5 ms·
Physicist here, don't worry if you don't understand that PDF, it is a pretty terrible explanation.
by xchip 9y ago
Physicist 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.
- edflsafoiewq 9y agoIf v is an eigenvector of T with eigenvalue λ, then (T - bI)v = λv - bv = (λ - b)v. The image is a rescaling of v (and in particular has the same eigenvalue). Therefore (T-λ_2) ... (T-λ_m) v_k = (T-λ_2) ... (T-λ_(m-1)) (λ_k - λ_m) v_k = (T-λ_2) ... (T-λ_(m-2)) (λ_k - λ_(m-1)) (λ_k - λ_m) v_k = ... = (λ_k-λ_2) ... (λ_k - λ_(m-1)) (λ_k - λ_m) v_k Now if k is not 1, then the factor (λ_k - λ_k) appears in the above product, so the term drops outs. So the only term left is the v_1 one. The eigenvalues are all distinct by hypothesis: "Non-zero eigenvectors corresponding to distinct eigenvalues...".
- T_D_K 9y agoGreat explanation, thanks. And the "distinct eigenvalues" part is obvious in hindsight. For some reason my brain thought that we were adding them, not subtracting.
- xchip 9y agoThere are more intuitive ways to explain all that. I saw once a webpage explaining all that with just graphics but I cannot find it anymore. If I find I will update this comment with it.
- tptacek 9y agoIs that what it's supposed to be?
- deleted 9y ago[deleted]
- tptacek 9y agoI don't mind the votes, but just to be clear, this isn't snark. My understanding of this paper is that it's an appeal to other linear algebra educators, not a conceptual introduction.
- xchip 9y agoIt is not an appeal, all the stuff it is describing it is pretty simple and obvious but it is done in the most pedantic way.