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Problem 2. Given nЄ Z and σ Є Sn, let Qo denote the n x n matrix defined by¹ n Qo = Σ Eo(j),j. j=1 Here Eij denotes the (i, j)-th matrix unit. Its entries are all zero except there is a 1 in the i-th row and j-th column. In particular, one has EijEkl = 8jkEil, where §jk is 1 if j = k and 0 otherwise. 2.1. Show that, for each σ, Є Sn, one has (Qo) = Qo-1 and QoQx = Qoy. Conclude that the assignment (σ) = Qo defines an injective group homomorphism Þ : Sn → On(R). 2.2. Now let det: On (R) → R* denote the determinant group homomorphism. Since orthogonal matrices have determinant ±1, det has image contained in %2 = {±1}. Define E det o Sn → Z2. For σ € Sn, the number (σ) is called the sign of σ. Show that if σ = Sn is a k-cycle, then ε(0) = (−1)k+1. Hint. For 2.2, it may be helpful to proceed as follows: ... Step I. First explain why (a1a2 ak) = (a1 ak)(a₁ ak−1) ··· (a₁ a2). Step II. Show that ɛ(b₁ b₂) = −1. To do this, use that det (I) ... = 1 and that if a matrix A is obtained from a matrix B by interchanging two rows or columns, then det(A) = = det(B). Step III. Combine the above two steps with the fact that is a group homomorphism to compute € (α1 a2 ... ak). 2.3. The kernel Ker() is called the alternating group of degree n, and denoted An. Using 2.2 and cycle decomposition, determine which of the following two hich of the following two permutations in S7 belong to A7: 1 2 3 4 5 6 7 1 2 3 4 5 6 7 σ = Υ == 2 5 4 6 1 3 7 6 5 4 3 2 1

Problem 2. Given nЄ Z and σ Є Sn, let Qo denote the n x n matrix defined by¹ n Qo = Σ Eo(j),j. j=1 Here Eij denotes the (i, j)-th matrix unit. Its entries are all zero except there is a 1 in the i-th row and j-th column. In particular, one has EijEkl = 8jkEil, where §jk is 1 if j = k and 0 otherwise. 2.1. Show that, for each σ, Є Sn, one has (Qo) = Qo-1 and QoQx = Qoy. Conclude that the assignment (σ) = Qo defines an injective group homomorphism Þ : Sn → On(R). 2.2. Now let det: On (R) → R* denote the determinant group homomorphism. Since orthogonal matrices have determinant ±1, det has image contained in %2 = {±1}. Define E det o Sn → Z2. For σ € Sn, the number (σ) is called the sign of σ. Show that if σ = Sn is a k-cycle, then ε(0) = (−1)k+1. Hint. For 2.2, it may be helpful to proceed as follows: ... Step I. First explain why (a1a2 ak) = (a1 ak)(a₁ ak−1) ··· (a₁ a2). Step II. Show that ɛ(b₁ b₂) = −1. To do this, use that det (I) ... = 1 and that if a matrix A is obtained from a matrix B by interchanging two rows or columns, then det(A) = = det(B). Step III. Combine the above two steps with the fact that is a group homomorphism to compute € (α1 a2 ... ak). 2.3. The kernel Ker() is called the alternating group of degree n, and denoted An. Using 2.2 and cycle decomposition, determine which of the following two hich of the following two permutations in S7 belong to A7: 1 2 3 4 5 6 7 1 2 3 4 5 6 7 σ = Υ == 2 5 4 6 1 3 7 6 5 4 3 2 1

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter9: Systems Of Equations And Inequalities
Section9.8: Determinants
Problem 34E
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Problem 2. Given nЄ Z and σ Є Sn, let Qo denote the n x n matrix defined by¹ n Qo = Σ Eo(j),j. j=1 Here Eij denotes the (i, j)-th matrix unit. Its entries are all zero except there is a 1 in the i-th row and j-th column. In particular, one has EijEkl = 8jkEil, where §jk is 1 if j = k and 0 otherwise. 2.1. Show that, for each σ, Є Sn, one has (Qo) = Qo-1 and QoQx = Qoy. Conclude that the assignment (σ) = Qo defines an injective group homomorphism Þ : Sn → On(R). 2.2. Now let det: On (R) → R* denote the determinant group homomorphism. Since orthogonal matrices have determinant ±1, det has image contained in %2 = {±1}. Define E det o Sn → Z2. For σ € Sn, the number (σ) is called the sign of σ. Show that if σ = Sn is a k-cycle, then ε(0) = (−1)k+1. Hint. For 2.2, it may be helpful to proceed as follows: ... Step I. First explain why (a1a2 ak) = (a1 ak)(a₁ ak−1) ··· (a₁ a2). Step II. Show that ɛ(b₁ b₂) = −1. To do this, use that det (I) ... = 1 and that if a matrix A is obtained from a matrix B by interchanging two rows or columns, then det(A) = = det(B). Step III. Combine the above two steps with the fact that is a group homomorphism to compute € (α1 a2 ... ak). 2.3. The kernel Ker() is called the alternating group of degree n, and denoted An. Using 2.2 and cycle decomposition, determine which of the following two hich of the following two permutations in S7 belong to A7: 1 2 3 4 5 6 7 1 2 3 4 5 6 7 σ = Υ == 2 5 4 6 1 3 7 6 5 4 3 2 1
Transcribed Image Text:Problem 2. Given nЄ Z and σ Є Sn, let Qo denote the n x n matrix defined by¹ n Qo = Σ Eo(j),j. j=1 Here Eij denotes the (i, j)-th matrix unit. Its entries are all zero except there is a 1 in the i-th row and j-th column. In particular, one has EijEkl = 8jkEil, where §jk is 1 if j = k and 0 otherwise. 2.1. Show that, for each σ, Є Sn, one has (Qo) = Qo-1 and QoQx = Qoy. Conclude that the assignment (σ) = Qo defines an injective group homomorphism Þ : Sn → On(R). 2.2. Now let det: On (R) → R* denote the determinant group homomorphism. Since orthogonal matrices have determinant ±1, det has image contained in %2 = {±1}. Define E det o Sn → Z2. For σ € Sn, the number (σ) is called the sign of σ. Show that if σ = Sn is a k-cycle, then ε(0) = (−1)k+1. Hint. For 2.2, it may be helpful to proceed as follows: ... Step I. First explain why (a1a2 ak) = (a1 ak)(a₁ ak−1) ··· (a₁ a2). Step II. Show that ɛ(b₁ b₂) = −1. To do this, use that det (I) ... = 1 and that if a matrix A is obtained from a matrix B by interchanging two rows or columns, then det(A) = = det(B). Step III. Combine the above two steps with the fact that is a group homomorphism to compute € (α1 a2 ... ak). 2.3. The kernel Ker() is called the alternating group of degree n, and denoted An. Using 2.2 and cycle decomposition, determine which of the following two hich of the following two permutations in S7 belong to A7: 1 2 3 4 5 6 7 1 2 3 4 5 6 7 σ = Υ == 2 5 4 6 1 3 7 6 5 4 3 2 1
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