Talk:Determinant of Matrix Product

I do not think that the elementary operations on rows converting a square matrix to a triangular one can introduce a minus sign before the determinant. In other words, multiplying one row by a constant and subtracting it from another row does not change the value of the determinant, and this is how the square matrix is converted to a triangular matrix. The second stated proof is therefore robustly correct. The proof of this concept may be found in elementary math analysis texts.

You are correct. The method used in Square Matrix Row Equivalent to Triangular Matrix can be adapted to avoid the type 3 elementary operations. I will hopefully get to this later today. --Lord_Farin 11:51, 3 November 2011 (CDT)

Thx guys. I will amend the proof as appropriate. --prime mover 15:17, 3 November 2011 (CDT)

The proof of mentioned statement has been adapted. --Lord_Farin 17:33, 3 November 2011 (CDT)

missing step

From the proof 1, it is clear than:

$\det \left({\mathbf T_A \mathbf T_B}\right) = \det \left({\mathbf A}\right) \det \left({\mathbf B}\right)$

But I think we still have to proof that:

$\det \left({\mathbf T_A \mathbf T_B}\right) = \det \left({\mathbf A \mathbf B}\right)$

Could anybody clarify? Abaca

We have $\det \left({\mathbf T_A}\right) = \det \left({\mathbf A}\right)$ and $\det \left({\mathbf T_B}\right) = \det \left({\mathbf B}\right)$ from a bit further up.

Then we have $\det \left({\mathbf T_A \mathbf T_B}\right) = \det \left({\mathbf T_A}\right)\det \left({\mathbf T_B}\right)$.

Unless I'm also missing something obvious. BTW, please sign your talk page posts by pressing the signature icon (it's the one with scribble in it) from the row of icons above the edit pane. Thx. --prime mover 19:08, 14 March 2012 (EDT)

The point is that it is not immediately clear from the current phrasing that $\mathbf T_A \mathbf T_B$ is row equivalent to $\mathbf{AB}$. Probably that's immediate from the definition, but it deserves a link at the very least. --Lord_Farin 05:59, 15 March 2012 (EDT)

Exactly, that is my doubt. On the other hand, is the triangular matrix provided by Square Matrix Row Equivalent to Triangular Matrix unique? If it is, we may prove that ${\mathbf T_A \mathbf T_B}$ is the triangular matrix equivalent to $ {\mathbf A \mathbf B}$. Thanks (now with the signature) --Abaca 16:28, 15 March 2012 (EDT)

Yes, you're right, I understand now. That does indeed seem to be a missing step. Anyone able to plug it? --prime mover 16:37, 15 March 2012 (EDT)