6 ms·

Determinants have a very basic intuition behind them: it's the stretch factor of the n-volume of a linearly transformed unit n-cube (area of a linearly transfor

Determinants have a very basic intuition behind them: it's the stretch factor of the n-volume of a linearly transformed unit n-cube (area of a linearly transformed unit square in 2D, volume of a linearly transformed unit cube in 3D, etc.) Why would one want to banish them from linear algebra?

whatshisface 9y ago
The author alludes to that by pointing out that they're needed for jacobians, which are essentially also stretch factors. However, a nice geometric interpretation does not nice math make. If you remember their definition, the one with the sub-products and the alternating +/-es, and then imagine trying to prove that it has that geometric interpretation - you've arrived at a huge pain for undergrad math students that are just being introduced to matrices.
- jessriedel 9y ago
  > then imagine trying to prove that it has that geometric interpretation Exactly. We leaned heavily on determinants in my freshman linear algebra class but I went at least an additional year before I even heard the interpretation, much less could prove it from the standard terrible definition.
- enriquto 9y ago
  for me the definition is precisely the volume
- vanderZwan 9y ago
  > However, a nice geometric interpretation does not nice math make. I keep seeing proponents of the Clifford Algebra as a much more elegant type of maths that supersedes linear algebra and also makes it much more intuitive, but I haven't really found a clear source on it yet.
zodiac 9y ago
Yes they are intuitive but it is pretty hard to define them rigorously in a way that captures that intuition. Also, every way I've seen to define them requires choosing a basis (or even an orthonormal basis), which isn't that nice. If you read the pdf you'd notice that Axler's proofs completely avoids relying on the fact that every finite-dimensional vector space has a basis, or choosing one.
- Sniffnoy 9y ago
  Defining the determinant definitely does not require choosing a basis (though I don't know how you'd define it without at least knowing that a basis exists, for reasons you can see below). Let's say you've got a linear transformation T:V->V, and suppose V is n-dimensional. Consider exterior powers of V, Λ^k(V); for each one we naturally get a linear transformation Λ^k(T):Λ^k(V)->Λ^k(V). In particular take k=n, so we get a linear transformation Λ^n(T):Λ^n(V)->Λ^n(V). But Λ^n(V) is one-dimensional, so Λ^n(T) must be multiplication by some constant factor. That factor is the determinant. That's basically the most "natural" definition of the determinant (notice how multiplicativity immediately falls out of it, assuming of course you already know that Λ^k is a functor). You need the idea of a basis in order to (a) make sense of the statement "V is n-dimensional" and (b) prove that Λ^n(V) is 1-dimensional, but that's it, and you certainly never need to choose one in order to define the determinant. Probably not a great definition for beginning students of linear algebra, but I do have to correct this idea that defining the determinant requires choosing a basis...
  - contravariant 9y ago
    I think you get a basis of an n-dimensional vector space by the definition of dimension (if e.g. the dimension is the maximum size of a set of linearly independent vectors). It's a bit trickier when the dimension is infinite, but again most definitions of dimension require there to be a basis of a particular size, the difficult part is proving that this makes sense (i.e. the dimension is unique and defined).
    - zodiac 9y ago

whatshisface 9y ago

The author alludes to that by pointing out that they're needed for jacobians, which are essentially also stretch factors. However, a nice geometric interpretation does not nice math make. If you remember their definition, the one with the sub-products and the alternating +/-es, and then imagine trying to prove that it has that geometric interpretation - you've arrived at a huge pain for undergrad math students that are just being introduced to matrices.

jessriedel 9y ago
> then imagine trying to prove that it has that geometric interpretation Exactly. We leaned heavily on determinants in my freshman linear algebra class but I went at least an additional year before I even heard the interpretation, much less could prove it from the standard terrible definition.
enriquto 9y ago
for me the definition is precisely the volume
vanderZwan 9y ago
> However, a nice geometric interpretation does not nice math make. I keep seeing proponents of the Clifford Algebra as a much more elegant type of maths that supersedes linear algebra and also makes it much more intuitive, but I haven't really found a clear source on it yet.

zodiac 9y ago

Yes they are intuitive but it is pretty hard to define them rigorously in a way that captures that intuition. Also, every way I've seen to define them requires choosing a basis (or even an orthonormal basis), which isn't that nice. If you read the pdf you'd notice that Axler's proofs completely avoids relying on the fact that every finite-dimensional vector space has a basis, or choosing one.

Sniffnoy 9y ago
Defining the determinant definitely does not require choosing a basis (though I don't know how you'd define it without at least knowing that a basis exists, for reasons you can see below). Let's say you've got a linear transformation T:V->V, and suppose V is n-dimensional. Consider exterior powers of V, Λ^k(V); for each one we naturally get a linear transformation Λ^k(T):Λ^k(V)->Λ^k(V). In particular take k=n, so we get a linear transformation Λ^n(T):Λ^n(V)->Λ^n(V). But Λ^n(V) is one-dimensional, so Λ^n(T) must be multiplication by some constant factor. That factor is the determinant. That's basically the most "natural" definition of the determinant (notice how multiplicativity immediately falls out of it, assuming of course you already know that Λ^k is a functor). You need the idea of a basis in order to (a) make sense of the statement "V is n-dimensional" and (b) prove that Λ^n(V) is 1-dimensional, but that's it, and you certainly never need to choose one in order to define the determinant. Probably not a great definition for beginning students of linear algebra, but I do have to correct this idea that defining the determinant requires choosing a basis...
- contravariant 9y ago
  I think you get a basis of an n-dimensional vector space by the definition of dimension (if e.g. the dimension is the maximum size of a set of linearly independent vectors). It's a bit trickier when the dimension is infinite, but again most definitions of dimension require there to be a basis of a particular size, the difficult part is proving that this makes sense (i.e. the dimension is unique and defined).
  - zodiac 9y ago