If \(\sigma\) is a permutation of degree \(n\) and \(A\) is an \(n \times n\) matrix over the field \(F\) with row vectors \(\alpha_1, \ldots, \alpha_n\), let \(\sigma(A)\) denote the \(n \times n\) matrix with row vectors \(\alpha_{\sigma 1}, \ldots, \alpha_{\sigma n}\).
Prove that \(\sigma(AB) = \sigma(A)B\), and in particular that \(\sigma(A) = \sigma(I)A\).
If \(T\) is the linear operator of Exercise 9, prove that the matrix of \(T\) in the standard ordered basis is \(\sigma(I)\).
Is \(\sigma^{-1}(I)\) the inverse matrix of \(\sigma(I)\)?
Is it true that \(\sigma(A)\) is similar to \(A\)?
Solution
Let the rows of \(A\) be \(\alpha_1,\ldots,\alpha_n\).
For (a), the \(i\)-th row of \(\sigma(AB)\) is the \(\sigma(i)\)-th row of \(AB\), namely \(\alpha_{\sigma(i)}B\). This is also the \(i\)-th row of \(\sigma(A)B\). Hence \[ \sigma(AB)=\sigma(A)B. \] Taking \(A=I\) in the identity \(\sigma(IA)=\sigma(I)A\) gives \[ \sigma(A)=\sigma(I)A. \]
For (b), the operator of Exercise 9 sends \[ (x_1,\ldots,x_n)\longmapsto(x_{\sigma(1)},\ldots,x_{\sigma(n)}). \] For a column vector \(x\), the \(i\)-th coordinate of \(\sigma(I)x\) is precisely \(x_{\sigma(i)}\). Therefore its standard matrix is \(\sigma(I)\).
For (c), apply (a): \[ \sigma^{-1}(\sigma(I))=\sigma^{-1}(I)\sigma(I). \] The left side is \(I\), since applying the inverse row permutation undoes \(\sigma\). Thus \(\sigma^{-1}(I)\sigma(I)=I\); similarly the reverse product is \(I\). Hence \(\sigma^{-1}(I)=\sigma(I)^{-1}\).
For (d), not in general. Take \(A=I\) and any nonidentity permutation \(\sigma\). Then \[ \sigma(A)=\sigma(I)\ne I. \] But every matrix similar to \(I\) equals \(I\), since \(P^{-1}IP=I\). Hence \(\sigma(A)\) need not be similar to \(A\).