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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!

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Is this the sort of proof your book is giving? proofwiki.org/wiki/… –  AppliedSide Feb 18 '11 at 4:01
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If $\det(A)=0$, then you have $\det(A)\det(B) = 0 = \det(AB)$, so there is nothing to do. So all you need to do is consider the case where $A$ is invertible, i.e., where $\det(A)\neq 0$.

To show the result holds if $B$ is not invertible (not necessary, but you ask about it) notice that if $B$ is not invertible then neither is $AB$ (since there exists a nontrivial solution to $B\mathbf{x}=\mathbf{0}$, and this nontrivial solution is also a nontrivial solution to $AB\mathbf{x}=\mathbf{0}$, so $AB$ is not invertible either). So $\det(AB)=0$, and equality holds.

The key to the argument the book uses is that if $A$ is invertible, then it can be written as a product of elementary matrices: $A = E_1\cdots E_t$. These elementary matrices correspond to the elementary row operations performed to get rrefA (in the order of the indices). But when you left-multiply any matrix by an elementary matrix, then you perform the corresponding elementary row operation on the matrix.

Now, what happens if $A$ is just an elementary matrix, and you look at $\det(AB)$? If $A$ is an elementary matrix corresponding to a swap of rows, then $AB$ is the result of swapping rows of $B$, so $\det(AB) = -\det(B)$. But if $A$ is an elementary matrix corresponding to a swap of rows, then $\det(A) = -1$. So in this case, $\det(AB) = -\det(B) = \det(A)\det(B)$.

If $A$ corresponds to adding a multiple of one row to another row, then $\det(A)=1$ and $\det(AB) = \det(B) = \det(A)\det(B)$.

Finally, if $A$ corresponds to multiplying a row by $k$, then $AB$ is the result of multiplying a row of $B$ by $k$, so by the properties of the determinant you have that $\det(AB) = k\det(B)$; but again, $\det(A) = k$, so $\det(AB)=k\det(B)$ = \det(A)\det(B)$.

So the result works if $A$ is an elementary matrix.

Now, how do the $E_i$ correspond to the operations you do when you compute the rref of $A$? They are the opposite of what you do to $A$ to get the rref: because if $F_1,\ldots,F_t$ are what you do to $A$ to get the rref of $A$, then $F_t\cdots F_1 A = I$, so $A = (F_t\cdots F_1)^{-1} = F_1^{-1}\cdots F_t^{-1}$.

The inverse of an elementary matrix corresponding to a swap of rows is an elementary matrix corresponding to a swap of rows; the inverse of an elementary matrix corresponding to adding a multiple of one row to another is an elementary matrix corresponding to adding a multiple of one row to another; and the inverse of an elementary matrix that corresponds to dividing a row by $k$ is the elementary matrix that corresponds to multiplying a row by $k$.

So, to see how $E_1$, $E_2,\ldots,E_t$ affect $\det(B)$ when you compute $\det(E_1\cdots E_tB)$, you only need to keep track of (i) the number of times you swapped rows; and (ii) the constants you divided the rows of $A$ by to get the rref of $A$. Each row swap multiplies the determinant by $-1$, and dividing a row of $A$ by $k$ multiplies the determinant by $k$. So, if $s$ is the number of times you swapped rows, and $k_1,\ldots,k_r$ are the constants by which you divided rows of $A$, then $$\det(AB) = \det(E_1\cdots E_t B) = (-1)^sk_1\cdots k_r\det(B),$$ which is the step you were having trouble with.

I hope that helps.

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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.)

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