Given two square matrices A and B, how do you show that \det(AB) = \det(A)\det(B) where \det(\cdot) is the determinant of the matrix?

**Answer**

Let’s consider the function B\mapsto \det(AB) as a function of the columns of B=\left(v_1|\cdots |v_i| \cdots | v_n\right). It is straight forward to verify that this map is multilinear, in the sense that

\det\left(A\left(v_1|\cdots |v_i+av_i’| \cdots | v_n\right)\right)=\det\left(A\left(v_1|\cdots |v_i| \cdots | v_n\right)\right)+a\det\left(A\left(v_1|\cdots |v_i’| \cdots | v_n\right)\right). It is also alternating, in the sense that if you swap two columns of B, you multiply your overall result by -1. These properties both follow directly from the corresponding properties for the function A\mapsto \det(A).

The determinant is completely characterized by these two properties, and the fact that \det(I)=1. Moreover, any function that satisfies these two properties must be a multiple of the determinant. If you have not seen this fact, you should try to prove it. I don’t know of a reference online, but I know it is contained in Bretscher’s linear algebra book.

In any case, because of this fact, we must have that \det(AB)=c\det(B) for some constant c, and setting B=I, we see that c=\det(A).

For completeness, here is a proof of the necessary lemma that any a multilinear, alternating function is a multiple of determinant.

We will let f:\mathbb (F^n)^n\to \mathbb F be a multilinear, alternating function, where, to allow for this proof to work in characteristic 2, we will say that a multilinear function is alternating if it is zero when two of its inputs are equal (this is equivalent to getting a sign when you swap two inputs everywhere except characteristic 2). Let e_1, \ldots, e_n be the standard basis vectors. Then f(e_{i_1},e_{i_2}, \ldots, e_{i_n})=0 if any index occurs twice, and otherwise, if \sigma\in S_n is a permutation, then f(e_{\sigma(1)}, e_{\sigma(2)},\ldots, e_{\sigma(n)})=(-1)^\sigma, the sign of the permutation \sigma.

Using multilinearity, one can expand out evaluating f on a collection of vectors written in terms of the basis:

f\left(\sum_{j_1=1}^n a_{1j_1}e_{j_1}, \sum_{j_2=1}^n a_{2j_2}e_{j_2},\ldots, \sum_{j_n=1}^n a_{nj_n}e_{j_n}\right) = \sum_{j_1=1}^n\sum_{j_2=1}^n\cdots \sum_{j_n=1}^n \left(\prod_{k=1}^n a_{kj_k}\right)f(e_{j_1},e_{j_2},\ldots, e_{j_n}).

All the terms with j_{\ell}=j_{\ell’} for some \ell\neq \ell’ will vanish before the f term is zero, and the other terms can be written in terms of permutations. If j_{\ell}\neq j_{\ell’} for any \ell\neq \ell’, then there is a unique permutation \sigma with j_k=\sigma(k) for every k. This yields:

\begin{align}\sum_{j_1=1}^n\sum_{j_2=1}^n\cdots \sum_{j_n=1}^n \left(\prod_{k=1}^n a_{kj_k}\right)f(e_{j_1},e_{j_2},\ldots, e_{j_n}) &= \sum_{\sigma\in S_n} \left(\prod_{k=1}^n a_{k\sigma(k)}\right)f(e_{\sigma(1)},e_{\sigma(2)},\ldots, e_{\sigma(n)}) \\ &= \sum_{\sigma\in S_n} (-1)^{\sigma}\left(\prod_{k=1}^n a_{k\sigma(k)}\right)f(e_{1},e_{2},\ldots, e_{n}) \\ &= f(e_{1},e_{2},\ldots, e_{n}) \sum_{\sigma\in S_n} (-1)^{\sigma}\left(\prod_{k=1}^n a_{k\sigma(k)}\right). \end{align}

In the last line, the thing still in the sum is the determinant, although one does not need to realize this fact, as we have shown that f is completely determined by f(e_1,\ldots, e_n), and we simply define \det to be such a function with \det(e_1,\ldots, e_n)=1.

**Attribution***Source : Link , Question Author : Learner , Answer Author : Aaron*