Where can I download these videos in 480p? I checked the ocw website, the download links are for 360p or less quality. And, thank you for this awesome course.

That just also removed many comments under the original video.

Sounds like a git commit message

@@biao9957 😂

@@shubhamtalks9718 youtube-dl

Lecture 18 has 200k views. The fall is still big, but I believe this lecture has 20k for the last 6 months after the video was replaced.

Hello, this video replaces one from 2009 because the original video had a delay problem, if you look you can see that it was posted only recently in September of 2019. So unlike all the other videos in the series, it didn't have 10 years to accumulate views, but to be honest, seeing that 20 000 people have come this far in the course in only 6 months is quite astonishing!

chill dude why does it matter

calm down bruh, we get it. You're studying lmao.

Well as long as students have to deal with linear Algebra, they will meet Gilbert Strang - lol.

This should have more likes

@glyn hodges no you wouldn't wanna watch pythagoras's videos

His legacy is already tainted from being a Nazi unfortunately

Yeah ehm, he ruined it all by himself.

Thank you, i was really thinking how he arrived at R.

Thanks a lot!!!

Won't we get the transpose of R then? The elements will be q1ta1, q1ta2 and so on, right?

you are the savior of all mankind

Q^T isn't quite the inverse of Q. Q may not have an inverse (if it is not square) instead think: A = QR (multiply both sides by Q^T from the left) -> Q^T * A = Q^T * Q * R -> Q^t A = R ( because Q^t Q = I by definition of Q) From here you can see that R is just the dot product between q1, q2,q3 ... and a, b, c... So, q2 dot a = 0 (because q2 is just b - the component of b in a's direction)

Αλέξανδρε το πέρασες τελικά?

@@theodorei.4278 Δυστυχώς οχι,το ξαναδεινω φέτος ομως σε συγκριση με τον καθηγητή του πανεπιστημίου έμαθα πολυ περισσότερα απο αυτά τα βίντεο. Edit: Για λιγο δεν περασα,0.25 που ηταν να γραψω 5 μοναδες παραπανω στο τελικο, ανεξάρτητα ομως με βοηθησε πολυ

​@@azrael6882 Σπουδάζεις μαθηματικά, φυσική, μηχανική ή επιστήμη υπολογιστών;

@@kkounal974 επιστήμη υπολογιστών

its been a while... did you end up passing your final??

calm down girl hes not interested 😂

Only to finish him of "maybe thats what Schmidt did - he, brilliant Schmidt - thought okay, divide by the length. Okay, that - is Smiths contribution :') Absolute favorite part.

@@jan-willemgmeligmeyling8393 Schmidt was a Nazi sympathizer.

2023 still here😆

University was always 95% self-education. That's not really what the professors are there for.

got my degree from there but i disagree. you have plenty of great mathematicians in other universities too

you are supposed not to know the determinants

@Avi Aaron if you are refering to the Q from the top it is indeed a rotation

It's both. The cool thing about math is that you can think about things in different ways.

The readings are assigned in: Strang, Gilbert. Introduction to Linear Algebra. 4th ed. Wellesley-Cambridge Press, 2009. ISBN: 9780980232714. See the course on MIT OpenCourseWare for more info at: ocw.mit.edu/18-06S05. Best wishes on your studies!

I also have the same query??

He place it wrong

actually, since the A here is treated like a n dimentional vector, i.e. a nX1 matrix, where the properties of vectors and matrix algebra is the same.

Since A' just a row, A'b a scalar, the switch is fine

Multiplying by transpose A is the idea that was used in projection when vector b is not in the column space of A. Also, Transpose A * A makes it a square matrix which has unique solution or infinitely many solutions.

A further elaboration of 강병우 's comment. Ax = b has no solution when b is not in the column space of A, meaning no non-zero combination of columns of A gives b. The best solution \hat{x} expresses the projection of b onto the column space, p = Pb in the column space, so that p = A \hat{x}. This means b is actually decomposed into p and the error vector e in the subspace that is perpendicular to the column space of A (left nullspace), i.e., b = e +p. This error vector has been dropped, but its length is minimized. 'Best' is in this sense. As the error vector is perpendicular to the column space, A'e=0-->A'(b-A\hat{x})=0, thus we have A'Ax=A'b. Recall the projection matrix P = A (A'A)^-1A', which has an inverse when A has independent columns. Alternatively, multiplying A' to both sides of A\hat{x} = Pb, we also get A'Ax=A'b. To solve this new equation, just use the inverse of ( A'A), so that \hat{x} = (A'A)^-1A'b.

@@angfeng9601 that looks like some serious LaTeX lover!

When A has independent columns AtA is invertible , actually the nullspace of A is identically the same as of AtA

How would that work? Why project first?

It doesn't matter because A is a vector so (A^T b)/(A^T A) is a constant

Placement of A does not matter in case where A is a vector, but does matter in case where A is an nxn matrix. In latter case, projection matrix is P = A(A^T A)^-1 A^T, where projection matrix P acts on some input (vector b, for example). Full illustration of P acting on vector b is: A(A^T A)^-1 A^T b

Thanks!

Oh, he did.

We know that we choose 'A' (the first ORTHOGONAL vector) as 'a1'. Later, we see that 'q1' is just 'a1/||a1||'. This means that 'q1' is the unit vector of a1. Now, in the equation, the first column of the LHS is 'a1'. Similarly in the RHS, the first column in the first matrix is 'q1'. Since q1 and a1 are the just same vectors with different directions, all you need is q1 to be scaled by a constant to become a1. This means that the q2 vector is not required. Not required naturally means the coefficient is 0 (remember that a1 and q2 are orthogonal vectors and their dot product gives zero ). Since the second matrix in the RHS is a 2x2 matrix, the cell at row =2 and column = 1 is 0, the matrix becomes upper triangular. That's all.

@CoeusQuantitative SO If I suppose a,b are basis of matrix A, then the error part, e, is a linear combination of a and b. that's why column space of Q is same as column space of A. Is this what are you trying to say?

@CoeusQuantitative ok, wait, I am figuring it out..... anyway I am really appreciate your replies. Live long and Prosper.😄👍🙏

@CoeusQuantitative what a great picture!! I finally fully understand the picture. The problem was I was confused vector 'b' becasue of merely overused terminology........ In the previous projection and least sqare chapter, vector 'b' was out of C(A) and we tried to project vector 'b' onto C(A). That is Ax=b, the right side. However, in orthonomal and Gram process(pf.Gilbert doesn't like SChmidt so...😅) chapter, vector 'b' is just one of the basis of matrix A. That's why I was so confused 😥. My thought was the error vector 'e' should be in the left null space, and how can vector 'e' be in C(A)?? Just using a1, a2, a3 would be much easier to understand....I figured this out by solving some questions in the text book. And I watched your video, that was awesome. you made it for me😉. I want to say thx again.👍

the vector 'b' in orthonormal bases chapter is the one which is in the right side of eqation. It's not the basis. Be careful guys.

From the lecture on projection: for projection in two-dimensional space, you use the following formula: A * A^t / (A^t * A) * b, which is the projection matrix times the projected vector b: p = P * b. You can rewatch the video on how he derived it. Basically, you find out what is x(hat), substitute it in P = A * x(hat), and you'll get P = A * A^t / (A^t * A) for the two-dimensional space.

@@TheSalosful not quite

Good point. Thanks

But in higher diminsions, how do you gaurantee that the cross product gives the vector that is in the space spanned by the 3 vectors that we started with (a,b,c).

As he had showed initially Q transpose times Q = Identity and if Q is square it will have inverse (Since all columns are independent). So we can show Q inverse = Q transpose. Now A inverse times A = A times A inverse = Identity If A is an invertible matrix. Using this we can show Q times Q transpose = Q tranpose times Q = Identity. When Q is not square then we don't have an inverse. So this is not possible and we are left with Q times Q transpose. Correct me if I am wrong!

@@helikthacker413 It seems right in this formula way, but when I wanna explain it in multiplication way, sth is wrong there... When Q times Q^T we got the (1,1) item in the result from the row1 in Q times col 1 in Q^T, it desnt get 1.... Hope you could help with this puzzle. Thx

@@dianel.9238 What Q are you talking about? A square one or a non-square one. In square taking a few examples you will get Identity matrix. But in non-square it is expected that it will not be an identity matrix and so you wont get a 1 in diagonal. This is my understanding, please correct me if I am wrong.

@@helikthacker413 It is the columns in Q are orthogonal ,right? not the rows. In this case, the (1,1) item in Q*Q^T is the result of row1 in Q times col1 in Q^T ,ie row1 in Q... how come that gives 1 (in this case square or not doent seem matter)?

@@dianel.9238 kzfaq.info/get/bejne/ZrOkqtSZrt3Ion0.html This shows if Q is square, Q^T = Q^-1 since Q*Q^T = Identity. So after that it is clear that Q^-1 can be left multiplied or right multiplied to Q to get Identity. I don't know the proof but taking any Q and right multiplying by Q^T gives Identity.

to make column vector orthonormal.

Because he wants it to be an orthonormal matrix. Which means all columns are PERPENDICULAR to each other and of UNIT LENGTH.He shapes his matrix in such a form that the 2 columns , when dotted with each other ,return 0 , which is what we would expect of perpendicular vectors.However They do not have unit length. The length of each column vector would be sqrt(1²+1²) = sqrt(2). So if we wanna "normalize" the vector , meaning turn it into a unit length vector , then we just divide by the length. Since everything iin the matrix is divided by the same number , the dot product remains 0 (you can just factor the divisor out),and now the column vectors have length 1

@@DeadPool-jt1ci thank you so much! This is helpful!

Unit vectors | Matrix transformations | Linear Algebra | Khan Academy kzfaq.info/get/bejne/oreeasmc2MeooWQ.html

Converting to unit vectors(it’s already 4months I hope u get it already lol

To normalize

It is said A^T*Q=R, but it should be Q^T*A=R, am I wrong, is that what you are saying?

*65

what dont you get? a lot of this is computational.