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The best, by far, book on Linear Algebra that elegantly teaches it from Vector Spaces and Linear Operators point of view is Paul Halmos "Finite-Dimensional Vec

by lr1970 6y ago

The best, by far, book on Linear Algebra that elegantly teaches it from Vector Spaces and Linear Operators point of view is Paul Halmos "Finite-Dimensional Vector Spaces" For instance, the way Halmos introduces the determinant of a matrix (or an operator) is the most consistent, elegant and simple way I ever encountered. OTOH, in Kenneth Kuttler's LinAlg books the determinant is pulled out of the thin air like in 1000+ other similar books.

cpp_frog 6y ago
Thanks, I'll look it up. The best textbook from which I studied (operators) was Elon Lima's Algebra Linear. Sadly the only physical copies are sold in Brazil.
- spekcular 6y ago
  If you want to see the matrix point of view done well, there's Linear Algebra Done Wrong: https://www.math.brown.edu/streil/papers/LADW/LADW.html https://www.math.brown.edu/streil/papers/LADW/LADW.html. You can read a bit about the motivation for doing it that way on that website. The title is a reference to a somewhat well-known book, Linear Algebra Done Right, which avoids using determinants to develop the theory (resulting in a somewhat novel/cleaner presentation). It's unfortunately not freely available online (published by Springer – I would suspect most university students can get it freely through their library's website, however).
  - Tomte 6y ago
    LADR was freely available at Springer at least at some point in time. Under their open access program. I couldn't find it again in a few minutes' search, so it may be gone.
- in9 6y ago
  Ah yes, a fellow Brazillian. I've alwys found Elon's book on Linear Algebra a masterpiece. Coupled with the exercises book, going through it is an eye opening experience.
enriquto 6y ago
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.
deleted 6y ago
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ABeeSea 6y ago
I find both the geometric and algebraic definitions quoted here unsatisfying. What is a “volume” spanned by a vector space of polynomials or co-tangent functionals? *A* determinate function (not the) is simply a skew symmetric n-linear map into the underlying field. Done. Now we get the volume interpretation when it’s appropriate, the wedge product interpretation, and the generalization to finitely generated projective modules (if a determinate function exists, there are additional conditions needed for the existence.)

by lr1970 6y ago

cpp_frog 6y ago

Thanks, I'll look it up. The best textbook from which I studied (operators) was Elon Lima's Algebra Linear. Sadly the only physical copies are sold in Brazil.

spekcular 6y ago
If you want to see the matrix point of view done well, there's Linear Algebra Done Wrong: https://www.math.brown.edu/streil/papers/LADW/LADW.html https://www.math.brown.edu/streil/papers/LADW/LADW.html. You can read a bit about the motivation for doing it that way on that website. The title is a reference to a somewhat well-known book, Linear Algebra Done Right, which avoids using determinants to develop the theory (resulting in a somewhat novel/cleaner presentation). It's unfortunately not freely available online (published by Springer – I would suspect most university students can get it freely through their library's website, however).
- Tomte 6y ago
  LADR was freely available at Springer at least at some point in time. Under their open access program. I couldn't find it again in a few minutes' search, so it may be gone.
in9 6y ago
Ah yes, a fellow Brazillian. I've alwys found Elon's book on Linear Algebra a masterpiece. Coupled with the exercises book, going through it is an eye opening experience.

enriquto 6y ago

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.

deleted 6y ago

[deleted]

ABeeSea 6y ago

I find both the geometric and algebraic definitions quoted here unsatisfying. What is a “volume” spanned by a vector space of polynomials or co-tangent functionals? *A* determinate function (not the) is simply a skew symmetric n-linear map into the underlying field. Done. Now we get the volume interpretation when it’s appropriate, the wedge product interpretation, and the generalization to finitely generated projective modules (if a determinate function exists, there are additional conditions needed for the existence.)