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