# Determinant of matrix times elementary matrix

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!

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:

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

• Brilliant thanks very much! – jake walsh Jan 1 '19 at 2:41

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.