Why is the signature of a transposition -1? I'm going through Linear Algebra by Peter Lax and I'm completely stumped as to why the signature of the transpose is -1.  The signature is through the discriminant, which is defined as
$$P(x_1,...,x_n)=\prod_{i<j}(x_i-x_j)$$
and a permutation is expressed as
$$p(x_1,\dots,x_n)$$
and represents a bijective map of $x_1,\dots,x_n$ onto themselves.
The signature $\sigma$ for a permutation $p$ is defined as
$$P(p(x_1,...,x_n))=\sigma(p)P(x_1,...,x_n)$$
where $\sigma(p)=\pm1$.  The book then defines a transposition as
$$p(i)=i \qquad \text{for}\: i\neq j \:\text{or}\: k$$
$$p(j)=k \qquad p(k)=j$$
How I've been approaching this so far is by considering the number of changes where $(x_i-x_k)\rightarrow(x_i-x_j)$ where $x_j>x_i$, since these will introduce an extra -1 term in addition to the -1 term introduced by $(x_j-x_k)\rightarrow(x_k-x_j)$.  However, it always seems as though whatever solution I come up with is either true only for an odd or even number of said exchanges.  How can I approach this differently?
 A: We need to count the number $N$ of pairs $(r,s)$ where $r<s$ but $p(r)>p(s)$.
The sign of $p$ is then $(-1)^N$. Call such a pair an inversion of $p$.
Suppose $j<k$. Then the inversions of the permutation $p$ swapping $j$ and $k$
are $(j,k)$ and also both $(j,l)$
and $(l,k)$ where $j<l<k$. There are $k-j-1$ of these $l$ and so overall
$N=2(k-j-1)+1$, an odd number, so $(-1)^N=-1$.
A: Let $p$ be the transposition that swaps $k$ and $l$ (where $k<l$).
Then let us compare
$$\tag1 \prod_{i<j}(x_i-x_j)$$
and
$$\tag2 \prod_{i<j}(x_{p(i)}-x_{p(j)})$$
That is, for $i<j$, let us check whether $x_i-x_j$ or $-(x_i-x_j)$ occurs in $(2)$:

*

*If $\{i,j\}\cap\{k,l\}=\emptyset$, then $x_i-x_j$ occurs in $(2)$

*If $i<k$ and $j=l$, then $x_i-x_j=x_{p(i)}-x_{p(k)}$ occurs in $(2)$

*If $i<k$ and $j=k$, then $x_i-x_j=x_{p(i)}-x_{p(l)}$ occurs in $(2)$

*If $i=k$ and $j>l$, then $x_i-x_j=x_{p(l)}-x_{p(j)}$ occurs in $(2)$

*If $i=l$ and $j>l$, then $x_i-x_j=x_{p(k)}-x_{p(j)}$ occurs in $(2)$

*If $i=k<j<l$, then $-(x_i-x_j)=x_{p(j)}-x_{p(l)}$ occurs in $(2)$

*If $k<i<j=l$, then $-(x_i-x_j)=x_{p(l)}-x_{p(i)}$ occurs in $(2)$

*If $i=k$, $j=l$, then $-(x_i-x_j)=x_{p(k)}-x_{p(l)}$ occurs in $(2)$
The first five bullet points are irrelevant, the sixth and seventh bullet point always cancel (each point occurs $l-k-1$ times).  Remains the last bullet point, i.e., we have $(2)=-(1)$.
