Proof $\det(AB)=\det(A)\det(B)$

Question

I have read the following proof , in here:

Why can we go from the first line to the second one?

why $Det(E^k)\cdot Det(E^m)=Det(E^k\cdot E^m)$?

is it because $\det(E^k)\in \mathbb{F}$ for all $k$?

Have you read the three lines above, starting by "But for any matrix $\mathbf{D}$"? — Clement C., Aug 20 '15 at 15:34
The idea of the proof is to show that elementary matrices satisfy the multiplicative property for each case of elementary matrices, and then to use that fact to prove it generally. — Tae Hyung Kim, Aug 20 '15 at 15:35
You can go from the first line to the second because they've already established that the multiplicative property holds for elementary matrices. — Jahan Claes, Aug 20 '15 at 15:38
@gbox what do you mean by $\det(E^k) = \Bbb F$? Do you actually mean $\det(E^k) \in \Bbb F$? — Ben Grossmann, Aug 20 '15 at 15:39
@thkim1011 why does elementary matrices satisfy the multiplicative property? — gbox, Aug 20 '15 at 15:42
@gbox Have you read this article? It links to the three proofs for how elementary matrices change the determinant of a matrix after multiplication. — eigenchris, Aug 20 '15 at 15:44
@eigenchris yes, the are changing the determinate by scalar, that is why it is true? — gbox, Aug 20 '15 at 15:46
@gbox The three proofs are linked under the "proof" section. — eigenchris, Aug 20 '15 at 15:48
@gbox I don't know what else to tell you... the proof for the question you are asking is linked right in the article I gave you... — eigenchris, Aug 20 '15 at 15:53
@eigenchris my question is why $Det(E^k)\cdot Det(E^m)=Det(E^k\cdot E^m)$ — gbox, Aug 20 '15 at 16:15
@gbox Have you read these three articles? What is it about them you don't understand? — eigenchris, Aug 20 '15 at 16:27
@eigenchris yes, maybe it is trivial from those three articles, but I am trying to understand why $Det(E^k)\cdot Det(E^m)=Det(E^k\cdot E^m)$ — gbox, Aug 20 '15 at 16:31

eigenchris · Accepted Answer · 2015-08-24T14:40:11.883

The property $\det(E^iE^j) = \det(E^i)\det(E^j)$ doesn't just hold for two elementary matrices $E^i,E^j$, it holds for any elementary matrix $E$ and any matrix $A$:

$$\det(EA) = \det(E)\det(A)$$

Proof:

There are three types of elementary matrices:

row-scaling
row-swapping
row-adding

If I have some matrix $A$ with determinant $\det(A)$, and I multiply $A$ by some elementary matrix $E$ to get $EA$, then...

if $E$ is a row-scaling matrix which scales a row of $A$ by a constant $c$, we can easily see that $\det(E) =c$ since $E$ is the identity matrix, but with one diagonal element changed to $c$. We can also show that $\det(EA) = c\det(A) = \det(E)\det(A)$ [proof]
if $E$ is a row-swapping matrix, then $det(E) = -1$ since it is simply the identity matrix with two rows swapped. We can show that $\det(EA) = -\det(A) = \det(E)\det(A)$ [proof]
if $E$ is a row-adding matrix which adds two rows of $A$ together, with one of the rows scaled by a constant $k$, then $\det(E) = 1$ since it is simply the identity matrix with one non-zero off-diagonal element $k$. We can also show that $\det(EA) = \det(A) = \det(E)\det(A)$ [proof]

score 1 · Answer 2 · answered Aug 20 '15 at 19:08

1

The determinant is a multiplicative function on $M_n(R)$, where $R$ is the base (commutative) ring: $$\det(AB)=\det A\cdot\det B$$ for any $n\times n$ matrices with coefficients in $R$, and not only when one is an elementary matrix.

answered Aug 20 '15 at 19:08

Bernard

175,478

Thanks, but I have yet learned about rings – gbox Aug 20 '15 at 19:09
2

Replace ring by field, or even $\mathbf R$ or $\mathbf C$: I wanted to point this is a very general property. – Bernard Aug 20 '15 at 19:13

Proof $\det(AB)=\det(A)\det(B)$

2 Answers2

Linked