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1 Math 121A Linear Algebra (Friday, Week 0) 5 to 7 quizzes (homework problems) 2 midterms 1 final exam Chapters 1 to 4 Solutions, practice exams, and schedule on website. Website: http://math.uci.edu/~swise/classes/math_121a.html Email: swise(at)math(dot)uci(dot)edu Office Hours: Monday 3pm – 4pm, Wednesday 1pm – 2pm. TA: Andrew-David Bjork Office: RH 420B Office Hours: Tu 12pm – 2pm, W 10am – 12pm. Email: abjork(at)math(dot)uci(dot)edu Linear Algebra About vector space, denoted V and W, and about linear transformation (operation) on vector spaces. Suppose V and W are vector spaces. A transformation (operator) L: V W is linear if and only if L(u + v) = L(u) + L(v) u, v V and “scalars” and . Example: Let , g C[a, b] and , . ba ( + g) dx = ba dx + ba g dx A linear transformation Arrows in Real Two-Space x = (a1, a2) 2 y = (b1, b2) 2 Addition of arrows at the origin Parallelogram Law (textbook 1.1) Addition of arrows at a point P O Apply Parallelogram Law at P. (a1, a2) + (b1, b2) = (a1 + b1, a2 + b2) Scalar Vector Multiplication t and x 2 t(a1, a2) = (ta1, ta2) 2 Example: A(–2, 0, 1) and B(4, 5, 3) C is a vector emanating from the origin and having the same direction from A to B. (4, 5, 3) – (–2, 0, 1) = (6, 5, 2) The equation of line through A and B is x = (–2, 0, 1) + t(6, 5, 2). Example: A(1, 0, 2), B(–3, –2, 4), and C(1, 8, –5). A vector from A to B: (–3, –2, 4) – (1, 0, 2) = (–4, –2, 2) A vector from A to C: (1, 8, –5) – (1, 0, 2) = (0, 8, –7) The equation of plane containing A, B, and C is x = (1, 0, 2) + s(–4, –2, 2) + t(0, 8, –7) x = (a1, a2) 2, ai y = (b1, b2) 2, bi x + y = (a1 + b1, a2 + b2) tx = (ta1, ta2) –x = (–1)x = (–a1, –a2) x – y = x + (–1)y = (a1 – b1, a2 – b2) 3 Lecture (Monday, Week 1) Properties of Arrows A1. x, y 2 x + y = y + x commutativity of addition A2. x, y, z 2 (x + y) + z = x + (y + z) associativity of addition A3. x 2 0 2 s. t. x + 0 = x, where 0 = (0, 0) A4. x 2 y 2 s. t. x + y = 0 If x = (a1, a2), then y = (–a1, –a2). A5. x 2 1x = x A6. a, b x 2 (ab)x = a(bx). A7. a x, y 2 a(x + y) = ax + ay. A8. a, b x 2 (a + b)x = ax + bx. Fields Examples: , Q, and C. x2 + 1 = 0 x = i is not algebraically closed. Properties of Fields F1. a + b = b + a and ab = ba commutativity of addition and multiplication F2. (a + b) + c = a + (b + c) and (ab)c = a(bc) associativity of addition and multiplication F3. 0 + a = a and 1a = a existence of additive and multiplicative identity F4. a F b 0 F c, d F s. t. a + c = 0 and bd = 1 existence of additive and multiplicative inverse F5. a(b + c) = ab + ac distributivity of multiplication over addition We denote a field by F. Define a vector space Example: 2, 3, 4, …, n, where n Z+ . Definition: A vector space V over a field “F” consists of a “set” on which two operations “addition and scalar multiplication” are defined so that x, y V ! x + y V and a F x V !ax V such that VS1. x, y V x + y = y + x. VS2. x, y, z V (x + y) + z = x + (y + z). VS3. x V 0 V s. t. x + 0 = x. (What does 0 look like?) VS4. x V y V s. t. x + y = 0, where y is called the additive inverse. (What does y look like?) VS5. x V 1x = x, where 1 is the multiplicative identity of F. VS6. a, b F x V (ab)x = a(bx). VS7. a F x, y V a(x + y) = ax + ay. VS8. a, b F x V (a + b)x = ax + bx. is replaced by F and 2 by V. 4 Elements of F are called scalars. Elements of V are called vectors. Definition: Let ai F for i = 1, 2, …, n, where n Z+. Then x = (a1, a2, …, an) is an n-tuple of scalars from F. Two n-tuples x = (a1, a2, …, an) and y = (b1, b2, …, bn) are equal if and only if ai = bi for i = 1, 2, …, n. Two zero n-tuple is the element 0 = (0, 0, …, 0). xi = ai for i = 1, 2, …, n. yi = bi for i = 1, 2, …, n. x + y = (a1 + b1, a2 + b2, …, an + bn) (x + y)i = xi + yi = ai + bi for i = 1, 2, …, n. Let t F. tx = (ta1, ta2, …, tan) (tx)i = txi = tai for i = 1, 2, …, n. Example: Fn = {x: x = (a1, a2, …, an), ai F for i = 1, 2, …, n} Together with the field F and addition and scalar multiplication as defined is a vector space. i.e. Fn = V Proof of VS2: WTS: (x + y) + z = x + (y + z) x, y, z V. xi = ai, yi = bi, zi = ci for i = 1, 2, …, n. (x + y) + z = ((a1 + b1) + c1, (a2 + b2) + c2, …, (an + bn) + cn) = (a1 + (b1 + c1), a2 + (a2 + c2), …, a3 + (b3 + c3)) by F2 = x + (y + z) 5 Lecture (Wednesday, Week 1) Homework: Section 1.2 #7, 8, 9, 10, 12, 20, 21 Definition: An m n matrix, denoted A, over the field F is a rectangular array of the form a11 a12 … a1n a21 a22 … a2n A= where aij F, 1 i m, 1 j n. … … Am1 am2 … amn Definition: The diagonal elements are those aij such that i = j. a11 a12 a13 Example: A = is a 2 3 matrix with aij F. a21 a22 a23 a31 a32 a33 Definition: The set of all m n matrices over the field F is denoted Mmn(F), n Z+ = {1, 2, …}. Notation: Z* = {0, 1, 2, 3, …} Z– = {-1, -2, -3, …} The components of A are Aij and Aij = aij. Example: Aij = i + j The ith row of A is ri = (ai1, ai2, …, ain) , a “row” vector from Fn. a1j The jth column of A is cj = a2j , a “column” vector from Fn. … Amj Zero matrix is the element 0, where all aij = 0. Two matrices are equal when all of their components are equal. i.e. Aij = Bij aij = bij Addition and Scalar Multiplication: Given A, B Mmn(F), the components of A + B are (A + B)ij = Aij + Bij = aij + bij. 6 a11 a12 a13 Example: A + B = a21 a22 a23 b11 b12 b13 + a31 a32 a33 b21 b22 b23 = b31 b32 b33 a11 + b11 a12 + b12 a13 + b13 a21 + b21 a22 + b22 a23 + b23 a31 + b31 a32 + b32 a33 + b33 Given A Mmn(F) and t F, the components of tA are (tA)ij = tAij = taij. Example: V = (Mmn(F), +, ) is a vector space. VS6 WTS a, b F x V (ab)x = a(bx). Let xij = aij. [(ab)x]ij = (ab)xij = (ab)aij [a(bx)]ij = a(bxij) = a(baij) By F2, (ab)aij = a(baij) ij Definition: A polynomial (x) with coefficients from F is an “expression” of the form (x) = anxn + an-1xn-1 + … + a1x + ao, where ak F, k = 0, 1, 2, …, n. Degree of the polynomial (x) is the largest value k such that ak 0, denoted deg() = k. Addition: Let be a polynomial of degree m. Let g be a polynomial of degree n. (x) = amxm + am-1xm-1 + … + a1x + ao g(x) = bnxn + bn-1xn-1 + … + b1x + bo Without loss of generality, if n m, bm, bm-1, …, bn+1 = 0. g(x) = bmxm + … + bnxn + … + b1x + bo (x) + g(x) = (am + bm)xm + … + (an + bn)xn + … + (a1 + b1)x + (ao + bo) Scalar Multiplication: Let t F and be a polynomial of degree m. Then t(x) = tamxm + tam-1xm-1 + … + ta1x + tao. The zero polynomial is the expression z(x) = 0. z(x) = mxm + m-1xm-1 + … + 1x + o , where k = 0 for k = 0, 1, …, m. deg(z) = –1 Two polynomials are equal if and only if all their coefficients are equal. [(x) + g(x)]k = (ak + bk)xk [tf(x)]k = takxk Example: Pn(F) = {(x): (x) = amxm + am-1xm-1 + … + a1x + ao, ak F for k = 0, 1, …, m} We do not require am 0. deg(f) n V(Pn(F), +, ) is a vector space. 7 Theorem: (1.1) If x, y, z V and x + z = y + z, then x = y. Proof: z V v V such that z + v = 0 (VS4) x = x + 0 = x + (z + v) = (x + z) + v = (y + z) + v = y + (z + v) = y + 0 = y 8 Lecture (Friday, Week 1) Homework: Section 1.3 #3, 4, 5, 6, 7, 12, 20, 28 Section 1.3 Subspaces of Vector Spaces Definition: A subset W of a vector space V over a field F is called a subspace of V if and only if W itself is a vector space over F with the same addition and multiplication as defined on V. Remark: We find that if W is a vector subspace, then the properties VS1, 2, 5, 6, 7, 8 are automatically inherited. i.e. (VS1) x, y V x + y = y + x Four things are not automatically inherited: (VSS1) x, y W x + y W (closure under addition). (VSS2) a F x W ax W (closure under multiplication). (VSS3) x W 0’ W such that x + 0’ = x. (VSS4) x W y’ W such that x + y’ = 0’. Theorem: (1.3) Let V be a vector space and W V. W is a vector subspace of V the following three conditions hold: (a) 0 W, where 0 is the zero vector from V. (b) x + y W x, y W. (c) cx W c F x W. Note: This theorem allows us to check only three properties for subspaces. Proof: Suppose V is a vector space and W V. () Assume W is a vector subspace of V. VSS1 – 4 are satisfied, hence (b) and (c) automatically hold. Since W is a vector space, x W 0’ W such that x + 0’ = x. Since x W, x V, hence 0 V such that x + 0 = x. x + 0’ = x + 0 0’ = 0 W by Theorem 1.1. () Assume (a), (b), and (c). (b) VSS1, (c) VSS2, and (a) VSS3. Let x W. Then (–1)x W by (c). (Here, since 1 F, –1 F.) x + (–1)x W by (a). x + (–1)x = (1 + (–1))x = 0x = 0, where 0 F. Thus we have shown VSS4. 9 Definition: The transpose of an m n matrix A is an n m matrix At whose components satisfy (At)ij = Aji. Remark: Only square matrices (m = n) can be symmetric. Example: A = 1 2 2 1 At = 3 2 2 3 A = At Example: Snn = {A Mmn(F): m = n, A = At} Snn is a vector subspace of Mnn(F). Snn Mnn(F) is clear. Check A, B Snn A + B Snn. (A + B)ij = Aij + Bij = Aji + Bji = (A + B)ji A + B = (A + B)t A + B Snn. The rest of proof is similar. Theorem: (1.4) Any intersection of subspaces of a vector space V is a vector space. Proof: Let W1, W2 be vector subspaces of vector space V. Let W = W1 W2. Then W W1 V and W W2 V. Use Theorem 1.3 Check (a), (b), and (c). (a) 0 W1, W2 0 W (b) x, y W x, y W1, W2 x + y W1, W2 x + y W (c) Let c F and x W x W1, W2 cx W1, W2 cx W Definition: Let A, B Mmn(F). Trace (A) = tr(A) = A11 + A22 + … + Ann (all diagonal elements). Exercise 6: TZnn = {A Mmn(F): tr(A) = 0} TZnn is a vector subspace of Mmn(F). 10 Lecture (Monday, Week 2) Homework: Section 1.4 #10, 12, 13, 14 Quiz 1: Thursday in discussion (Covering homework 1.2 – 1.4) Today: Linear Combination (section 1.4) Definition: Let V be a vector space over a field F and S V. A vector v V is a linear combination of vectors of S if and only if a finite number of vectors u1, …, un S and a1, …, an F such that v = a1u1 + … + anun = i=1n a1u1. In this case, we say v is a linear combination of u1, …, un (or {u1, …, un}). Example: Let u1 = (1, 2), u2 = (1, 3), u3 = (2, 4). Let a1 = 1, a2 = 0, a3 = –1. i=13 aiui = (1, 2) – (2, 4) = (–1, –2) Key Problem: Given v V and a subset S V, V a vector space, is v a linear combination of vectors from S? Example: v = (2, 6, 8) u1 = (1, 2, 1), u2 = (–2, –4, –2), u3 = (0, 2, 3), u4 = (2, 0, –3), u5 = (–3, 8, 16). Is v a linear combination of u1, …, u5? In other word, is v = a1u1 + a2u2 + a3u3 + a4u4 + a5u5? 2 = a1 – 2a2 + 0a3 + 2a4 – 3a5 6 = 2a1 – 4a2 + 2a3 + 0a4 + 8a5 8 = a1 – 2a2 + 3a3 – 3a4 + 16a5 … Row-Reduced Echelon Form a1 – 2a2 + 5a5 = –9 a3 + 3a5 = 7 a4 – 2a5 = 3 Solution set: a1 = 2a2 – a5 – 4 a3 = –3a5 + 7 a4 = 2a5 + 3 (a2, a5) a solution v is a linear combination of (u1, …, u5). Solution: Let a2 = a5 = 0 a1 = –4, a3 = 7, a4 = 3 v = i=15 aiui v is also a linear combination of u1, u3, u4. (2, 6, 8) = –4(1, 2, 1) + 7(0, 2, 3) + 3(2, 0, –3) 11 Example: 3x3 – 2x2 + 7x + 8 is not a linear combination of 1(x) = x3 – 2x2 – 5x – 3, 2(x) = 3x3 – 5x2 – 4x – 9. We want to show that there is no a1, a2 such that (x) = a11(x) + a22(x). Suppose it is true that a1, a2 such that (x) = a11(x) + a22(x). 3x3 – 2x2 + 7x + 8 = (a1 + 3a2)x3 + (–2a1 – 5a2)x2 + (–5a1 – 4a2)x + (–3a1 – 9a2) 3 = a1 + 3a2 –2 = –2a1 – 5a2 7 = –5a1 – 4a2 8 = –3a1 – 9a2 Row-Reduced Echelon Form: a1 + 3a2 = 3 a2 = 4 11a2 = 22 0 = 17 contradiction! Definition: Let S V, V a vector space, S nonempty. Then the span of S denoted span(S) is the set of all linear combination of the vectors in S. Define span() = {0}. Theorem: (1.5) The span of any subset S of a vector space V is a vector subspace of V. Moreover, any vector subspace T that contains S must also contain span(S). Proof: Use Theorem 1.3. a) 0 span(S) b) x + y span(S) x, y span(S) c) cx span(S) x span(S) c F Trivially, 0 span(S). Let x, y span(S) and c F. Suppose S = {u1, …, un}. a1, …, an such that x = i=1n aiui. b1, …., bn such that y = i=1n biui. x + y = i=1n (ai + bi)ui, where ai + bi F x + y span(S) cx = i=1n (cai)ui, where cai F cx span(S) 12 Discussion (Tuesday, Week 2) Tutoring Center: RH 414, W 10am – 12pm Even Functions Closure under addition Let f, g be even functions. Let x . ( + g)(–x) = (–x) + g(–x) = (x) + g(x) because , g even = ( + g)(x) Hence + g is a even function. Closure under scalar multiplication Let be even, , and x . ()(–x) = (–x) = (x) = ()(x) Hence is even. because even 13 Section 1.3 #28 If Mt = –M, then we call M Mnn(F) skew-symmetric. Let W1 = {M: Mt = –M} Mnn(F). WTS W1 is a subspace. a) 0 = (0) 0t = (0) = –(0) Hence 0 W1. b) Let A, B W1. (A + B)t = At + Bt by problem #3 = –A + (–B) = –(A + B) Hence A + B W1. c) Let A W1 and F. (A)t = At by problem #3 = (–A) = –(A) Hence A W1. Let W2 = {M Mnn(F): Mt = M}. If F has characteristic > 2, then Mnn(F) = W1 W2. Let Eij be n n matrix whose entries are all zero except for having value 1 in the i th row and jth column. B = {Eij: 1 i, j n} is the standard basis for Mnn(F). A W1, B W2 Aij = 1/2, Aji = –1/2, and all zero except for the other entries. Bij = 1/2, Bji = 1/2, and all zero except for the other entries. Eij = Aij + Bij Let M Mnn(F). Since B is a basis, there are ij F such that M = i,jn ijEij = i,jn ij(Aij + Bij) = i,jn ijAij + i,jn ijBij, where i,jn ijAij W1 and i,jn ijBij W2. Hence M W1 W2. 14 Lecture (Wednesday, Week 2) Homework: Section 1.5 #5, 9, 14, 16 Example: In 3, Let S = {(1, 0, 0), (0, 1, 0), (0, 0, 1)}. Let (a, b, c) 3. Then (a, b, c) = a(1, 0, 0) + b(0, 1, 0) + c(0, 0, 1). Definition: A subset S V, V a vector space, is called linearly dependent if and only if a finite number of vectors u1, …, un S and a finite number of scalars a1, …, an F (not all zero) such that a1u1 + … + anun = 0, a non-trivial representation of the zero vector. Remark: Trivially, if ai= 0 i = 1, …, n, then i=1n aiui = 0. Example: Consider S = {(1, 3, -4, 2), (2, 2, -4, 0), (1, -3, 2, 4), (-1, 0, 1, 0)}. i=14 aiui = a1(1, 3, -4, 2) + a2(2, 2, -4, 0) + a3(1, -3, 2, 4) + a4(-1, 0, 1, 0) = (0, 0, 0, 0) a1 + 2a2 + a3 – a4 = 0 3a1 + 2a2 – 3a3 + 0a4 = 0 – 4a1 – 4a2 + 2a3 + a4 = 0 0a1 + 2a2 + 4a3 + 0a4 = 0 a1 = 4, a2 = -3, a3 = 2, a4 = 0 S is linearly dependent. Definition: A subset S V is called linearly independent if and only if S is not linearly dependent. finite collection of vectors u1, …, un S and a1, …, an F, i=1n aiui 0 unless (of course) ai = 0 i. Example: S = {(1, 0, 0, -1), (0, 1, 0, -1), (0, 0, 1, -1), (0, 0, 0, 1)} i=14 aiui = a1(1, 0, 0, -1) + a2(0, 1, 0, -1) + a3(0, 0, 1, -1) + a4(0, 0, 0, 1) = (0, 0, 0, 0) a1 = 0 a2 = 0 a3 = 0 – a1 – a2 – a3 + a 4 = 0 a4 = 0 S is linearly independent. Example: S = {(1, 0, 0), (0, 1, 0), (-2, -1, 0)} i=13 aiui = a1(1, 0, 0) + a2(0, 1, 0) + a3(-2, -1, 0) = (0, 0, 0) a1 – 2a3 = 0 a1 = 2a3 a2 + a 3 = 0 a2 = -a3 Clearly, a1 = 2, a2 = -1, a3 = 1 is a solution (actually, infinitely many solutions). S is linearly dependent. 15 Theorem: (1.6) Let V be a vector space and S1 S2 V. If S1 is linearly dependent, then S2 is also linearly dependent. Proof: (Exercise 12) If S1 is linearly dependent, u1, …, un S1 and a1, …, an F such that i=1n aiui = 0 and not all ai = 0. Since S1 S2, u1, …, un S2 and a1, …, an F. Thus u1, …, un S2 and a1, …, an F such that i=1n aiui = 0, where not all ai = 0. S2 is linearly dependent. Corollary: S1 S2 V. If S2 is linearly independent, then S1 is also linearly independent. Proof: (Exercise 12) Proof by contra-positive: (A B) (B’ A’) (S1 is linearly dependent) (S2 is linearly dependent) Not(S2 is linearly dependent) Not(S1 is linearly dependent) (S2 is linearly independent) (S1 is linearly independent) Theorem: (1.7) Let S be a linearly independent subset of a vector space V and let v V such that v S. Then S {v} is linearly dependent if and only if v span(S). Proof: () If v span(S), then v is a linear combination of vectors from S. i.e. u1, …, un S and a1, …, an F such that v = a1u1 + … + anun. a1u1 + … + anun + (-1)v = 0. Set an+1 = -1 and un+1 = v. i=1n+1 aiui = 0, where not all ai = 0 because an+1 0. () If S {v} is linearly dependent, then u1, …, un S and a1, …, an, an+1 F such that a1u1 + … + anun + an+1v = 0, where not all ai = 0. a1u1 + … + anun = (-1)an+1v v = (–a1/an+1)u1 + … + (–an/an+1)un, where (–ai/an+1) F i. v is a linear combination of vectors u1, …, un from S. v span(S) 16 Lecture (Friday, Week 2) Homework: Read the 1st half of section 1.6. Section 1.6 Bases and Dimension Definition: A basis for a vector space V is a linearly independent subset of V that generates V (span() = V). If is a basis for V, we also say that the vectors of form a basis for V. Example: span() = {0} and is linearly independent by definition. Thus is a basis for {0}, the zero vector space. Example: In Fn, let e1 = (1, 0, 0, …, 0), e2 = (0, 1, 0, …, 0), …, en = (0, 0, …, 0, 1). S = {e1, e2, …, en} is linearly independent span(S) = Fn Therefore S is a basis for Fn; called the standard (or canonical) basis. Example: In Pn(F), the set {1, x, x2, …, xn} is a basis for Pn(F). (x) = anxn + … + a1x + ao = {1, x, x2, …, xn} is called the standard basis for Pn(F). Example: P(F) is the set of all polynomials of any degree. The set {1, x, x2, x3, …} is a basis for P(F). Theorem: (1.8) Let V be a vector space and = {u1, …, un} be a subset of V. Then is a basis v V ! linear combination of vectors from . i.e. ! linear scalars a1, …, an F such that v = i=1n aiui. Proof: () Let be a basis for V span() = V Let v V. Since span() = V, a1, …, an F such that v = i=1n aiui. Suppose another set of scalars b1, …, bn F such that v = i=1n biui. v – v = i=1n (ai – bi)ui ai – bi = 0 ai = bi () Exercise 19 Let v V and suppose !ai, …, an F such that v = i=1n aiui. We know that span() = V. Uniqueness implies that if v = i=1n biui, then ai = bi for i = 1, …, n. In particular, 0 V. Thus ! linear combination i=1n ciui = 0. If di = 0 for i = 1, …, n, then we trivially have 0 = i=1n diui. ci = di for i = 1, …, n. is linearly independent. 17 Theorem: (1.9) If a vector space V is generated by a finite set S (span(S) = V), then some subset of S is a basis for V. Hence, V has a finite basis. Proof: S = and S = {0} are trivial. Algorithm: S contains a nonzero vector u1. Set 1 = {u1}. Then 1 is linearly independent. Pick v2 0 such that 2 = {v1, v2} and 2 is linearly independent. Continue until k = {v1, v2, …, vk}. Stop when vk+1 = 0 or vk+1 0 but Bk {vk+1} is linearly dependent. Claim: = k = {u1, …, uk} is a basis for V. Proof: is linearly independent. It remains to check if span() = V. We know S V span (S) = V. If = S, then span() = span(S) = V. Now, suppose S. Let v V. Then a1, …, ak, b1, …, bm F, m > 0 such that v = i=1k aiui + j=1m bjvj, where vj are the vectors in S\B. B {vj} is linearly dependent j = 1, …, m. a1, …, ak+1 F such that i=1k aiui + ak+1vj = 0, not all ai = 0. In particular, ak+1 0. i=1k aiui = – ak+1vj i=1k (–ai/ak+1)ui = vj Let –ai/ak+1 = ai(j). Then vj = i=1k ai(j)ui. For vector v V, v = i=1k aiui + j=1m bj(i=1k ai(j)ui), where ui . v V can be written as a linear combination of vectors from . span() = V. 18 Lecture (Monday, Week 3) Homework: Section 1.6 #5, 6, 9, 14, 20, 21, 28 Disregard Lagrange Interpretation Polynomial. Example: S = {(1, 0, 0), (0, 1, 0), (2, 0, 0), (0, 0, -3)} span(S) = V = F3 = {(1, 0, 0), (0, 1, 0), (0, 0, 1)} = {e1, e2, e3} is a canonical basis for F3. ’ = {(2, 0, 0), (0, 1, 0), (0, 0, -3)} = {2e1, e2, -3e3}. A basis is {a1e1, a2e2, a3e3} for any nonzero scalars ai F. Theorem: (1.10) Let V be a vector space that is generated by G V, where #G = n. And let L V be linearly independent, where #L = m. Then m n and H G such that #H = n – m and L H generates V. Remark: We cannot find a linearly independent set of # larger than n. Example: S = {(1, 0, 0), (0, 1, 0), (0, 0, 1), (0, -3, 0)} span(S) = V = F3 S is not linearly independent. = {(1, 0, 0), (0, -3, 0), (0, 0, 1)} is a linearly independent subset of V. Also, ’ = {(1, 0, 0), (0, -3, 0)} is a linearly independent subset of V. ” = {(2, 0, 0), (0, -1, 0) is a linearly independent set of V. L is not a subset of G necessarily. Replacement Theorem Corollary 1: Let V be a vector space having a finite basis (# < ). If is another basis for V, then # = #. Proof: Let n = #, m = #. If m > n, we can select a subset S such that #S = n + 1. S is linearly independent because is a basis (hence linearly independent). Since generates V and S V, by the Replacement Theorem, n + 1 n. Thus m n. If we assume m < n, then it leads to n m. Therefore m = n. Definition: A vector space V is called finite-dimensional if and only if it has a basis such that # < . The unique number # = n is called the dimension of V, denoted dim(V) = n. Example: The vector space Pn(F) has dimension n + 1 = {1, x, x2, …, xn} 19 Example: The vector space P(F) is infinite-dimensional = {1, x, x2, …} Corollary 2: Let V be a vector space and dim(V) = n. a) If G V, #G < , and span(G) = V, then #G n. In particular, if #G = n, then G is a basis. b) If L V is linearly independent and #L = n, then L is a basis. c) Proof: If L V s linearly independent and #L < n, then L may be extended to a basis for V. Let V be a vector space and dim(V) = n. a) Suppose #G = m < and span(G) = V. By theorem 1.9, H G such that H is a basis. By Corollary 1, #H = n. Thus #G n. If #G = n, then H = G. Thus G is a basis. Theorem: (1.11) Let W be a subspace of a finite-dimensional vector space V. Then W is finite-dimensional and dim(W) dim(V). If dim(W) = dim(V), then W = V. Corollary: If W is a subspace of a vector space V, V finite-dimensional, then any basis for W can be extended to a basis for V. Example: W = {v = {v1, v2, v3} F3: v3 = 0} F3 (like F2) S = {(1, 0, 0), (0, 1, 0)} is a basis for W. = S {(0, 0, 1)} is a basis for F3. 20 Discussion (Tuesday, Week 3) Section 1.5 #16 S V is linearly independent every finite subset of S is linearly independent. () Assume S is linearly independent. Let S’ be a finite subset of S. Let u1, …, un S’ and a1, …, an F such that a1u1 + … + anun = 0. Since S’ S, u1, …, un S. Since S is linearly independent, a1 = … = an = 0. () Assume every finite subset of S is linearly independent. Let v1, …, vm S and a1, …, am F such that a1v1 + … + amvm = 0. But S’ = {v1, …, vm} S is finite and linear independent. Hence, a1 = … = am = 0. 21 Section 1.6 #21 V is an infinite dimensional vector space V contains an infinite linearly independent subset. () Proof by contra-positive If V is finite dimensional vector space, then V does not contain an infinite linearly independent subset. Assume dim(V) = n < . Let S V be linearly independent. Then #S n by Replacement Theorem. Hence S is finite. () Construct a family of subsets of V in the following way. Construct S1: Since V is infinite dimensional v1 V such that v1 0 Let S1 = {v1}. Inductive step: Suppose we have constructed Sn = {v1, …, vn} linearly independent. Since span(Sn) V (because dim(V) > n), vn+1 V such that {v1, …, vn+1} is linearly independent. Let Sn+1 = {v1, …, vn+1} Now, we have constructed {Sn: n }. Set S = n Sn. Claim 1: S V. Proof 1: Let v S = n Sn. no such that v Sn_o V. Hence v V. Claim 2: #S = Proof 2: Assume not; S is finite. N such that sm = SN m N. Consider SN+1. By the construction, {v1, …, vn+1} is linearly independent. Hence vn+1 vi for i n. SN+1 SN. Contradiction! Claim 3: S is linearly independent. Proof 3: S is linearly independent every finite subset is linearly independent. Let S’ S, where #S’ < . x S’ n(x) such that x = vn(x). Let M = maxxS’{n(x)}. Hence S’ SM is linearly independent by the construction. S’ is linearly independent. 22 Lecture (Wednesday, Week 3) Section 2.1 Linear Transformation Definition: Let V, W be two vector spaces over F. We call T: V W a linear transformation (linear operator) if and only if x, y V and c F, a) T(x + y) = T(x) + T(y) b) T(cx) = cT(x) Properties: (Exercise 7) a) If T is linear, then T(0V) = 0W. T(0V) = T(x – x) = T(x) + T(–x) = T(x) + (–1)T(x) = T(x) – T(x) = 0W b) T is linear if and only if T(cx + y) = cT(x) + T(y) x, y V c F. c) If T is linear, then T(x – y) = T(x) – T(y) x, y V. d) T is linear if and only if x1, …, xn V a1, …, an F T(i=1n aixi) = i=1n aiT(xi). Example: T: 2 2 T((a1, a2)) = (a1, –a2) Reflection Transformation. Let x = (a1, a2), y = (b1, b2). T(cx + y) = T(c(a1, a2) + (b1, b2)) = T(ca1 + b1, ca2 + b2) = (ca1 + b1, –(ca2 + b2)) = c(a1 – a2) + (b1 – b2) = cT(x) + T(y) This shows that T is a linear transformation. Example: Let V = C() = {(x) : (x) is continuous x }. Consider T() = ba (t) dt. Let (x), g(x) C() and c F. T(c + g) = ba [c(t) + g(t)] dt = ba c(t) dt + ba g(t) dt = cba (t) dt + ba g(t) dt = cT() + T(g) This shows that T is a linear transformation. Definition: IV: V V defined as IV(v) = v v V is called the identity transformation. T0: V W defined as T0(v) = 0W v V is called the zero transformation. Both of these are linear transformations. Definition: Let V, W be vector spaces and let T: V W be linear. The null space (or kernel) is defined as N(T) = {x V: T(x) = 0W}. The range (or image) is R(T) = {y W: x V such that T(x) = y} = {T(x): x V}. 23 Theorem: (2.1) Let V, W be vector spaces and T: V W be linear. Then N(T) is vector subspace of V and R(T) is a vector subspace of W. Proof: Let 0V = 0 V, 0W = 0 W. To show G V is a vector subspace, need to show (Theorem 1.5) a) 0G b) x + y G x, y G c) cx G x G c F a) T(0V) = 0W Hence 0V N(T) and 0W R(T). b) Let x, y N(T). T(x) = T(y) = 0W T(x + y) = T(x) + T(y) = 0W + 0W = 0W Hence x + y N(T). Let x, y R(T). Then x’, y’ V such that T(x’) = x and T(y’) = y. Set z = x’ + y’. Then z V. T(z) = T(x’ + y’) = T(x’) + T(y’) = x + y z such that T(z) = x + y. Hence x + y R(T). c) Let c F and x N(T). Then T(x) = 0W T(cx) = cT(x) = c0W = 0W Hence cx N(T). Let c F and y R(T). x V such that T(x) = y. T(cx) = cT(x) = cy. Set cx = z. z V such that T(z) = T(cx) = cy. Hence cy R(T). 24 Lecture (Friday, Week 3) Homework: Section 2.1 #7, 8, 13 Midterm: 5th week, Friday Theorem: (2.2) Let V, W be vector spaces and T: V W be linear. If = {v1, …, vn} is a basis for V, then R(T) = span(T()), where T() = {T(v1), …, T(vn)}. In other words, the set of image of is the range. Example: Consider T: P2() M22(). T((x)) = f (1) f (2) 0 is linear. f (0) 0 = {1, x, x2} is basis for P2(). R(T) = span(T()) = span({T(1), T(x), T(x2)}) 0 0 = span 0 0 = a1 0 1 , 1 0 0 1 + a2 1 0 0 0 = span 0 1 , 1 0 0 3 , 0 0 0 0 0 3 + a3 0 0 0 0 0 0 0 Since these two matrices are linearly independent, 0 0 1 , 1 0 0 is a basis of R(T). 0 dim(R(T)) = 2 Definition: Let V and W be vector spaces, T: V W be linear. Then nullity(T) = dim(N(T)) and rank(T) = dim(R(T)). Remember N(T) is a vector subspace of V and R(T) is a vector subspace of W. Back to the previous example: dim(R(T)) = rank(T) = 2 What is dim(N(T))? N(T) = {(x) P2(): T((x)) = 0} = {(x) P2(): (1) – (2) = 0 and (0) = 0} (0) = ao + a1(0) + a2(0)2 (0) = 0 ao = 0 (1) = ao + a1(1) + a2(1)2 = a1 + a2 (2) = ao + a1(2) + a2(2)2 = 2a1 + 4a2 25 (1) – (2) = 0 –a1 – 3a2 = 0 a1 = –3a2 a2 = (–1/3)a1 (x) = ao + a1x + a2x2 = a1x – (1/3)a1x2 = {x – (1/3)x2} dim(N(T)) = 1 nullity(T) = 1, rank(T) = 2, dim(V) = 3 nullity(T) + rank(T) = dim(V) Theorem: (2.3) “Dimension Theorem” Let V and W be vector spaces and T: V W be linear. If V is finite-dimensional, then nullity(T) + rank(T) = dim(V). Proof: Suppose dim(V) = n, dim(N(T)) = k n. Remember that N(T) is a vector subspace of V. Let N = {v1, …, vk} be a basis for N(T). By the corollary to theorem 1.11 (page 51), N may be extended to = {v1, …, vk, …, vn} for V. Claim: S = {T(vk+1), …, T(vn)} is a basis for R(T). WTS 1) span(S) = R(T). WTS 2) S is linearly independent. 1) Use theorem 2.2 R(T) = span(T()) = span({T(v1), …, T(vk), …, T(vn)}) = span({0, …, 0, T(vk+1), …, T(vn)}) since v1, …, vk N(T) = span({T(vk+1), …, T(vn)}) = span(S) 2) Suppose i=k+1n biT(vi) = 0 for bk+1, …, bn F. Since T is linear, T(i=k+1n bivi) = 0. i=k+1n bivi N(T). ! C1, …, ck F such that i=k+1n bivi = i=1k civi. i=k+1n (–ci)vi + i=k+1n bivi = 0, where v1, …, vk, vk+1, …, vn , a basis for V. Since = {v1, …, vk, vk+1, …, vn} is linearly independent, ai = 0 and bi = 0. Hence, S = {T(vk+1), …, T(vn)} is linearly independent. #S = n – k dim(R(T)) = rank(T) = n – k nullity(T) = k, rank(T) = n – k, and dim(V) = n. Hence, nullity(T) + rank(T) = k + (n – k) = n = dim(V). 26 Lecture (Monday, Week 4) Homework: Section 2.1 #14, 16, 17, 18, 20 Midterm: Section 1.1 – 2.3 Definition: Let T: V W be a mapping from a set V to a set W. Then T is called one-to-one y R(T) ! x V such that T(x) = y. T(x) = T(y) x = y x y T(x) T(y) Definition: T is onto R(T) = W. Theorem: (2.4) Let V and W be vector spaces and T: V W be linear. Then T is one-to-one N(T) = {0V} Proof: () Assume T is one-to-one. Let x N(T). Then T(x) = 0W. Now, it is always true that T(0V) = 0W. T(x) = T(0V) x = 0v () Assume N(T) = {0V}. Let x, y V such that T(x) = T(y). 0W = T(x) – T(y) = T(x – y) = T(0V) x – y = 0V Thus, x = y. Hence T is one-to-one. Theorem: (2.5) Let V and W be vector spaces of equal finite dimension. Let T: V W be linear. Then the followings are equivalent. (a) T is one-to-one. (b) T is onto. (c) rank(T) = dim(V). (d) Proof: From the Dimension Theorem, nullity(T) + rank(T) = dim(V) = dim(W). (a) (b) Suppose T is one-to-one. By theorem 2.4, N(T) = {0V}. nullity(T) = dim(N(T)) = dim({0V}) = 0 rank(T) = dim(R(T)) = dim(V) = dim(W) Since R(T) is a vector subspace of W and dim(R(T)) = dim(W), R(T) = W. Hence, T is onto. (b) (c) Suppose T is onto. Then R(T) = W. rank(T) = dim(W) = dim(V) 27 (c) (a) Suppose rank(T) = dim(V) = dim(W). Then nullity(T) = 0 N(T) = {0V} By theorem 2.4, T is one-to-one. Example: T(a1, a2) = (a1 + a2, a2) T: F2 F2 is linear. Claim: N(T) = {0} = {(0, 0)} {0} N(T) is trivial. WTS N(T) {0}. Let x N(T). Then x = (a1, a2). T(x) = T(a1, a2) = (a1 + a2, a2) = (0, 0) a1 + a2 = 0 and a2 = 0 a1 = a2 = 0 x = (a1, a2) = (0, 0) = 0 F2 x {0}. Theorem: (2.6) Let V and W be vector spaces over F, and suppose = {v1, …, vn} is a basis of V. For (arbitrary) vectors w1, …, wn W, ! linear transformation T: V W s.t. T(vi) = wi, i = 1, …, n. Idea of Proof: Let x V. ! a1, …, an F such that x = i=1n aivi. Define T: V W by T(x) = i=1n aiwi. We need to show 3 things. (1) T is linear. (2) T(vi) = wi If x = vi, then aj = 0, j i, ai = 1. T(vi) = T(x) = j=1n ajwj = wj (3) Uniqueness of T Corollary: Let V and W be vector spaces, and suppose V has a finite basis = {v1, …, vn}. If T: V W and U: V W are linear and T(vi) = U(vi), for i = 1, …, n, then T = U. Proof: Let v V, and suppose T(v) = w W. Since = {v1, …, vn} is a basis for V, ! a1, …, an F such that v = i=1n aivi. T(v) = T(i=1n aivi) = i=1n aiT(vi) = i=1n aiU(vi) = U(i=1n aivi) = U(v) Hence, T = U. 28 Discussion (Tuesday, Week 4) T: V W over F N(T) = {v V: T(v) = 0} R(T) = {w W: v V such that T(v) = w} T is one-to-one N(T) = {0} T is onto R(T) = W Section 2.1 Problem #14: Let V and W be vector spaces and T: V W be linear. a) Prove that T is one-to-one if and only if T carries linearly independent subsets of V onto linearly independent subsets of W. i.e. T is one-to-one If A V is linearly independent, then T(A) = {w W: v A such that T(v) = w} W is linearly independent. () Assume T is one-to-one. Let A V be linearly independent. Let w1, …, wn T(A) and a1, …, an F such that i=1n aiwi = 0. WTS ai = 0 i = 1, …, n. Since i = 1, …, n vi A such that T(vi) = wi, i=1n aiwi = i=1n aiT(vi). i=1n aiwi = i=1n aiT(wi) = T(i=1n aivi) = 0 Since T is one-to-one, N(T) = {0}. T(i=1n aivi) = T(0) = 0 i=1n aivi = 0 ai = 0 i= 1, …, n because A is linearly independent. () T is one-to-one N(T) = {0} (T(v) = 0 v = 0) Suppose T(v) = 0. Let be basis for V. v1, …, vm and a1, …, am F such that v = i=1m aivi. T(v) = T(i=1m aivi) = i=1m aiT(vi) = 0, where T(vi) T(). Since is linearly independent, T() is linearly independent. Hence, ai = 0 i = 1, …, m. v = i=1m aivi = i=1m 0vi = 0 b) Suppose that T is one-to-one and that S is a subset of V. Prove that S is linearly independent if and only if T(S) is linearly independent. () Always true by part a) (). () Assume T(S) is linearly independent. Let s1, …, sn S and a1, …, an F such that i=1n aisi = 0. T(i=1n aisi) = T(0) = 0 Since T is linear, i=1n aiT(si) = 0 Since T(si) is linearly independent, ai = 0 i = 1, …, n. 29 c) Suppose = {v1, …, vn} is a basis for V and T is one-to-one and onto. Prove that T() = {T(v1), …, T(vn)} is a basis for W. T() is linearly independent by part b) (). It remains to show that T() spans W. Let w W. Since T is onto, v V such that T(v) = w. a1, …, an F such that v = i=1n aivi. w = T(v) = T(i=1n aivi) = i=1n aiT(vi), where T(vi) T() i = 1, …, n. Hence, T() spans W. Section 2.1 Problem #18: Give an example of linear transformation T: 2 2 such that N(T) = R(T). Note that T(N(T)) = 0. T(N(T)) = T(R(T)) = 0 T2 = 0 Consider a 2 2 upper triangular matrix, where aij = 0 for i = j. 0 0 a 0 0 0 a 0 0 0 0 0 0 N(T) R(T) and R(T) N(T) N(T) = R(T) 30 Lecture (Wednesday, Week 4) Homework: Section 2.2 #4, 6, 8, 9, 13 Definition: An ordered basis for V is a basis for V endowed with a specific ordering. Example: Consider F3. = {e1, e2, e3} where e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1). = {e2, e3, e1} as an ordered basis. = {ei}i=1n is an ordered basis for Fn. We call {ei}i=1n the standard ordered basis for Fn. Example: {xi}i=1n is the standard ordered basis for Pn(F). Definition: Let = {u1, …, un} = {ui}i=1n be an ordered basis for V. x V let a1, …, an F be the unique scalars such that x = i=1n aiui. a1 The coordinate vector of x relative to is [x] = ... . an Notice [ui] = ei. The mapping T: V Fn defined by T(x) = [x] is linear. Definition: Let V and W be vector spaces with ordered basis = {vi}i=1n and = {wj}j=1m, respectively. Let T: V W be linear. ! aij F, 1 i m such that T(vj) = i=1m aijwi, 1 j n. Why? Where? T(v) = i=1m aiwi for some j, T(vj) = i=1m aijwi. The mn matrix A Mmn(F) defined via Aij = aij is called matrix representation of T in the ordered basis , . We write A = [T]. If V = W and = , then we write A = [T]. Example: T: 2 3 T((a1, a2)) = (a1 + 3a2, 0, 2a1 – 4a2) = {(1, 0), (0, 1)} is a basis for 2. = {(1, 0, 0), (0, 1, 0), (0, 0, 1)} is a basis for 3. Now, T(e1) = T(1, 0) = (1, 0, 2) = 1e1 + 0e2 + 2e3 a11 = 1, a21 = 0, a31 = 2. T(e2) = T(0, 1) = (3, 0, –4) = 3e1 + 0e2 + (–4)e3 a12 = 3, a22 = 0, a32 = –4. 1 [T] = 0 2 3 0 4 31 Definition: Let T: V W and U: V W be transformation from a vector space V to a vector space W. Let a F. Then we define a) T + U: V W by (T + U)(x) = T(x) + U(x) x V. b) aT: V W by (aT)(x) = aT(x) x V. Theorem: (2.7) Let V and W be vector spaces over F, and let T: V W and U: V W be linear. Then a) a F aT + U is a linear transformation. b) Using (+) and () defined above, the set of all linear transformations from V to W is a vector space. To = zero transformation is the zero vector. Definition: That vector space is called L(V, W). If V = W, we write L(V). Theorem: (2.8) Let V and W be finite-dimensional vector spaces with ordered bases and , respectively. Let T: V W and U: V W be linear. Then a) [T + U] = [T] + [U] Mmn(F). b) [aT] = a[T] Mmn(F). Proof: Let = {vi}i=1n and = {wj}j=1m. a) ! aij, bij F, 1 i m, 1 j n such that T(vj) = i=1m aijwi and U(vj) = i=1m bijwi. Hence, (T + U)(vj) = T(vj) + U(vj) = i=1m aijwi + i=1m bijwi = i=1m (aij + bij)wi Thus, ([T + U])ij = aij + bij = ([T] + [U])ij [T + U] = [T] + [U] b) (aT)(vj) = aT(vj) = ai=1m aijwi = i=1m aaijwi. ([aT])ij = aaij = (a[T])ij [aT] = a[T] 32 Lecture (Friday, Week4) Section 2.3 Composition of Linear Transformations and Matrix Multiplication Theorem: (2.9) Let V, W, Z be vector spaces. Let T: V W and U: W Z be linear. Then UT is linear. Remark: UT(v) = U(T(v)) Z, where T(v) W. Proof: UT(ax + y) = U(T(ax + y)) = U(aT(x) + T(y)) because T is linear. = aU(T(x)) + U(T(y)) because U is linear. = aUT(x) + UT(y) Theorem: (2.10) Let V be a vector space. Let T, U1, U2 L(v). Then (a) T(U1 + U2) = TU1 + TU2 (U1 + U2)T = U1(T) + U2(T) (b) T(U1U2) = (TU1)U2 [= TU1U2] (c) TI = IT = T, where I(v) = v v V. (d) A(U1U2) = (aU1)U2 = U1(aU2) a F. Proof: (Exercise 8) (a) Let v V. T(U1 + U2)(v) = T((U1 + U2)(v)) = T(U1(v) + U2(v)) = T(U1(v)) + T(U2(v)) = TU1(v) + TU2(v) = (TU1 + TU2)(v) T(U1 + U2) = TU1 + TU2 33 Let T: V W and U: W Z be linear. Let V, W, Z be vector spaces. Let = {vi}i=1n, = {wi}i=1m, = {zi}i=1p be ordered bases for V, W, Z, respectively. Let A = [U] Mpm. Let B = [T] Mmn. Define C Mpn by Cij = k=1m AikBkj, 1 i p, 1 j n. Definition: C Mpn is the unique matrix representation of [UT]. UT(vj) = i=1p Cijzi UT(vj) = U(T(vj)) = U(k=1m Bkjwk) = k=1m BkjU(wk) = k=1m Bkj(i=1p Aikzi) = i=1p (k=1m AikBkj) zi = i=1p Cijzi Thus ([UT])ij = Cij. Definition: A Mpm and B Mmn. Then the product of A and B is the p n matrix C whose components are Cij = k=1m AikBkj, 1 i p, 1 j n. Example: If A M32 and B M22, then C M32. 34 Lecture (Monday, Week5) Friday: Midterm Exam I (sections 1.1 – 1.6 & 2.1 – 2.2) Today (after 5pm): Practice exam on website Theorem: Suppose A Mmn, B Mnp. Then (AB)t = BtAt. Remark: AB = C Mmp and BtAt = Ct Mpm Proof: Recall (At)ij = Aji and (Bt)ij = Bij. [(AB)t]ij = (AB)ji = k=1n AjkBki = k=1n (At)kj(Bt)ik = k=1n (Bt)ik(At)kj = (BtAt)ij Hence, (AB)t = BtAt. Theorem: (2.4) Let V, W, Z be vector spaces of finite dimension with ordered bases , , , respectively. Let T: V W and U: W Z be linear. Then [UT] = [U][T]. Proof: Done in the previous lecture. Definition: We define the kronecker delta ij by ij = 1, for i = j. ij = 0, for i j. The n n identity matrix In is defined by (In)ij = ij, for 1 i, j n. 1 0 0 Example: I3 = 0 1 0 multiplicative (matrix product) identity 0 01 0 0 0 03 = 0 0 0 the additive identity 0 0 0 35 Theorem: (2.12) Let A Mmn, B, C Mnp, and D, E Mqm. Then a) A(B + C) = AB + AC (D + E)A = DA + EA b) a(AB) = (aA)B = A(aB) a F c) ImA = A = AIn Mmn d) If dim(V) = n, where V is a vector space with ordered basis , then [Iv] = In. Iv(v) = v V. Proof: a) [A(B + C)]ij = k=1n AikMkj = k=1n Aik(Bkj + Ckj) = k=1n AikBkj + AikCkj = k=1n AikBkj + k=1n AikCkj = (AB)ij + (AC)ij Hence, A(B + C) = AB + AC. [(D + E)A]ij = k=1m MikAkj, 1 i q, 1 j n. = k=1m (Dik + Eik)Akj = k=1m DikAkj + EikAkj = k=1m DikAkj + k=1n EikAkj = (DA)ij + (EA)ij Hence, (D + E)A = DA + EA. b) – d) See the book & Exercise 5. Corollary: Let A Mmn, B1, …, Bk Mnp, C1, …Ck Mqm, and a1, …, ak F. Then Ai=1k (aiBi) = i=1k aiABi and i=1k (aiCi)A = i=1k aiCiA Definition: Let A Mmn, k {0, 1, 2, 3, …}. Ak = AA … A (k times), for k > 0. Ak = In, for k = 0 36 Lecture (Wednesday, Week 5) Theorem: (2.13) Let A Mmn(F) and B Mnp(F). ( AB ) 1j and v = For each j, 1 j p, let uj = ... j ( AB ) mj a) B1 j ... . Then B nj uj = Avj. b) vj = Bej. Remark: Avj is a matrix vector multiplication defined by assuming v j is an n 1 matrix. n A1k B kj ( AB ) B1 j 1j k 1 Proof: a) uj = ... = A ... = Avj = ... ( AB ) B mj n mj Amk B kj k 1 b) See Exercise 6. Theorem: (2.14) Let V and W be vector spaces of finite-dimension with ordered bases and , respectively. Let T: V W be linear. Then u V, [T(u)] = [T][u]. Proof: See the book. Example: Let T: P3() P2() defined by T((x)) = ’(x). = {xi}i=03 and = {xi}i=02 are basis for P3() and P2(). 0 1 0 0 Let A = 0 0 2 0 . 0 0 0 3 Claim: [T] = A. T(vj) = i=1m aijwi T(1) = 01 + 0x + 0x2 T(x) = 11 + 0x + 0x2 T(x2) = 01 + 2x + 0x2 T(x3) = 01 + 0x + 3x2 Consider p(x) = 2 – 4x + x2 + 3x3. T(p(x)) = q(x) = p’(x) = – 4 + 2x + 9x2 2 4 and [T(p(x))] = [q(x)] = [p(x)] = 1 3 4 2 9 37 [T][p(x)] = [T(p(x))] 0 1 0 0 0 0 2 0 0 0 0 3 2 4 4 = 2 1 9 3 Definition: Let A Mmn(F). Denote LA : Fn Fm by LA(x) = Ax x Fn. It is called the left matrix multiplication transformation. Remark: (m n)(n 1) = (m 1) Fm. Example: A = 1 21 0 1 2 M23() LA: 3 2 Ax = b M21() 1 1 21 LA 3 = 0 1 2 1 1 3 = 6 1 1 Theorem: Let A Mmn(F). Then LA : Fn Fm is linear. If B Mmn(F), = {ei}i=1n, = {ei}i=1m, then a) [LA] = A b) LA = LB A = B i.e. Ax = Ab a = B c) LA+B = LA + LB i.e. (A + B)x = Ax + Bx d) If T: Fn Fm is linear, then ! C Mmn such that T = LC and C = [T]. i.e. T(x) = Cx e) If E Mmp(F), then LAE = LALE. f) If m = n, LI_n = IF^n i.e. Inx = x = IF^n(x) Theorem: (2.16) Let A, B, C be matrices such that A(BC) is well-defined (compatible). Then A(BC) = (AB)C associativity of matrix products. Proof: 1) From the definition of AB. 2) Use properties e) and b) in the previous theorem. 38 Lecture (Monday, Week 6) Homework: Read Appendix B and section 2.4. Today: Section 2.4 Definition: Let V and W be vector spaces and T: V W be linear. A function U: W V is called an inverse of T if and only if TU = Iw and UT = Iv. If T has an inverse, it is called invertible. Remark: If T is invertible, then the inverse of T is unique and is denoted T -1 (Appendix B). Remark: The following hold for invertible function T and U. (1) (TU)-1 = U-1T-1 TUU-1T-1 = TIT-1 = TT-1 = I (2) (T-1)-1 = T T-1T = T and TT-1 = T (3) Let T: V W be linear and let dim(V) = dim(W) = n be finite. Then T is invertible if and only if rank(T) = dim(V). (It follows from theorem 2.5.) Theorem: (2.17) Let V and W be vector spaces and T: V W be linear and invertible. Then T-1: W V is linear. Proof: Let y1, y2 W and c F. Since T is one-to-one and onto, ! x1, x2 V such that T(x1) = y1 and T(x2) = y2. Thus T-1(y1) = x1 and T-1(y2) = x2. WTS: T-1(cy1 + y2) = cT-1(y1) + T-1(y2) T-1(cy1 + y2) = T-1[cT(x1) + T(x2)] = T-1[T(cx1 + x2)] = cx1 + x2 = cT-1(y1) + T-1(y2) Definition: Let A Mnn. Then A is invertible if and only if B Mnn such that AB = BA = In. Remark: If A Mnn is invertible, and AB = BA = In for some B Mnn, then B is the unique inverse of A. We write A-1 = B. Proof: Suppose C Mnn such that AC = CA = In. C = CIn = C(AB) = (CA)B = InB = B. 39 Lemma: Let V and W be vector spaces, and T: V W be linear and invertible. Then dim(V) is finite if and only if dim(W) is finite. In this case, dim(V) = dim(W). Proof: () Suppose dim(V) = n. Let = {vi}i=1n be an ordered basis for V. By theorem 2.2 (page 68), span(T()) = R(T) = W. Hence dim(W) n by theorem 1.9 (page 44). () T-1: W V. Let = {wi}i=1m be an ordered basis for W. Then span(T-1()) = R(T-1) = V. Thus dim(V) m = dim(W). Since dim(V) dim(W) and dim(W) dim(V), dim(V) = dim(W). Theorem: (2.18) Let V and W be vector spaces and dim(V) and dim(W) be finite. Let and be ordered bases for V and W, respectively. Let T: V W be linear. Then T is invertible if and only if [T] is invertible. Furthermore, [T -1] = ([T])-1. Proof: () Suppose T is invertible. dim(V) = dim(W) = n is finite by the previous lemma. Thus [T] Mnn. Let T-1: W V denote the inverse transformation. By definition, TT-1 = Iw and T-1T = Iv. By theorem 2.4, In = [Iv] = [T-1T] = [T-1][T]. Similarly, [T][T-1] = In. ([T])-1[T][T-1] = ([T])-1In [T-1] = ([T])-1 () Continued on Wednesday. 40 Lecture (Wednesday, Week 6) Homework: section 2.4 #4, 5, 9, 10, 12, 16, 17, 20 Corollary 1: Let V be a finite dimension vector space with ordered basis and T: V V be linear. Then T is invertible if and only if [T] is invertible. Furthermore, [T-1] = ([T])-1. Corollary 2: Let A Mnn(F). Then A is invertible if and only if LA is invertible. Furthermore, (LA)-1 = LA-1. Recall: LA: Fn Fm, where LA(x) = Ax. Definition: Let V and W be vector spaces. We say that V is isomorphic to W if and only if T: V W that is invertible. T is called an isomorphism from V onto W. Example: T: F2 P1(F) defined via T((a1, a2) = a1 + a2x is linear and invertible. Thus F2 is isomorphic to P1(F). In general, Fn is isomorphic to Pn-1(F) where T((a1, a2, a3, …, an)) = a1 + a2x + a3x2+ … anxn-1. Theorem: (2.19) Let V and W be finite dimensional vector spaces over F. Then V is isomorphic to W if and only if dim(V) = dim(W). Proof: () Assume V is isomorphic to W. This means T: V W and T is invertible. By lemma (page 101), dim(v) = dim(W) because T is invertible. () Suppose dim(V) = dim(W) = n. Let = {vi}i=1n and = {wi}i=1n be ordered bases for V and W, respectively. By theorem 2.6 (page 72), T: V W such that T is linear and T(vi) = wi, 1 i n. By theorem 2.2 (page 68), R(T) = span(T()) = span() = W. Hence T is onto. From theorem 2.5 (page 71), T must be one-to-one because dim(V) = dim(W). Hence T is invertible. Thus V is isomorphic to W. Corollary: Let V be a vector space over F. Then V is isomorphic to Fn if and only if dim(V) = n. Theorem: (2.20) Let V and W be finite dimensional vector spaces over F where dim(V) = n and dim(W) = m. Let and be ordered bases for V and W, respectively. Then : L(V, W) Mmn(F) defined by (T) = [T] for T L(V, W) is an isomorphism. In other word, L(V, W) is isomorphic to M mn(F). Corollary: Let V and W be vector spaces, where dim(V) = n and dim(W) = m. Then dim(L(V, W)) = m n. 41 Definition: Let be an ordered basis for V where dim(V) = n. The standard representation of V with respect to is denoted by : V Fn and defined by (x) = [x]. Recall: Let = {vi}i=1n. Then x has the unique representation, x = a1v1 + a2v2 + … anvn. a1 a [x] = 2 Fn. ... an Theorem: (2.21) For any vector space V, where dim(V) = n and ordered basis , is an isomorphism of V onto Fn. 42 Lecture (Friday, Week 6) Homework: Section 2.5 #6 (a, b, c), 10, 13 Change of Basis Example: Consider the relation R: 2x2 – 4xy + 5y2 = 0 x = (2/5)x’ – (1/5)y’ y = (1/5)x’ + (2/5)y’ R’: (x’)2 + 6(y’)2 = 1 ellipse 2 1 , 1/5 } 1 2 x 2 1 [x] = = (1/5) y 1 2 ’ = {(1/5) x' = Q[x]’ y' Q = [I2]’ , I2: 2 2: (’) () [x] = Q[x]’ = [I]’[x]’ Theorem: (2.22) Let and ’ be two ordered bases for V, a vector space with dim(V) = n < . Let Q = [Iv]’. Then a) Q is invertible. b) v V [v] = Q[v]’. Proof: a) Since Iv is invertible, Q is invertible by theorem 2.18 (page 101). b) By theorem 2.14 (page 91), T: V W: () (), u V [T(u)] = [T][u]. Let T = Iv, W = V, = ’, and = . Then [u] = [Iv]’[u]’. Definition: The matrix Q = [Iv]’ is called a change of coordinate matrix. Q changes ’ coordinates into coordinates. Remark: Q-1 changes coordinates into ’. Q-1[u] = [u]’ Remark: If = {x1, …, xn} and ’ = {x1’, …, xn’}, then I: V V: (’) (), I(v) = v v V. xj’ = i=1n Qijxi, 1 j n The jth column of Q is [xj’]. a1 Recall: If v = i=1n aixi, then [v] = ... . an 43 Theorem: (2.23) Let T: V V be linear, V a vector space with dim(V) = n < . Let and ’ be ordered bases for V. Let Q = [Iv]’. Then [T]’ = Q-1[T]Q. Proof: Let Iv: V V. Then T = IvT = TIv. Q[T]’ = [Iv]’[T]’’ = [IvT]’ = [TIv]’ = [T][Iv]’ = [T]Q Hence, [T]’ = Q-1[T]Q. 44 Example: T: 2 2 defined via T((a, b)) = (3a – b, a + 3b). = {(1, 1), (1, -1)} and ’ = {(2, 4), (3, 1)}. T((1, 1)) = (2, 4) = a11(1, 1) + a21(1, -1) a11 + a21 = 2 a11 – a21 = 4 a11 = 3, a21 = -1 T((1, -1)) = (4, -2) = a12(1, 1) + a22(1, -1) a12 + a22 = 4 a12 – a22 = -2 a12 = 1, a22 = 3 [T] = 3 1 1 3 Consider Q = [I2]’. I2((2, 4)) = (2, 4) = a11(1, 1) + a21(1, -1) a11 + a21 = 2 a11 – a21 = 4 a11 = 3, a21 = -1 I2((3, 1)) = (3, 1) = a12(1, 1) + a22(1, -1) a12 + a22 = 3 a12 – a22 = 1 a12 = 2, a22 = 1 1 2 3 2 -1 1 1 Q = (1/5) 1 3 4 1 [T]’ = Q-1[T]Q = =B 2 2 Hence, Q = [I2]’ = Check: T((2, 4)) = (2, 14) = a11(2, 4) + a21(3, 1) 2a11 + 3a21 = 2 4a11 + a21 = 14 a21 = -2, a11 = 4 T((3, 1)) = (8, 6) = a12(2, 4) + a22(3, 1) 2a12 + 3a22 = 8 4a12 + a22 = 6 a22 = 2, a12 = 1 45 Corollary: Let A Mnn(F) and let be an ordered basis for Fn. Then [LA] = Q-1AQ. Recall LA(x) = Ax. Proof: Let = {ei}i=1n. Set Q = [IFn]. Recall A = [LA]. [LA] = Q-1[LA]Q = Q-1AQ Definition: Let A, B Mnn(F). We say that B is similar to A if and only if an invertible Q Mnn(F) such that B = Q-1AQ. 46 Lecture (Monday, Week 7) Homework: section 3.1 #2, 3, 4, 5, 6. Quiz: covers section 2.4, 2.5, and 3.1. Definition: Let A Mmn(F). Any one of the following 3 operations on rows [column] of A is called an elementary row [column] operation: Type (1): interchanging any two rows [columns] of A. Type (2): multiplying any row [column] by a nonzero scalar. Type (3): adding any scalar multiple of a row [column] to another row [column]. 1 Example: A = 2 4 2 3 1 1 0 1 4 3 2 1 1 2 3 0 1 3 1 6 3 0 1 3 4 2 2 A: (r1 r2, r2 r1) B = 1 4 2 A: (3c2 c2) C = 1 4 17 A: (4r3 + r1 r1) M = 2 4 2 1 0 3 4 2 12 1 3 1 2 7 Facts: Row or column operations can be reversed (undone). The reverse operation is the same type as the original. 47 Definition: An n n elementary matrix is one obtained by performing an elementary row [column] operation on In. The elementary matrix is of type (1), (2), or (3) according to the type of elementary operation performed on In. 0 Example: I3: (r1 r2, r2 r1) E = 1 0 1 I3: (–2r3 + r1 r1) E = 0 0 1 Fact: I3: (–2c1 + c3 c3) E = 0 0 0 0 2 0 1 Type (3) elementary matrix 0 1 1 Type (1) elementary matrix 0 0 0 0 0 1 1 2 0 1 Theorem: (3.1) Let A Mmn(F) and suppose that B is obtained from A by performing an elementary row [column] operation. Then an elementary matrix E where E Mmm(F) [E Mnn(F)] such that B = EA [B = AE]. In fact, E is obtained form Im [In] by performing the same elementary row [column] operation as that which was performed on A to obtain B. Conversely, If E is an elementary m m [n n] matrix, then EA [AE] is the matrix obtained form A by performing the same elementary row [column] operation as that which produces E from I m [In]. 1 Example: A = 2 4 2 3 1 1 0 1 4 3 2 2 A: (r1 r2, r2 r1) B = 1 4 0 EA = 1 0 1 0 0 0 2 0 1 1 4 1 1 2 3 0 1 1 1 2 3 0 1 3 4 and I3: (r1 r2, r2 r1) E = 2 3 1 4 = 2 4 2 2 3 1 1 0 1 0 1 0 1 0 0 0 0 1 4 3 2 Theorem: (3.2) Elementary matrices of all types are invertible, and the inverse of an elementary matrix is an elementary matrix of the same type. 48 Lecture (Wednesday, Week 7) Quiz: covers section 2.4 and 2.5 only. Today: section 3.2 Homework: read along section 3.2. A = [LA], = {ei}i=1n, LA: Fn Fm, LA(x) = Ax. Definition: If A Mmn(F), then rank(A) = rank(LA). Remark: rank(LA) = dim(R(LA)) Theorem: (3.3) Let T: V W be linear, V and W finite-dimensional vector spaces with ordered basis and , respectively. Then rank(T) = rank([T]). Proof: exercise 2.4.20 Theorem: (3.4) Let A Mmn(F). If P Mmm(F) and Q Mnn(F) invertible, then (a) rank(AQ) = rank(A) (b) rank(PA) = rank(A) (c) rank(PAQ) = rank(A) Proof: (a) rank(AQ) = rank(LAQ) = dim(R(LAQ)) R(LAQ) = R(LALQ) = LALQ(Fn) = LA(LQ(Fn)) = LA(Fn). Note: since Q is onto, LQ: Fn Fn is onto, hence LQ(Fn) = Fn. R(LAQ) = LA(Fn) = R(LA) rank(AQ) = dim(R(LAQ)) = dim(R(LA)) = rank(LA) = rank(A) (b) By definition, rank(PA) = rank(LPA). LPA = LPLA: Fn Fm rank(LPA) = rank(LPLA) R(LPLA) = LPLA(Fn) = Lp(LA(Fn)) = LP(R(LA)) rank(PA) = rank(LPA) = rank(LPLA) = dim(R(LPLA)) = dim(LP(R(LA))) = dim(R(LA)) By exercise 2.4.17 because LP is an isomorphism. rank(PA) = dim(R(LA)) = rank(LA) = rank(A) Corollary: Elementary row [column] operations are rank-preserving. Idea of Proof: Since E is invertible, B = EA rank(B) = rank(EA) B = AE rank(B) = rank(AE) 49 Theorem: (3.5) The rank of any matrix equals the maximum number of linearly independent columns. Proof: For any matrix A Mmn(F), rank(A) = rank(LA) = dim(R(LA)). Let = {ei}i=1n be the standard basis for Fn. By theorem 2.2 (page 68), R(LA) = span(LA()). span(LA()) = span({LA(e1), …, LA(en)}) = span({Ae1, …, Aen}) = span({a1, …, an}), where aj is the jth column of A. rank(A) = rank(LA) = dim(R(LA)) = dim(span({a1, …, an})) 1 Example: A = 1 1 1 B = 1 0 1 C = 0 0 1 D = 0 0 1 E = 0 0 1 3 2 2 0 1 1 0 3 = E1A 1 1 2 2 2 1 0 2 1 0 2 1 1 2 = E2E1A 1 1 2 = E2E1AE3 1 0 2 = E2E1AE3E4 1 By the previous corollary, rank is preserved through elementary row [column] operations. Hence, rank(A) = rank(E) = 2 50 Lecture (Monday, Week 8) Today: Finish section 3.2 Homework: section 3.2 #2 abc, 5 bcd, 8, 13, 15, 16 Midterm 2: Wednesday, Week 9 Practice Midterm: Friday, this week Quiz: covers sections 3.1 and 3.2 Theorem: (3.6) Let A Mmn(F) and rank(A) = r (= rank(LA)). Then r m and r n, and by means of elementary row or column operations, A may be transformed into D = Ir O2 O . O 1 3 Corollary 1: Let A Mmn(F) and rank(A) = r. Then invertible matrices B Mmm(F) and C Mnn(F) such that D = BAC Mmn(F). Corollary 2: Let A Mmn(F). Then (a) rank(At) = rank(A) (b) rank(A) is equal to the number of linearly independent rows. (c) The rows and columns of any matrix generate subspaces of the same dimensions, numerically equal to rank(A). Proof: (a) By corollary 1, B, C invertible such that D = BAC where D = Dt = Ir O1 Ir O2 O O 1 Mmn(F). 3 O O 2 3 Dt = (BAC)t = CtAtBt Since B and C are invertible, Bt and Ct are invertible by exercise 2.4.5. By theorem 3.4, rank(At) = rank(Dt) = r = rank(A). Corollary 3: Every invertible matrix is a product of elementary matrices. Proof: Let A Mnn(F) be invertible. Note that rank(A) = rank(LA) = dim(R(LA)) = dim(Fn) = n, where LA: Fn Fn. By corollary 1, B, C Mnn(F) which are invertible such that D = BAC = In. B = EpEp-1…E1 and C = G1G2…Gq where Ei and Gj are elementary matrices for 1 i p and 1 j q. D = EpEp-1…E1AG1G2…Gq = In A = E1-1…Ep-1InGq-1…Gq-1 = E1-1…Ep-1Gq-1…Gq-1 51 Theorem: (3.7) Let T: V W and U: W Z be linear transformations on finite-dimensional vector spaces. Let A and B be matrices such that AB is defined. Then (a) rank(UT) rank(U) (b) rank(UT) rank(T) (c) rank(AB) rank(A) (d) rank(AB) rank(B) Proof: (a) Note that R(T) is a vector subspace of W. R(UT) = U(T(V)) = U(R(T)) U(W) = R(U). rank(UT) = dim(R(UT)) dim(R(U)) = rank(U). (c) By proof (a), rank(AB) = rank(LAB) = rank(LALB) rank(LA) = rank(A) (d) By proof (c) and corollary 2 (page 158), rank(AB) = rank((AB) t) = rank(BtAt) rank(Bt) = rank(B). Matrix Inversions Definition: Let A and B be m n and m p matrices, respectively. By the augmented matrix (A B), we mean the m (n + p) matrix (AB). Let A Mnn(F) be an invertible matrix. Consider C = (A In) Mn2n(F). By exercise 15, A-1C = (A-1A A-1In) = (In A-1) A-1 = Ep…E1, where Ei is elementary matrix for 1 i p. Ep…E1(A In) = A-1C = (In A-1) 52 Lecture (Wednesday, Week 8) Today: Section 3.3 Linear Systems of Equations a11x1 + a12x2 + … + a1nxn = b1 a21x1 + a22x2 + … + a2nxn = b2 … am1x1 + am2x2 + … + amnxn = bm aij, xj, bi F for 1 i m and 1 j n. a11 a Ax = b, where A = 21 ... a m1 a a 12 a 22 m2 ... 2n , x = ... a mn ... a a 1n x1 x 2 , and b = ... xn b1 b 2 ... bm A Mmn(F), x Fn, and b Fm. LA: Fn Fm, where LA(x) = Ax = b. b is known and x is unknown. s1 s Definition: A solution of (S) is any n-tuples S = 2 Fn such that AS = b. ... sm The set of all solutions of (S) is called the solution set of (S). = {S Fn: AS = b} is the solution set and Fn. The system (S) is called consistent if and only if ; otherwise (S) is called inconsistent. Example: (1b) 2x1 + 3x2 + x3 = 1 x1 – x2 + 2x3 = 6 Ax = b 2 1 x1 3 1 1 x 2 = 1 2 6 x3 6 S1 = 2 and S2 = 7 8 4 3 53 Definition: A system Ax = b of m linear equations in n unknowns is said to be homogeneous if and only if b = 0; otherwise the system is called inhomogeneous (non-homogeneous). Theorem: (3.8) Let Ax = 0 where A Mmn(F) and x Fn. Let KH = {S Fn: AS = 0} = ker(A). Then KH = N(LA). Hence, KH is a subspace of Fn of dimension n – rank(LA) = n – rank(A) = nullity(LA). [dim(KH) = n – rank(A)] Proof: Recall from the Dimension Theorem that dim(V) = rank(T) + nullity(T), T: V W. Recall LA: Fn Fm, where LA(x) = Ax. n = dim(Fn) = rank(LA) + nullity(LA) = rank(A) + nullity (LA) nullity(LA) = n – rank(A) N(LA) = {x Fn: LA(x) = 0 Fm} = KH dim(N(LA)) = nullity(LA) = dim(KH) Definition: KH = kernel of A = ker(A) Example: Consider A = 1 1 1 and x F3. 1 1 2 What are the solution to Ax = 0 F2. rank(A) = # of linearly independent rows or column = 2 dim(KH) = n – rank(A) = 3 – 2 = 1 Equation 1: x1 + 2x2 + x3 = 0 Equation 2: x1 – x2 – x3 = 0 3x2 – 2x3 = 0 Let x3 = t F. x2 = – (2/3)t x1 + 2x2 + x3 = 0 x1 – 2(2/3)t + t = 0 x1 – (1/3)t = 0 x1 = (1/3)t 1/ 3 1/ 3 (1 / 3)t St = (2 / 3)t = KH = {x F3: x = t 2 / 3 , where t F} = span({ 2 / 3 }) 1 1 t 1/ 3 Hence, = { 2 / 3 } a the basis for KH. 1 54 Lecture (Friday, Week 8) Theorem: (3.9) Let K be the solution set of Ax = b. Let KH be the solution set of Ax = 0. Then for any solution s to Ax = b, K = {s} + KH = {s + sH: sH KH}. Proof: Fix s, a solution to Ax = b. Let w K. Then Aw = b. Hence, A(w – s) = Aw – As = b – b = 0. Thus w – s KH. sH KH such that w – s SH. w = s + sH w {s} + KH. Hence, K {s} + KH. Conversely, suppose w {s} + KH. Then w = s + sH for some sH KH. However, Aw = A(s + sH) = As + AsH = b + 0 = b. w K. Hence, {s} + KH K. Remark: Note that we always have A0n = 0m for 0n Fn and 0m Fm. Hence, 0n KH. Suppose that for a given b Fm s Fn such that As = b. Further, suppose KH = {0n}. Then s is the only solution of Ax = b because K = {s} + KH = {s} + {0n} = {s}. Theorem: (3.10) Let Ax = b be a system of n linear equations in n unknowns. (i.e. LA: Fn Fn, A Mnn(F)) Then A is invertible if and only if the system has exactly one solution. Proof: () Suppose that A is invertible. Then A-1 exists such that AA-1 = A-1A = In. Let s = A-1b. Then As = A(A-1b) = (AA-1) = Inb = b. Let s1 and s2 be any two solutions of Ax = b. Then As1 = b and As2 = b. LA(s1 – s2) = A(s1 – s2) = 0 Since LA is one-to-one and onto, s1 – s2 = 0 s1 = s2 () Suppose there is only one solution to Ax = b. Let s be the solution. Then K = {s} = {s} + KH. Thus KH = {0}. N(LA) = KH = {0} LA is invertible A is invertible. 55 Theorem: (3.11) Let Ax = b be a system of linear equations. Then the system is consistent if and only if rank(A) = rank(Ab). Proof: () Suppose the system is consistent. Then s Fn such that As = b. This implies b R(LA) Fm. R(LA) = span({a1, …, an}), where ai is a column of A. Thus b span({a1, …, an}). span({a1, …, an}) = span({a1, …, an, b}) dim(span({a1, …, an})) = dim(span({a1, …, an, b})) Hence, rank(A) = rank(Ab) () Since the proof of () is an equivalent argument, () holds automatically.