Vtyjkyuk

I see this in my textbook:

I don't really understand the proof or why elementary matrices are involved. What is going on here?

First, let me give you an slight outline of the approach of your textbook: You first establish the result only for elementary matrices, i.e. that $|EB|=|E|cdot |B|$. You then generalizes this for arbitrary finite products of elementary matrices, i.e. $|E_1cdots E_nB|=|E_1|cdots|E_n||B|$. Why this fondling with elementary matrices? The results above are easy to establish and regular matrices have a representation as a product of elementary matrices. From there on, you can generalize the result.

I want to give you a slightly different approach for a proof: I'm not sure how you have defined the determinant in your textbook, so let me give a convenient definition for my layout. Let $mathbbF$ be a field.

The determinant function of dimension $n$ over $mathbbF$, $det:(mathbbF^n)^ntomathbbF$, is the unique such function satisfying

1. multi-linearity, i.e. $det$ is linear in each argument.


2. anti-symmetry, i.e. $a_i=a_jRightarrow det(a_1,dots,a_n)=0$


3. normalization, i.e. $det(e_1,dots,e_n)=1$ for the unit vectors $e_1,dots,e_n$ of $mathbbF^n$

That this function is unique actually has to be proven, but I skip this. The key thing is to define the determinant directly in the columns of a matrix instead on the whole matrix object. Thus, for a matrix $AinmathbbF^(n,n)$, the expression $det(A)$ per se has no meaning, but we define $det(A):=det(a_1,dots,a_n)$ for $A=(a_1,dots,a_n)$ being the corresponding column vector representation, as a an abuse of notation.

Let $A,BinmathbbF^(n,n)$ and let $C=AB$ for $A=(a_1,dots,a_n), B=(b_1,dots,b_n)$ and $C=(c_1,dots,c_n)$ as their column vector representation.

By the rules of matrix multiplication, we have that $c_j=Ab_j$:

$$c_j=Ce_j=sum_i=1^nc_ije_i=ABe_j=Ab_j=A(sum_i=1^nb_ije_i)=sum_i=1^nb_ijAe_i=sum_i=1^nb_ija_i$$

Thus, you we obtain:

$$det(C)=det(c_1,dots,c_n)=det(Ab_1,dots,Ab_n)=det(sum_k=1^nb_k1a_k,dots,sum_k=1^nb_kna_k)=sum_k_1=1^nsum_k_2=1^ndotssum_k_n=1^ndet(b_k_11a_k_1,dots,b_k_nna_k_n)$$

Now, in this nested sum, all these states of the summation indices $k_1,dots,k_n$ vanish for which the map $imapsto k_i$ is not a permutation(bijection) on $1,dots,n$ as in these cases anti-symmetry turns the determinant values to $0$. Thus, we can see every remaining such map as a permutation $sigma$ in the set of all permutation $S_n$. The nested sums vanish and are replaced by one large sum over all permutation, where we additionally pull out(by multi-linearity) the scalar values of $B$. Thus, we have

$$det(C)=sum_sigmain S_nb_sigma(1)1cdots b_sigma(n)ndet(a_sigma(1),dots,a_sigma(n))=sum_sigmain S_nmathrmsign(sigma)b_sigma(1)1cdots b_sigma(n)ndet(a_1,dots,a_n)=det(a_1,dots,a_n)cdotsum_sigmain S_nmathrmsign(sigma)b_sigma(1)1cdots b_sigma(n)n=det(A)cdotdet(B)$$

In the last steps, I've used two facts about determinants and permutations. First, I've used the following lemma:

Lemma: For any anty-symmetric function $f:(mathbbF^n)^ntomathbbF$, $sigmain S_n$, any $a_1,dots,a_ninmathbbF^n$: $$f(a_sigma(1),dots,a_sigma(n))=mathrmsign(sigma)f(a_1,dots,a_n)$$

where $mathrmsign(sigma)$ is a scalar value, the sign of the permutation $sigma$. Secondly, I've used that

$$sum_sigmain S_nmathrmsign(sigma)b_sigma(1)1cdots b_sigma(n)n=det(B)$$

This is known as the Leibniz-formula for determinants.

EDIT: Since this is a long post and an extensive topic I've tried to cover with the least text possible, any continuing questions or just pointing out errors are/is well-appreciated.