People who say things like you do just don't understand how to teach. Teaching in a class and teaching on TH-cam are fundamentally two different things.
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)
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.
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 :)
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
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
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?
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!!
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.
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.
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.
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
@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!!
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?
√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.
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.
This 10 minute explanation is better than 2 hour lecture in class.
People who say things like you do just don't understand how to teach. Teaching in a class and teaching on TH-cam are fundamentally two different things.
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)
@@nkwdtwg1352 how so? 🧐
So true
Legit
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!!
😂
It’s incredible how hard lectures can explain math and how poorly they do. Thank you for explaining it really helped
Looked through several videos describing Gram-Schmidt and this was the most clear, thank you!
hello lovely Jessica , I like to conversesion with you
@@blockchaineenthusiast3932 lol
@@anonymous9217w2 lol
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.
Simplicity is a great demonstration of genius. Thank you!
your videos are saving me hours of studying time for my upcoming Linear Algebra exams 😭 thank you
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.
Best, easiest & most visual explanation of the Gram-Schmidt Process. Thank you Professor Dave :)
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 :)
Prof. Dave is a genius in education. We NEED you!
It's 11 pm. I have my final exam in linear algebra tomorrow and have not had time to learn this process yet
Best explanation I saw on TH-cam. Keep uploading such useful content. Thank You.
bro in just one day I finished more than 20 hours lecture just by using your videos. thank alot man you saved me ❤
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
understood 2 days worth of lecture in 2 minutes due to this video
Great explanation... Much better than my 1hr college class.
Keep Going🔥🔥❤
Linear final exam prep time
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
Dave is just that guy, he is that guy
I loved how the explanation is so simple and clear, thank you professor :)
Sir you are very understanding. Keep it up.
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
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?
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!!
@@Arman-v9f4iwell done bruhh
this guy deserve more views
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?
My wrist is numb just from working through both problems, but I understand it. Thanks Dave!
I will send the physio bill shortly.
You sir, are a LEGEND !
Thanks, looked several videos on this but your explanation is the simplest.❤
Terse and concise explanation! Thank you
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.
Had to implement it in python, was very easy thanks to your tutorial!
your answer (2/3) / sqrt(2/3) = 1/sqrt(6) is incorrect, the correct one is sqrt(6)/3. great video
Thank you so much from Berkeley grad student
Very simple and helpful explanation Thank you Sir 🙏🏻🙏🏻
High five Prof Dave. Thanks to u now am all set for my exam tomorrow 😊
Crystal clear explaination
You're a savior
You are a beast! Thank you Prof!! Differential Equation section next??
Yes as soon as I find someone who can write the scripts!
@@ProfessorDaveExplains Can't wait !!
Love from India 🇮🇳🇮🇳❤❤
I haven't never seen the lecture like this.....awesome..
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.
Sir, great. They teach only evaluation in class
Thank you x100000 I really appreciate your videos!!!!
These videos save me. Thank you!
thx you saved my life
I think you just saved my exam 😂
dude explanation is excellent
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.
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
Thanks a bunch! Short and well explained.
Is the answer provided for the exercise correct?
nice tutorial
Great explanation man
Thanks a lot for this series!!
what does it mean for vectors to be orthogonal?
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 ....
best and awesome explanation!!
thanks professor dave!
Very nicely explained sir thank you so much❤
Many thanks Professor Dave, it is very helpful
Bestest video👍 May Allah bless you❤
This has been really helpful to me. Thank you very much
@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!!
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?
this video is super helpful... thank you professor for your awesome explanation
Can you do a playlist on statistics? Thank you for time and effort.
what is the expression for orthonormal basis? just orthogonal basis/length?
thank you very much
simple and clear..👍
This is sooo amazing. Thanks buddy. I would love to join the channel but my local card ain't accepted.
Real professor 🌺
Thank you😍
great explanation ...
answer at 9:33 is orthogonal basis not the final answer.
Good for pencil and paper calculations but i need better procedure if i want to write program for orthogononalization because of numerical reasons
Am sure it will also help me figure out
Thanks dude
Thank you so much for this
thanks for this video
I have a final in 2 days, learning this for the first time😈
Thank You.
At 10:06, U2 1st component and 3rd component is not coming out as 1/√6. I get 1/3*√3/√2 = √6/6 .
√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.
is there dot prod mult for TI-84?
Thanku sir....its soo helpful
Thank you so much!
Excellent as always Dave.
Wowwwwww!! The procedure is that too simple
An amazing and clear breakdown of the process, thank you so much
Are the starting basis vectors assumed to be sorted by magnitude? If not, does that mean v1 is arbitrary?
I guess so
answer for u3 on the trial question is wrong
great content, thnx
Thanks
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.
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.
very much useful
Thankyou Math Jesus
nice cut
Grateful from VietNam 🤩🤩
Why 1/√6 8:15.
Bro how did u take u1. U1=3