Suppose \(L:V\mapto V\) is a linear operator and consider the tensor product map \(L^{\otimes r}=L\otimes\dots\otimes L:T^r(V)\mapto T^r(V)\). Then clearly \(L^{\otimes r}\circ A=A\circ L^{\otimes r}\) so that \(L^{\otimes r}|_{\Lambda^r(V)}:\Lambda^r(V)\mapto\Lambda^r(V)\). This restriction is typically denoted \(L^{\wedge p}\). Now, as we’ve already observed, if \(V\) is an \(n\)-dimensional vector space, then \(\dim\Lambda^n(V)=1\). So any \(L^{\wedge n}\) is multiplication by a scalar. Choosing a basis, \(\{e_i\}\), of \(V\), then \(e_1\wedge\dots\wedge e_n\) is the single basis element of \(\Lambda^n(V)\), and if we write, \(Le_i=L_i^je_j\), then
\begin{equation}
L^{\wedge n}(e_1\wedge\dots\wedge e_n)=d_Le_1\wedge\dots\wedge e_n,
\end{equation}
where \(d_L\) is some scalar. But we also have,
\begin{equation}
L^{\wedge n}(e_1\wedge\dots\wedge e_n)=L_1^{i_1}\cdots L_n^{i_n}e_{i_1}\wedge\dots\wedge e_{i_n}.
\end{equation}
Now, the right hand side here is only non-zero when the set of indices \(\{i_1,\dots,i_n\}\) is precisely \(\{1,2,\dots,n\}\) and in this case
\begin{equation}
L_1^{i_1}\cdots L_n^{i_n}e_{i_1}\wedge\dots\wedge e_{i_n}=\sum_{\sigma\in S_n}\sgn(\sigma)L_1^{\sigma_1}\cdots L_n^{\sigma_n}e_1\wedge\dots\wedge e_n,
\end{equation}
in which we see precisely our original definition of the determinant,
so that \(d_L=\det L\).