Determinant of matrix times elementary matrix

Question

I have been reading Linear Algebra Done Wrong and came across this Lemma:

Lemma 3.6. For a square matrix A and an elementary matrix E (of the same size)

     det(AE) = (det A)(det E)

Proof: The proof can be done just by direct checking: determinants of special matrices are easy to compute; right multiplication by an elementary matrix is a column operation, and effect of column operations on the determinant is well known.

This can look like a lucky coincidence, that the determinants of elementary matrices agree with the corresponding column operations, but it is not a coincidence at all.

Namely, for a column operation the corresponding elementary matrix can be obtained from the identity matrix I by this column operation. So, its determinant is 1 (determinant of I) times the effect of the column operation. And that is all! It may be hard to realize at first, but the above paragraph is a complete and rigorous proof of the lemma!

I am having a very hard time understanding this proof any help would be great, thanks!

Answer 1

OK, so this is a weird proof, but there's really only two parts of this proof that actually matter here. First:

[...] right multiplication by an elementary matrix is a column operation, and effect of column operations on the determinant is well known.

What this means is that when you multiply $A$ by $E$ on the right, you are doing a column operation on $A$, whether that be switching two columns, doing a column addition, or multiplying a column by a scalar. All of these operations will multiply the determinant by some known number, $k$ (i.e. switching two columns multiplies determinant by $k=-1$, column addition multiplies determinant by $k=1$, and multiplying a column by the scalar $c$ multiplies determinant by $k=c$). Thus, for any matrix $A$, the determinant of $AE$ will be $k$ times the original determinant of $A$. We can write this in an equation as follows:

$$\det AE=k(\det A)$$

Second:

[...] for a column operation the corresponding elementary matrix can be obtained from the identity matrix I by this column operation. So, its determinant is 1 (determinant of I) times the effect of the column operation.

Now, this is really confusing at first, but it can be understood in terms of our $\det AE=k(\det A)$ above. See, this equation works for any matrix $A$, which means we could also substitute the identity matrix $I$ for $A$ into this equation. Therefore, we get:

$$\det IE=k(\det I)\rightarrow \det E=k$$

(Note that this uses the fact that $IE=E$ and $\det I=1$.)

Therefore, we now know the value of $k$ is $\det E$. Thus, we can substitute that back into our original equation $\det AE=k(\det A)$ to get:

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

Since scalar multiplication is commutative, we can switch the right side around to get the final lemma:

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

Answer 2

The text defines the determinant in terms of column operation and there is an elementary matrix for each column operation. They spend a lot of time building the determinant based on column operations and the payoff is that this lemma is essentially true by construction.