Below are two well-known statements regarding the determinant function:
- When $A$ is a square matrix, $\det(A)$ is the signed volume of the parallelepiped whose edges are columns of $A$.
- When $A$ is a square matrix, $\det(A) = \det(A^T)$.
It appears to me that 1 is how $\det$ got invented in the first place. The multilinearity and alternating property are natural consequences. Cramer's rule can also be easily understood geometrically. However, 1 does not immediately yield 2.
Of course the proof of 2 can be done algebraically, and it's not wrong to say that the algebra involved does come from 1. My question is, can 2 be viewed more geometrically? Can it feed us back a better geometric interpretation?
[How would one visualize the row vectors of $A$ from the column vectors? And how would one think, geometrically, about the equality of the signed volume spanned by two visually different parallelepipeds?]
Edit: If you are going to use multiplicativity of the determinant, please provide a geometric intuition to that too. Keep reading to see what I mean.
Here is one of my failed attempts. I introduce a different interpretation:
- $\det(A)$ is the ratio of the volume of an arbitrary parallelepiped changed by applying $A^T$ (left multiplication), viewed as a linear transformation.
The reason I say applying $A^T$ instead of applying $A$ is that I want the transpose to appear somewhere, and coincidentally the computation of $A^T u$ can be thought of geometrically in terms of columns of $A$: $A^T u$ is the vector whose components are dot products of columns of $A$ with $u$. (I'll take for granted the validity of the word arbitrary in 3 for the moment.) What is left to argue is the equality of 1 and 3, which I have not been able to do geometrically.
Is This a Duplicate Question? In a sense yes, and in a sense no. I did participate in that question a long time ago. There was no satisfactory answer. The accepted answer does not say what a "really geometric" interpretation of (*) should be. Yes you can always do certain row and column operations to put a matrix into a diagonal form without affecting the determinant (with an exception of a possible sign flip), and by induction, one can prove $\det(A) = \det(A^T)$. This, however, seems to me more algebraic than geometric even though each step can be thought of as geometric. I am aware that the property of an interpretation being "geometric" or "algebraic" is not a rigorous notion and is quite subjective. Some may say that this is as far as "geometry" can assist us. I just want to know if that is really the case. That is why I tag this as a soft question. I also secretly hope that maybe one such interpretation would serve as a simple explanation of Cauchy-Binet formula.