0

I am looking for some intuitive explanation based on the geometric transformation in 2D or 3D space.

In general, for two matrices $A$ and $B$, $AB \neq BA$, i.e., the matrix product isn't commutative. This is sort of intuitive because the end result of two successive transformations, in general, depends on the order.

When it comes to determinants, $det(AB) = det(A).det(B)$ Since a determinant is a scalar, $det(A).det(B)=det(B).det(A)=det(BA)=det(AB)$

As a matrix can be regarded as a transformation of space and its associated determinant gives the factor by which an area changes due to the transformation,

Is there an intuitive explanation for why should two successive transformations, which when put in reverse order do not necessarily give the same outcome, must end up with the same factor by which an area changes due to the successive transformations irrespective of the order of application?

P.S. Why I am asking for an intuitive answer is: successive transformations don't always give the same result when their order of application is changed. The fact that the factor by which an area changes due to the successive transformations irrespective of the orders of application, tells us that the different results of the successive transformations with different orders of application are mutually constrained. Can someone at least state that constraint in explicit mathematical equations?

ThePhysicist
  • 189
  • 9
  • Does this make sense to you if you restrict the transformations to uniform scaling? How about to scaling along one dimension only? What if you scale in one direction and then another—does the order matter? – amd Sep 11 '19 at 06:15
  • 2
    I'm not sure what can get more intuitive than the fact that one transformation multiples the area by $\det A$ and the other multiplies the area by $\det B$. –  Sep 11 '19 at 06:40
  • It may be easy to explain this with the "method of physics", which is intuition based on formulas that are observed in experiments. The key word is change of volume. Start with $e_1,e_2,\dots, e_n$ a / the base of $\Bbb R^n$. Or just only with a "box". Apply $A$ on the "box" and observe the multiplication factor for the (signed) volume. It is (the absolute value of) the determinant of $A$. Then apply $B$ on the "deformed box". It is a linear game, we multiply with $\det B$. This experiment is same with the one where we apply $BA$ directly. Note: Mathematically they build $\bigwedge^n A$ – dan_fulea Sep 11 '19 at 06:50
  • Why I am asking for an intuitive answer is: successive transformations don't always give the same result when their order of application is changed. The fact that the factor by which an area changes due to the successive transformations irrespective of the orders of application, tells us that the different results of the successive transformations with different orders of application are mutually constrained. Can someone at least state that constraint in explicit mathematical equations? – ThePhysicist Sep 12 '19 at 03:48
  • the second answer in this link is the proof i prefer for this result. i think it just requires some basic knowledge about multi linear forms to understand – C Squared Jun 11 '21 at 11:30

2 Answers2

1

The first transformation changes the area or volume by a factor $f$, and the second transformation changes the area or volume by a factor $g$. The combined effect changes the area/volume by a factor of $fg$. These factors are scalars, and scalar multiplication is commutative.

Brian Borchers
  • 10,563
1

@dan_fuela commented It may be easy to explain this with the "method of physics", which is intuition based on formulas that are observed in experiments. The key word is change of volume. Start with $e_1,e_2,…,e_n$ the basis of $\mathbb{R}^n$. Or just only with a "box". Apply $A$ on the "box" and observe the multiplication factor for the (signed) volume. It is (the absolute value of) the determinant of $A$. Then apply $B$ on the "deformed box".

Building on @dan's comment, it's because the multilinearity of the determinant means that it doesn't matter what "box" you look at, if you look at a box under a linear transformation, the ratio between the volumes of the box and it's image is fixed for all such boxes. This is, in my head, the best understanding of the determinant, but it can be understood by just looking at the standard basis box with the volume being the same by multilinearity. So, we start with the standard $e_1, e_2, ..., e_n$. Applying $A$ we get some new box $Ae_1, Ae_2, ..., Ae_n$, and applying $B$ to that some other box $BAe_1, BAe_2, ..., BAe_n$. So, the only thing that matters are the successive ratio of the volumes, and those don't depend on where the basis vectors are mapped to. So the ratios just multiply. Because it doesn't matter where the new box is, only it's volume matters. So $\det(AB) = \det A \det B$, and the same applies for the other order of transformation. So, the "boxes" themselves change depending on what the preimage of the boxes are, but because of linearity, the ratio between the volumes is fixed.

Cade Reinberger
  • 2,396