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

Question

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

2 Answers

Impact of this question

2589 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-09-14T06:28:46

See explanation...

Explanation:

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

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

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

The $n$ $n$ 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 $M$ $M$ .

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

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

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

Answer 2 · 2017-09-14T16:04:28

We seek to prove that:

$det (A B) = det (A) det (B)$ $det(bb(AB)) = det(bb(A)) det(bb(B))$

Consider, as a specific case, a general $2 \times 2$ $2xx2$ system

$bb(A) = ( (a_11, a_12), (a_21, a_22) )$ and $bb(B) = ( (b_11, b_12), (b_21, b_22) )$

Evaluation of the LHS:

$det( bb(A) ) = a_11 a_22 - a_21 a_12$
$det( bb(B) ) = b_11 b_22 - b_21 b_12$

And the product is:

$det( bb(A) )det( bb(B) ) = (a_11 a_22 - a_21 a_12)(b_11 b_22 - b_21 b_12)$

$" " = a_11 a_22b_11 b_22 - a_11 a_22b_21 b_12 -$
$" " a_21 a_12b_11 b_22 + a_21 a_12b_21 b_12$

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

$bb(AB) = ( (a_11b_11+a_12b_21,a_11b_12+a_12b_22), (a_21b_11+a_22b_21, a_21b_12+a_22b_22) )$

Leading to:

$det(bb(AB)) = (a_11b_11+a_12b_21)(a_21b_12+a_22b_22) -$
$" " ( a_11b_12+a_12b_22) (a_21b_11+a_22b_21)$

$" " = a_11b_11a_21b_12 + a_12b_21a_21b_12 +$
$" " a_11b_11a_22b_22 + a_12b_21a_22b_22 -$
$" " a_11b_12a_21b_11 - a_12b_22a_21b_11 -$
$" " a_11b_12a_22b_21 - a_12b_22a_22b_21$

$" " = a_11b_11a_22b_22 +a_12b_21a_21b_12-$
$" " a_11b_12a_22b_21 -a_12b_22a_21b_11$

$" " det( bb(A) )det( bb(B) )$

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

Science

Math

Humanities

... and beyond

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

2 Answers

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

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

2 Answers

Explanation:

Related questions

Impact of this question

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