TITLE: Two questions with respect to the determinants
QUESTION [5 upvotes]: I have got a proof of $det(AB)$=$(detA)(detB)$ in my book.
It goes as follows (for invertible A): we know that rref[A|AB]=[$I_{n}$|B]
We also know that det(A)=$(-1)^{s}k_{1}k_{2}...k_{r}$ where s is the number of row swaps needed to get to the rrefA, and $k_{i}$ are coefficients by which we divide rows of A to get to rrefA. Hence, the book concludes, det(AB)=$(-1)^{s}k_{1}k_{2}...k_{r}$(detB)=(detA)(detB), but I don't see how we go from det(AB) to $(-1)^{s}k_{1}k_{2}...k_{r}$(detB). Could you please explain the logic behind this step? (I see how (detA)(detB)=$(-1)^{s}k_{1}k_{2}...k_{r}$(detB), obviously).
The book then also uses the fact that if A is not invertible, neither will be AB (because image of AB is contained in A), so (detA)(detB)=0(detB)=0=det(AB). My question is, how would we prove that this equation holds if B is non-invertible, and hence detB=0? I could think of saying that since B is not invertible then it can't be represented as a product of elementary matrices, while A can, so AB can't be represented as such either, but that sounds hand-wavy to me.
Thanks a lot!

REPLY [2 votes]: First, in case anyone else, like me, doesn't know the abbreviation "rref": it means "reduced row echelon form".
The structure of the proof in the book isn't entirely clear from your exposition. It is clear that the first part assumes that $A$ is invertible. (You didn't mention that assumption.) It's not quite clear to me whether the first part also assumes that $B$ is invertible. (If it does not, your question in the second part wouldn't affect the validity of the proof.)
I) To the first part of the question ("Could you please explain the logic behind this step?"):
a) If we assume that $B$ is invertible, we can proceed as follows: Since $\mathrm{rref}[A|AB]=[I_n|B]$, we know that applying row operations that transform $A$ into reduced row echelon form to $AB$ yields $B$. But that means that whatever row operations we perform to get $B$ into reduced row echelon form, we can perform the same operations on $AB$ after first performing $s$ row swaps and dividing rows of $AB$ by the $k_i$. Thus, applying $\det(A)=(-1)^{s}k_{1}k_{2}...k_{r}$ to $AB$, we see that this product for $AB$ must be the corresponding product for $A$ multiplied by the corresponding product for $B$.
b) If we don't assume that $B$ is invertible in the first part, I think we need a bit more than what you've written. For instance, it would be enough to know how row operations affect the determinant, namely that a row swap multiplies the determinant by $-1$, dividing a row by $k_i$ divides the determinant by $k_i$, and adding a multiple of a row to another row leaves the determinant unchanged. Applying this to $\mathrm{rref}[A|AB]=[I_n|B]$ yields the desired equation, since $AB$ is transformed into  $B$ by $s$ row swaps and divisions by the $k_i$.
II) To the second part of the question ("how would we prove that this equation holds if B is non-invertible"):
As mentioned, this is only required if we assume in the first part that $B$ is invertible.
Your "hand-waving" proof seems fine to me -- more explicitly: We've dealt with the case where $A$ is not invertible. So assume that $A$ is invertible. Then if $AB$ were invertible, that would allow us to express both $A$ and $AB$ as a product of elementary matrices, which in turn would allow us to represent $B$ as such a product, which can't be if $B$ is not invertible.
Other ways of proving the same thing would be: If $A$ is invertible (assumed as above), we can write $B=(A^{-1}A)B=A^{-1}(AB)$. So if $AB$ were invertible, then $(AB)^{-1}A$ would be an inverse for $B$, which can't be if $B$ is not invertible. Or, in the same vein as what the book does in the case where $A$ isn't invertible: If $B$ isn't invertible, its kernel is not empty, but the kernel of $AB$ contains the kernel of $B$, and hence the kernel of $AB$ is not empty, and thus $AB$ isn't invertible. (Note that here we don't have to assume that $A$ is invertible.)