What is a vector space? -- Abstract Linear Algebra 6

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[hxc]

I've never seen a finite field drawn out like as a cartesian plane. That was really cool.

[synaestheziac]

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!

[beniborukhov]

Around 13:30, shouldn't it be (a1 + b1, a2+b2) ?

[NutziHD]

Nice Vid:) Cant wait for more advanced topics, keep it going!

[TimHaloun]

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.

[SurfinScientist]

So, a vector space over the field of complex numbers is in fact a vector space over a vector space.

[s]

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.

[shalvagang]

can you also try to make videos on Multilinear Algebra

[SoundVoltage]

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?

[lordstevenson]

Love the content, any idea when you’ll start complex analysis videos?

[bobdowling]

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.

[scottmiller]

I was really hoping the finite fields would tile nicely, but that doesn't seem to be the case.

[khaledjebari]

Where can i get your jacket of "let's do hard math"?
I'm big fan of your videos from Tunisia

[StanleyDevastating]

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.

[AbuMaxime]

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?

[MrRyanroberson]

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.

[scramah]

@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)

[iabervon]

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.