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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!

$\begin{align*}\det(AB)&=\det((AB)_1, (AB)_2, ..., (AB)_k) \text{ where } (AB)_i \text{denotes the } i^{th} \text{column of } AB\\ &=\det(\sum_{i=1}^k\sum_{j_1=1}^na_{i,j_1}b_{j_1,1}\cdot \hat{e}_i,\sum_{i=1}^k\sum_{j_2=1}^na_{i,j_2}b_{j_2,2}\cdot \hat{e}_i , ..., \sum_{i=1}^k\sum_{j_k=1}^na_{i,j_k}b_{j_k,k}\cdot \hat{e}_i ) \\ &=\sum_{i_1,i_2, ..., i_k=1}^k\det(\sum_{j_1=1}^na_{i_1,j_1}b_{j_1,1}\cdot \hat{e}_{i_1},\sum_{j_2=1}^na_{i_2,j_2}b_{j_2,2}\cdot \hat{e}_{i_2} , ..., \sum_{j_k=1}^na_{i_k,j_k}b_{j_k,k}\cdot \hat{e}_{i_k} ) \\ &=\sum_{i_1,i_2, ..., i_k=1}^k\sum_{j_1,j_2, ...,j_k=1}^n\det(a_{i_1,j_1}b_{j_1,1}\cdot \hat{e}_{i_1},a_{i_2,j_2}b_{j_2,2}\cdot \hat{e}_{i_2} , ..., a_{i_k,j_k}b_{j_k,k}\cdot \hat{e}_{i_k} ) \\ &=\sum_{j_1,j_2, ...,j_k=1}^n\sum_{i_1,i_2, ..., i_k=1}^k\det(a_{i_1,j_1}b_{j_1,1}\cdot \hat{e}_{i_1},a_{i_2,j_2}b_{j_2,2}\cdot \hat{e}_{i_2} , ..., a_{i_k,j_k}b_{j_k,k}\cdot \hat{e}_{i_k} ) \\ &=\sum_{j_1,j_2, ...,j_k=1}^n b_{j_1,1}b_{j_2,2}...b_{j_k,k}\sum_{i_1,i_2, ..., i_k=1}^k\det(a_{i_1,j_1}\cdot \hat{e}_{i_1},a_{i_2,j_2}\cdot \hat{e}_{i_2} , ..., a_{i_k,j_k}\cdot \hat{e}_{i_k} ) \\ &=\sum_{j_1,j_2, ...,j_k=1}^n b_{j_1,1}b_{j_2,2}...b_{j_k,k}\det(A(j_1, j_2, ..., j_k)) \\ \end{align*}$

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    $\begingroup$ Why do you think the last lines are wrong? They are correct. $\endgroup$ May 26, 2019 at 18:50
  • $\begingroup$ So, then does $\det(A(j_1, j_2, ..., j_k)):=\sum_{i_1,i_2, ..., i_k=1}^k\det(a_{i_1,j_1}\cdot \hat{e}_{i_1},a_{i_2,j_2}\cdot \hat{e}_{i_2} , ..., a_{i_k,j_k}\cdot \hat{e}_{i_k} )$, or does it equal that for another reason? I felt that was a problem, because I did not think those terms were equal. Answering that would answer my question. $\endgroup$
    – W. G.
    May 26, 2019 at 18:53
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    $\begingroup$ Yes. Note that $a_{i_1,j_1}$ is $A(J)_{i_1,1}$, $a_{i_2,j_2}$ is $A(J)_{i_2,2}$, etc. so this reduces to the definition of determinant. $\endgroup$ May 26, 2019 at 18:58
  • $\begingroup$ Ohhh! I just got it! Please feel free to put what you just wrote above to answer this question. It makes sense to me now. Thank you! $\endgroup$
    – W. G.
    May 26, 2019 at 19:10

1 Answer 1

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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.

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