The direct sum and product of vector spaces -- Abstract Linear Algebra 9

[@carlosvaldeonrincon799]

I have given several linear algebra courses but none have I enjoyed as much as this one, and free for everyone who wants to learn on youtube, thanks for doing this for math lovers... and amateur climbers!

[@lilyhayden5732]

yay category theory! the advantages of approaching linear algebra from this angle begin to be seen

[@BiscuitZombies]

\( ﾟヮﾟ)/ This is my first exposure to 'university level' Linear Algebra (in high school we just covered the basics like addition/subtraction, scaling of vectors, projections and dot products). However, starting next week I have a class on Linear Algebra (first year university, but approached from a 'proof' angle.) I look forward to seeing how this translates.

[@bobmvideos]

at 21:00 should it be: u element of W implies u1 = 0 ?

[@skibaa1]

11:13 why Tv and Tw have to be surjective? We can construct h: U->VxW for any (U, Tv, Tw). When this universal construction is defined on Vec category, I don't remember they demand epimorphisms. If I miss something, can you think of a counterexample of such (U, Tv, Tw) where Tv or Tw are not surjective so I will not be able to construct a proper h?

[@mrphlip]

There is one property of the direct sum that wasn't mentioned (or if it was, I missed it)... Say we have a vector x that is an element of V+W, we know by definition we can find v in V and w in W such that v+w=x. However, if V+W is a _direct_ sum, we also have that the choice of v and w is _unique_. I believe this property is necessary for the proof of the universal property at 25:45 to hold... without it, your proof is not depending on the fact that it's a direct sum, and the property definitely does not hold for non-direct sums...

[@goodplacetostop2973]

30:28

[@jeffreycloete852]

Hi Prof Penn..that was nice..universal properties are often challenging..but within the context of Linear Algebra they don't seem so intractable..looking forward to what you still have in store for your Abstract Linear Algebra course..Thanks Again for all the effort you put into your videos..much appreciated!!

[@harryh5666]

14:40 smallest? can someone define that a bit more precisely

[@epsilia3611]

11:00 I'm not getting something here. We have to prove the uniqueness of an object right here. So my question is, how do we know it exists at the first place ? The proof starts by assuming there exists one already ...

[@StratosFair]

At 13:20 we didn't use the surjectivity of T_V and T_W ?

[@StratosFair]

Same at 27:25 the injectivity wasn't used ?

[@mayabartolabac]

oooo a little bit of category theory in this video 👀

[@BiscuitZombies]

👀

[@nathanisbored]

some videos are missing from the playlist

[@roygreen8265]

Where can i check warm-up answers.

[@StanleyDevastating]

Not sure I really grasped what the point is of establishing that the linear transformations satisfying those particular composition requirements are unique? If we had different requirements couldn't we just define different functions pi and iota? Does it mean something?

[@cycklist]

First time I've ever seen iota used in maths!

[@schweinmachtbree1013]

We tend to use iota when we would otherwise use i, but this is already taken for the imaginary unit. here iota is being used for injective maps. when there are two injections I like to use ι for the first and a dotless j for the second, which I would pronounce as "j-ota"

[@mayabartolabac]

24:52 y o d a

[@bobmvideos]

This is how I say iota from now on

[@cletushumphrey9163]

why do T_v and T_w have to be surjective?

[@schweinmachtbree1013]

Good catch, they don't; that's a mistake. Similarly, in the universal property for the direct sum T_V and T_W don't have to be injective. Indeed this is quite a bad mistake because the whole point of the universal property (of, say, the direct product) is that there is a one-to-one correspondence between (k-)linear maps _f_ : _U_ → _V_ × _W_ and pairs of (k-)linear maps ( _g_ : _U_ → _V_ , _h_ : _U_ → _W_ ). Therefore writing Hom_k( _X_ , _Y_ ) for the collection of linear maps from a k-vectorspace _X_ to a k-vectorspace _Y_ , (this is a standard notation, which I wouldn't be surprised if it shows up in the coming videos) we have a bijection of sets Hom_k( _U_ , _V_ × _W_ ) Hom_k( _U_ , _V_ ) × Hom_k( _U_ , _W_ ). For anybody who is psyched about category theory, it is a nice exercise to show that Hom_k( _X_ , _Y_ ) is not only a set but actually a k-vectorspace in its own right, since we can add linear maps and scalar multiply them by elements of k in the "obvious" way. You can then show that the bijection of sets above is moreover an isomorphism of k-vectorspaces Hom_k( _U_ , _V_ × _W_ ) ≅ Hom_k( _U_ , _V_ ) × Hom_k( _U_ , _W_ )

[@StanleyDevastating]

@@schweinmachtbree1013 Thanks, I couldn't see where the injectivity condition came in to it.

[@Mr4thdimention]

Am I wrong here? It seems like the direct sum of two subspaces would be isomorphic to their product. If true it's kind of neat that products and co-products are isomorphic in the category of vector spaces.

[@larrye.goinesjr.1535]

Visualizing Quantum Logic Gates - Bloch Sphere X-State = (1 + -1 × X / √(X² + Y² + Z²)) / 2 Y-State = (1 + -1 × Y / √(X² + Y² + Z²)) / 2 Z-State = (1 + -1 × Z / √(X² + Y² + Z²)) / 2

[@NutziHD]

Cool Video:) Why exactly were T_V and T_W required to be surjective/injective, did I miss something?

[@ConManAU]

I think if they’re not surjective then h doesn’t need to be unique, because any vector not in the image of T_v or T_w isn’t affected by the equality requirement when you compose the functions together.

[@NutziHD]

@@ConManAU But if two functions f,g : A -> B satisfy f(a) = g(a) for all a in A, then they are the same, regardless whether f or g was surjective, what did I miss?

[@schweinmachtbree1013]

@@NutziHD you didn't miss anything, it's a mistake! (see my reply to Cletus' comment for an elaboration)

[@aashsyed1277]

Wow so amazing ...................

[@Alex_Deam]

At 4:15, you forgot to check inverses and commutativity. Not too difficult to see both hold though.

[@schweinmachtbree1013]

he says that he's not going to check all the axioms. and going through all of them wouldn't make for a good viewer experience anyway

[@Alex_Deam]

@@schweinmachtbree1013 Actually at 2:35 Michael implies he's going to check for inverses but then forgets after he finishes associativity. As for commutativity, he didn't mention it, not even to say "we won't check that". And yeah, checking all these axioms can get tedious, but I figured it's important someone mentioned what got left out, otherwise someone new to linear algebra or abstract algebra might get confused.

[@schweinmachtbree1013]

@@Alex_Deam You're right that he forgot to mention commutativity, but in my opinion it's clear that michael doesn't intend to check for inverses: he says "we need to check associativity and inverses, let's look at associativity" (and he completely fills up the space on the board checking associativity). If the viewer is paying attention they can see that inverses we're not presented and figure it out on their own if they need to. (maybe you would have been happy if he had said "one needs to check associativity and inverses", but to be honest this discussion is pretty pedantic and silly).

[@Alex_Deam]

@@schweinmachtbree1013 Normally Michael specifically says something like this is HW rather than leaving it vague. I'm sure some people can figure it out, but not every learner is the same or has the same level of confidence. And much of maths is just organised pedantry lol.

[@scp3178]

Inspiring. The semidirect product of groups could have been noticed here (abstract algebra like 😉). Thanks Michael for your constantly good videos. Chris

[@ratandmonkey2982]

I'm going to have to go back to books. You lost me in the last 5 minutes

[@artificialresearching4437]

I might be wrong, but this could work: kzbin.info/www/bejne/hqe4o5WmqL2Ngrc P.S. A piece of advice: make video 1.5 times faster, I speak very slowly)