Square matrices as a product of elementary matrices,

Question

I am trying to prove det(A) = det($A^T$), starting with the idea that every square matrix is the product of elementary matrices.

Is this true, even for the non-invertible square matrices?

So, I'd like to start with the identity matrix and multiply it with elementary matrices until I get my desired matrix. Then use the fact that $(AB)^T$ = $B^T$$A^T$.

(The proof using the definition of determinant with the signum function and permutation stuff is a little difficult and not instructive at all, in my opinion.)

Thanks,

Answer 1

The elementary matrices are invertible, so any product of them is also invertible. However, invertible matrices are dense in all matrices, and determinant and transpose are continuous, so if you can prove that $\det(A) = \det(A^T)$ for invertible matrices, it follows that this is true for all matrices.

Answer 2

@Robert Israel is right that the usual elementary matrices are all invertible, and hence that products of them all have nonzero determinant.

But if you augment the set of elementary matrices with ones that allow you to multiply a row (or column) by zero, then you can create every matrix from (enlarged) elementary operations on the identity; if you show this, then you've got a proof.

Showing that every mat arises from a product of ordinary elementary matrices and these new ones that allow multiplication by zero in some row or column...is not so bad. If you do ordinary Gaussian elimination, each step corresponds to an (ordinary) elementary matrix multiply. If you get to the point where you've got a missing pivot, as in

1 * * * * 
0 0 * * *
0 0 * * *
0 0 * * *
0 0 * * *

you should put the next pivot on the diagonal, as in

1 * * * * 
0 0 * * *
0 0 1 * *
0 0 * * *
0 0 * * *

and then use it to clear out entries above and below, to get

1 * 0 * * 
0 0 0 * *
0 0 1 * *
0 0 0 * *
0 0 0 * *

You'll then end up with some zero-rows, which can be created from the identity by mutiplication by the "new" elementary ops.