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- เผยแพร่เมื่อ 13 ก.ย. 2024
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That hat notation is so much nicer than writing in the i-1 and i+1 terms!
amazing ............... th-cam.com/video/CEz8tYePuhs/w-d-xo.html
I hope you can do this full time in the future. This is top tier quality education
perhaps it should be pointed out that the theorem at 19:00 only applies to finite-dimensional vectorspaces, whereas almost all of the other theorems so far can be adapted to possibly-infinite-dimensional vectorspaces (using the definitions that the span of an infinite subset is the set of all *finite* linear combinations, or equivalently the union of the spans of all its finite subsets, and that an infinite subset is linearly independent if all of its finite subsets are linearly independent)
@UCgll1K4epLc49jcBDVB8MJQ yes but the other theorems so far are stated similarly when they actually hold more generally. I was just pointing out that this theorem in particular is one of the few that only holds for finite-dimensional spaces. The series is titled "abstract linear algebra" so this (doing linear algebra in high generality) is likely of interest to people in the comments.
amazing ............... th-cam.com/video/CEz8tYePuhs/w-d-xo.html
The proof of the final implication (3) => (2) in the first theorem is incorrect. the proof of the first implication (1) => (3) is also strange (maybe not incorrect, but definitely strange) - one doesn't need to bother with the alphas; one can just consider B U {v} from the start and follow ones nose (B U {v} is dependent so there is a relation lambda_1 v_1 + ... + lambda_n v_n + lambda v = 0. lambda must be non-zero so we can divide by it and rearrange to find that v in span(B) as was to be shown)
There are better treatments of bases out there. Prof not uniformly wonderful, but he is good.
@@get2113 I think having this kind of stuff in video format is extremely valuable despite mistakes every now and then - we as the viewers can always point out errata in the comments :)
amazing ............... th-cam.com/video/CEz8tYePuhs/w-d-xo.html
Why is it incorrect? There is a slightly confusing bit where he appears to write w as a linear combination of vectors in B, but actually it's a linear combination of vectors in S and since S a subset of B all the basis vectors are in B. However, apart from that I don't see what the problem is with the argument? He has constructed a w in V, that is not in the span (S). So no strict subset of B can span V. So B is a minimal spanning set.
@@StanleyDevastating yes that is what I was referring to; he should have written v_i ϵ S not v_i ϵ B, and then it follows that v_1, ..., v_n, w ϵ B
The proof of theorem 19:00 is nice, but a bit tedious. Wasn't there a way to argue by contradiction and using what we have shown before about minimal spanning sets ?
Great lecture! For what it's worth it seems like TH-cam maths loves dimensions and infinite cardinalities, could clickbait with that in the title.
amazing ............... th-cam.com/video/CEz8tYePuhs/w-d-xo.html
@4:46 first proof could be as simple as this: if w is other than zero, then B isn't maximal; contradiction and done.
I don't understand the point of the theorem at @19:00 . We already proved that a set being maximally linearly independent was equivalent to a set being the minimal spanning set. So we already know that a set of linearly independent vectors cannot have more members than any set that spans the vector space, right?
Nvm, I see what I have missed. A minimal spanning set, A , only rules out subsets of itself spanning V, it does not rule out the existence of another smaller spanning set, B, where B is not a subset of A. Similarly for a maximal linearly independent set. Therefore, if we take a maximal linearly independent set, C, and a minimal spanning set, D, where neither is a subset of the other, we need to prove that C cannot have more elements than D.
Basics of bases
So what is the basis of the vector space defined by x_1 + x_2 + x_3 = 1?
It isn't a vector space. It's a coset of the vector space where x1 + x2 + x3 = 0
I'm taking this under assumption that your set is a subspace of some k^n.
To show that, let's suppose that a zero vector z exists in our set, which we'll call V'.
Then for all v in V', z+v = v+z = v.
With normal vector addition in k^n, we have that for a,b in k^n:
a + b = (a1+b1) +... + (ai+bi) +...+ (an+bn) for 1
@@cinnabun-ysera Thanks for your reply. Looks like I need a quotient space then.
@@SurfinScientist I'd recommend taking a look at affine spaces and constructing affine bases.
@@cinnabun-ysera Thanks a lot! I will take a look at affine spaces and bases.
Hi Prof Penn..Thanks for your continued efforts to bring maths to a larger audience..could you please do a video on EXACT SEQUENCES..maybe ..SHORT EXACT SEQUENCES..in the same vein as your wonderful and very popular video on Tensors?
amazing ............... th-cam.com/video/CEz8tYePuhs/w-d-xo.html
I might be wrong, but this could work: th-cam.com/video/PvUrbpsXZLU/w-d-xo.html
P.S. A piece of advice: make video 1.5 times faster, I speak very slowly)