{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

u4 - Unit IV Determinants 1 Permutations Definition A...

This preview shows pages 1–2. Sign up to view the full content.

Unit IV: Determinants 1. Permutations Definition: A permutation of a set S is a function from S to S that is one-to-one and onto. If S has n elements, then there are n ! permutations of S . The notation π : { 1 , . . . , n } 7→ { k 1 . . . k n } refers to the permutation π of the integers 1 , . . . , n for which π (1) = k 1 , π (2) = k 2 , and so on, to π ( n ) = k n . Example: There are two permutations of the set { 1 , 2 } , namely the identity { 1 , 2 } 7→ { 1 , 2 } and the switch { 1 , 2 } 7→ { 2 , 1 } . There are six permutations of the set { 1 , 2 , 3 } . They are the identity { 1 , 2 , 3 } 7→ { 1 , 2 , 3 } , two circular shifts { 1 , 2 , 3 } 7→ { 2 , 3 , 1 } and { 1 , 2 , 3 } 7→ { 3 , 1 , 2 } , and three switches { 1 , 2 , 3 } 7→ { 2 , 1 , 3 } , { 1 , 2 , 3 } 7→ { 3 , 2 , 1 } , and { 1 , 2 , 3 } 7→ { 1 , 3 , 2 } . The composition σ π of two permutations σ and π is a permutation. Each permutation π has an inverse permutation π - 1 , defined so that π - 1 π = π π - 1 = the identity permutation. We will refer to a permutation that simply switches two values as a switch . (The math- ematical word is transposition .) Any permutation is a composition of switches. For in- stance, the circular shift { 1 , 2 , 3 } 7→ { 2 , 3 , 1 } is a composition of two switches, the switch { 1 , 2 , 3 } 7→ { 2 , 1 , 3 } of the first two values, followed by the switch { 1 , 2 , 3 } 7→ { 3 , 2 , 1 } of the first and third values. The inversion number of a permutation π : { 1 , . . . , n } 7→ { π (1) , . . . π ( n ) } is the number of pairs ( j, k ) such that j < k but π ( j ) > π ( k ). Example: The identity permutation of { 1 , 2 , 3 } has inversion number 0, the circular shifts have inversion number 2, and the switches have inversion numbers 1 or 3. For instance, the circular shift { 1 , 2 , 3 } 7→ { 2 , 3 , 1 } has inversion number 2, corresponding to the inversions π (1) = 2 > 1 = π (3) and π (2) = 3 > 1 = π (3). The switch { 1 , 2 , 3 } 7→ { 2 , 1 , 3 } has inversion number 1, corresponding to the inversion π (1) = 2 > 1 = π (2). Definition: An even permutation is a permutation whose inversion number is even, and an odd permutation is a permutation whose inversion number is odd. The parity of a per- mutation is even or odd according to whether its inversion number is even or odd. Definition: The sign of a permutation is +1 if the permutation is even and - 1 if the permutation is odd. We denote the sign of a permutation π by sgn( π ). Theorem 1. If we switch two values of a permutation, the sign changes. Proofsketch. If we switch the values of adjacent integers, say the values π ( j ) and π ( j +1), the inversion number of the new permutation changes by +1 or - 1, depending on whether π ( j ) < π ( j + 1) or π ( j ) > π ( j + 1). Switching two values of arbitrary integers, say π ( j ) and π ( k ) can be done through a succession of an odd number of switches of values of adjacent integers. Procedure. The theorem provides an easy way to determine the parity of a permutation. Count the number of switches you need to get from the identity permutation to π . If you can do it with an even number of switches, then π is even. If you can do it with an odd number of switches, then π is odd.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern