Vector Space - Yes or No? Example 1

[@rohanramani6243]

How does this one man know how to do niche problems from every calc class. He knows Calc I, Calc II, Infinite Series, Three dimensional and Four dimensional mathematics and Matrices. What a guy.

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[@gatlinthatcher7651]

@Kendrick Colin No problem =)

[@riobenlopega2619]

Why can't all of my math teacher be like you? You're so great at teaching! God Bless you. :)

[@CooperCable]

I just learned more in 8 mins from you than in 2 hours of lectures from my professor. Wow.

[@upamanyudixit8298]

i am seriously telling your video is still helping people after 5 years

[@chadandersen2774]

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[@MrFatperson]

you are my justification to skipping class

[@hayamramadan3610]

I just got 100 on my calculus exam because of u and here comes the algebra ... U r a true hero professor❤❤

[@kimberlin6568]

Thanks! You helped me through a completely different problem with your easy step by step way of proving vector spaces!

[@patrickjmt]

excellent :)

[@androidrobot8809]

you are a life saver. Plz keep up the good work

[@رانيافهمى-ظ3ق]

dude I love you, keep doing those life saving videos please

[@brydenknowshisgun8486]

you write matrices so well for a left handed person... good job...

[@swiftplanner]

really thanks alot I was stuck at this for about two days and you got that. Love you. thanks alot.

[@kariiiiim]

Thanks for the help. Minor detail on prop. 8. Just a typo, you end up getting cU + dU not cU +cU.

[@tamalmazumdar9831]

Sir your lectures are still helping many people in 2024. Keep it up 🎉🎉

[@patrickjmt]

ill try!

[@karlchrisfordrequioma4908]

prop. 8 is should be cu+du??? amazing

[@sultanaljohani7359]

At 7.22 i think it should be (cU+dU) ,am i right ? and think you very much)

[@8kigana]

sultan aljohani I saw it 7:10 and some change, but yes you may be on to something, he definitely said going through these was tedious and sped through that error.

[@anthonypascual9347]

Yeah I noticed that as well. I believe you are correct, but I'm learning as you are so it's possible we, ourselves, skimmed over something.

[@suleymanalishah]

Yeah cu + du definitely makes sense because he was proving that (c + d)u = cu + du

[@lingchiu4121]

Thanks for making this, it's very helpful. The only problem is every time I want to take a look, you hand is going to block me...I wish you were right handed:))

[@ammanagrawal762]

u explained it so nicely. Thank u brother

[@mankotshephang8588]

I have a question if that kind of question comes in test should I do all 10 conditions of vector space?

[@ridafatima1739]

No, you would first check the three conditions that is the closure under addition and scalar multiplication and if any of those fails to hold then it won't be a vector space. Moreover, if the odds are in their favor then check for additive identity.

[@kollyman2005]

You are the best man.

[@shahhildanial3525]

do you have the softcopy of those notes? I mean your hand written notes that u show in the video??

[@BankruptGreek]

x belongs in R, pretty sure that's necessary. the vector could exist in R² but X could be only positive or only negative which wouldn't be a Vector Space

[@ahnekah5261]

thank you for explaining point 3

[@avlogs3223]

Mayn you are a lifesaver.

[@noranhosny1598]

By The Way I guess there is calculation mistake in property 8 Everything is right except for the last step it is supposed to be cu+cd instead of cu+cv But this does not prevent from being an amazing video

[@matthewhuebsch13]

I disagree. If I was understanding the concept really well up to that point and something is said that is incorrect and makes me confused, then I would say that one mistake can ruin the whole video. The fact that videos can be edited is the greatest advantage over in person instruction. If the investment is made to do it right, the video can save the viewers time and in the end is far more beneficial.

[@ousheshharadhun3773]

@patrickJMT do you happen to have your cheat sheet.

[@nandinisingh9570]

let p2(R) the vector space of polynomial in x of degree at most 2 with real coefficients.Let M2(R) be the vector space of 2×2 real matrices.If a linear transformation T:p2(R) to M2(R) is defined as T(f)= [f(0)-f(2) 0 0 f(1)] then T is one one or onto or range(T)=span{[0 0 ,[-2 0 0 1] 0 1]} or Null(T)= span{x^2-2x,1-x}???

[@Paulovrish7334]

God bless, what is the book you used?

[@ben0857123]

what if you only have one vector. Does that count as a vector space?

[@agapaymarjoriem.9018]

Do you have powerpoint presentation for this example?

[@edmondmalepane3141]

don't you have a long video for vector spaces and examples

[@marthaeugeniamaldonadorodr3837]

Hi, you help me a lot! with all my thanks

[@MahmoodShafi0]

7:15 I think the right answer is cu+du NOT cu+cv

[@hiddenarray2315]

No, he let u be the transpose of the vector [x x].

[@matthewhuebsch13]

How does that change the u into a v?

[@anthonypascual9347]

Actually, if you keep watching for axiom 9, you can see he does replace du for cv, so it was a minor mistake.

[@matthewhuebsch13]

How do you get a v in number 8?

[@user-yu3ub6oy7t]

Keep going 👏🏻👏🏻

[@ერეკლეგუჯეჯიანი]

7:12 he was supposed to write cu + du

[@metahadiprabowo7884]

why the vektor must use the alphjabet ?

[@majesticbarrack5506]

its too hard to test all proterties can we skip some or any short method?

[@timothyolenyo2473]

+h.m.ashhad ashhad no can not you have to show all properties

[@patrickmoloney672]

1-10 are all Axioms, not properties. We don't have to prove them, they are just a set of rules. But we have to test all of them for them to be a Vector Space, so that none of the rules are broken.

[@tcpjh]

Yes, you totally can! Just make use of the subspace criteria :)

[@mohamedmoustafa5098]

If the first one doesn't work in the question, then stop and say: This is not a vector space, but if all of the axioms you tested are valid then go on till you finish then, maybe you find that it's not a vector space and maybe be a vector space

[@saveUyghurs]

So even if just one of the axioms don't hold, then it's not a vector space?

[@firaseljerdy1244]

bengalifob2 yes

[@mohamedmoustafa5098]

Yes

[@santoryu8940]

how will i know if the vectors are in the vector space?

[@santoryu8940]

also what does "closure" under addition means? can anyone rephrase this for me :/

[@deonpersaud9540]

it means if you add say two elements in the set then the answer must be an element of that set as well. for example 3 and 4 are in the set of real numbers. 3 + 4 = 7. 7 is also a real number. the set of real numbers is closed under addition.

[@santoryu8940]

dammmm thanks deon

[@looc1612]

where the hell is third one ?

[@alimuhammadsamnani1857]

How to know whether zero vector is present in vector space or not? If someone can explain

[@कत्यूषा]

satisfy zero in the equation if equation satisfy then zero belong to subspace vector and it is non empty

[@lavenderoluoch3375]

I thought vector v =component (y, y) in axiom 8

[@timothyolenyo2473]

+Lavender Oluoch yeah it is

[@lavenderoluoch3375]

Thanks

[@timothyolenyo2473]

welcome

[@sarthakdhiman381]

wanted to tell u , u r still helping me'

[@jameyatesmauriat6116]

This is mathematics for fetuses

[@syavv7514]

Thank you sir :)

[@heyysan9860]

Thank you!

[@sarmadnadeem_]

No doubt he's best at teaching. but everyone is a good teacher, it just depends on us when we focus on the lecture. :)

[@l3aIIin23]

thank you

[@njabulomahlalela2912]

thanks a lot!

[@sikujuasturridge4301]

Its amazing....what about giving You a question and give us the solution?

[@user-yu3ub6oy7t]

Thank yooou 😭😭

[@brianombongi14]

Thank so much my all time professor. You are the best

[@metahadiprabowo7884]

i'm sorry sir,but my head confuse to lesson the vektor

[@metahadiprabowo7884]

apik tenan pak,,nanging aku ra mudeng blas (Good ,sir ,but i dont understand, what you talking about

[@md.sadeknayem7913]

thanks

[@misterbguy123]

Cx

[@zuest696]

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[@VenomFrogPwned]

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i was look for this comment xDDDD

[@sultanaljohani7359]

in properaty 8

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Flesh bro show me more

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jai mata di

[@fireflies15]

truly tedious

[@bakhtawarsarwar666]

GOOD but boring.