Linear algebra (Osnabrück 2024-2025)/Part I/Lecture 18/refcontrol
- Permutations
In this lecture, we provide another description for the determinant with the help of permutations.
For a set , we call the set
of all bijective mappingsMDLD/bijective mappings
on the automorphism group or the permutation group of .The operation is the composition of mappings, therefore, it is associative, the identity is the neutral element. The inverse element for a bijective mapping is just the inverse mapping. Hence, this is a group. A bijective mapping is also called a permutation.
For the finite set , we also write
A permutation of a finite set can be described with a (complete) value table or with an arrow diagram.
Let be a finite set and let be a permutationMDLD/permutation (finite) on . Then is called a cycle of order , if there exists a subset , containing elements and such that is on the identity and such that commutes the elements of in a cyclic way. If , then we write
We consider the permutation
We can write this as the product of the two cyclesMDLD/cycles (permutation) and .
An element with is called a fixed pointMDLD/fixed point of the permutation. The (action) scope of a permutation is the set of points from which are not fixed points. For a cycle, the set is the scope. We mention without proof that every permutation is a product of cycles. Such a product representation is called a cycle representation.
For a natural number , one puts
Let be a finite set with elements. Then the permutation groupMDLD/permutation group
Let . For , there are possible images, for , there are possible images remaining, for , there are possible images remaining, etc. Therefore, there are altogether
possible permutations.
- Transpositions
A transposition on a finite set is a permutationMDLD/permutation
on which swaps two elements and leaves all other elements on their place.A transposition is a cycle of length .
Every permutationMDLD/permutation on a finite set can be written as a product of
transpositions.MDLD/transpositionsWe proof the statement by induction over the cardinality of the set . For , there is nothing to show, so let . The identity is the empty product of transpositions. So suppose that is not the identity, and let . Let be the transposition which swaps and . Then is a fixed pointMDLD/fixed point of , and we can consider as a permutation on . By the induction hypothesis, there exist transpositions on such that on . This does also hold on , and we get .
- The sign of a permutation
Let and let be a permutationMDLD/permutation on . Then we call the number
The sign is or , because in the numerator and in the denominator, up to sign, the same differences occur. Thus, for the sign, there are only two possible values. For , we say that is an even permutation, and for , we say that is an odd permutation.
Let and let be a permutationMDLD/permutation on . We call an index pair
an inversion (of ), if
holds.Let and let be a permutationMDLD/permutation on . Let denote the number of inversionsMDLD/inversions (permutation) of . Then the signMDLD/sign (permutation) of equals
We write
because, after this reordering, we have in the numerator as well as in the denominator the product of all positive differences.
We consider the permutation
with the cycle representation
The inversions are
so there are of those. The sign is due to fact, and the permutation is odd.
The sign is a group homomorphism in the sense of the following definition.
Let and denote groups.MDLD/groups A mappingMDLD/mapping
is called group homomorphism, if the equality
holds for all
.Let two permutations and be given. Then
Let and let be a permutationMDLD/permutation on . Let
be written as a product of transpositions.MDLD/transpositions Then the signMDLD/sign (permutation) can be described as
Suppose that the transposition swaps the numbers . Then
The last equation follows from the fact that, in the first and the second product, all numerators and denominators are positive, and the fact that, in the third and in the forth product, the numerators are negative and the denominators are positive. Therefore, as the index sets of the third and the fourth product coincide, all the signs cancel each other.
The statement follows from the case of a transposition via the homomorphism property.
Let be an arbitrary set with elements, but without an ordering, and let be a permutation on . Then we can not talk about inversions,MDLD/inversions and the definition of signMDLD/definition of sign via products of differences is not directly applicable. However, we can look at fact in order to define the sign in this slightly more general situation. For this, we write as a product of transpositions and define
To see that this is well-defined, we consider a bijection
The permutation on defines on the permutation . Let be a representation as a product of transpositions on . Then
where . These are also transpositions, so that the parity of is determined by the sign of .
- The Leibniz formula for the determinant
We do induction over , the base case is clear, so let . The set of permutations can be split up, by sorting along , and considering the bijective mappings
as a permutation on , by both sets in an order-preserving way with . This yields a bijection , where denotes the set of permutations on which send to . Between the signs, there is the relation
since we need transpositions to put the -th place to the first place. Altogether, there is a natural bijection
Hence, we get
Here, is the submatrix in which the first row and the -th column is omitted. For the penultimate equation, we use the induction hypothesis, and the last equation rests on Laplace expansion with respect to the first row.
<< | Linear algebra (Osnabrück 2024-2025)/Part I | >> PDF-version of this lecture Exercise sheet for this lecture (PDF) |
---|