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.
Best Answer
Write $A$ as a product of the form $E_1E_2\cdots E_kDF_1F_2\cdots F_\ell$, where each $E_i$ (resp. $F_j$) is an elementary matrix corresponding to row (resp. column) switching or row (resp. column) addition, and $D$ is a diagonal matrix. You are done if you can explain why
But these are essentially what motivate the definition of determinant: we want to define the determinant as the volume of the parallelepiped generated by the columns of a matrix. Shears do not change the volume (thus the determinant of an elementary matrix for row/column addition is $1$) and scaling in one direction changes the volume in the same proportion (thus $\det D$ is the product of $D$'s diagonal entries). Since we want to define the volume as zero in the degenerate case, it follows that together with the previous two requirements, the volume must be signed and by flipping two columns, the sign flips too (thus the determinant of an elementary matrix for row/column switching is $-1$).