The Gram-Schmidt Process
ฝัง
- เผยแพร่เมื่อ 5 ก.ค. 2024
- We know about orthogonal vectors, and we know how to generate an orthonormal basis for a vector space given some orthogonal basis. But how do we generate an orthogonal or orthonormal basis given some other basis comprised of vectors that are not orthogonal? We can use the Gram-Schmidt Process! It looks tricky at first glance, but it's not too bad. Let's give it a try!
Script by Howard Whittle
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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!!
😂
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
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.
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.
your videos are saving me hours of studying time for my upcoming Linear Algebra exams 😭 thank you
Best explanation I saw on TH-cam. Keep uploading such useful content. Thank You.
Simplicity is a great demonstration of genius. Thank you!
Thanks, looked several videos on this but your explanation is the simplest.❤
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 :)
I loved how the explanation is so simple and clear, thank you professor :)
Prof. Dave is a genius in education. We NEED you!
Great explanation... Much better than my 1hr college class.
Keep Going🔥🔥❤
Terse and concise explanation! Thank you
this video is super helpful... thank you professor for your awesome explanation
High five Prof Dave. Thanks to u now am all set for my exam tomorrow 😊
Had to implement it in python, was very easy thanks to your tutorial!
Thanks a lot for this series!!
I haven't never seen the lecture like this.....awesome..
This has been really helpful to me. Thank you very much
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
bro in just one day I finished more than 20 hours lecture just by using your videos. thank alot man you saved me ❤
Thanks a bunch! Short and well explained.
Many thanks Professor Dave, it is very helpful
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
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!!
Very nicely explained sir thank you so much❤
Sir you are very understanding. Keep it up.
best and awesome explanation!!
You sir, are a LEGEND !
My wrist is numb just from working through both problems, but I understand it. Thanks Dave!
I will send the physio bill shortly.
Crystal clear explaination
These videos save me. Thank you!
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?
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
I'm paying $7k for the class I'm in but this is free and much clearer. Life is strange.
An amazing and clear breakdown of the process, thank you so much
Great explanation man
Dave is just that guy, he is that guy
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.
dude explanation is excellent
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 !!
Can you do a playlist on statistics? Thank you for time and effort.
Excellent as always Dave.
great explanation ...
Thank you so much for this
what is the expression for orthonormal basis? just orthogonal basis/length?
Thank you so much from Berkeley grad student
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.
simple and clear..👍
Good for pencil and paper calculations but i need better procedure if i want to write program for orthogononalization because of numerical reasons
thanks professor dave!
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.
this guy deserve more views
Thank you so much!
Thanku sir....its soo helpful
thanks for this video
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.
Is the answer provided for the exercise correct?
hello, what if i have 2 vectors that are perpendicular but one is not?
thx you saved my life
great content, thnx
This is sooo amazing. Thanks buddy. I would love to join the channel but my local card ain't accepted.
thank you very much
Are the starting basis vectors assumed to be sorted by magnitude? If not, does that mean v1 is arbitrary?
I guess so
very much useful
Thank You.
Real professor 🌺
Thank you😍
is there dot prod mult for TI-84?
Am sure it will also help me figure out
Thanks dude
Sir, great. They teach only evaluation in class
Thanks
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.
@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!!
DO U HAVE THE SOFT COPY OF YOUR PRESENTATION?
answer at 9:33 is orthogonal basis not the final answer.
nice cut
I think you just saved my exam 😂
Thenx
Shouldn't U3 be equal to (-1/2,0,-1/2)?
Yes they're the same
thanxs
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 ....
Bro how did u take u1. U1=3
Thank you so much Sir.../\
Linear final exam prep time
Grateful from VietNam 🤩🤩
Bestest video👍 May Allah bless you❤
Damn I just studied this 15 minutes before my test and guess what I could solve it :)
Wowwwwww!! The procedure is that too simple
Legend
answer for u3 on the trial question is wrong
I wish school could be like thing instead of the factory line approach it is 😢
Да,это жестко!
I have a final in 2 days, learning this for the first time😈
Perhaps it would have been better to explain with diagrams comprising vectors v1, v2, v3 --> u1, u2, u3 which leads towards building this formula.
ooooof professor dave is hotter with shorter hair
p.s his vids are really commendable!
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