Given the following formulation of the determinant with Levi-Civita permutation symbols, show that $\det(AB)=\det A \det B$.

$$\det A = \sum\limits_{ij\cdots l}\epsilon_{ij\cdots l} A_{i1}A_{j2}\cdots A_{ln}\,,\,\,\,\,\,\textrm{where A is an }n\times n \textrm{ matrix}$$

I have been trying to show this for so long, but I can't seem to get past a certain point. Here is my work so far.

$$\begin{align*} \det (AB)&=\sum\limits_{ij\cdots l}\epsilon_{ij\cdots l}(AB)_{i1}(AB)_{j2}\cdots(AB)_{ln}\\ &=\sum\limits_{ij\cdots l}\epsilon_{ij\cdots l}\left(\sum\limits_{k_1}A_{ik_1}B_{k_11}\right)\left(\sum\limits_{k_2}A_{jk_2}B_{k_22}\right)\cdots \left(\sum\limits_{k_n}A_{lk_n}B_{k_nn}\right)\\ &=\sum\limits_{ij\cdots l}\epsilon_{ij\cdots l}\left(\sum\limits_{k_1,k_2,\cdots k_n} A_{ik_1}A_{jk_2}\cdots A_{lk_n}B_{k_11}B_{k_22}\cdots B_{k_nn}\right)\\ &=??? \end{align*} $$

Any tips on how to proceed? Have I made a mistake anywhere?

  • $\begingroup$ That one must start from that definition doesn't means that one must get tortured with manipulating the sums and the permutations. You can show from that definition that the determinant is invariant by the rows transformations and then reduce it to the case of triangular matrices. $\endgroup$
    – user218170
    Feb 22, 2015 at 5:42
  • 1
    $\begingroup$ Still I don't understand why we must use this complicated identity instead of directly using the definition. $\endgroup$
    – Vim
    Feb 22, 2015 at 5:44
  • $\begingroup$ math.rutgers.edu/~zeilberg/mamarimY/DM85.pdf gives a proof along the lines the OP suggested. A more modern way to do the proof would be to use the multilinearity repeatedly to reduce the problem to a case where $A$ has at most one nonzero entry per column and $B$ has at most one nonzero entry per row, and then casebash according to whether these entries form permutations or not. $\endgroup$ Feb 22, 2015 at 5:45
  • $\begingroup$ @Vim Because that identity is a definition and if you are writing a book or teaching a course and begin there you need to prove the remaining properties from the definition. $\endgroup$
    – user218170
    Feb 22, 2015 at 5:47
  • $\begingroup$ Oh yes it is.Just now looking at your post I thought mistakenly that your $ A_{i1}A_{j2}$ etc are Matrices since they are in capital form. In my textbook elements are all in lower case. $\endgroup$
    – Vim
    Feb 22, 2015 at 5:49

1 Answer 1


$$ \begin{align*}\det(AB) &= \sum_{ij\cdots l} \epsilon_{ij\cdots l} \sum_{k_1,\dots,k_n} A_{1k_1}B_{k_1i} A_{2k_2}B_{k_2j} \cdots A_{nk_n}B_{k_nl} \\ &= \sum_{k_1,\dots,k_n} A_{1k_1} A_{2k_2} \cdots A_{nk_n} \sum_{ij\cdots l} \epsilon_{ij\cdots l} B_{k_1i}B_{k_1j} \cdots B_{k_nl} \\ &= \sum_{k_1,\dots,k_n} A_{1k_1} A_{2k_2} \cdots A_{nk_n} \det(C_{k_1k_2\cdots k_n}) ,\end{align*}$$ where $C_{k_1k_2\cdots k_n}$ is the matrix whose first row is the $k_1$th row of $B$, whose second row is the $k_2$th row of $B$, ..., and whose $n$th row is the $k_n$th row of $B$.

Hence $$\det(C_{k_1k_2\cdots k_n}) = \epsilon_{k_1k_2\cdots k_n} \det(B).$$ Therefore $$ \det(AB) = \sum_{k_1,\dots,k_n} \epsilon_{k_1k_2\cdots k_n} A_{1k_1} A_{2k_2} \cdots A_{nk_n} \det(B) = \det(A) \det(B) .$$

  • $\begingroup$ Wow, how nice. The solution was staring me right in the face. Thank you so much! $\endgroup$ Feb 22, 2015 at 17:38
  • $\begingroup$ Wait, I just noticed something. How come in the first line (right after $\det (AB)$) you have the subscripts for $A$ as $1k_1,2k_2,\cdots nk_n$ and the subscripts for $B$ as $k_1i,k_2j,\cdots k_nl$, when in the last part that I left off at in my problem had them the other way around? $\endgroup$ Feb 23, 2015 at 14:37
  • $\begingroup$ What allowed you to switch those? $\endgroup$ Feb 23, 2015 at 14:37
  • $\begingroup$ Given that the result is correct, I'm guessing that any switching of subscripts (so long as it is maintained) is fine. $\endgroup$ Feb 23, 2015 at 15:18

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