Let $A$ be a $k\times n$ matrix and $B$ be a $n\times k$ matrix.
$\textbf{Goal:}$ I'm trying to show $\begin{equation}\det(AB)=\sum_{j_1,j_2, …,j_k=1}^n b_{j_1,1}b_{j_2,2}…b_{j_k,k}\det(A(j_1, j_2, …, j_k)) \end{equation}$ as shown in this proof for the Cauchy Binet Formula.
Note: I was using a linear map for determinants which can be found here on Wikipedia, and $\hat{e}_j$ denotes a single column with all zero's except the $j^{th}$ row which is a one.
$\textbf{Question:}$ Would the proof below work? I feel like I went wrong somewhere because I believe the last couple lines are wrong. Any help would be appreciated!
$\begin{align*}\det(AB)&=\det((AB)_1, (AB)_2, …, (AB)_k) \text{ where } (AB)_i \text{denotes the } i^{th} \text{column of } AB\\
&=\det(\sum_{i=1}^k\sum_{j_1=1}^na_{i,j_1}b_{j_1,1}\cdot \hat{e}_i,\sum_{i=1}^k\sum_{j_2=1}^na_{i,j_2}b_{j_2,2}\cdot \hat{e}_i , …, \sum_{i=1}^k\sum_{j_k=1}^na_{i,j_k}b_{j_k,k}\cdot \hat{e}_i ) \\
&=\sum_{i_1,i_2, …, i_k=1}^k\det(\sum_{j_1=1}^na_{i_1,j_1}b_{j_1,1}\cdot \hat{e}_{i_1},\sum_{j_2=1}^na_{i_2,j_2}b_{j_2,2}\cdot \hat{e}_{i_2} , …, \sum_{j_k=1}^na_{i_k,j_k}b_{j_k,k}\cdot \hat{e}_{i_k} ) \\
&=\sum_{i_1,i_2, …, i_k=1}^k\sum_{j_1,j_2, …,j_k=1}^n\det(a_{i_1,j_1}b_{j_1,1}\cdot \hat{e}_{i_1},a_{i_2,j_2}b_{j_2,2}\cdot \hat{e}_{i_2} , …, a_{i_k,j_k}b_{j_k,k}\cdot \hat{e}_{i_k} ) \\
&=\sum_{j_1,j_2, …,j_k=1}^n\sum_{i_1,i_2, …, i_k=1}^k\det(a_{i_1,j_1}b_{j_1,1}\cdot \hat{e}_{i_1},a_{i_2,j_2}b_{j_2,2}\cdot \hat{e}_{i_2} , …, a_{i_k,j_k}b_{j_k,k}\cdot \hat{e}_{i_k} ) \\
&=\sum_{j_1,j_2, …,j_k=1}^n b_{j_1,1}b_{j_2,2}…b_{j_k,k}\sum_{i_1,i_2, …, i_k=1}^k\det(a_{i_1,j_1}\cdot \hat{e}_{i_1},a_{i_2,j_2}\cdot \hat{e}_{i_2} , …, a_{i_k,j_k}\cdot \hat{e}_{i_k} ) \\
&=\sum_{j_1,j_2, …,j_k=1}^n b_{j_1,1}b_{j_2,2}…b_{j_k,k}\det(A(j_1, j_2, …, j_k)) \\
\end{align*}$
Best Answer
Note that $$ A_{i_m,j_m}=A(J)_{i_m.m}\quad \text{for all }m=1,2,\dots,k $$ since the $k\times k$ matrix $A(J)$ is constructed by picking $j_1,j_2,j_3,\dots$ columns of $A$ as the first, second, third, ... columns of $A(J)$. So $$ \det(a_{i_1,j_1}\cdot\hat e_{i_1},a_{i_2,j_2}\cdot\hat e_{i_2},\dots,a_{i_l,j_k}\cdot\hat e_{i_k}) =\det(A(J)_{i_1,1}\cdot\hat e_{i_1},A(J)_{i_2,2}\cdot\hat e_{i_2},\dots,A(J)_{i_k,k}\cdot\hat e_{i_k}) $$ and thus the sum over all $i_1,i_2,\dots,i_k$ is $\det A(J)$ by definition.