[proofplan] We compute the determinant of $P_\sigma$ by comparing it with the determinant of matrices attached to transpositions. First we record that the assignment $\sigma\mapsto P_\sigma$ respects composition. Then we observe that a transposition matrix is obtained from the identity matrix by swapping two columns, so its determinant is $-1$. Finally, we choose a [transposition decomposition](/theorems/777) of $\sigma$ and compare the resulting determinant with the signature formula from [citetheorem:7876]. [/proofplan] [step:Show that permutation matrices respect composition] Let $\operatorname{id}_{\mathbb{R}^n}: \mathbb{R}^n\to\mathbb{R}^n$ denote the identity [linear map](/page/Linear%20Map), and let $(e_1,\dots,e_n)$ denote the standard ordered basis of $\mathbb{R}^n$. For each $\rho\in S_n$, the matrix $P_\rho$ is characterized by the condition $P_\rho e_i=e_{\rho(i)}$ for all $i\in\{1,\dots,n\}$. Let $\sigma,\pi\in S_n$. For every $i\in\{1,\dots,n\}$, matrix multiplication gives \begin{align*} P_\sigma P_\pi e_i=P_\sigma e_{\pi(i)}=e_{\sigma(\pi(i))}=P_{\sigma\pi}e_i. \end{align*} Since two linear maps $\mathbb{R}^n\to\mathbb{R}^n$ agreeing on the standard basis are equal, we have \begin{align*} P_{\sigma\pi}=P_\sigma P_\pi. \end{align*} In particular, $P_{\operatorname{id}}=I_n$, where $I_n\in\mathbb{R}^{n\times n}$ is the identity matrix. [/step] [step:Compute the determinant of a transposition matrix] Let $\tau\in S_n$ be a transposition. Choose distinct indices $a,b\in\{1,\dots,n\}$ such that $\tau(a)=b$, $\tau(b)=a$, and $\tau(i)=i$ for every $i\notin\{a,b\}$. The $i$th column of $P_\tau$ is $P_\tau e_i=e_{\tau(i)}$. Thus $P_\tau$ is obtained from $I_n$ by interchanging its $a$th and $b$th columns and leaving all other columns fixed. Since the determinant is alternating in the columns and $\det(I_n)=1$, interchanging two distinct columns multiplies the determinant by $-1$. Therefore \begin{align*} \det(P_\tau)=-\det(I_n)=-1. \end{align*} [guided] We need the determinant of a single transposition matrix because arbitrary permutations can be built from transpositions. Let $\tau\in S_n$ be a transposition. This means that there are distinct indices $a,b\in\{1,\dots,n\}$ such that $\tau(a)=b$, $\tau(b)=a$, and every other index is fixed by $\tau$. By the definition of $P_\tau$, the $i$th column of $P_\tau$ is the vector $P_\tau e_i=e_{\tau(i)}$. For $i\notin\{a,b\}$, this column is $e_i$, exactly as in the identity matrix $I_n$. The $a$th column is $e_b$, and the $b$th column is $e_a$. Hence $P_\tau$ is precisely the identity matrix with the $a$th and $b$th columns exchanged. The determinant is an alternating multilinear function of the columns, normalized by $\det(I_n)=1$. The alternating property says that swapping two distinct columns changes the sign of the determinant. Applying this to the two exchanged columns of $I_n$, we obtain \begin{align*} \det(P_\tau)=-\det(I_n)=-1. \end{align*} This is the determinant analogue of the fact that a transposition has odd signature. [/guided] [/step] [step:Decompose the permutation into transpositions] We use the standard fact that every element of $S_n$ is a product of transpositions. Thus there exist an integer $m\ge 0$ and transpositions $\tau_1,\dots,\tau_m\in S_n$ such that \begin{align*} \sigma=\tau_1\tau_2\cdots\tau_m. \end{align*} If $m=0$, this means $\sigma=\operatorname{id}$ and the empty product of matrices is $I_n$. [/step] [step:Multiply the transposition determinants] Using the composition identity for permutation matrices repeatedly, we obtain \begin{align*} P_\sigma=P_{\tau_1\tau_2\cdots\tau_m}=P_{\tau_1}P_{\tau_2}\cdots P_{\tau_m}. \end{align*} By multiplicativity of the determinant on square matrices, \begin{align*} \det(P_\sigma)=\det(P_{\tau_1})\det(P_{\tau_2})\cdots\det(P_{\tau_m}). \end{align*} Each $\tau_i$ is a transposition, so the previous step gives $\det(P_{\tau_i})=-1$ for every $i\in\{1,\dots,m\}$. Therefore \begin{align*} \det(P_\sigma)=(-1)^m. \end{align*} [/step] [step:Identify the result with the signature] The chosen decomposition writes $\sigma$ as a product of $m$ transpositions. By [citetheorem:7876], \begin{align*} \operatorname{sgn}(\sigma)=(-1)^m. \end{align*} Combining this with the determinant computation gives \begin{align*} \det(P_\sigma)=(-1)^m=\operatorname{sgn}(\sigma). \end{align*} This proves the claimed identity. [/step]