The determinant product rule states that for any two square matrices A and B, the determinant of their product equals the product of their determinants: det(AB) = det(A)det(B). This can be proven using the definition of the determinant involving permutations and the properties of the sign function. The proof involves expressing the determinant of the product AB in terms of the determinants of A and B, ultimately leading to the conclusion that the determinant of the product is indeed the product of the determinants.
PREREQUISITESMathematicians, students of linear algebra, and anyone interested in understanding the properties of determinants and their applications in various mathematical contexts.
MOG_Manuel said:
Fredrik said:The definition of the determinant is
\det A=\sum_\sigma (\operatorname{sgn}\sigma)A^1_{\sigma(1)}\cdots A^n_{\sigma(n)}
where the sum is over all permutations of the set {1,2,...,n}, sgn σ is =1 when the permutation is even and =-1 when it's odd, and A^i_j denotes the entry on row i, column j. With this notation, you can do it as a fairly straightforward calculation:
<br /> \begin{align*}<br /> \det(AB) &=\sum_\sigma (\operatorname{sgn}\sigma)(AB)^1_{\sigma(1)}\cdots (AB)^n_{\sigma(n)}=\sum_\sigma (\operatorname{sgn}\sigma)\Big(\sum_{i_1}A^1_{i_1}B^{i_1}_{\sigma(1)}\Big)\cdots \Big(\sum_{i_n}A^1_{i_n}B^{i_n}_{\sigma(n)}\Big)\\<br /> &=\sum_{i_1,\dots,i_n}A^1_{i_1}\cdots A^n_{i_n}<br /> \underbrace{\sum_\sigma (\operatorname{sgn}\sigma) B^{i_1}_{\sigma(1)}\cdots B^{i_n}_{\sigma(n)}}_{=0\text{ unless }(i_1,\dots,i_n)\text{ is a permutation of }(1,\dots,n).}\\<br /> &=\sum_\tau A^1_{\tau(1)}\cdots A^n_{\tau(n)}\sum_\sigma (\operatorname{sgn}\sigma) B^{\tau(1)}_{\sigma(1)}\cdots B^{\tau(n)}_{\sigma(n)}\\<br /> &=\sum_\tau A^1_{\tau(1)}\cdots A^n_{\tau(n)}\sum_\sigma \underbrace{(\operatorname{sgn}\sigma)}_{=(\operatorname{sgn}\tau)(\operatorname{sgn}(\tau^{-1}\circ\sigma))} B^{1}_{\tau^{-1}\circ\sigma(1)}\cdots B^{n}_{\tau^{-1}\circ\sigma(n)}\\<br /> &=\sum_\tau(\operatorname{sgn}\tau)A^1_{\tau(1)}\cdots A^n_{\tau(n)}\underbrace{\sum_{\tau^{-1}\circ\sigma}(\operatorname{sgn}(\tau^{-1}\circ\sigma))B^{1}_{\tau^{-1}\circ\sigma(1)}\cdots B^{n}_{\tau^{-1}\circ\sigma(n)}}_{=\det B}\\<br /> &=(\det A)(\det B)<br /> \end{align*}<br />
but you will probably have to stare at this for a while before you understand all the steps.
\tau and \sigma are both permutations of {1,2,...,n}, i.e. they are bijections from that set onto itself. In the revised calculation below, \rho represents a permutation of {1,2,...,n} too.etnad179 said:Just to check the \tau is the set of permutation of the elements {i_1,...i_n}?
I think I did that step wrong. This is how I would like to handle the product B^{\tau(1)}_{\sigma(1)}\cdots B^{\tau(n)}_{\sigma(n)} today: First use the fact that real numbers commute, to rearrange the factors so that the row indices (the upper indices) appear in the order (1,2,...,n). I'll write asterisks instead of the column indices until we have figured out what we should write in those slots.etnad179 said:And how can we get from this permutation \tau of the row is equal to the inverse \tau^{-1} of the column for matrix element B^{\tau(1)}_{\sigma(1)}