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Down with Determinants (1995)
- tzahola 9y agoI had the same uneasy feeling about determinants when I was studying linear algebra at the university. Years later I found Sheldon Axler’s “Linear Algebra done right”, and I loved it!
- experiment0 9y agoThis is a great paper. As a counterpoint, one place where determinants are incredibly useful is in Hartree-Fock theory, where they effective encode the Pauli exclusion principle (or anti-symmetry requirements) of atomic orbitals. https://en.wikipedia.org/wiki/Hartree–Fock_method https://en.wikipedia.org/wiki/Hartree–Fock_method
- nategri 9y agoAlso: Cross products.
- marcelluspye 9y agoBased on this article, I would venture to guess that Axler is also not a fan of the cross product, though, so he wouldn't feel that much is lost. Cross products are practically a vector-calculus hack that lack generality and risk obscuring intuition at the mathematical level.
- ajkjk 9y agoI'm sure it's out there somewhere, but it would be an appropriate corollary to the OP if there was a "Down with Cross Products", which argues that multivectors and wedge products should be taught instead of cross products in multivariable calculus. Then, determinants are "the wedge product of N linearly independent vectors", and cross products are "the wedge product of 2 vectors in 3 dimensions", which gives a bivector and trivially encodes their pseudovector properties. (Also, surface normals in integrals are bivectors, the 'i' of complex analysis is the bivector resulting from wedge product x^y, and e^(i theta) is the exponential map applied to the i operator, and (del wedge vector-function f) is the (bivector-valued) curl while (del wedge bivector-function g) is the (scalar valued) divergence (and that's why del(del(f)) = 0).) (But differential forms should probably be omitted in a first course, because they get hairy quickly and are hard to wrap one's head around. It's enough to know that dxdy in integrals is actually dx^dy, and therefore the Jacobian appears when changing variables because of the factor that appears from dx'^dy' = dx'(x,y)^dy'(x,y).)
- zodiac 9y agoWhich books would you recommend to learn this wedge product / differential forms approach to linear algebra and complex numbers?
- selimthegrim 9y agoDoesn't Spivak get into them?
- wolfgke 9y agoMichael Spivak has written multiple books. Which one do you mean?
- tgb 9y agoCalculus on Manifolds. I'd recommend Hubbard and Hubbard over that as it's a little easier read with the same material.
- wolfgke 9y ago> I'd recommend Hubbard and Hubbard over that as it's a little easier read with the same material. John Hamal Hubbard, Barbara Burke Hubbard - Vector Calculus, Linear Algebra and Differential Forms: A Unified Approach
- Koshkin 9y agoOne problem with cross product, it only exists in three dimensions.
- wolfgke 9y agoNo, it exists in dimension 3 and 7: > https://en.wikipedia.org/wiki/Seven-dimensional_cross_product https://en.wikipedia.org/wiki/Seven-dimensional_cross_produc... EDIT: Nore precisely: A common way to axiomatize the cross product yields a cross products exactly in dimension 3 and 7.
- sixo 9y agoA wedge product exists in all dimensions but only in 3 & 7 is it identifiable with a unique vector.
- pacala 9y ago> A complex number λ is called an eigenvalue of T if T −λI is not injective. Uhm, what the intuition behind _that_?
- zodiac 9y agoIt is equivalent to "A complex number λ is called an eigenvalue of T if Tx = λx for some nonzero x"
- pacala 9y agoThanks! Looking up wikipedia, "In mathematics, an injective function or injection or one-to-one function is a function that preserves distinctness". From that definition, "T −λI is not injective" means that there exists 2 vectors, y and z, such that (T −λI)y = (T −λI)z. Giving T(y - z) = λI(y - z). Via x = (y - z), I get to your (usual) definition. The question I have is _why_ use the injective-based definition instead of the usual well-known one? Is there some further insight down the road?
- Koshkin 9y agoThe merit in using the purely functional language is in that it opens the door for application of powerful methods and concepts of Category Theory. (Note also that even the very notion of a function being injective can be defined in a purely functional manner, i.e. without any reference to sets or their elements.)
- zodiac 9y agoLinear Algebra is a great place to get exposed to some category-theoretical concepts (maybe not when first learning linear algebra, though). My linear algebra prof used the category-theoretical definitions of subspace and quotient space (eg S is a subspace of L if there exists an injective linear map from S to L, a quotient space if there is an injective linear map) and tensor product (universal property)
- zodiac 9y ago
- hexane360 9y agoIf anyone's looking for a (high level) overview of linear algebra, I'd highly recommend 3blue1brown's video series: https://www.youtube.com/watch?v=kjBOesZCoqc&list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab https://www.youtube.com/watch?v=kjBOesZCoqc&list=PLZHQObOWTQ... It's mostly graphical, and is really helpful in forming and cementing an intuition for linear algebra.
- InclinedPlane 9y agoI really hate the way that mathematics is taught in most school systems, it's all machinery with none of the beauty and underlying understanding. I was lucky enough to maintain an interest in mathematics sufficient to allow me to basically teach myself, with a heavy reliance on an underlying conceptual understanding. A lot of things like calculus and linear algebra are mostly just about understanding the basic concepts and then building up some experience working with them. There's no need to memorize gazillions of formulae or anything like that, you just need to actually know what you're doing when you do the work, but you will inevitably rely on tables and references except for very simple work. Which is to say, I could not recommend 3blue1brown's videos more highly, they are an invaluable aid to learning linear algebra and actually helping you understand what is you're doing when you're doing these various operations to "solve problems".
- Izkata 9y agoRan across these videos a few months ago; at one video a night (~15-20 minutes a day over two weeks), it gave me a better intuitive understanding of linear algebra than a full semester in college. These videos should be required before diving into the math in class, I'd've definitely done better had I seen these beforehand.
- CalChris 9y agoAxler seems to have won that battle. His textbook Linear Algebra Done Right is widely used at 308 universities including Berkeley, Stanford and MIT. He has a PDF available without proofs, videos, etc. 3blue1brown likes the book. http://linear.axler.net/ http://linear.axler.net/ I suffered through determinants.
- nextos 9y agoAxler's book is fantastic. But sadly Springer altered the typesetting on the 3rd edition. A really classic and clear LaTeX layout got turned into something much less clear. This freaked me out. Look inside and compare: * Second edition: https://www.amazon.com/dp/0387982582 https://www.amazon.com/dp/0387982582 * Third edition: https://www.amazon.com/dp/3319307657 https://www.amazon.com/dp/3319307657 I wonder whether widespread adoption of his book pushed editors to make it look flashier and watered down. The contents are the same though.
- colechristensen 9y agoOn this topic, I love older textbooks that read closer to prose than whatever passed for flashy textbooks (at least 10 years ago, I can only imagine things have gotten worse)
- CalChris 9y agoAlong these lines, I prefer black and white matte rather than glossy color.
- nilkn 9y agoThis was one of my favorite books as a math undergraduate. I'm sad to see the highly legible and clear layout has been replaced by something so gratuitous and distracting. For better or worse, the 3rd edition formatting and styling is something I've come to mentally associate with low-quality cash-grab big-lecture-hall tomes designed and written by committee over the course of a dozen editions. I wonder if I would have written off the book when I was a student if I'd seen it in such a form.
- xchip 9y agoPhysicist here, don't worry if you don't understand that PDF, it is a pretty terrible explanation.
- benrbray 9y agoWhy?
- CamperBob2 9y agoFor one thing, it uses extensive mathematical jargon that won’t make any sense to beginners... or even to advanced students other than math majors.
- zodiac 9y agoYeah, I think it's written for an audience of other math lecturers, to convince them to not use determinants in their own classes
- wolfgke 9y ago> For one thing, it uses extensive mathematical jargon that won’t make any sense to beginners In Germany this is the usual style for lectures for absolute beginners from 1st semester on - even commonly for people who don't major in math. This style is even not uncommon for 1st semester math lectures for student who don't major in mathematics or physics. Hardly any faculty has a problem with this - they love it that the math departments weed out "unsuitable" students in their lectures so they don't have to If you don't believe me and know a little German, here are two common German textbooks about linear algebra covering about 1.5 semesters of linear algebra for math majors: - Gerd Fischer - Lineare Algebra: Eine Einführung für Studienanfänger (note the title "Linear Algebra: An introduction for freshmen" - I am really not kidding) - Siegfried Bosch - Lineare Algebra Even more: I know a lecturer from Hungary who had very direct words about how relaxing he considers the curriculum for math majors in Germany (he is used to a Sowjet-Russian-style-inspired math program).
- T_D_K 9y agoI got lost in 2.2, I can't work out how applying the transformation leads to the result. Which is frustrating since it's the only non-trivial line in the proof, lol. Also, after applying the transformation, the author states that "a1(λ1 − λ2)(λ1 − λ3)...(λ1 − λm)v1 = 0" => "a1 = 0". But he never says why we know "λa != -λb for all a, b in 1..m" -- that seems non-obvious to me.
- omazurov 9y agoDeterminants 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 agoThe 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 agofor 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 agoYes 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.
- forkandwait 9y agoAxlers book is lovely, but my (amateur) opinion is that determinants are pretty damn intuitive and useful in the applied world. They appear quite naturally in the systems of equations I have worked with. Furthermore, some would argue that mathematics has lost its way as it becomes dedicated to abstraction alone.
- hidenotslide 9y agoI don't understand, can you give an example? For most of the classical applications determinants are computationally terrible compared to factorization methods, e.g. for matrix inverse elimination is O(n^3) and Cramer's rule is something like O(n!).
- tgb 9y agoI think it's false to equate determinants with "determinants computed by cofactor expansion". One can compute determinants efficiently through Gauss elimination, too.
- hidenotslide 9y agoFair, but I'm still not aware of any practical applications for "systems of equations" like the person I responded to mentioned. If you know any please share. The determinant intuition for me is the signed volume factor for a change of basis. I've seen the combinatorial lattice path application and I'm sure there are more in other fields. But not much reason I can see to have them feature so prominently in an intro linear algebra class. Better to spend more time with SVD for instance, which wasn't even covered in the first linear algebra class I took.
- CamperBob2 9y agoMaybe I need a little more handholding than the average linear algebra student, but if the language in that paper made any sense to me at all, I probably wouldn’t need any instruction on determinants.
- zodiac 9y agoUnfortunately I think the paper is written for an audience of other math lecturers, to convince them to not use determinants in their own classes, and not for beginning linear algebra students
- adamnemecek 9y agoObligatory mention that geometric algebra, an alternative to linear algebra, doesn't need determinants while maintaining all the power.
- enriquto 9y agoI never understood this kind of racism against determinants. They are very useful and intuitive, especially in 2D and 3D, where they represent areas and volumes. For example, they give an intuitive meaning to the notion of linear independence of 3 spatial vectors: they are independent when they span a non-zero volume.
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