Linear Algebra. Paul Yiu. Department of Mathematics Florida Atlantic University. Fall PDF Free Download

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

Linear Algebra Paul Yiu Department of Mathematics Florida Atlantic University Fall 2011 1A: Vector spaces

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

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

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

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

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

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

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

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

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

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

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 2 + + a n e n. 12

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

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

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

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

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

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

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

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

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

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 2 + + 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

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

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

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

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

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

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

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

Linear Algebra Paul Yiu Department of Mathematics Florida Atlantic University Fall 2011 1D: Sums of vector spaces

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

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

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

Linear Algebra Paul Yiu Department of Mathematics Florida Atlantic University Fall 2011 1E: Finite fields

Characteristic of a finite field Let F be a finite field. There is a least positive integer n for which n 1 = } 1 + 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

Characteristic of a finite field Let F be a finite field. There is a least positive integer n for which n 1 = 1 + 1 + + 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

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

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 1 1 + α 1 + α 1 α 5

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

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. 1 6 5 0 2 4 3 7

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

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