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- เผยแพร่เมื่อ 29 ส.ค. 2024
- ⭐Highly Suggested Linear Algebra books⭐
Linear Algebra, an introduction to abstract mathematics: amzn.to/3rkp4Wc
Linear Algebra Done Right: amzn.to/3rkp4Wc
The Manga Guide to Linear Algebra: amzn.to/3HnS59o
A First Course in Linear Algebra: linear.ups.edu/
Linear Algebra Done Wrong: www.math.brown...
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I've never seen a finite field drawn out like as a cartesian plane. That was really cool.
I especially liked the geometric interpretation
Around 13:30, shouldn't it be (a1 + b1, a2+b2) ?
yes
I really like that you include vector spaces over finite fields. I’ve gone through at least five linear algebra courses and I don’t think I’ve seen any other professors even mention those!
13:33 Okay
47:51Good Place To Stop
Is (-a+bi)/(ai^2+bi^2) a complex conjugate of a+bi?
Nice Vid:) Cant wait for more advanced topics, keep it going!
Best math channel
So, a vector space over the field of complex numbers is in fact a vector space over a vector space.
technically the reals are a vector space over the rationals, so...
I'd love to see somewhere like a venn diagram or a growing se of circles starting with a set then going outwards to magma group ring field vector space, inner product space algebra etc. I have trouble keeping them all distinct in my head since I'm not intimately familiar with it all yet.
I would say you don't have to bother remembering what a magma is because they are almost never used.
The inclusions for "group-like structures" are semigroup ⊃ monoid ⊃ group ⊃ abelian group, the inclusions for "ring-like structures" are ring ⊃ commutative ring ⊃ integral domain ⊃ field, the inclusions for "vectorspace-like structures are" vectorspace ⊃ normed vectorspace ⊃ inner product space (and there are some more), and finally the inclusions for "algebra-like structures" are (associative) algebra ⊃ commutative (associative) algebra ⊃ commutative (associative) division algebra (again there are some more).
Instead of categorising things by their structure (i.e. their operations, as above) one can think about adding structure - in this case one has things like set ⇐ abelian group ⇐ vectorspace ⇐ algebra, and set ⇐ abelian group ⇐ ring ⇐ algebra. feel free to ask if any of this is confusing.
Rather than just looking at someone else's breakdown, try drawing it out yourself. It will be easier to remember and also encoded in a way that makes the most sense to you.
Mistakes at 13:30 and 19:30
19:12 ergo, βa₁ = αb₁ and βa₂ = αb₂ (of course I am aware that was a mistake)
The link for "Linear Algebra Done Right" is pointing to the wrong book. Otherwise a nice presentation.
I had some fun, investigating the statement in the video that certain vector-spaces inherit their vector-space structure from being a ring.
It seems like a necessary condition is every element of the field, k, has to be in the ring. Then scalar multiplication needs to be defined as a special case of the already existing multiplication operation (take ring multiplication r*s, but r is restricted to be from the subset of the ring that is also in the field). Making those two assumptions I was able to demonstrate it.
I'm really enjoying your new channel, and the Abstract Linear Algebra series in particular. In college I was a math major, and while I really enjoyed the abstract algebra classes, I have to admit I never really _got_ linear algebra. I'm coming back now (a couple decades later..) and doing the Coursera "Math for Machine Learning" class at the same time as going through your series - the two really complement each other well. Thanks for the great videos.
One thing: it looks like #6 & #7 of this series are not in the playlist?
Interestingly: the set {0, 2, 4} modulo 6 also forms a field.
2 + 4 = 0, therefore they are one another's inverses
2 * 2 = 4 * 4 = 4 mod 6
2 * 4 = 4 * 2 = 2 mod 6
multiplication is well-defined and commutative and zero (the additive identity) is the only element for which its multiplicative inverse is undefinable.
2^-1 = 2, since 2 * 2 * 2 = 4 * 2 = 2; 4 * 2 * 2 = 2 * 2 = 4
4^-1 = 4, since 2 * 4 * 4 = 2 * 4 = 2; 4 * 4 * 4 = 4 * 4 = 4
which makes 4 the multiplicative identity, which means the following behavior-preserving bijection exists: {0, 2, 4} mod 6 === {0, 2, 1} mod 3.
I only noticed that actually while writing that line.
can you also try to make videos on Multilinear Algebra
i 2nd this!
Lets quit fooling around and move on to elliptic algebra.
@@deltalima6703 but what are your thoughts about stuffs like Multilinear Algebra and geometric algebra
Geometric would be interesting to me. Multilinear sounds like slogging with not much interesting stuff tbh.
@@deltalima6703 YOU ARE NOT TOTALLY WORNG BUT NOT TOTALLY CORRECT YOU ARE SAYING THAT GEOMETRIC ALGEBRA IS QUITE INTERESTING BIT BUT THE THING IS THAT THE ALGEBRAIC ASPECT OF GEOMETRIC CONTAIN MUCH ABOUT THINGS IN THE TOPIC OF Multilinear Algebra LIKE MULTILINEAR FORM ,VECTOR LIKE BIVECTOR TRIVECTOR AND MUCH MORE ... YOU CAN ALSO SAY LIKE GEOMETRIC ALGEBRA IS A EXTENSION OF LINEAR ALGBERA AND COMES IN APLLICATION OF MULTILINEAR ALGBERA
These videos are a godsend
Your example of a finite discrete lattice V= F3 x F3 over the field F3 brings the case of the infinite discrete lattice such as V = Z x Z. This would be the square lattice in surface crystallography for example. Now the question is over which field could such a vector space be built? Because Z itself is not a field (element 2 does not have a multiplicative inverse in Z for instance). We can't take R because scalar multiplication of an element of V by pi, for instance, wouldn't be in V. Can we drop the requirement that k is a field, and use only a ring for the scalar space?
I think you should include a few of the stranger examples of vector spaces to demonstrate that the abstraction covers way more ground than the easily visualisable examples you have used. For example: the Reals as a vector space over the Rationals, etc.
I would argue that he has already done this with vectorspaces of functions and especially vectorspaces over finite fields. and R over Q is kind of already covered by the last warm-up exercise (a field is a vectorspace over any of its subfields. in fact a *ring* is a vectorspace over any of its subfields: for example the dual numbers (see e.g. MIchael's video on his main channel from yesterday) are a vectorspace over the real numbers)
actually I wouldn't be surprised if Michael does mention R over Q once he gets to dimension, so stay tuned I guess :D
Love the content, any idea when you’ll start complex analysis videos?
I was really hoping the finite fields would tile nicely, but that doesn't seem to be the case.
Where can i get your jacket of "let's do hard math"?
I'm big fan of your videos from Tunisia
I wonder if it would be useful to consider the vector space whose elements are elements of a field with "m" written after them, like they're distances in meters. In obviously doesn't affect the math, but it might help to clarify when something is a scaler and when it's a vector, and that you aren't going to get mismatches and when you're working in the field versus the vector space.
Good! Let's go!
@6:15 , when you are providing definitions of how your operations work, is the 0+0=0 statement ALSO part of the definition? Or is the only definition that is being offered the alpha scalar multiplication 0 = 0 statement? Otherwise, it is not clear to me why the (V,+) statement is satisfied...as you have not defined how to compute the vector addition.
Edit: If you go to @8:11, it is clarified that 0+0=0 IS, in fact, part of the definition...where the + on the left hand side is describing vector addition (not scalar addition)