This is actually (hopefully) the first post in a series, and I’ll talk about this way of looking at the determinant in a subsequent post. It actually generalizes past vector spaces over R if you do it in the appropriate way.
The problem is that the equivalence of this to the usual determinant is not easy to prove unless you have some machinery at your disposal already. It’s “obvious” that volume should be invariant under skew translations, for example, but the easiest proof I know of this simply goes through the determinant of a skew translation matrix and shows it’s equal to 1.
If you have the time, try thinking of how you would prove that the characterization you give here is actually equivalent to the characterization of the determinant in the post—alternating and multilinear map that’s equal to 1 on the identity matrix. The “multilinear” part turns out to be rather tricky to establish properly.
Thinking about how to prove the multilinearity of the volume of a parallelepiped definition I like this sketched approach:
The two dimensional case is a “cute” problem involving rearranging triangles and ordinary areas (or you solve this case in any other way you want). The general case then follows from linearity of integrals (you get the higher dimensional cases by integrating the two dimensional case appropriately).
So, this is not exactly a rigorous proof, but off the top of my head, I would justify/remember the properties like: identity has determinant 1 because it doesn’t change the size of any volumes, determinant is alternating because swapping two axes of a parallelpiped is like reflecting those axes in a mirror, changing the orientation. Multilinearity is equivalent to showing that the volume of a parallelpiped A=a1∧a2∧(...)∧an is linear in all the ais. But this follows since the volume of A is equal to the volume of a2∧(...)∧an multiplied by the component of a1 projected onto the axis orthogonal to a2,...,an, which is clearly linear in a1. This last fact is a bit weird, to justify intuitively I imagine having a straight tower of blocks which you then ‘slant’ by pulling the top in a given direction without changing the volume, this corresponding to the components of a1 not orthogonal to a2,...,an.
This is actually (hopefully) the first post in a series, and I’ll talk about this way of looking at the determinant in a subsequent post. It actually generalizes past vector spaces over R if you do it in the appropriate way.
The problem is that the equivalence of this to the usual determinant is not easy to prove unless you have some machinery at your disposal already. It’s “obvious” that volume should be invariant under skew translations, for example, but the easiest proof I know of this simply goes through the determinant of a skew translation matrix and shows it’s equal to 1.
If you have the time, try thinking of how you would prove that the characterization you give here is actually equivalent to the characterization of the determinant in the post—alternating and multilinear map that’s equal to 1 on the identity matrix. The “multilinear” part turns out to be rather tricky to establish properly.
Thinking about how to prove the multilinearity of the volume of a parallelepiped definition I like this sketched approach:
The two dimensional case is a “cute” problem involving rearranging triangles and ordinary areas (or you solve this case in any other way you want). The general case then follows from linearity of integrals (you get the higher dimensional cases by integrating the two dimensional case appropriately).
So, this is not exactly a rigorous proof, but off the top of my head, I would justify/remember the properties like: identity has determinant 1 because it doesn’t change the size of any volumes, determinant is alternating because swapping two axes of a parallelpiped is like reflecting those axes in a mirror, changing the orientation. Multilinearity is equivalent to showing that the volume of a parallelpiped A=a1∧a2∧(...)∧an is linear in all the ais. But this follows since the volume of A is equal to the volume of a2∧(...)∧an multiplied by the component of a1 projected onto the axis orthogonal to a2,...,an, which is clearly linear in a1. This last fact is a bit weird, to justify intuitively I imagine having a straight tower of blocks which you then ‘slant’ by pulling the top in a given direction without changing the volume, this corresponding to the components of a1 not orthogonal to a2,...,an.
Yeah, that’s what I mean by saying it’s “obvious”. Similar to the change of variables theorem in that way.