Well said.

one is 9 yrs ago and the other one is 9 months ago lol @@tianjoshua4079

May glory come to him.

given how old he looks now I suppose you're right

zhen de ei

domain expansion typa shit

+YOUWEI QIN I thought the same thing. There's just like infinite sliding blackboards stacked on top of each other :)

quantity?

MIT has everything of a very high quality! Its such a pity that you only noticed the blackboards!

Agree, no body mentions the chalks.

lol

i was noticing them

Why don't we ask them?

Makes you feel like you're in the classroom even more...

Exactly haha they’re in most of the lectures

This is exactly what got me confused at first. I really appreciate professor Strang and MIT for making this gem of a lecture available online, but I felt he presented a few tricks like the block matrix one for finding the spanning set of the null space of a linear map (a linearly independent one*, too, because of that I block) in a rather hand-wavy manner. Perhaps a better way to visualize it is as follows: 1. Draw the RREF matrix as staircases with pivots, preferably with interlaced free columns for generality. 2. If there are any all-zero rows at the bottom of the RREF matrix, trim off that part. 3. Pluck out a free column from the staircase, then turn it sideways (90 degrees counterclockwise.) 4. Multiply each component in the free column by -1, to reverse their signs. (This is for building the -F block) 5. Insert the "selector" (coefficient of 1) component at the same index as the index of the extracted free column in the RREF matrix. 6. Insert "N/A" (coefficient of 0) components at the same indices as the indices of the rest of the free columns in the RREF matrix. 7. Now turn the free column back to its original position (90 degrees clockwise) 8. Put the finished column in the "special solutions" matrix. 9. Do the same with the rest of the free columns in the RREF. 10. In the special cases where F is NOT interspersed with the I in the original RREF matrix, what you get is a matrix with the -F block stacked on top of the I block. P.S. The point of plucking out a free column and then laying it on its side is to make the step 5 and 6 easier to visualize. *If only one column vector has a non-zero entry at a specific row index in a set of columns, then there is no linear combination of the rest of columns in the set that is equal to that column. That is why the special solutions matrix built this way always contains a linearly independent set of columns.

This really bothered me. The block presentation wasn't exactly blocked. But I'll stick with it.

yes, I'm a bit confused what to do with matrices which rreff is something like [ 1 * 0 * 0 [ 0 0 1 * 0 [ 0 0 0 0 1 they are clearly not [I F I will check this answer

Truly agree. It's quite impressing how MIT has so many sliding boards... the # of blackboards in MIT is INFINITE. LOL

readap427 Or Spanish film. Both from Latin

fin means the end

Indeed! It's like a story - with characters, and plot, and plot twists...Mr Strang is a shining example of what education should be - accessible, engaging and with the sense of disccovery!

Yes! this is what I felt as I was watching! And I felt that I was as happy as I would be watching a favourite movie.

definitely more than a couple. I dont know why people are saying its magical

26:31 even the ghost gets mindblowned

hahahahahah me with linear algebra 2 years ago. will get a 10 now easy

Any updates?

same issue with me

Dr. Strang wanted us to realize that the reduced row echelon form of the original matrix consisted of the identity matrix (when only looking at the pivot columns) and some other matrix, which he called F, when only looking at the free columns. He generalized this notion by defining the matrix R using placeholders I and F for the identity matrix (I) and the matrix formed by the free columns (F), with possible rows of 0s beneath. Since he was generalizing, he wrote R as a block matrix (where I and F represent matrices). We know I has dimensions r x r (since I is the identity matrix formed by the pivot columns, and the number of pivot columns = number of pivot variables = rank = r) We know F has dimensions n - r x n - r (since F is the matrix formed by the free columns, and we know there are n - r free columns). So our original Ax = 0 can be rewritten -- throughout the whole process of his lecture -- as Rx = 0. He then wonders what would the solution of this matrix equation would be. Well, since defined R generally using I and F, he unintentionally (I am assuming given how pleasantly surprised he sounded) was defining R as a block matrix, he decided to find all the special solutions at once in which he called a null space matrix N. This N would solve the Rx=0 equation, i.e., would make RN = 0 true. Well, knowing how matrix multiplication works, N needs to be a matrix that, when multiplied with the row(s) of R, would produce 0's. Since the first row of R is I F, what linear combination of I F would = 0? We would need to multiply I by -F, and F by I (because then we'd have -F + F = 0). This is how to look at it pure algorithmically. Dr. Strang actually uses wonderful logic. If the first row of R = [I F], then of course we want I in the free variable row (the second row of N) in order to preserve it, and to cancel it out, of course we would need -F in the identity row (the first row of N) in order to cancel out the F in the free variable block of R. This is how he knows the null space matrix N is always going to be [-F I] (obviously written as a column, but I can't type that out in this comment). He then goes further to show us how this actually is not surprising. Going back to Rx = 0, remember that R (as a block matrix) = [I F]. x = [x_pivot, x_free] (as a column matrix). If we actually did the matrix multiplication we would have: I * x_pivot + F * x_free = 0. Solving for x_pivot we get: x_pivot = -F * x_free So, if in our solution we make our free variables the identity (remember when Dr. Strang said "hey, these are free variables. Let's make them whatever we want. Let's make x_2 = 1 and x_4 = 0" and later he said "hey, let's make x_2 = 0 and x_4 = 1"), then by the above equation, of course x_pivot HAS to be -F.

@@toanvo2829 F has dimensions r x n - r (NOT n - r x n -r), no?

thank you so much for clearing this doubt.

Yes. So the dimension of the identity matrix in R is not the same as the dimension of the identity matrix in N. And the sum of the dimensions of these identity matrix should be equal to the number of columns in A.

lol would be annoying af if sittin behind em.

+hoan huynh See his department page for contact information: www-math.mit.edu/~gs/

Thank you! This helped enormously! And 9 years later!

I know!!😂 He's the best professor I know.