User:Anghel/Sandbox

Proof
This proof assumes that $\mathbf A$ and $\mathbf B$ are $n \times n$-matrices over a commutative ring with unity $\left({R, +, \circ}\right)$.

Let $\mathbf C = \left[{c}\right]_n = \mathbf A \mathbf B$.

From Square Matrix Row Equivalent to Triangular Matrix, it follows that $\mathbf A$ can be converted into a upper triangular matrix $\mathbf A'$ by a finite sequence of elementary row operations $\hat o_1, \ldots, \hat o_{m'}$.

Let $\mathbf C'$ denote the matrix that results from using $\hat o_1, \ldots, \hat o_{m'}$ on $\mathbf C$.

From Elementary Row Operations Commutes with Matrix Multiplication, it follows that $\mathbf C' = \mathbf A' \mathbf B$.

Effect of Sequence of Elementary Row Operations on Determinant shows that there exists $c_1 \in R$ such that:


 * $c_1 \det \left({\mathbf A'}\right) = \det \left({\mathbf A}\right)$
 * $c_1 \det \left({\mathbf C'}\right) = \det \left({\mathbf C}\right)$

Let $\mathbf B^t$ be the transpose of $B$.

From Transpose of Matrix Product, it follows that $\left({\mathbf C'}\right)^t = \left({\mathbf A' \mathbf B}\right)^t = \mathbf B^t \left({\mathbf A'}\right)^t$

From Square Matrix Row Equivalent to Triangular Matrix, it follows that $\mathbf B^t$ can be converted into a lower triangular matrix $\left({\mathbf B^t}\right)'$ by a finite sequence of elementary row operations $\hat p_1, \ldots, \hat p_{m''}$.

Let $\mathbf C$ denote the matrix that results from using $\hat p_1, \ldots, \hat p_{m}$ on $\left({\mathbf C'}\right)^t$.

From Elementary Row Operations Commutes with Matrix Multiplication, it follows that $\mathbf C'' = \left({\mathbf B^t}\right)' \left({\mathbf A'}\right)^t$.

Effect of Sequence of Elementary Row Operations on Determinant shows that there exists $c_2 \in R$ such that:


 * $c_2 \det \left({\left({\mathbf B^t}\right)'}\right) = \det \left({\mathbf B^t}\right)$
 * $c_2 \det \left({\mathbf C''}\right) = \det \left({ \left({\mathbf C'}\right)^t }\right)$

From Transpose of Upper Triangular Matrix is Lower Triangular, it follows that $\left({\mathbf A'}\right)^t$ is a lower triangular matrix.

Then Product of Triangular Matrices shows that $\left({\mathbf B^t}\right)' \left({\mathbf A'}\right)^t$ is a lower triangular matrix whose diagonal elements are the products of the diagonal elements of $\left({\mathbf B^t}\right)'$ and $\left({\mathbf A'}\right)^t$.

From Determinant of Triangular Matrix, we have that $\det \left({\left({\mathbf A'}\right)^t }\right)$, $\det \left({\left({\mathbf B^t}\right)' }\right)$, and $\det \left({\left({\mathbf B^t}\right)' \left({\mathbf A'}\right)^t }\right)$ are equal to the product of their diagonal elements.

Combinining these results shows that


 * $\det \left({ \left({\mathbf B^t}\right)' \left({\mathbf A'}\right)^t }\right) = \det \left({\left({\mathbf B^t}\right)' }\right) \det \left({\left({\mathbf A'}\right)^t }\right)$

Then: