# How to show that det\det(AB) =\det(A) \det(B)?

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

Let’s consider the function $$B\mapsto \det(AB)B\mapsto \det(AB)$$ as a function of the columns of $$B=\left(v_1|\cdots |v_i| \cdots | v_n\right)B=\left(v_1|\cdots |v_i| \cdots | v_n\right)$$. It is straight forward to verify that this map is multilinear, in the sense that
$$\det\left(A\left(v_1|\cdots |v_i+av_i’| \cdots | v_n\right)\right)=\det\left(A\left(v_1|\cdots |v_i| \cdots | v_n\right)\right)+a\det\left(A\left(v_1|\cdots |v_i’| \cdots | v_n\right)\right).\det\left(A\left(v_1|\cdots |v_i+av_i'| \cdots | v_n\right)\right)=\det\left(A\left(v_1|\cdots |v_i| \cdots | v_n\right)\right)+a\det\left(A\left(v_1|\cdots |v_i'| \cdots | v_n\right)\right).$$ It is also alternating, in the sense that if you swap two columns of $$BB$$, you multiply your overall result by $$-1-1$$. These properties both follow directly from the corresponding properties for the function $$A\mapsto \det(A)A\mapsto \det(A)$$.

The determinant is completely characterized by these two properties, and the fact that $$\det(I)=1\det(I)=1$$. 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 $$\det(AB)=c\det(B)\det(AB)=c\det(B)$$ for some constant $$cc$$, and setting $$B=IB=I$$, we see that $$c=\det(A)c=\det(A)$$.

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

We will let $$f:\mathbb (F^n)^n\to \mathbb Ff:\mathbb (F^n)^n\to \mathbb F$$ 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 $$e_1, \ldots, e_ne_1, \ldots, e_n$$ be the standard basis vectors. Then $$f(e_{i_1},e_{i_2}, \ldots, e_{i_n})=0f(e_{i_1},e_{i_2}, \ldots, e_{i_n})=0$$ if any index occurs twice, and otherwise, if $$\sigma\in S_n\sigma\in S_n$$ is a permutation, then $$f(e_{\sigma(1)}, e_{\sigma(2)},\ldots, e_{\sigma(n)})=(-1)^\sigmaf(e_{\sigma(1)}, e_{\sigma(2)},\ldots, e_{\sigma(n)})=(-1)^\sigma$$, the sign of the permutation $$\sigma\sigma$$.

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

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

All the terms with $$j_{\ell}=j_{\ell’}j_{\ell}=j_{\ell'}$$ for some $$\ell\neq \ell’\ell\neq \ell'$$ will vanish before the $$ff$$ term is zero, and the other terms can be written in terms of permutations. If $$j_{\ell}\neq j_{\ell’}j_{\ell}\neq j_{\ell'}$$ for any $$\ell\neq \ell’\ell\neq \ell'$$, then there is a unique permutation $$\sigma\sigma$$ with $$j_k=\sigma(k)j_k=\sigma(k)$$ for every $$kk$$. 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} \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 $$ff$$ is completely determined by $$f(e_1,\ldots, e_n)f(e_1,\ldots, e_n)$$, and we simply define $$\det\det$$ to be such a function with $$\det(e_1,\ldots, e_n)=1\det(e_1,\ldots, e_n)=1$$.