Not subspace

Examples of sets that are not subspaces and showing why they’re not subspaces
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[@chinesecabbagefarmer]

You truly seem to enjoy what you're doing! It's admirable.

[@waleedharalkathiri1836]

And we truly enjoy what is he doing

[@chinesecabbagefarmer]

@@waleedharalkathiri1836 Absolutely!

[@ecmrn]

I've been watching linear alg videos for a month. Why oh why didn't I find your channel sooner, I would have done much better on my midterm. You explain everything the best out of all the instructors on KZbin

[@MardkoMBR]

Brilliant. Thank you for your contagious enthusiasm

[@adolfocarrillo248]

Doctor Peyote, this is the first Time that I do understand a tutorial from all your series of tutorials. Your students are so lucky to have you as their teacher. Recieve a big huggie on my behalf.

[@joshuakennedy5629]

I am learning about subspaces in linear algebra right now and your videos are really helping me understand the concepts of this chapter. Your students are very lucky to have you!

[@dhunt6618]

Thanks for showing the ways to drop out of subspace without (hopefully) breaking the warp engines!

[@ChristinaRichardsonFitness]

Thank you! Again and again and again! I am in a crummy grad level linear class so I ordered the book you use and I am going through your videos. I need this information. He uses your least favorite linear book. Lol. Thank you!!

[@drpeyam]

Thank you so much!!! And sorry to hear that your prof uses that other book 😥

[@ecmrn]

Bless you! I don't know which University you to teach in but your students are lucky

[@yogeetavensiani4958]

Thanks a lot! I will never forget this concept after watching this video!

[@waleedharalkathiri1836]

That is amazing Dr. Peyam really appreciated 👍👍

[@ibraschwan]

Thanks for the video ❤

[@SmileyHuN]

Great

[@JASMINEMICHAELASC]

what did he say at 5:34 between 'thats why its super important....' and '....to show something is false'?

[@icew0lf98]

I think you should make more videos like this, but there are two things I didn't like about this one : 1) at one point you said what will be the reason the example wont be a subspace, without letting the viewer think. 2) it was too easy, its fine to have easy non-examples but I think one needs to be hard.

[@neptunian6226]

In terms of definition, do we NEED to define that 0 is in the subspace or does constant multiplication suffice because our scalar field MUST have a 0 element, that implies that the zero vector is in it?

[@drpeyam]

Basically you just need it to be non empty, then 0 is in it because of what you said

[@Davidamp]

Actually, the theorem states that a subset S of the vector space V is a vector space if and only if: i) S is not the empty set ii) is closed in addition iii) is closed in s-multiplication. The zero vector (Ov) is included in the closure of s-multiplication because 0•v = Ov. Even though, most of the time in problems, I show the Ov is in the set to show it is not empty.

[@drpeyam]

Yep

[@orlandomoreno6168]

Then the non ex 1 is actually a subspace?

[@Davidamp]

@@orlandomoreno6168 No, because actually is not closed in the operations nor the Ov is in the set.

[@Davidamp]

Also, one of the Axioms of the Vector Space is that there exist the Ov in the space. So if it doesn't exist in the subset, it is not a vector space neither a subspace.

[@coreyplate1001]

I have a math challenge / video suggestion for you (one that I don't know the answer to myself): what wave function, when added to its derivative, gives a standing wave with a node at zero and, if possible, at nπ for all n∈Z

[@rounakpoonia8742]

Why are you so happy though

[@aneeshsrinivas9088]

Is it a good idea to respond to the sexond counter ezample:”no, (1,1) managed to GTFO from this set by scalar multiplication by -1, i suggest this set installs unclimbable walls to prevent those scalars”

[@RealEverythingComputers]

DR PEYAM THE GOAT STRIKES BACK

[@CornishMiner]

Super. Your videos get better and better :)

[@dgrandlapinblanc]

Thanks Mister.

[@cristiandavidmillergonzale710]

I will owe at least the half of my engineering degree to you

[@norman.20]

I wanted to ask if the following expressions and conclusion are true or not: Consider [x y 1]. we can think of it as Z=1 plane. This is not a subspace of R3 , then it cannot be written as span of some vectors in R3 It means there is no set of vectors in R3 which linear combination of them, can form the plane Z = 1 In other words, we cannot have a matrix that its corresponding column space generate Z = 1 Now It seems kind of limitation for manipulation of objects in Z = 1 with Linear tools. Is it true?

[@drpeyam]

True, but it’s what’s called an affine space, namely a vector (0,0,1) + a subspace (the xy plane), so you can still study it!

[@TheDiederikdehaan]

Dr Peyam, what about the following subspace: H is a square for which |x|=< 1 & |y|=

[@TheDiederikdehaan]

Nevermind after watching your previous video I found my answer - I retract my question :)

[@ElifArslan-l9g]

thank you!

[@luanvasco3572]

perfect !!!

[@IronLotus15]

"From the producer of" lol

[@Silver_G]

Is this series pre-recorded? If not, please adjust the camera's angle... This Linear Algebra series is epic but the angle is ruining the whole series😅

[@drpeyam]

Pre-recorded!

[@suhailmohamed3013]

@@drpeyam the camera angle is GREAT

[@SlipperyTeeth]

Trick question: Is R a subspace of C?

[@ramchandanibhavesh]

Of course

[@SlipperyTeeth]

@@ramchandanibhavesh I wasn't kidding about it being a trick question. The correct answer is "It depends". Specifically, it depends on what field you're using for your scalars. If you're using real numbers as your scalar field, then the answer is yes, of course. If you're using complex numbers as your scalar field, then the answer is no. What I mean is that if you look at the rule that says that for any u in the vector space and scalar c, cu also has to be in the vector space, it is not immediately obvious that c has to be a real number. More abstractly, we say that c just has to be a field. Some common fields are rational numbers, real numbers, and complex numbers. So, if you use complex numbers as your scalars, then i×r isn't in R for most r in R.

[@ramchandanibhavesh]

@@SlipperyTeeth Let V be a vector space over K. Then W is a vector subspace of V if for every a, b in K and for every, u, v in V, au + bv belongs to W. According to this, you can't change your inicial field as per your choice. Indeed, my first answer is incorrect, but "it depends" shouldn't be correct either. The answer is that R is not a vector subspace of C, because, as you pointed out, most of the scalars of the field C multiplied by a vector in R will result in a vector in C.

[@SlipperyTeeth]

@@ramchandanibhavesh You can say that C is a vector space over the reals. This is completely valid. It does depend on if when we say that C is a vector space, are we referring to it as a vector space over the reals, or are we referring to it as a vector space over the complex numbers. Using your letters in your previous comment, it is not necessary that for any u,v there be an a such that au=v. This is an additional restriction that I think you are subconsciously appending to the normal restrictions. For example, there is no such a for vectors in R^2 over the reals [take u=(1,0) and v=(0,1)]. You can say similar things between R and Q with the reals and rational numbers. Or C and Q with complex numbers and rational numbers.

[@ramchandanibhavesh]

@@SlipperyTeethI had understood C as vector space over complex number, didn't think about defining it over the real numbers. Thanks for your explanations. Although I don't see how did you deduce that au = w. I just meant that any linear combination of u and v should remain in the subspace W.

[@MatheusNasi]

"cu" means "asshole" in portuguese, really nice to watch you writing it lol

[@drpeyam]

Hahahaha, OMG 😂😂😂

[@dolphinlunggrin6594]

Reminds me of the days of ICQ messenger where we would sign off with cu which spelled out would sound like "see you". And thinking about it "I'm off, asshole" would have fit as well 90% of the time 😂

[@blackpenredpen]

dolphin lunggrin omg did you just mention ICQ?!!!! Oh mannnn that brings back memories!!!!!!!

[@Silver_G]

@@blackpenredpen "uh-oh"

[@thedoublehelix5661]

You don't NEED an explicit example to show something false. For instance, in non-example 2, you can show that c*u is not an element of h by case work. Since c is an element of R, we can use cases. case 1: c>=0, ... yada yada yada case 2: c=0, c x*c

[@drpeyam]

Yes you do! Try the same argument with the set x^2 + y^2 = 0

[@thedoublehelix5661]

Dr Peyam I’m confused. Is the set x^2+y^2=0 not a sub space? kzbin.info/www/bejne/f5nUiGiOpN-ImK8

[@chinesecabbagefarmer]

Iron that shirt!

[@asht750]

The content is good. However, try to write down the language of proof-writing instead of talking too much. If I were in your class, I would drop it because of the sloppy explanations.

[@drpeyam]

???