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While Halmos' book is lovely, I still prefer the geometric definition of determinant to the algebraic one: The determinant of a matrix is the signed volume (or

While Halmos' book is lovely, I still prefer the geometric definition of determinant to the algebraic one: The determinant of a matrix is the signed volume (or area) of the parallellepiped spanned by its columns. Equivalently, the determinant of a linear map is the volume of the image of a unit cube by that map (or any arbitrary shape of volume one, not necessarily a cube). All the algebraic properties of the determinant follow easily from the geometric definition (multi-linearity, anti-symmetry, etc). Really, I don't see what you like about Halmos definition of the determinant... I have just read it (page 99 of my copy) and he admits that it is a "somewhat roundabout procedure", just after giving the definition! There's other references that seem much cleaner (e.g. Spivak's calculus on manifolds, using exterior algebra).

lr1970 6y ago
> Really, I don't see what you like about Halmos definition of the determinant... Halmos shows (it is almost trivial) that the space of anti-symmetric n-forms Wn over L_n is 1-dimensional. Wn(Ae1,...,Aen) = const*Wn(e1,...,en). This scalar const is called determinant. It has all the properties you would ascribe to Volume like volume spanned by collinear column-vectors is zero. This is a nice bridge to geometry in Ln. Also, in a space of just one page (p.99) he introduces determinant and proves its main properties like det(A*B) = det(A)*det(B) and therefore det(A^-1) = 1/det(A).
- jacobolus 6y ago
  The determinant of n vectors {vi} relative to a particular basis {ei} in an n-dimensional vector space is the scalar-valued ratio: ( v1 ∧ v2 ∧ ··· ∧ vn ) / ( e1 ∧ e2 ∧ ··· ∧ en ) The signed volume per se is just the n-vector: v1 ∧ v2 ∧ ··· ∧ vn Generally working with the wedge product is more pleasant and conceptually clearer than working with determinants. Among other things we don't need to make an arbitrary choice of basis or unit n-vector. There's also no reason to limit ourselves to n terms. v1 ∧ v2 is also a reasonable quantity to use, etc.
  - 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.
The_suffocated 6y ago
If you define determinant as volume, how do you define volume? I agree that it's pedagogically sound to motivate the notion of determinant by the volume of a parallelepiped, but using volume as the definition of determinant just doesn't sound right. And strictly speaking, determinant is not volume because the former is dimensionless. It is the scaling factor of the volume when a geometric entity is transformed by a linear map.
- enriquto 6y ago
  > If you define determinant as volume, how do you define volume? How do you define "length" and "area"? I guess that if you don't have already a very firm grasp of these basic concepts, then there's no business for you (yet) in studying determinants. Much later, once you master thoroughly lengths, areas, volumes and hypervolumes; and also linear algebra and determinants (however they are defined), then you can embark in the elegant definitions using exterior algebra and the like. Notice that Halmos itself says that his treatment is appropriate for a *second* course in linear algebra, preparing the field for the later study of infinite-dimensional spaces. > And strictly speaking, determinant is not volume because the former is dimensionless. This really depends on the context. If you are working on euclidean space, you already have "units" and the determinant makes sense in itself, as the volume spanned by sets of vectors.

lr1970 6y ago

> Really, I don't see what you like about Halmos definition of the determinant... Halmos shows (it is almost trivial) that the space of anti-symmetric n-forms Wn over L_n is 1-dimensional. Wn(Ae1,...,Aen) = const*Wn(e1,...,en). This scalar const is called determinant. It has all the properties you would ascribe to Volume like volume spanned by collinear column-vectors is zero. This is a nice bridge to geometry in Ln. Also, in a space of just one page (p.99) he introduces determinant and proves its main properties like det(A*B) = det(A)*det(B) and therefore det(A^-1) = 1/det(A).

jacobolus 6y ago
The determinant of n vectors {vi} relative to a particular basis {ei} in an n-dimensional vector space is the scalar-valued ratio: ( v1 ∧ v2 ∧ ··· ∧ vn ) / ( e1 ∧ e2 ∧ ··· ∧ en ) The signed volume per se is just the n-vector: v1 ∧ v2 ∧ ··· ∧ vn Generally working with the wedge product is more pleasant and conceptually clearer than working with determinants. Among other things we don't need to make an arbitrary choice of basis or unit n-vector. There's also no reason to limit ourselves to n terms. v1 ∧ v2 is also a reasonable quantity to use, etc.
- 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.

The_suffocated 6y ago

If you define determinant as volume, how do you define volume? I agree that it's pedagogically sound to motivate the notion of determinant by the volume of a parallelepiped, but using volume as the definition of determinant just doesn't sound right. And strictly speaking, determinant is not volume because the former is dimensionless. It is the scaling factor of the volume when a geometric entity is transformed by a linear map.

enriquto 6y ago
> If you define determinant as volume, how do you define volume? How do you define "length" and "area"? I guess that if you don't have already a very firm grasp of these basic concepts, then there's no business for you (yet) in studying determinants. Much later, once you master thoroughly lengths, areas, volumes and hypervolumes; and also linear algebra and determinants (however they are defined), then you can embark in the elegant definitions using exterior algebra and the like. Notice that Halmos itself says that his treatment is appropriate for a *second* course in linear algebra, preparing the field for the later study of infinite-dimensional spaces. > And strictly speaking, determinant is not volume because the former is dimensionless. This really depends on the context. If you are working on euclidean space, you already have "units" and the determinant makes sense in itself, as the volume spanned by sets of vectors.