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8. Solving Ax = b: Row Reduced Form R
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- เผยแพร่เมื่อ 17 ส.ค. 2024
- MIT 18.06 Linear Algebra, Spring 2005
Instructor: Gilbert Strang
View the complete course: ocw.mit.edu/18-...
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8. Solving Ax = b: Row Reduced Form R
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oh.my.god. he is just TOO PRECIOUS!!!!!
every time he starts with his way too excited "OK!" I actually get excited too :)
Thanks Dr. Strang!
Excellent quality of teaching.. really appreciate the initiative taken by MIT to make these lectures available to everyone across the world for free
abhishek G.Y
MIT!! 🇺🇸🏆❤️
i got so invested in the lecture i was replying whenever he asked the class a question
Anxiously Waiting at the end for him to switch the r=m
Haha same, I get to the point where I'm like, "you heard me, right?" 😂😂 These lectures are just too good!
so did I
good one, made me laugh ^_^
Prof Strang is a blessing to humanity!
@@Parkertannerz we haven't really seen the messiah, but we have seen this man teach with all the passion in the world. Professor Strang's lectures are nothing short of blessings
to be honest, the blackboard is better than my uni
Mine either !!
Well can your uni represent four dimensions?
It’s better than mine too
You can see how passionate he is to the idea of making everyone see linear algebra as intuitive as he sees it. This is what being a professor should be all about. Absolutely fantastic.
Nice description 👍
I love his blinking single eye, literally. this man is great.
Xiang Zhang double the speed, double the blinks.
I don't think I'd pass linear algebra without your lectures
Reviewing these in grad school since I've let my linear algebra get rusty. I picked up the book from the library. I wish my linear algebra class was this good. I just purchased his "Linear Algebra and Learning from Data" as soon as I saw that there's also lectures
"lemme give an example, okay BRILLIANT example" That "brilliant" always gets me
Thank you professor Gilbert Strang and MiT for this absolutely fabulous series on linear algebra, for the benefit of students all over the world. This, for me, is USA at its very best. Greetings from Trondheim , Norway.
I"ve watched the 18.06 from lecture 1 to this section, suddently realized that all contents were in Prof. Strang's brain and he didn't utilize any auxiliary materials/tools to control the rhythm, so amazing... Thank you Sir for your great teaching!
i am working real hard on this course carefully reading the book and doing all the exercises hopefully i can become a real linear algebra professional thanks to professor Gilbert Strang
all the best Sir!!
Prof.Strang is very very cute . I love how he teaches . I feel a warmth in my heart seeing how he moves around the classroom trying to teach the students .It often feels like he is trying to teach himself . I can connent with him easily . It's so AWESOME!
all points of the nullspace are shifted by the vector x-particular
A true informative lecture by a brilliant professor! Can't say how much I enjoy the way he explains every concept; so simple, so elegant!
He is working magic here in his lectures. I am experiencing one of those "math is beautiful" moments with these concepts and ways of thinking about matrices.
I`m learning English watching these incredible classes, thank you so much Teacher Strang and MIT
Thank you Dr Strang for the amazing videos. I love the way you explain things so simply. Most of my teachers at University fail to explain simply things. They go into big terms while the basics is tough. With your videos I barely attend my classes but I am able to nail my exams. Thank you again!!
Thank you so much, what an amazing lecture! 23:50 - 24:12 is definitely going in my notes word by word!
Watching these in junior-year ME is incredible. He essentially condensed the past three years of studying into one simple mental picture.
@@brandonlawrence5851 sameee
this lecture really puts together the concepts of pivot ranks, null space and special solutions, and complete solutions to Ax=b. I think we've also covered row reduction to get the augmented identity matrix to find an inverse and transpose.
This video was recorded on a Saturday and I am watching it on a Saturday. The Professor caught me off guard when at the end, he said, "have a great weekend. See you on Monday"
These lectures are addictive.
There is an important piece of information missing in the lecture; The fact that any particular solution p_i is reachable from another particular solution p_j + some combination of nullspace. Now one can claim that all solutions are included in x_p+x_null
So basically, professor here assumed a scenario of just an unique particular solution? Please confirm!!
@@yt.abhibhav Yes. That's what professor said in 11:53.
You can set whatever you want in free variable, but setting them all to 0 makes the following computation easier.
? Ax = b has only one ONE particular solution given all free variables are set to zero.
The particular solution graphically is the vector joining the (0,0,0...) to the point where Ax = b is satisfied but all all free variable components are zero
The null space is then defined graphically by finding the ratio of each free variable with every pivot variables and finding the line which has all the same ratios with respect to variable magnitude
In this way we find n-r lines for n-r free variables
the linear combinations of all those lines give us the whole "plane" where Ax = b
The particular solution isn't unique; it is the vector found in one step of this method to find the complete solution to Ax = b
all other solutions p_j like you say are indeed x_p + x_n
Dr. Strang thank for helping me to relearn linear algebra and it's important concepts. When I took this class at the University of Maryland Baltimore County the professor did not care if I learn it or not.
One of my favorite MIT professor. May Allah Almighty bless with long life.
Amazing. Think about it, xparticular plus the nullspace gives final x in r4, is the same as y=a*x+b in r2! This really blows your mind. Great class
The taste is different with MIT people! You "feel" the science with them! I envy the ones who had the opportunity to sit in these class rooms!
omg that resonates with me so deep!
45:00
in the case r = m < n, then
rref (A) = [ Id. F ] , if we allow permutation of columns.
This part of the lecture is fantastic when he overwrites I and F!!
I'm proud to have found out that he had made a mistake before Mr Strang. It means I've actually undertood the previous lecture (which I had to watch several times to grasp)
He is able to be so wonderfully passionate because he cares that his students understand. So he is able to be in a one-on-one framework while standing and talking to an entire class.
he's a great professor. very rarely you see such a talented teacher in math.
Thank you Prof. Strang & thank you MIT.
Brilliant lectures, much appreciated.
46:37
I just record the timestamp in case I forget the conclusion of this great course in the future.
I actually passed my 3 Applied Linear Algebra courses with flying colors by just remembering all these results because the lecturer never tried to explaine what's happening behind the scene. So after almost 20 years I'm still here trying to figure out what's what. I shoulda gone to MIT.
Its amazing how the previous recitation perfectly syncs with the lecture at 16:17
this lecture NEVER GETS OLD.
랄뚜기가 왜 여기에..
thank you so much professor Gilbert Strang. You literally make my day.
Best way to learn -> Understand and understand how precious education opportunity we all here get just by generous decision by MIT and prof Strang. We never pay any(even ads), we even can do repeat as much as we want and stop for a while to think or write.
I do not know how to say thank you. I started loving and understanding linear algebra from when i started watching your lectures. Thank you so much. I really appreciate it. wish you the best.
Every time I see one of Professor Strang’s lectures I run and get my old Finite Math with Calculus book just to see how much I’ve forgotten 😉 He’s awesome to listen too🙏
It's so weird these videos are from 1999..
I was in elementary school learning the multiplication table in 1999.. and now I'm using these lectures to help my pass my University midterms.. woah
And here I am 10 years later trying to pass my finals. Btw how did they go?
@@ahmadabdullah262 Wow, I have no memory of those exams. I ended up dropping out of engineering after a couple years - it wasn't for me. I did some travelling, and went into architecture. Ten years later I just graduated with a master of architecture degree, and I'm transitioning from architecture into product design. Funny where life takes you... I have NO IDEA what is going on in this video
@@LukieRawr I changed from architecture to computer science. Guess you never know where life takes you huh
@@ahmadabdullah262 So how its going for you mate?
He always asks permission from some unseen student in the sky that only he can hear: “can I give an example here? OK, can I try this?” It’s a pleasure to take his course.
Gilbert Strang - a legend in the making...
Guess he's already made as a legend...
thank you MIT, all these linear algebra lectures are very helpful, and Prof. Gilbert Strang teach very well
thank you very much!!:))
21:47 those blackboards are very nice actually.
I also watched the Stanford physics lectures (by professor Leonard Susskind), and they also have very nice blackboards (whiteboards actually).
I found that top universities always have excellent blackboards, because those good professors always prefer to do the demonstrations and calculations by themselves on the board.
My uni was very mediocre and so were the teachers. after a whole semester they seldom wrote anything on the board, all they do is repeating what's on the textbook and playing the powerpoint with most calculations and results already written on it,
One exception was my Mathematical Physics professor, he was a brilliant guy and he always do the demonstrations on the board, and during classes we can often hear him complain about the size of the board and the poor quality of the chalk :)
The summary is brief and billiant!, Ax=b, cool!
So I have to draw a 4-D picture on this MIT cheap balckboard - Einstein could do it..:D
hypnoticpoisons I read this comment just as he said that!
Thanks a lot Dr. Strang, i'm really excited for the next lecture.
"ok.... diurghurghh uhg" - Gilbert Strang, 46:37
jokes aside, i love these classes, he is the best math teacher I have ever seen by a mile. Makes me really passionate about learning linear algebra compared to my uni classes
These videos finally make me realize that I wasn't stupid, my undergrad professor was!
On 20:30, Professor Strang mentions that X-particular is unique and can’t be multiplied by a constant (it is not a line space). So I went back to step one of X-particular where he said (12:00) - that setting all free variables to zero is one way of finding X-Particular. Why is X-p one guy if there are many ways to find different xp? thank you
Same doubt
You can specify the exact same set of "solution space" (i.e. set of vectors that solve the system) just by adding a vector from the null space to the Xp. The solution space may now look different on the surface, however rest assured that all possible Xp's are "reachable" just by adding a vector from the null space.
Bless this man and his family, he should be an idol for teachers
"What about the solution to Ax=b, what's the story on that one?" hahahah he's adorable
I should have been in his lectures instead of being 8 years old. 😞
Conclude the lecture: The solution will exist only when rank(A)=rank(A|b), which would be x=x(particular)+x(nullspace), x(nullspace) exists only when rank(A)
Here's something they didn't quite spell out to me in my Linear Algebra class:
Let E be the product of row operation matrices. Let F be its inverse, FE = I.
Obviously if Ax = b then EAx = Eb (because f(x) = f(y) for all f when x = y).
But also importantly and relevantly if EAx = Eb then Ax = b:
If EAx = Eb then Ax = IAx = (FE)Ax = F(EAx) = F(Eb) = (FE)b = Ib = b.
That is, divide by E (i.e. multiply by F) on both sides to recover Ax = b.
This is the justification for the algorithm: Compute E * [A b] where E is chosen such that solutions x can be read off easily. Solving E*[A b] means x solves EAx = Eb, but then Ax = b.
So the outputs of the algorithm are in fact solutions to Ax = b.
At 43:32, shouldn't Professor Strang have put *infinite* as opposed to *1* solutions to Ax = b. When r = m = n, all n (or m) - dimensional vectors can be obtained from linear combinations of A's columns because the column space is also n-dimensional.
"Me and the guys in the nullspace"
I have fallen for Dr. Strang. He is actually Dr. Strange. 💗 I don't know how he gives me some feeling through these videos. Only kudos....👏.
"...giving other self-learners in the world opportunities of learning from the best."
True enough. Of course, it would also be nice if they would release a few more lecture series from other subjects. I'm spoiled now... it is so frustrating walking into a course without already knowing it forward and back.
;)
Dr. strang is too energetic!!!
I nod my head to the screen like listening to rock music, while I am not.
Great videos! the only thing that i'm missing is "to rise my hand and ask him some questions"..
I really like his teaching. Perfect.
I teach some electronics courses at a local community college, and I am watching these as a refresher on linear algebra. I am taking notes on Dr. Strang's teaching style!
23:57 What a genius. I love this lessons.
"Einstein can do it." I love it when he said that :)
This information gives solution of pulling up and pulling down gravity operated on two planets orbiting the orbit bt its mass difference operated on 60 degrees in between them when the mass difference is much higher side. Now the matrics giving more informations by column vector by elemination.
When he says *OK* you know it's about to get good 😂😂
Thanks (2 years after) because I didn't understand it. Now it's obvious :D !
I think the main point of this lecture is not to find the solution at all, but to know when you can find it. So you can watch in perspective and answer when you can and when you cannot to find it. That's the reason for what he did'nt begun by putting numbers as results of the equations, but with literals, meaning they can be any number and the answer to that system just can be found if those numbers meet the condition of beeing a mix of every column in the matrix.
Why is the nullspace(A)= *0* ?
1. the reduced echelon form of A is R
2. From past lesson RN = *0*, where R = [ I F ] But here R = [ I ]
3. As there are no free variables, then I N = 0 implies N = [ 0 ] !
So, interestingly, for full rank matrices, there are no solutions coming from the null space (just c[0] i.e 0)
33:07 dude, there's something moving under the professor's desk.
I believe it’s vector trying to escape from the desk space.
At 28:11 ,How can we assume Nullspace to be 0,if free variables are 0,in Reduced echleon eqn=[-F I] to find nullspace F=0 as free variable is 0,But I can still write 1 in place of I to get Nullspace non zero???
Ayush Kumar I wondered the same, however I realised you are choosing the free variable to be the I. If you don't have a free variable then you can't choose that I in the first place. I would suggest only follow that template if you have free variables.
35:17 The lovely sound of science
rank tells you everything about the number of solutions that number of rank all the information except the exact entries in the solution
He got me when he said: "can I just right [ IF]"
21:55 My brain stopped working when he drew ℝ4 on 2 dim board! Woah!!
sorry....but really not sorry 38:32. Thank you Gilbert Strang!
i'm wondering that when he draw the picture of the complete solution , why doesn't the Null space pass the origin ?
thankyou professor!! thanku MIT!!
oh god, prof. strang is so funny. he is acting natural. like him so much.
I’m confused about how he explained adding a x particular to nullspace will not give a subspace. I think it this way. Nullspace should always contain the origin, so in this case it is a plane going through the origin, since x particular is a vector not in that plane, adding them together will never give a zero, so it is not a subspace.
At 36:40 every right hand side 'b' corresponds to the ones within A's column space, right?
When he plots the solutions x in R^4 at minute 23-24, he plots Xp, Xn and Xcomplete=X. Shouldn't Xn be drawn through the origin and Xcomplete be drawn away from the origin and at the 'plane' without going through the origin?
+Hiten Chojer The plane that he drew on the board wasn't Xn, it was Xcomplete, which is Xp+Xn. It's like Xn that has been shifted away from the origin by Xp, just like he said..
+Adam Levin You are right. I had the same thought as Chojer initially. But the man eventually made it clear that the plane is Xcomplete which is the Nullspace being shifted to go through Xp.
At 45:20, if we are talking about null space then indeed there would be an infinite number of solutions, but if we are talking about Ax=b then there needs to be a particular solution in the first place for the infinite solutions to be existing.
Am I right?
Language of column space , OMGGG
At t = 11:00, finding the complete solution to Ax = b.
I LOVE YOU MAN
when r< m, n, i.e. the last case in the summary board, can it have exactly one solution?
Xparticular is calculated by setting all the free variables to zero. But in case, when rank=n, there wont be any free variables. How to find the Xparticular in this case?
+adarsh k murthy I guess that when rank=n, there will be either one solution or NO solutions. In the case of one solution, Xcomplete = Xparticular.
+Sam Mao thank you very much
+adarsh k murthy
yes either 1 solution or nothing,then we go for solutions with errors-as in next lectures
The number of free variables gives you the number of dimensions of the solution. No free variables is a zero dimensional solution which is a point. Just for anyone reading this really.
Thank you MIT!!
Everything ok sir it is being thought during my school but where to use or apply this maths in sceintific or commuter making softwares.
Thank You, Prof Strang.
Thanks a million times!
Did he choose the b vector (1,5,6) at random, based on 0 = b3-b2-b1? EDIT: yes
Einstein couldn't do it, but... Gilbert can!
I can't understand why A(xp + xn) = 0 are all the possible solution we can have multiple xp how can we show that we can't comeup with an xp which is not xp+xn
Can’t I also choose other values for x2 and x4 to find other particular solutions? Would they not also shift the null space to another spot and cover regions which were not already defined? Or does the particular solution + null space contain the set of all possible solutions for Rx=b?
i believe it contains all solutions since if Ax = b => Ax + A(any vector in null space) = b + 0 = b , and since the null space contains all such vectors, any other solution you obtain would be this x + a vector in the null space.
God, I love this guy!!!!