# Linear algebra (Osnabrück 2024-2025)/Part I/Lecture 18

    Permutations
In this lecture, we provide another description for the determinant with the help of permutations.

## Definition

For a set $M$, we call the set
    $\operatorname {Aut} \,(M)=\operatorname {Perm} \,(M)={\left\{\varphi :M\longrightarrow M\mid \varphi {\text{ bijective}}\right\}}\,$
of all bijective mappings
$M$
$M$
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 $\varphi \colon M\rightarrow M$ is also called a permutation.
For the finite set $I=\{1,\ldots ,n\}$, we also write
    $S_{n}=\operatorname {Perm} \,(I)\,.$
A permutation of a finite set can be described with a (complete) value table or with an arrow diagram.

## Definition

Let $M$ be a finite set and let $\pi$ be a permutation on $M$. Then $\pi$ is called a cycle of order $r$, if there exists a subset $Z\subseteq M$, containing $r$ elements and such that $\pi$ is on $M\setminus Z$ the identity and such that $\pi$ commutes the elements of $Z$ in a cyclic way. If $Z=\{z,\pi (z),\,\pi ^{2}(z),\ldots ,\pi ^{r-1}(z)\}$, then we write
    $\pi =\langle z,\pi (z),\,\pi ^{2}(z),\ldots ,\pi ^{r-1}(z)\rangle \,.$

## Example

We consider the permutation
We can write this as the product of the two cycles $\langle 1,2\rangle$ and $\langle 3,5,4\rangle$.
An element $x\in M$ with $\pi (x)=x$ is called a fixed point of the permutation. The (action) scope of a permutation is the set of points from $M$ which are not fixed points. For a cycle, the set $Z$ is the scope. We mention without proof that every permutation is a product of cycles. Such a product representation is called a cycle representation.

## Definition

For a natural number $n$, one puts
    $n!:=n(n-1)(n-2)\cdots 3\cdot 2\cdot 1\,,$
$n$

## Lemma

Let $M$ be a finite set with $n$ elements. Then the permutation group
    $\operatorname {Perm} \,(M)\cong S_{n}\,$
$n!$

### Proof

Let $M=\{1,\ldots ,n\}$. For $1$, there are $n$ possible images, for $2$, there are $n-1$ possible images remaining, for $3$, there are $n-2$ possible images remaining, etc. Therefore, there are altogether
    $n(n-1)(n-2)\cdots 2\cdot 1=n!\,$
possible permutations.
$\Box$
    Transpositions

## Definition

A transposition on a finite set $M$ is a permutation
$M$
A transposition is a cycle of length $2$.

## Lemma

Every permutation on a finite set $M$ can be written as a product of

### Proof

We proof the statement by induction over the cardinality of the set $M$. For ${\#\left(M\right)}=1$, there is nothing to show, so let ${\#\left(M\right)}\geq 2$. The identity is the empty product of transpositions. So suppose that $\pi$ is not the identity, and let $\pi (x)=y\neq x$. Let $\tau$ be the transposition which swaps $x$ and $y$. Then $y$ is a fixed point of $\pi \tau$, and we can consider $\pi \tau$ as a permutation on $M'=M\setminus \{y\}$. By the induction hypothesis, there exist transpositions $\tau _{j}$ on $M'$ such that $\pi \tau =\prod _{j}\tau _{j}$ on $M'$. This does also hold on $M$, and we get $\pi ={\left(\prod _{j}\tau _{j}\right)}\tau$.
$\Box$
    The sign of a permutation

## Definition

Let $M=\{1,\ldots ,n\}$ and let $\pi$ be a permutation on $M$. Then we call the number
    $\operatorname {sgn} (\pi )=\prod _{i<j}{\frac {\pi (j)-\pi (i)}{j-i}}\,$
$\pi$
The sign is $1$ or $-1$, because in the numerator and in the denominator, up to sign, the same differences $\pm (j-i)$ occur. Thus, for the sign, there are only two possible values. For $\operatorname {sgn} (\pi )=1$, we say that $\pi$ is an even permutation, and for $\operatorname {sgn} (\pi )=-1$, we say that $\pi$ is an odd permutation.

## Definition

Let $M=\{1,\ldots ,n\}$ and let $\pi$ be a permutation on $M$. We call an index pair
    $i<j\,$
an inversion (of $\pi$), if $\pi (i)>\pi (j)$

## Lemma

Let $M=\{1,\ldots ,n\}$ and let $\pi$ be a permutation on $M$. Let $k={\#\left(F\right)}$ denote the number of inversions of $\pi$. Then the sign of $\pi$ equals
    $\operatorname {sgn} (\pi )=(-1)^{k}\,.$

### Proof

We write
    ${\begin{aligned}\operatorname {sgn} (\pi )&=\prod _{i<j}{\frac {\pi (j)-\pi (i)}{j-i}}\\&=\prod _{(i,j)\in F}{\frac {\pi (j)-\pi (i)}{j-i}}\prod _{(i,j)\not \in F}{\frac {\pi (j)-\pi (i)}{j-i}}\\&=(-1)^{k}\prod _{(i,j)\in F}{\frac {\pi (i)-\pi (j)}{j-i}}\prod _{(i,j)\not \in F}{\frac {\pi (j)-\pi (i)}{j-i}}\\&=(-1)^{k},\end{aligned}}$
because, after this reordering, we have in the numerator as well as in the denominator the product of all positive differences.
$\Box$

## Example

We consider the permutation
with the cycle representation
    $\langle 1,2,4,5,3,6\rangle .$
The inversions are
    $(1,6),\,(2,5),\,(2,6),\,(3,4),\,(3,5),\,(3,6),\,(4,5),\,(4,6),\,(5,6),$
so there are $9$ of those. The sign is $(-1)^{9}=-1$ due to fact, and the permutation is odd.
The sign is a group homomorphism in the sense of the following definition.

## Definition

Let $(G,\circ ,e_{G})$ and $(H,\circ ,e_{H})$ denote groups. A mapping
    $\psi \colon G\longrightarrow H$
is called group homomorphism, if the equality
    $\psi (g\circ g')=\psi (g)\circ \psi (g')\,$
holds for all
$g,g'\in G$

## Theorem

The sign mapping
    $S_{n}\longrightarrow \{1,-1\},\pi \longmapsto \operatorname {sgn} (\pi ),$
is a

### Proof

Let two permutations $\pi$ and $\rho$ be given. Then
    ${\begin{aligned}\operatorname {sgn} (\pi \circ \rho )&=\prod _{i<j}{\frac {(\pi \circ \rho )(j)-(\pi \circ \rho )(i)}{j-i}}\\&={\left(\prod _{i<j}{\frac {(\pi \circ \rho )(j)-(\pi \circ \rho )(i)}{\rho (j)-\rho (i)}}\right)}\prod _{i<j}{\frac {\rho (j)-\rho (i)}{j-i}}\\&={\left(\prod _{i<j,\,\rho (i)<\rho (j)}{\frac {\pi (\rho (j))-\pi (\rho (i))}{\rho (j)-\rho (i)}}\right)}{\left(\prod _{i<j,\,\rho (i)>\rho (j)}{\frac {\pi (\rho (j))-\pi (\rho (i))}{\rho (j)-\rho (i)}}\right)}\,\operatorname {sgn} (\rho )\\&={\left(\prod _{i<j,\,\rho (i)<\rho (j)}{\frac {\pi (\rho (j))-\pi (\rho (i))}{\rho (j)-\rho (i)}}\right)}{\left(\prod _{i<j,\,\rho (i)>\rho (j)}{\frac {\pi (\rho (i))-\pi (\rho (j))}{\rho (i)-\rho (j)}}\right)}\,\operatorname {sgn} (\rho )\\&=\prod _{k<\ell }{\frac {\pi (\ell )-\pi (k)}{\ell -k}}\operatorname {sgn} (\rho )\\&=\operatorname {sgn} (\pi )\operatorname {sgn} (\rho ).\end{aligned}}$
$\Box$

## Lemma

Let $M=\{1,\ldots ,n\}$ and let $\pi$ be a permutation on $M$. Let
    $\pi =\tau _{1}\cdots \tau _{r}\,$
be written as a product of $r$ transpositions. Then the sign can be described as
    $\operatorname {sgn} (\pi )=(-1)^{r}\,.$

### Proof

Suppose that the transposition $\tau$ swaps the numbers $k<\ell$. Then
    ${\begin{aligned}\operatorname {sgn} (\tau )&=\prod _{i<j}{\frac {\tau (j)-\tau (i)}{j-i}}\\&=\prod _{i<j,\,i,j\neq k,\ell }{\frac {\tau (j)-\tau (i)}{j-i}}\cdot \prod _{i<j,\,i=k,j\neq \ell }{\frac {\tau (j)-\tau (i)}{j-i}}\cdot \prod _{i<j,\,i\neq k,j=\ell }{\frac {\tau (j)-\tau (i)}{j-i}}\cdot \prod _{i<j,\,i=k,j=\ell }{\frac {\tau (j)-\tau (i)}{j-i}}\\&=\prod _{j>k,\,j\neq \ell }{\frac {j-\ell }{j-k}}\cdot \prod _{i\neq k,\,i<\ell }{\frac {k-i}{\ell -i}}\cdot {\frac {k-\ell }{\ell -k}}\\&=\prod _{j>\ell }{\frac {j-\ell }{j-k}}\cdot \prod _{i<k}{\frac {k-i}{\ell -i}}\cdot \prod _{k<j<\ell }{\frac {j-\ell }{j-k}}\cdot \prod _{k<i<\ell }{\frac {k-i}{\ell -i}}\cdot (-1)\\&=-1.\end{aligned}}$
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.
$\Box$

## Remark

Let $I$ be an arbitrary set with $n$ elements, but without an ordering, and let $\pi$ be a permutation on $I$. Then we can not talk about
inversions, and the 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 $\pi$ as a product of $r$ transpositions and define
    $\operatorname {sgn} (\pi )={\begin{cases}1\,,{\text{ in case }}r{\text{ is even}}\,,\\-1\,,{\text{ in case }}r{\text{ is odd}}\,.\end{cases}}\,$
To see that this is well-defined, we consider a bijection
    $\varphi \colon I\longrightarrow \{1,\ldots ,n\}.$
The permutation $\pi$ on $I$ defines on $\{1,\ldots ,n\}$ the permutation $\pi '=\varphi \pi \varphi ^{-1}$. Let $\pi =\tau _{1}\cdots \tau _{r}$ be a representation as a product of $r$ transpositions on $I$. Then
    $\pi '=\varphi \pi \varphi ^{-1}=\varphi \tau _{1}\cdots \tau _{r}\varphi ^{-1}=\varphi \tau _{1}\varphi ^{-1}\varphi \tau _{2}\varphi ^{-1}\varphi \cdots \varphi ^{-1}\varphi \tau _{r}\varphi ^{-1}=\tau _{1}'\tau _{2}'\cdots \tau _{r}'\,,$
where $\tau _{j}'=\varphi \tau _{j}\varphi ^{-1}$. These are also transpositions, so that the parity of $r$ is determined by the sign of $\pi '$.
    The Leibniz formula for the determinant

## Theorem

For the determinant of an $n\times n$-matrix
    $M={\left(a_{ij}\right)}_{ij}\,,$
    $\det M=\sum _{\pi \in S_{n}}\operatorname {sgn} (\pi )a_{1\pi (1)}\cdots a_{n\pi (n)}\,$

### Proof

We do induction over $n\geq 1$, the base case is clear, so let $n\geq 2$. The set of permutations $\pi \in S_{n}$ can be split up, by sorting along $\pi (1)$, and considering the bijective mappings
    $\pi {|}_{\{2,\ldots ,n\}}\colon \{2,\ldots ,n\}\longrightarrow \{1,\ldots ,n\}\setminus \{i\}$
as a permutation $\rho$ on $\{1,\ldots ,n-1\}$, by both sets in an order-preserving way with $\{1,\ldots ,n-1\}$. This yields a bijection $S_{n,i}\cong S_{n-1}$, where $S_{n,i}$ denotes the set of permutations on $\{1,\ldots ,n\}$ which send $1$ to $i$. Between the signs, there is the relation
    $\operatorname {sgn} (\pi )=(-1)^{i-1}\operatorname {sgn} (\rho )=(-1)^{i+1}\operatorname {sgn} (\rho )\,,$
since we need $i-1$ transpositions to put the $i$-th place to the first place. Altogether, there is a natural bijection
    $S_{n}=\biguplus _{i\in \{1,\ldots ,n\}}S_{n,i}=\biguplus _{i\in \{1,\ldots ,n\}}S_{n-1}\,.$
Hence, we get
    ${\begin{aligned}\sum _{\pi \in S_{n}}\operatorname {sgn} (\pi )a_{1\pi (1)}\cdots a_{n\pi (n)}&=\sum _{i=1}^{n}\sum _{\pi \in S_{n,i}}\operatorname {sgn} (\pi )\prod _{j=1}^{n}a_{j\pi (j)}\\&=\sum _{i=1}^{n}a_{1i}\sum _{\pi \in S_{n,i}}\operatorname {sgn} (\pi )\prod _{j=2}^{n}a_{j\pi (j)}\\&=\sum _{i=1}^{n}a_{1i}\sum _{\rho \in S_{n-1}}(-1)^{i+1}\operatorname {sgn} (\rho )\prod _{k=1}^{n-1}(M_{1i})_{k\rho (k)}\\&=\sum _{i=1}^{n}(-1)^{i+1}a_{1i}\det M_{1i}\\&=\det M.\end{aligned}}$
Here, $M_{1i}$ is the submatrix in which the first row and the $i$-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.
$\Box$