Basis of a vector space Let V be a vector space over F. A set A V satisfying any of these conditions is called a basis of V. (A) A is a minimal spanning set of V. (B) A is a maximal linearly independent set in V. (C) A is a linearly independent spanning set of V. 6

24 Example Consider the standard n-dimensional vector space F n. The set of n vectors e 1 = (1, 0, 0..., 0), e 2 = (0, 1, 0,..., 0),. e n = (0, 0, 0,..., 1), is a spanning set since for every vector a = (a 1, a 2,..., a n ) F n, It is also linearly independent since a = a 1 e 1 + a 2 e a n e n. 0 = a 1 e 1 +a 2 e 2 + +a n e n = (a 1, a 2,..., a n ) = a 1 = a 2 = = a n = 0. The set (e 1, e 2,..., e n ) is a basis of F n, often called the standard (or canonical) basis of F n. 7

25 Linear Algebra Paul Yiu Department of Mathematics Florida Atlantic University Fall C: Dimension

26 Steinitz exchange theorem Theorem. Let V be a vector space over F, with a linearly independent set A of m elements, and a spanning set B of n elements. Then, (i) m n, and (ii) there is a subset B B such that A B is a basis of V. 1

27 Proof of Steinitz exchange theorem Proof. Let A = {a 1, a 2,..., a m } be a linearly independent set, and B = {b 1, b 2,..., b n } a spanning set of V. Since the set {a m, b 1,..., b n } is not minimally spanning, it must be linearly dependent. Let i be the largest index for which b i is linearly dependent on its predecessors in this list. Removing b i we have a new spanning set of n vectors. Continue inserting, one at a time, the vectors a m 1,..., a 1 to the beginning of the list of spanning vectors, and removing a vector from B dependent on its predecessors (which cannot be any of the vectors from A), we obtain a sequence of generating sets, each consisting of n vectors. If n < m, this process will replace the spanning set B by a proper subset of A, which is a linearly independent spanning set. This proper subset of A is a maximal linearly independent set, a contradiction. 2

28 Proof of Steinitz exchange theorem Proof. Let A = {a 1, a 2,..., a m } be a linearly independent set, and B = {b 1, b 2,..., b n } a spanning set of V.... If n < m,... contradiction. Therefore, m n, and we eventually replace m of the vectors of B by the m vectors of A. The remaining n m vectors in B form a subset B 1 such that A B 1 is a spanning set of V. By further removing elements from B 1, if any, which are linearly dependent on their predecessors, we obtain a subset B B 1 B so that A B is a minimal spanning set, a basis of V. 3

29 Dimension The notion of the dimension of a vector space is justified by the following corollary. Corollary. If a vector space V has a basis of n vectors, then every basis of V has exactly n vectors. Proof. If V has a basis of n vectors, then by the Steinitz exchange theorem, no basis of V can be infinite. 4

30 Dimension The notion of the dimension of a vector space is justified by the following corollary. Corollary. If a vector space V has a basis of n vectors, then every basis of V has exactly n vectors. Proof. If V has a basis of n vectors, then by the Steinitz exchange theorem, no basis of V can be infinite. Let A be a basis of n vectors, and B an infinite basis. By the Steinitz exchange theorem, we replace n of the vectors of B by A, and obtain a basis without removing any further vectors from B (since none is dependent on its predecessors in B). This contradicts the maximality of A as a linearly independent set. 5

31 Dimension The notion of the dimension of a vector space is justified by the following corollary. Corollary. If a vector space V has a basis of n vectors, then every basis of V has exactly n vectors. Proof. If V has a basis of n vectors, then by the Steinitz exchange theorem, no basis of V can be infinite. Let A and B be finite bases of m and n elements respectively. Each is a linearly independent, spanning set. By the Steinitz exchange theorem, m n and n m, showing that m = n. The cardinality of a basis of a vector space V over F is called its dimension, usually denoted dim F V. 6

32 Basis extension theorem Corollary. Let X V be a subspace. Every basis of X can be extended to a basis of V. Proof. Given a basis A of X, we take an arbitrary basis B of V and apply the Steinitz extension theorem to obtain a basis of V consisting of A and appropriate members from B. 7

33 Linear Algebra Paul Yiu Department of Mathematics Florida Atlantic University Fall D: Sums of vector spaces

34 The sum of two subspaces Let X and Y be two subspaces of a vector space V. The subset X + Y := {x + y : x X, y Y } is a subspace of V. We call this the sum of X and Y. 1

35 The sum of two subspaces Let X and Y be two subspaces of a vector space V. The subset X + Y := {x + y : x X, y Y } is a subspace of V. We call this the sum of X and Y. Proposition. dim(x + Y ) + dim(x Y ) = dimx + dimy. Proof. Let m = dimx, n = dimy, and p = dim(x Y ). Begin with a basis A of X Y, and extend it to a basis B of X with m p extra vectors, and a basis C of Y with n p extra vectors. Then B C is a basis of X + Y consisting of p + (m p) + (n p) = m + n p vectors. 2

36 The sum of two subspaces Let X and Y be two subspaces of a vector space V. The subset X + Y := {x + y : x X, y Y } is a subspace of V. We call this the sum of X and Y. Proposition. dim(x + Y ) + dim(x Y ) = dimx + dimy. Proof. Let m = dimx, n = dimy, and p = dim(x Y ). Begin with a basis A of X Y, and extend it to a basis B of X with m p extra vectors, and a basis C of Y with n p extra vectors. Then B C is a basis of X + Y consisting of p + (m p) + (n p) = m + n p vectors. Corollary. If X Y = {0}, then dim(x + Y ) = dimx + dimy. If X Y = {0}, we write X Y (instead of X + Y ), and call this the internal direct sum of X and Y. 3

37 External direct sum Let X and Y be vector spaces over F. The external direct sum X Y is the Cartesian product X Y endowed with a vector space structure: (x 1, y 1 ) + (x 2, y 2 ) = (x 1 + x 2, y 1 + y 2 ), λ(x, y) = (λx, λy). 4

38 External direct sum Let X and Y be vector spaces over F. The external direct sum X Y is the Cartesian product X Y endowed with a vector space structure: (x 1, y 1 ) + (x 2, y 2 ) = (x 1 + x 2, y 1 + y 2 ), λ(x, y) = (λx, λy). Proposition. If dimx = m and dimy = n, then dim(x Y ) = m + n. Proof. If (x 1,..., x m ) is a basis of X and (y 1,..., y n ) a basis of Y, then ((x 1, 0),...,(x m, 0), (0, y 1 ),..., (0, y n )) is a basis of X Y. 5

39 Linear Algebra Paul Yiu Department of Mathematics Florida Atlantic University Fall E: Finite fields

40 Characteristic of a finite field Let F be a finite field. There is a least positive integer n for which n 1 = } {{ + 1 } = 0. n times This is called the characteristic of the field F, usually denoted char F. Lemma. The characteristic of a finite field is a prime number. 1

41 Characteristic of a finite field Let F be a finite field. There is a least positive integer n for which n 1 = (n times) = 0. This is called the characteristic of the field F, usually denoted char F. Lemma. The characteristic of a finite field is a prime number. Proof. Let char F = n. If n is not a prime, then n = ab for 1 < a, b < n. Since 0 = n 1 = (a 1)(b 1), one of a 1 and b 1 is zero. In any case, this contradicts the minimality of n. Therefore, n must be a prime. 2

42 Theorem. The cardinality of a finite field is a prime power. Proof. Let char F = p. F is a finite dimensional vector space over the field F p of p elements. If dim Fp F = n, then F has p n elements. 3

43 Construction of finite fields Let p be a prime and q = p n. If f(x) F p is an irreducible polynomial of degree n, then F = F p [x]/(f(x)) is a field. It has exactly q = p n elements since it is a vector space over F p, with basis 1, [x],..., [x n 1 ]. Every finite field can be constructed in this way. Any two finite fields with the same cardinality are isomorphic. We write F q for the field with q = p n elements. 4

44 Construction of finite fields Let p be a prime and q = p n. If f(x) F p is an irreducible polynomial of degree n, then F = F p [x]/(f(x)) is a field. It has exactly q = p n elements since it is a vector space over F p, with basis 1, [x],..., [x n 1 ]. Every finite field can be constructed in this way. Any two finite fields with the same cardinality are isomorphic. We write F q for the field with q = p n elements. Example. Construction of F 4 : The only irreducible quadratic polynomial over F 2 is x 2 + x + 1. Therefore, F 4 = F 2 [x]/(x 2 + x + 1). Let α = [x]. Then, F 4 = {0, 1, α, 1 + α}. Here is the multiplication table of the nonzero elements of F 4 : 1 α 1 + α 1 1 α 1 + α α α 1 + α α 1 + α 1 α 5

45 The projective space of a vector space Let V be a vector space over F. The relation on V \ {0} defined by u v if and only if u = λv for some λ F is an equivalence relation. The quotient, denoted by P(V ), is called the projective space of V. 6

46 Construction of finite projective planes Let F = F q for a prime power q. If we apply this construction to V = F 3, we obtain a finite projective plane of order q: P(V ) consists of q3 1 q 1 = q2 + q + 1 points. It also contains q 2 + q + 1 lines. Each point lies on q + 1 lines and each line contains q + 1 points

47 Construction of finite projective planes Let F = F q for a prime power q. If we apply this construction to V = F 3, we obtain a finite projective plane P(V ) of order q: P(V ) consists of q3 1 q 1 = q2 + q + 1 points. It also contains q 2 + q + 1 lines. Each point lies on q + 1 lines and each line contains q + 1 points. If q is not a prime power, the existence problem of a finite projective plane of order q is unsolved, except for nonexistence in the cases of q = 6 (Euler) and 10 (Lam, 1988). 8