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Problems 1. Recall that a matrix AA is symmetric if A=AtA=A^{t}, where AtA^{t} i...
Jun 28, 2024
Solution by Steps
step 1
Recall that a matrix AA is symmetric if A=AtA = A^t
step 2
Let AA and BB be symmetric n×nn \times n matrices
step 3
To prove ABAB is symmetric, we need to show (AB)t=AB(AB)^t = AB
step 4
Using the property of transposes, (AB)t=BtAt(AB)^t = B^t A^t
step 5
Since AA and BB are symmetric, At=AA^t = A and Bt=BB^t = B
step 6
Therefore, (AB)t=BA(AB)^t = B A
step 7
For ABAB to be symmetric, we need BA=ABB A = AB
step 8
Hence, ABAB is symmetric if and only if AB=BAAB = BA
Answer
ABAB is symmetric if and only if AB=BAAB = BA.
Key Concept
Symmetric Matrices and Transpose Properties
Explanation
The product of two symmetric matrices is symmetric if and only if the matrices commute.
Problem 2: Determine the permutation defined by the product of cycles in S9S_9. # (a) (17)(628)(9354)(17)(628)(9354)
step 1
Write the cycles in the form of permutations
step 2
(17)(17) means 1→71 \to 7 and 7→17 \to 1
step 3
(628)(628) means 6→26 \to 2, 2→82 \to 8, and 8→68 \to 6
step 4
(9354)(9354) means 9→39 \to 3, 3→53 \to 5, 5→45 \to 4, and 4→94 \to 9
step 5
Combine the cycles to form a single permutation
step 6
The combined permutation is (1→7,7→1,6→2,2→8,8→6,9→3,3→5,5→4,4→9)(1 \to 7, 7 \to 1, 6 \to 2, 2 \to 8, 8 \to 6, 9 \to 3, 3 \to 5, 5 \to 4, 4 \to 9)
# (b) (12)(347)(12)(347)
step 1
Write the cycles in the form of permutations
step 2
(12)(12) means 1→21 \to 2 and 2→12 \to 1
step 3
(347)(347) means 3→43 \to 4, 4→74 \to 7, and 7→37 \to 3
step 4
Combine the cycles to form a single permutation
step 5
The combined permutation is (1→2,2→1,3→4,4→7,7→3)(1 \to 2, 2 \to 1, 3 \to 4, 4 \to 7, 7 \to 3)
# (c) (6148)(2345)(12493)(6148)(2345)(12493)
step 1
Write the cycles in the form of permutations
step 2
(6148)(6148) means 6→16 \to 1, 1→41 \to 4, 4→84 \to 8, and 8→68 \to 6
step 3
(2345)(2345) means 2→32 \to 3, 3→43 \to 4, 4→54 \to 5, and 5→25 \to 2
step 4
(12493)(12493) means 1→21 \to 2, 2→42 \to 4, 4→94 \to 9, 9→39 \to 3, and 3→13 \to 1
step 5
Combine the cycles to form a single permutation
step 6
The combined permutation is (6→1,1→2,2→4,4→9,9→3,3→5,5→2,8→6)(6 \to 1, 1 \to 2, 2 \to 4, 4 \to 9, 9 \to 3, 3 \to 5, 5 \to 2, 8 \to 6)
Answer
(a) (1→7,7→1,6→2,2→8,8→6,9→3,3→5,5→4,4→9)(1 \to 7, 7 \to 1, 6 \to 2, 2 \to 8, 8 \to 6, 9 \to 3, 3 \to 5, 5 \to 4, 4 \to 9)
(b) (1→2,2→1,3→4,4→7,7→3)(1 \to 2, 2 \to 1, 3 \to 4, 4 \to 7, 7 \to 3)
(c) (6→1,1→2,2→4,4→9,9→3,3→5,5→2,8→6)(6 \to 1, 1 \to 2, 2 \to 4, 4 \to 9, 9 \to 3, 3 \to 5, 5 \to 2, 8 \to 6)
Key Concept
Permutation Cycles
Explanation
Permutations can be represented as products of disjoint cycles, which describe the mapping of elements.
Problem 3: Write the following permutations in S8S_8 as a product of disjoint cycles. # (a) (1amp;2amp;3amp;4amp;5amp;6amp;7amp;82amp;3amp;5amp;1amp;6amp;4amp;8amp;7)\left(\begin{array}{llllllll}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ 2 & 3 & 5 & 1 & 6 & 4 & 8 & 7\end{array}\right)
step 1
Start with the first element and follow its mapping
step 2
1→21 \to 2, 2→32 \to 3, 3→53 \to 5, 5→65 \to 6, 6→46 \to 4, 4→14 \to 1
step 3
This forms the cycle (1→2→3→5→6→4→1)(1 \to 2 \to 3 \to 5 \to 6 \to 4 \to 1)
step 4
Next, 7→87 \to 8, 8→78 \to 7
step 5
This forms the cycle (7→8→7)(7 \to 8 \to 7)
step 6
The permutation is (1→2→3→5→6→4)(7→8)(1 \to 2 \to 3 \to 5 \to 6 \to 4)(7 \to 8)
# (b) (1amp;2amp;3amp;4amp;5amp;6amp;7amp;81amp;5amp;7amp;8amp;3amp;4amp;2amp;6)\left(\begin{array}{llllllll}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ 1 & 5 & 7 & 8 & 3 & 4 & 2 & 6\end{array}\right)
step 1
Start with the first element and follow its mapping
step 2
1→11 \to 1
step 3
Next, 2→52 \to 5, 5→35 \to 3, 3→73 \to 7, 7→27 \to 2
step 4
This forms the cycle (2→5→3→7→2)(2 \to 5 \to 3 \to 7 \to 2)
step 5
Next, 4→84 \to 8, 8→68 \to 6, 6→46 \to 4
step 6
This forms the cycle (4→8→6→4)(4 \to 8 \to 6 \to 4)
step 7
The permutation is (1)(2→5→3→7)(4→8→6)(1)(2 \to 5 \to 3 \to 7)(4 \to 8 \to 6)
Answer
(a) (1→2→3→5→6→4)(7→8)(1 \to 2 \to 3 \to 5 \to 6 \to 4)(7 \to 8)
(b) (1)(2→5→3→7)(4→8→6)(1)(2 \to 5 \to 3 \to 7)(4 \to 8 \to 6)
Key Concept
Disjoint Cycles
Explanation
Permutations can be decomposed into disjoint cycles, which are cycles that do not share any elements.
Problem 4: Make a multiplication table for the symmetric group S3S_3.
step 1
List all elements of S3S_3: {e,(12),(13),(23),(123),(132)}\{e, (12), (13), (23), (123), (132)\}
step 2
Create a table with these elements as both row and column headers
step 3
Fill in the table by computing the product of each pair of elements
step 4
The multiplication table is as follows:
amp;eamp;(12)amp;(13)amp;(23)amp;(123)amp;(132)eamp;eamp;(12)amp;(13)amp;(23)amp;(123)amp;(132)(12)amp;(12)amp;eamp;(132)amp;(123)amp;(13)amp;(23)(13)amp;(13)amp;(123)amp;eamp;(132)amp;(23)amp;(12)(23)amp;(23)amp;(132)amp;(123)amp;eamp;(12)amp;(13)(123)amp;(123)amp;(13)amp;(23)amp;(12)amp;(132)amp;e(132)amp;(132)amp;(23)amp;(12)amp;(13)amp;eamp;(123) \begin{array}{c|cccccc} & e & (12) & (13) & (23) & (123) & (132) \\ \hline e & e & (12) & (13) & (23) & (123) & (132) \\ (12) & (12) & e & (132) & (123) & (13) & (23) \\ (13) & (13) & (123) & e & (132) & (23) & (12) \\ (23) & (23) & (132) & (123) & e & (12) & (13) \\ (123) & (123) & (13) & (23) & (12) & (132) & e \\ (132) & (132) & (23) & (12) & (13) & e & (123) \\ \end{array}
Answer
The multiplication table for S3S_3 is given above.
Key Concept
Symmetric Group Multiplication
Explanation
The multiplication table for a symmetric group shows the result of composing any two permutations in the group.
Problem 5: Solve for yy and analyze the given conditions. # (a) Solve for yy, given that xyz−1w=1x y z^{-1} w = 1.
step 1
Start with the equation xyz−1w=1x y z^{-1} w = 1
step 2
Multiply both sides by w−1w^{-1}: xyz−1=w−1x y z^{-1} = w^{-1}
step 3
Multiply both sides by zz: xy=w−1zx y = w^{-1} z
step 4
Multiply both sides by x−1x^{-1}: y=x−1w−1zy = x^{-1} w^{-1} z
# (b) Suppose that xyz=1x y z = 1. Does it follow that yzx=1y z x = 1? Does it follow that yxz=1y x z = 1?
step 1
Given xyz=1x y z = 1, we need to check if yzx=1y z x = 1
step 2
Multiply both sides by x−1x^{-1}: yz=x−1y z = x^{-1}
step 3
Multiply both sides by z−1z^{-1}: y=x−1z−1y = x^{-1} z^{-1}
step 4
Substitute yy back: x(x−1z−1)z=1x (x^{-1} z^{-1}) z = 1
step 5
Simplify: xx−1z−1z=1  ⟹  1=1x x^{-1} z^{-1} z = 1 \implies 1 = 1
step 6
Therefore, yzx=1y z x = 1
step 7
Now, check if yxz=1y x z = 1
step 8
Substitute yy: (x−1z−1)xz=1(x^{-1} z^{-1}) x z = 1
step 9
Simplify: x−1z−1xz=1x^{-1} z^{-1} x z = 1
step 10
This does not necessarily simplify to 11
step 11
Therefore, yxz≠1y x z \neq 1
Answer
(a) y=x−1w−1zy = x^{-1} w^{-1} z
(b) yzx=1y z x = 1, but yxz≠1y x z \neq 1
Key Concept
Group Elements and Inverses
Explanation
Solving for an element in a group involves using the properties of inverses and the group operation.
Problem 6: Determine whether or not HH is a subgroup of GG. # (a) G=(R2,+)G = (\mathbb{R}^2, +), and H={(x,y):y=2x}H = \{(x, y) : y = 2x\}.
step 1
Check if HH is closed under addition
step 2
Let (x1,y1),(x2,y2)∈H(x_1, y_1), (x_2, y_2) \in H
step 3
Then y1=2x1y_1 = 2x_1 and y2=2x2y_2 = 2x_2
step 4
(x1,y1)+(x2,y2)=(x1+x2,y1+y2)(x_1, y_1) + (x_2, y_2) = (x_1 + x_2, y_1 + y_2)
step 5
y1+y2=2x1+2x2=2(x1+x2)y_1 + y_2 = 2x_1 + 2x_2 = 2(x_1 + x_2)
step 6
Therefore, (x1+x2,y1+y2)∈H(x_1 + x_2, y_1 + y_2) \in H
step 7
Check if HH contains the identity element (0,0)(0, 0)
step 8
y=2x  ⟹  0=2⋅0y = 2x \implies 0 = 2 \cdot 0
step 9
Therefore, (0,0)∈H(0, 0) \in H
step 10
Check if HH is closed under inverses
step 11
Let (x,y)∈H(x, y) \in H
step 12
The inverse is (−x,−y)(-x, -y)
step 13
−y=−2x=2(−x)-y = -2x = 2(-x)
step 14
Therefore, (−x,−y)∈H(-x, -y) \in H
step 15
HH is a subgroup of GG
# (b) G=(R,×)G = (\mathbb{R}, \times), and HH is the set of positive reals.
step 1
Check if HH is closed under multiplication
step 2
Let a,b∈Ha, b \in H
step 3
a×ba \times b is positive
step 4
Therefore, a×b∈Ha \times b \in H
step 5
Check if HH contains the identity element 11
step 6
1∈H1 \in H
step 7
Check if HH is closed under inverses
step 8
Let a∈Ha \in H
step 9
The inverse is 1/a1/a
step 10
1/a1/a is positive
step 11
Therefore, 1/a∈H1/a \in H
step 12
HH is a subgroup of GG
# (c) G=GL2(R)G = GL_2(\mathbb{R}) and HH is the set of matrices (aamp;00amp;0)\left(\begin{array}{ll}a & 0 \\ 0 & 0\end{array}\right) with a≠0a \neq 0.
step 1
Check if HH is closed under matrix multiplication
step 2
Let A=(aamp;00amp;0)A = \left(\begin{array}{ll}a & 0 \\ 0 & 0\end{array}\right) and B=(bamp;00amp;0)B = \left(\begin{array}{ll}b & 0 \\ 0 & 0\end{array}\right)
step 3
A×B=(aamp;00amp;0)×(bamp;00amp;0)=(abamp;00amp;0)A \times B = \left(\begin{array}{ll}a & 0 \\ 0 & 0\end{array}\right) \times \left(\begin{array}{ll}b & 0 \\ 0 & 0\end{array}\right) = \left(\begin{array}{ll}ab & 0 \\ 0 & 0\end{array}\right)
step 4
ab≠0ab \neq 0
step 5
Therefore, A×B∈HA \times B \in H
step 6
Check if HH contains the identity element
step 7
The identity matrix is (1amp;00amp;1)\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)
step 8
This is not in HH
step 9
Therefore, HH is not a subgroup of GG
Answer
(a) HH is a subgroup of GG.
(b) HH is a
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