# Question about the proof of the Cauchy Binet Formula

Let $$A$$ be a $$k\times n$$ matrix and $$B$$ be a $$n\times k$$ matrix.

$$\textbf{Goal:}$$ I'm trying to show $$$$\det(AB)=\sum_{j_1,j_2, ...,j_k=1}^n b_{j_1,1}b_{j_2,2}...b_{j_k,k}\det(A(j_1, j_2, ..., j_k))$$$$ 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*}

• Why do you think the last lines are wrong? They are correct. – user10354138 May 26 at 18:50
• 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. – W. G. May 26 at 18:53
• 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. – user10354138 May 26 at 18:58
• 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! – W. G. May 26 at 19:10

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.