# Prove that  det(bb(AB)) = det(bb(A)) det(bb(B)) ?

Sep 14, 2017

See explanation...

#### Explanation:

The $n$ dimensional measure polytope is the $n$ dimensional analogue of the cube. So the unit measure polytope is a line segment, square, cube, hypercube, etc with sides of length $1$.

Suppose $M$ is any $n \times n$ matrix.

Multiplication by $M$ is a linear operator from $n$ dimensional space to itself.

The $n$ dimensional measure (length, area, volume, hypervolume, etc.) of the image of the unit measure polytope (unit line segment, unit square, unit cube, unit hypercube, etc.) under this linear transform is the determinant of $M$.

For example, in $2$ dimensional real space, a real $2 \times 2$ matrix $M$ with non-zero determinant $d$ maps the unit square to a parallelogram of area $d$.

Since matrix multiplication is linear, applying $M$ to any $n$ dimensional object of measure $x$ will result in an object of measure $x \det \left(M\right)$

So if we have two matrices $A$, $B$ and apply one then the other to the unit measure polytope, the measure of the resulting polytope will be $\det \left(A B\right) = \det \left(A\right) \cdot \det \left(B\right)$

Sep 14, 2017

We seek to prove that:

$\det \left(\boldsymbol{A B}\right) = \det \left(\boldsymbol{A}\right) \det \left(\boldsymbol{B}\right)$

Consider, as a specific case, a general $2 \times 2$ system

$\boldsymbol{A} = \left(\begin{matrix}{a}_{11} & {a}_{12} \\ {a}_{21} & {a}_{22}\end{matrix}\right)$ and $\boldsymbol{B} = \left(\begin{matrix}{b}_{11} & {b}_{12} \\ {b}_{21} & {b}_{22}\end{matrix}\right)$

Evaluation of the LHS:

$\det \left(\boldsymbol{A}\right) = {a}_{11} {a}_{22} - {a}_{21} {a}_{12}$
$\det \left(\boldsymbol{B}\right) = {b}_{11} {b}_{22} - {b}_{21} {b}_{12}$

And the product is:

$\det \left(\boldsymbol{A}\right) \det \left(\boldsymbol{B}\right) = \left({a}_{11} {a}_{22} - {a}_{21} {a}_{12}\right) \left({b}_{11} {b}_{22} - {b}_{21} {b}_{12}\right)$

$\text{ } = {a}_{11} {a}_{22} {b}_{11} {b}_{22} - {a}_{11} {a}_{22} {b}_{21} {b}_{12} -$
$\text{ } {a}_{21} {a}_{12} {b}_{11} {b}_{22} + {a}_{21} {a}_{12} {b}_{21} {b}_{12}$

And if we look at the matrix product, we have:

$\boldsymbol{A B} = \left(\begin{matrix}{a}_{11} {b}_{11} + {a}_{12} {b}_{21} & {a}_{11} {b}_{12} + {a}_{12} {b}_{22} \\ {a}_{21} {b}_{11} + {a}_{22} {b}_{21} & {a}_{21} {b}_{12} + {a}_{22} {b}_{22}\end{matrix}\right)$

$\det \left(\boldsymbol{A B}\right) = \left({a}_{11} {b}_{11} + {a}_{12} {b}_{21}\right) \left({a}_{21} {b}_{12} + {a}_{22} {b}_{22}\right) -$
$\text{ } \left({a}_{11} {b}_{12} + {a}_{12} {b}_{22}\right) \left({a}_{21} {b}_{11} + {a}_{22} {b}_{21}\right)$

$\text{ } = {a}_{11} {b}_{11} {a}_{21} {b}_{12} + {a}_{12} {b}_{21} {a}_{21} {b}_{12} +$
$\text{ } {a}_{11} {b}_{11} {a}_{22} {b}_{22} + {a}_{12} {b}_{21} {a}_{22} {b}_{22} -$
$\text{ } {a}_{11} {b}_{12} {a}_{21} {b}_{11} - {a}_{12} {b}_{22} {a}_{21} {b}_{11} -$
$\text{ } {a}_{11} {b}_{12} {a}_{22} {b}_{21} - {a}_{12} {b}_{22} {a}_{22} {b}_{21}$

$\text{ } = {a}_{11} {b}_{11} {a}_{22} {b}_{22} + {a}_{12} {b}_{21} {a}_{21} {b}_{12} -$
$\text{ } {a}_{11} {b}_{12} {a}_{22} {b}_{21} - {a}_{12} {b}_{22} {a}_{21} {b}_{11}$

$\text{ } \det \left(\boldsymbol{A}\right) \det \left(\boldsymbol{B}\right)$

In the general $n \times n$ case the same technique can be used but it is easier to use sums of columns and row vectors along with their inner products.