The Gram-Schmidt Process

[@omyrazeem2571]

This 10 minute explanation is better than 2 hour lecture in class.

[@nkwdtwg1352]

People who say things like you do just don't understand how to teach. Teaching in a class and teaching on KZbin are fundamentally two different things.

[@holyshit922]

I agree that it is explained well In my opinion decomposition the vector into its components could be the introduce to this method (decomposition the vector into its components is for Gram Schmidt like searching max for sorting)

[@nathanjeong]

@@nkwdtwg1352 how so? 🧐

[@Vishal-by3vr]

So true

[@asmundhommepaulsen945]

Legit

[@brownviking8609]

Saw you destroy flat earthers earlier in the year and now when I’m taking linear algebra in university, you come through for me again. Great vid!!

[@vizviz72405]

😂

[@jessicapeng6270]

Looked through several videos describing Gram-Schmidt and this was the most clear, thank you!

[@blockchaineenthusiast3932]

hello lovely Jessica , I like to conversesion with you

[@anonymous9217w2]

@@blockchaineenthusiast3932 lol

[@KeshavYadavgggvds]

​@@anonymous9217w2 lol

[@maxxie878]

Wow. I love it. It's easy to get mixed up when you're using it, so it's great so see this explained in simple steps.

[@bioscience6320]

I really, appreciate your way of presentation ( short, precise and to the point). I never understand my lectures without you. Thanks a lot professor Dave.

[@Atlas_Enderium]

your videos are saving me hours of studying time for my upcoming Linear Algebra exams 😭 thank you

[@abnerandreymartinezzamudio3366]

Simplicity is a great demonstration of genius. Thank you!

[@MadGermanX_X]

It’s incredible how hard lectures can explain math and how poorly they do. Thank you for explaining it really helped

[@YTjugal]

Best, easiest & most visual explanation of the Gram-Schmidt Process. Thank you Professor Dave :)

[@NicolaMigone-f8f]

I really have to thank you. I discovered your channel in my junior year when I went abroad to the US and studied Physics 1, when I already studied Physics 3 in Italy, my homeland. Obviously, your videos were very basic because the content was such. I'm now in my freshman year in university, and have rediscovered your channel through these videos about linear algebra and calculus BC, and am really glad I did, you have helped me a lot. My partial went amazing, all exercises were flawless (although I have to study theorems better), and it's also thanks to you :)

[@AkhtarMehraj]

Best explanation I saw on KZbin. Keep uploading such useful content. Thank You.

[@braydenchan138]

Prof. Dave is a genius in education. We NEED you!

[@tomatrix7525]

At 4:12, to get tye length of the vector, don’t we need to sq rt. the dot product? On your illustration I only see you performing the dot product operation without the root...Am I missunderstanding something? Edit: noticed thr square outisde the vector length, thus undoing the root. Makes sense now. Sorry about that silly question

[@norawallberg1345]

It's 11 pm. I have my final exam in linear algebra tomorrow and have not had time to learn this process yet

[@BlackCat-ov4tc]

I loved how the explanation is so simple and clear, thank you professor :)

[@kasra-ir5io]

bro in just one day I finished more than 20 hours lecture just by using your videos. thank alot man you saved me ❤

[@dhirdeepakmichael2977]

understood 2 days worth of lecture in 2 minutes due to this video

[@vardan_kashyap]

I am watching your video's related to linear algebrs in nov. 2024 this is so helpfull for me..❤😊 From India 🇮🇳 (University of Delhi) in 1st year

[@navagharkiran5769]

sir, will u make vidoes about Real Analysis , since its complicated in maths.. i Know u will explain it better to us plz . a lot of Indians not understand it due to lack of correct explaination in college

[@enthusiasticsimple901]

Great explanation... Much better than my 1hr college class. Keep Going🔥🔥❤

[@brianzhu4221]

Dave is just that guy, he is that guy

[@vaibhavkhobragade9773]

Terse and concise explanation! Thank you

[@bhattialdo3208]

Sir you are very understanding. Keep it up.

[@swaroopsalvi47]

Thanks, looked several videos on this but your explanation is the simplest.❤

[@yogitajindal3149]

Very simple and helpful explanation Thank you Sir 🙏🏻🙏🏻

[@arty7926]

Had to implement it in python, was very easy thanks to your tutorial!

[@akosdanielbollok]

In the comprehension, no matter what I do, as an orthogonal vector I always get [10/3 5/2 10/3] and when I divide it by it's length I get a totally different answer... what could I have been soing wrong?

[@Arman-v9f4i]

Do the calculations again, You'll get it. Here is the answer: So you get the orthogonal vector as [10/3 5/3 10/3]. You take out 5/3 as common from the orthogonal vector, you get 5/3*[2 1 2]. Calculate the length of this vector, you get √(10/3*10/3 + 5/3*5/3 + 10/3*10/3) = √(225/9) = 15/3. Then finding the orthonormal vector i.e., u2/(length of u2). (1 / (15/3) ) * 5/3*[2 1 2] = 3/15 * 5/3 * the vector which results 1/3 * [ 2 1 2] which also can be written as [ 2/3 1/3 2/3] . I hope you understood this and this would have helped you!!

[@aravindkannan.s3503]

​@@Arman-v9f4iwell done bruhh

[@KM-di8zl]

You sir, are a LEGEND !

[@mirai8274]

Thanks a bunch! Short and well explained.

[@DeliriumzzZ]

My wrist is numb just from working through both problems, but I understand it. Thanks Dave! I will send the physio bill shortly.

[@funnysareen]

You are a beast! Thank you Prof!! Differential Equation section next??

[@ProfessorDaveExplains]

Yes as soon as I find someone who can write the scripts!

[@funnysareen]

@@ProfessorDaveExplains Can't wait !!

[@professor_gurt]

Linear final exam prep time

[@tempurawizzo1880]

Thank you x100000 I really appreciate your videos!!!!

[@aminaahmed1997]

High five Prof Dave. Thanks to u now am all set for my exam tomorrow 😊

[@pugazhmathi.r3133]

I haven't never seen the lecture like this.....awesome..

[@reywilli8892]

Thank you so much from Berkeley grad student

[@reamabdulsalam524]

Hi thanks Dave really you made my life much easier you don’t know how long I have spent on this topic online to understand! Hours and weeks with no any benefit but with your 10 minutes explains it was really easy and now I can understand it should my Dr in university watch your clip thanks please do explain as well how to get the projection matrix onto a line ax=-by thanks

[@Lila-lc7wd]

Thanks a lot for this series!!

[@aviadlampert5956]

These videos save me. Thank you!

[@ryanjackson0x]

Isn't dividing by the (total) length of a vector only one way of normalizing a vector? How you should do it depends on the norm you are using, is my understanding.

[@akeemeviota426]

I have a question Professor Dave to the answers at the end of the video. Why is u1 = [1/3 2/3 -2/3]? Isn't v1 = u1?

[@daveleung8081]

this guy deserve more views

[@tiodharmawan4905]

Many thanks Professor Dave, it is very helpful

[@ditya3548]

This has been really helpful to me. Thank you very much

[@bobbykuhikar5527]

best and awesome explanation!!

[@ekleanthony7997]

this video is super helpful... thank you professor for your awesome explanation

[@gauravsaxena7075]

dude explanation is excellent

[@sugandhm2666]

Crystal clear explaination

[@sujitkale7808]

Very nicely explained sir thank you so much❤

[@Jehannum2000]

At first I couldn't understand how you could just remove bits of vectors willy-nilly (ad hoc). I then realised that these are not equations where if you take something off one side you have to add it to the other side - they're definitions of new vectors.

[@Youngballer739]

Great explanation man

[@randymartin5500]

At 10:06, U2 1st component and 3rd component is not coming out as 1/√6. I get 1/3*√3/√2 = √6/6 .

[@willjohnston2959]

√6/6 is same as 1/√6 so either answer is okay. Prove the ratios are equivalent using cross products, or multiply second fraction by √6/√6 to get first one.

[@t_n_t_o6519]

@Professor Dave Explains Shouldn't it be like that 5:23 -> (1 0 1)^T - 2/3 * (1 -1 1)^T = (1 0 1)^T - (-2/3 2/3 -2/3)^T = (5/3 -2/3 5/3)^T Because of the (-(2/3) * 1, (-2/3) * (-1) and (-2/3) * 1) -> (- & - equals + and - & + equals -) (I'm using the transpose of the the row vectors ^^ hope it is clear :D) It just confused me a little bit but else I love your videos seriously I'm a CS student and those videos of yours are better then my lectures!!

[@averywilliams2140]

The notation part in the definition you used a determinant notation '|u|' for the norm instead of the ||u|| notation. Just posted for edification. Good video also.

[@JanNowacki-t2o]

thx you saved my life

[@trashhollowknightplayer4199]

I think you just saved my exam 😂

[@RahulSharma-oc2qd]

Can we say that original basis and orthonormal basis, both can be shown as linear combination of the other? if I consider {e1,e2,...en) as the orthonormal basis of the vector space V and {g1,g2....gn} original basis of the vector space, then ei= Einstein summation(scalar).gi (where i=1,2....n)? Am I missing something at 4:10?

[@eittorres]

Can you do a playlist on statistics? Thank you for time and effort.

[@erinmeyers-t9w]

thanks professor dave!

[@mohammadidreesbhat1109]

Sir, great. They teach only evaluation in class

[@PetBuddies]

Is the answer provided for the exercise correct?

[@AK-lu8ml]

simple and clear..👍

[@easymath9783]

nice tutorial

[@browniesnofrownies4843]

Thank you so much for this

[@holyshit922]

Good for pencil and paper calculations but i need better procedure if i want to write program for orthogononalization because of numerical reasons

[@julianavarela4936]

You're a savior

[@JulissaMejiaPlaza]

what does it mean for vectors to be orthogonal?

[@sumitkumarshakya1525]

Love from India 🇮🇳🇮🇳❤❤

[@tubanexsenpai4174]

Why 1/√6 8:15.

[@sankhadas1080]

Thanku sir....its soo helpful

[@omyrazeem2571]

great explanation ...

[@novinnouri764]

thanks for this video

[@chamithdilshan3547]

Thank you so much!

[@MT.632]

Bestest video👍 May Allah bless you❤

[@bouyancyyang943]

what is the expression for orthonormal basis? just orthogonal basis/length?

[@darcash1738]

is there dot prod mult for TI-84?

[@thegeekypai]

Thank you😍

[@nuwagabaedgar6580]

Am sure it will also help me figure out

[@payaparapasu]

thank you very much

[@AsadKhan-ki2in]

Real professor 🌺

[@bouyancyyang943]

answer at 9:33 is orthogonal basis not the final answer.

[@transfo47]

Excellent as always Dave.

[@Rajath_2001]

Why is it always that, i don't understand, no matter how hard I try, in class. But, understand it completely in a 10 min video ....

[@alaadusokey262]

great content, thnx

[@abdullahyto]

Thanks dude

[@iamrising7176]

This is sooo amazing. Thanks buddy. I would love to join the channel but my local card ain't accepted.

[@manishverma6135]

Thank You.

[@taabostore]

your answer (2/3) / sqrt(2/3) = 1/sqrt(6) is incorrect, the correct one is sqrt(6)/3. great video

[@trashpanda70]

An amazing and clear breakdown of the process, thank you so much

[@GAMIDATA]

very much useful

[@ryanjackson0x]

Are the starting basis vectors assumed to be sorted by magnitude? If not, does that mean v1 is arbitrary?

[@hana29971]

I guess so

[@sachinrajpandey5242]

I am getting different answer for u3 for the question asked in checking comprehension, will you please tell me I am right or wrong. I tried a lot but my answer is different, for u3, from that given in the video.

[@jerrycorner225]

same question in u3 i come up orthormalization of -1/square root of 2,0 and 1/sqaure root of 2..is this also correct.

[@miminaechali7058]

Wowwwwww!! The procedure is that too simple

[@AnhLe-qw7yq]

Grateful from VietNam 🤩🤩

[@lochansanju6757]

Bro how did u take u1. U1=3

[@stanleymainastan]

answer for u3 on the trial question is wrong

[@linguafranca7834]

Thanks

[@krizh289]

I have a final in 2 days, learning this for the first time😈

[@RadoHudran]

nice cut

[@OsamaShehzad1995]

Perhaps it would have been better to explain with diagrams comprising vectors v1, v2, v3 --> u1, u2, u3 which leads towards building this formula.