2 ms·

The beauty of Halmos' derivation, which is similar but not identical to exterior algebra (wedge product), is that his approach is basis independent. A determina

by lr1970 6y ago

The beauty of Halmos' derivation, which is similar but not identical to exterior algebra (wedge product), is that his approach is basis independent. A determinant by his definition is scalar invariant over all bases. It is very geometrical in nature.

jacobolus 6y ago
The determinant inherently involves a basis (or at the very least a choice of unit n-vector). Or if you like you can think of the determinant as a function of a square matrix (grid of numbers), rather than a function of a collection of vectors. When you take the basis out, that's the wedge product, which inherently includes the orientation. Conveniently, there is only one degree of freedom for n-vectors in n-dimensional space. When we take the quotient of two n-vectors in n-dimensional space we therefore get a scalar.
- sannee 6y ago
  Let me sketch a way to get the determinant basis-free: Say we live in an n-dimensional vector space V and have an endomorphism f : V -> V. Now, we consider the pullback [1] f* : Λⁿ(V) -> Λⁿ(V) induced by f on the vector space of n-linear alternating forms Λⁿ(V) on V. This is just an endomorphism on Λⁿ(V). However, Λⁿ(V) is one-dimensional, hence necessarily invariant under f*. This means f* has an eigenvalue (!). This eigenvalue is what we usually call the determinant of f. This is completely independent of any choice of basis, orientation, or an inner product. [1] That is, given an element w ∈ Λⁿ(V) and an arbitrary n-tuple v₁, ..., vₙ of vectors from V, we have (f*w)(v₁, ..., vₙ) = w(f(v₁), ..., f(vₙ))
  - jacobolus 6y ago
    > and have an endomorphism f And the "outermorphism" f̱ of your linear transformation, when limited to considering its application to an arbitrary pseudoscalar, returns another pseudoscalar which necessarily has the same orientation, making that a scaling operation. So what we could say in that case is that f̱(p) / p = d (some scalar, the "determinant" of f), where p is any pseudoscalar p = v1 ∧ v2 ∧ ··· ∧ vn. This turns out to be about the same as what I wrote a few comments upthread. We are just dealing with f̱( v1 ∧ v2 ∧ ··· ∧ vn ) / ( v1 ∧ v2 ∧ ··· ∧ vn ) = d = ( f(v1) ∧ f(v2) ∧ ··· ∧ f(vn) ) / ( v1 ∧ v2 ∧ ··· ∧ vn ) instead of ( v1 ∧ v2 ∧ ··· ∧ vn ) / ( e1 ∧ e2 ∧ ··· ∧ en ) = d And now we are talking about a property of a linear transformation instead of a property of a collection of n vectors. In many practical situations, an oriented quantity like v1 ∧ v2 ∧ ··· ∧ vn is more useful than a scalar ratio d though.

by lr1970 6y ago

jacobolus 6y ago

The determinant inherently involves a basis (or at the very least a choice of unit n-vector). Or if you like you can think of the determinant as a function of a square matrix (grid of numbers), rather than a function of a collection of vectors. When you take the basis out, that's the wedge product, which inherently includes the orientation. Conveniently, there is only one degree of freedom for n-vectors in n-dimensional space. When we take the quotient of two n-vectors in n-dimensional space we therefore get a scalar.

sannee 6y ago
Let me sketch a way to get the determinant basis-free: Say we live in an n-dimensional vector space V and have an endomorphism f : V -> V. Now, we consider the pullback [1] f* : Λⁿ(V) -> Λⁿ(V) induced by f on the vector space of n-linear alternating forms Λⁿ(V) on V. This is just an endomorphism on Λⁿ(V). However, Λⁿ(V) is one-dimensional, hence necessarily invariant under f*. This means f* has an eigenvalue (!). This eigenvalue is what we usually call the determinant of f. This is completely independent of any choice of basis, orientation, or an inner product. [1] That is, given an element w ∈ Λⁿ(V) and an arbitrary n-tuple v₁, ..., vₙ of vectors from V, we have (f*w)(v₁, ..., vₙ) = w(f(v₁), ..., f(vₙ))
- jacobolus 6y ago
  > and have an endomorphism f And the "outermorphism" f̱ of your linear transformation, when limited to considering its application to an arbitrary pseudoscalar, returns another pseudoscalar which necessarily has the same orientation, making that a scaling operation. So what we could say in that case is that f̱(p) / p = d (some scalar, the "determinant" of f), where p is any pseudoscalar p = v1 ∧ v2 ∧ ··· ∧ vn. This turns out to be about the same as what I wrote a few comments upthread. We are just dealing with f̱( v1 ∧ v2 ∧ ··· ∧ vn ) / ( v1 ∧ v2 ∧ ··· ∧ vn ) = d = ( f(v1) ∧ f(v2) ∧ ··· ∧ f(vn) ) / ( v1 ∧ v2 ∧ ··· ∧ vn ) instead of ( v1 ∧ v2 ∧ ··· ∧ vn ) / ( e1 ∧ e2 ∧ ··· ∧ en ) = d And now we are talking about a property of a linear transformation instead of a property of a collection of n vectors. In many practical situations, an oriented quantity like v1 ∧ v2 ∧ ··· ∧ vn is more useful than a scalar ratio d though.