This is the problem I am given:
$\sigma \in S_{n}$. Define the $n \times n$ metrix $A = (\delta_{i \sigma^{-1}(j)})$.
(i). If $n=3, \sigma(1)=3,\sigma(2)=1,\sigma(3)=2,$ write out A explicitly
(ii). Prove $A \in O_{n}(\mathbb{R})$
(iii). Prove $A \in SO_{n}(\mathbb{R})$ if and only if $sgn(\sigma) = +1$
So far what I have (if ye could double check my work, that'd be greatly appreciated!):
(i).
$$\begin{array}
\verb A=\left(\begin{array}{@{}ccc@{}}
0 & 0 & 1 \\
1 & 0 & 0 \\
0 & 1 & 0
\end{array}\right)
\end{array}
$$
(ii).
$$\begin{array}
\verb A^{T}=\left(\begin{array}{@{}ccc@{}}
0 & 1 & 0 \\
0 & 0 & 1 \\
1 & 0 & 0
\end{array}\right)
\end{array} = A^{-1}
$$ Hence $A \in O_{3}(\mathbb{R})$
I am not sure how to prove the general case for this.
As for (iii) I am unsure how to go about it, any help would be appreciated! Thanks
Hint: The $i$-line of $A$ is $\delta_{i\sigma^{-1}(1)}..\delta_{i\sigma^{-1}(n)}$,
the $j$-column of the transpose of $A$ is the $j$-ligne of $A$, so it is $\delta_{j\sigma^{-1}(1)}..\delta_{j\sigma^{-1}(n)}$.
The $(i,j)$ term of $AA^T$ is the $\delta_{i\sigma^{-1}(1)}\delta_{j\sigma^{-1}(1)}+...+\delta_{i\sigma^{-1}(n)}\delta_{j\sigma^{-1}(n)}$
$\delta_{i\sigma^{-1}(k)}\delta_{j\sigma^{-1}(k)}$ is not zero if and only if $\delta_{i\sigma^{-1}(k)}=1$ and $\delta_{j\sigma^{-1}(k)}=1$ which is equivalent to $\sigma^{-1}(k)=i$, $k=\sigma(i)=\sigma(j)$ which implies that $i=j$. We deduce that $AA^T=Id_n$.