Let \(n\) be a positive integer and \(F\) a field. If \(\sigma\) is a permutation of degree \(n\), prove that the function \[ T(x_1, \ldots, x_n) = (x_{\sigma 1}, \ldots, x_{\sigma n}) \] is an invertible linear operator on \(F^n\).
Solution
1. \(T\) is linear.
1.1. \(T(x + y) = T(x) + T(y)\).
the \(i\)-th coordinate of \(T(x+y)\) is \((x+y)_{\sigma i} = x_{\sigma i} + y_{\sigma i}\), which is the \(i\)-th coordinate of \(T(x) + T(y)\).
1.2. \(T(cx) = c\,T(x)\) for \(c \in F\).
the \(i\)-th coordinate of \(T(cx)\) is \((cx)_{\sigma i} = c\, x_{\sigma i}\).
2. \(T\) is invertible.
2.1. \(T\) is the linear map with permutation matrix \(P_\sigma\) (the matrix with \(1\) in position \((\sigma i, i)\) and \(0\) elsewhere).
\(T\) permutes the standard basis vectors: \(T(e_i) = e_{\sigma i}\).
2.2. \(T^{-1} = T_{\sigma^{-1}}\), the map \(y \mapsto (y_{\sigma^{-1}1}, \ldots, y_{\sigma^{-1}n})\).
\(T_{\sigma^{-1}} \circ T(x) = (x_{\sigma(\sigma^{-1}1)}, \ldots) = (x_1, \ldots, x_n) = x\), and similarly the other way.
2.3. Hence \(T\) is invertible.
2 exhibits an inverse.
3. Q.E.D.
1 and 2.