0
$\begingroup$

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!

$\endgroup$
4
$\begingroup$

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:

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

$\endgroup$
  • $\begingroup$ Brilliant thanks very much! $\endgroup$ – jake walsh Jan 1 at 2:41
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.