4 ms·
Determinants are easy to use but very hardly to grasp intuitively, this is not a minority point of view. See countless of StackOverflow questions begging for a
by ak_111 2mo ago
Determinants are easy to use but very hardly to grasp intuitively, this is not a minority point of view. See countless of StackOverflow questions begging for a conceptual exposition of determinants.
The easiest conceptual handle is geometric: volume expansion, but seeing how this is related to the combinatorial sum over all permutations, or how those two point of views are related to the algebraic one (that a set of equations having a solution or not), is not easy to see even in the 2D case.
I won't be surprised if math professors don't have this issue like you said (especially if someone is comfortable with wedge products), but the vast majority of newcomers who are interested in understanding why something works rather than just how to use it struggle all the time with determinants.
- traes 2mo agoI think the volume explanation is one of the most intuitive pieces of math in existence, personally! Uninvertibility of a tranformation corresponds to a volume of zero because the transformation must squish two dimensions together, leaving them impossible to differentiate, det(AB) = det(A)det(B) because applying two transformations applies their scaling successively, det(A^-1) = 1/det(A) because you have to undo the scaling to invert a transformation etc. I don't think the permutation definition is even strictly necessary; if I recall correctly Linear Algebra Done Wrong defines the determinant in terms of its geometric definition and develops its formula from the properties it must have. I think that the concept is well worth the investment of initial confusion. Axler disagrees, however.
- ak_111 2mo agoBut if you see the geometric view then the permutation sum, for certain obsessive learners they want to see why they are equivalent and thats the hard part.
- abecedarius 2mo agoIt seems to me more like determinants are often badly motivated. The way I remember it from Apostol's Calculus was like "a volume multiplier would be very useful; it needs to be a signed volume for linearity, which implies antisymmetry. Here are axioms collecting these requirements. They're uniquely satisfied by the determinant. Proof: ..." Agreed that it should help if you got to learn wedge products first (I didn't).
- ak_111 2mo agoThis is indeed a good concise description of how to connect the two, but even making peace with this, there is something still magical in how the permutations in the sum cancel neatly (in an inclusion-exclusion kind of a way) to get the volume.
- fn-mote 2mo agoI found that “properties of the determinant uniquely determine this formula that I guessed” approach to determinants to be extremely unconvincing when I was learning linear algebra.
- abecedarius 2mo agoThis might come down to details of how you explain it: iirc Apostol took those basic moves (axioms) and calculated what the formula would have to be, rather than starting with a formula and checking that it has the properties of a signed volume. But I don't know, it's been a very long time for me. It's good to have a variety of approaches to the subject.
- impossiblefork 2mo agoYes, but that doesn't prove that the determinant is actually the real signed volume multiplier. I think this is what led me to feel unhappy about determinants when I was a first-year university student. You need to actually prove that the determinant is the volume of the N-dimensional parallelepiped, and the axiomatic proof doesn't do that. So you need basically two extra lines after proving those things so that people can say "okay, the determinant eats ignores all input vector non-orthogonality so that it gives volume".