So, a standart school second language class?

Except if high schooler taking Linear Algebra course

Wow, that's deep!

Not communication.. It's connection

Wow, that's exactly what is like, very giod analogy

This video is life changing for me

haaah 6 years since i left my 12years of education. And whenever I feel like something interesting to watch I occasionally come to this channel.

I doubt it took you 20 years to understand matrix multiplication

Then congrats, I must say?

cool

Absolutely.

There is now "Thanks" option in every youtube video. Donate some amount !

rly? i didnt understand shiet what his talking it was so fast iam still trying toprocess what he said in the beginning like wtf

@@simply6162 This is a great benefit of the video. You can replay it. Pause it. Think. Repeat until you understand. Can't do this in a lection!

Same

That's because engineers think "what can we use this for" instead of "why does this work or what does it mean".

Well as an engineer, I really, really wanted to know this, but you know, sometimes teachers are not good and when I asked how did anyone found out about matrices, their properties, how they work, why they work, etc. the teacher could not answer anything concrete, and the book on this also was really confusing and vague.

I have A BSEE and I am really loving this. My Linear algebra courses consisted of Appendixes in the back of my text books that summarized linear algebra in 4 pages. I never actually understood any of it but had lots of disconnected factoids about linear algebra.

@@brentlocher5049 true2. I know how the numbers add up. But never why, no fundamentals. Shear and rotation are new to me.

Personally, I find it a bit sad that engineers often learn math for 'practical application' without actually understanding the math. This is coming from a 3rd year engineering student.

Thanks for adding this, it's a really nice way to phrase something I should have communicated better.

+3Blue1Brown No no, thank you for doing such a good job in making these! I saw someone on my Facebook who just graduated from software engineering link this with the caption "I learnt more linear algebra in 30mins than I did in 5 years university" and decided to give them a watch. Admittedly, I already know this stuff, but these are entertaining to watch in and of themselves just for the gorgeous animations ahah if only you'd surfaced last semester when students in my mechanics class were like "What does a matrix have to do with stress transformations" and our engineering prof was like "EVERYTHING! YOUR MATH PROFS DESERVE TO GO TO JAIL" xD Keep up the awesome work!

I disagree with this explanation. You say, "Of course, the overall effect of BC is equivalent to applying C then B based on what he explained earlier in the video." Unfortunately, I can only assume you mean the section from 3:40 to 3:50, but in that very section, he simply drops some parentheses as if associativity has already been proven. So the argument, as far as I can tell, remains circular, and is not fixed by your comment.

By the way, I'm REALLY enjoying this video series. But this was my first major disappointment, with the claim that "This seems like cheating, but it's not; this is an honest to goodness proof." Actually, I think it is cheating. :( That's one of the dangers of visualization, not noticing the hidden assumptions necessary inherent in making the visualization in the first place. Visualizations are WONDERFUL for intuition, but can be very tricky as a deductive system. If this associativity confusion ends up being a circular proof (as I claim above) I hope that the end of the video can be redone so as not to misinform your (impressive number of) viewers.

He doesn't drop the parentheses to imply any sort of associativity. In that part of the video, he's actually still trying to define matrix multiplication. Essentially, he says that matrix multiplication should be defined in such a way that if you want to multiply two matrices A and B, then the resulting matrix C should be the one that transforms all vectors v the same way that A(Bv) does. i.e. we look for the matrix C such that A(Bv) = Cv for all v, in which case we say AB = C. By defining matrix multiplication this way, there turns out to be one and only one way to multiply matrices algebraically, and it's that funky little dance that you learn in a linear algebra course (proving this is not so easy). Using this definition, the proof of associativity is as straightforward as you saw in the video.

❤❤

learning through a personal tutor with an understandable answer to nearly every question is what we'll enter soon with AI

@@orang1921I hope AI will use that video as a basis for teaching

@@orang1921 ChatGPT is already helping me understand math much more than teachers ever did in school or university...

Yes....same thoughts. He is a perfectionist.

I came to the comments section after seeing this genius at 4:25, ad hoping to add the comment if I didn't see it. Also, the attention to detail added that Composition first has the ~teal of Rotation, which is the first transformation and then the pink of Shear. Just genius.

also a rotation and another rotation

Khan is a shill. This on the other side, is quality.

No doubt 3B1B was far too good to be compared with Khan. 3B1B is quality. Khan is quality.

Calxius those are strong words. Explain yourself.

It says in the videos thiss guy created the calc lectures for khan so lets take it easy. Khan is the man, hands down. Of course there will be people who can explain the information in a more digesable way that allows for deeper understanding, such as 3blue1brown, but Sal Khan has put out mucho content on his own that has gotten hundreds if not thousand of people through the first 2 years of their STEM degree. He even tries to relay a more intuitive undestanding as well.

right, as I say below khan has personally helped hundreds if not thousands of kids through their first 2 years of a stem degree

This is a great explanation, I was kinda stumped over the way he "proved" the associativity rule, but this way got through to me. No disrespect to 3b1b's explanations, they're great, but the way he said "C... then B then A" when referring to (AB)C felt a little funky.

Great explanation, I was also a little confused by the way it was explained in the video

Apparently this is a great explanation. It definitely doesn’t help me. Composition???

I don't know, I personally think there's still something missing from his proof as there is in this one. Say AB = Q. It seems a bit of a jump to me to assume that applying Q after C is equivalent to applying A after B after C. I feel some proof is missing from this. Similarly in the above proof, it seems like a jump to me when you assume q(c(x)) is equal to a(z(x)). If anyone has any way of explaining these gaps I would love to hear!

The transformation is always happening right to left , so the way I understand is (AB)C is apply Transformation C then B then A; even A(BC) says apply C then B then A; the brackets only change the order of multiplication but we are not changing the order of transformation

you are lucky!

Some really do have all the fun

Doing the same

I am doing it right now and I hope this will be of much help when I start my linear algebra course in a few weeks :)

@@3drws314 how did that go? I am doing the same thing.

maybe math is not so wanted content these days :)

Because the majority of the 7 billion people dont care for understanding the universe(through math in this case) and care about primal instincts like sex, food and money more. They are on a lower level of Maslow hierarchy. If at least 20% of people really cared about science we'd be on Mars and Titan already.

It is quite well known among math majors and math grad students, as far as I can tell.

Jo Kah he has 2.42 million subscribers, that’s a lot compared to other KZfaq channels. I’d say he is doing very well

It has 1.5 million views.

real and alive

I had to disagree with that part actually. I came back to this video to see how he proves associativity again and realized he just said the translations are in the same order. That's just explaining what associativity is. lol That's what we're questioning and seeking to actually prove. Obviously you can imagine translations in the same order. That's not what we're asking. We're asking whether it's associative. Which is about different orders. Only A(BC) is "C then B then A"... (AB)C does "B then A" first (producing a whole new translation) which means we're doing the math in a different order, hence the word "associative" exists. Because we're smart enough to realize some things may not be. Or we can all act like everything's associative just cause we can imagine them being the same order we want them to be. lol With something like associativity, this is the pure example for symbolic proofs. You can't rely on your "good explanation". You wouldn't even begin to start explaining something until you've actually proven it. If you're someone who likes to actually understand, you want solid proof. You don't want some simple "it makes sense so just accept it" so called "proof". And then you look back with hindsight and tell others it just makes sense and ask if they can see why. lol

A(BC) is "C then B then A" is the same as (AB) C because you're doing "B then A" then add C to the front ending with "C then B then A"

Exactly at 7.37 you told that ""Take a shear which fixes i-hat and smooshes j-hat over to the right and then Rotate 90 degrees"" At 7.45 you first did the shear fixing i hat and rotated 90 degree. clear and perfect... But at 7.55 you first rotated 90 degree and took a sheer ""Fixing J-HAT"" instead of fixing i-hat. So,you end up having different results. I want a clarification whether that's right or wrong??? Thank You.. By the way you are the best in the business for explaining mathematics.

@@DlcEnergy there is another comment regarding this which actually makes this proof quite rigourous. You may like to check that.

@@nikhilnagaria2672 i'm interested. can you link me to it?

I know right!!!! This gave me the intuition behind all of matrices.

This is a million times more interesting than just learning the formula.

"I've never been thought" What an ironic typo.

+Zardo Dhieldor ooops! thanks for spotting that one out. =)

Jérôme J. This should become a new figure of speech: "being thought sth." Only, I don't know what it would mean.

"Muuuahahah... Try *that* with us, you fools!!!"

+SafetySkull Atleast when the right matrix is non-squared, you can consider it as just a (linear transform) x (vector) multiplication. Giving you a transformed vector.

THIS, I should've scrolled down earlier, I was so confused about vectors losing dimensions, unexplained by any simple geometric transformations.

non-square matrices can be represented as square matrices by substituting 0's on missing dimensions - the video still makes sense on that count

AB,BA are only possible only when B,A are square matrices

When you get it for the Basic vectors [1 0] and [0 1] it should be easy for every other vectors. Its a rotation of 90° and then a vertical flip. Two visualize use your fingers: Raise left hand. Index finger up. Middle finger to the right. The transformation M2: Middle finger must be where the index finger is now. Index finger must be where the middle finger is now (and must double - but let's just imagine that). We can do this by lifting the elbow to the left. This was the M2 transformation (for the base vectors) No matter where your fingers are now - if you make this rotation (and imagine that your index finger becomes twice as long) then you perform the M2 transformation.

Bruh !!! ... I watch you videos. Glad to see your interest in maths.

5m subs :0

MAGMA?!?!?!?! WHATTTTTTT

Lm

YES FINALLY

lmao

Knew I would find this comment

Yeah! But it has its own disadvantages. Like I always read comic styled meme in the same way it gets spoiled.

@@AnuragKumar-xh3wc a small price to pay for salvation

Very good point, I'll try to keep that in mind in the future. For many of the videos already made in this series, though, the green/red is already kind of locked in, and I wouldn't want to be inconsistent.

Additionally, It's worth noting that you've reversed the standard color conventions for X and Y. X is almost represented by red and Y is almost always represented by green. When in 3d, Z is usually blue. In fact, between various 3d applications, there's more agreement over this color convention than there is over whether Y or Z is the vertical axis. (And I maintain Z should always be vertical)

Z going into or out of the board/screen also makes more sense for Z-indexing and Z-buffering.

***** I'm not a board or paper person. I'm a 3d modeling and graphics person. Putting Z up makes the most sense to me and it's the convention used for aerospace engineering and 3d printing. This REALLY wasn't the point of my comment though. I was just trying to point out the standard color convention, which I feel is fairly important. The axis orientation comment was more of an aside.

I will say that I am red-green colorblind and I don't have any trouble with the colors in these videos. Of course, colorblindness has a lot of variation between people, so I'm not representative of everyone.

Thanks Noah! The song is just a short little made up thing. There's not really a full song to it, just enough to sandwich the videos.

Acttually, it's not a song at all, since those actually have words.

+John David moonlight sonata doesn't have words either

+Yunis Yilmaz We don't call it a song for that reason.

I second this +1. Kudus to the author(s) of these videos and I wish them the best

Yes but you need symbolic proof to ensure correctness.

@@purefatdude2 no you don't, symbolic proof can only be proven on case by case basis (2D), here is a better proof. I will use A' for the inverse of A (easier to type, A(BC)=(AB)C, apply A' on both sides, A'A(BC)=A'(AB)C, A'A=I, BC=A'(AB)C, apply B' on both sides, B'BC=B'(AB)C, B'B=I, C=B'A'(AB)C, but B'A'=(AB)', C=(AB)'(AB)C, (AB)'(AB)=I, for this to be true, the associativity rule must be true. It is consistent with 3b1b's transformation rule

@@tehyonglip9203 I wasn't referring to this specific problem. I was referring to mathematics in general.

Teh Yong Lip The symbolic proof isn't just taking 3 2x2 matrixes and doing the multiplication. You just take 3 generic matrixes for which the product is defined (the number of columns of the first must be equal to the number of lines of the second) and then apply the definition

@@tehyonglip9203 lol this proof even without thinking about not square matrices, existence of square matrices, that have'nt any inverse matrice is wrong Mistake is in first step apply A' on both sides and A'A(BC)=A'(AB)C is already wrong and the biggest mistake is in not wanting to write * operation symbol(or other symbol for multiplication of matrices) Because it should be like this A'*(A*(B*C))=A'*(A*B)*C ,but when you are aware of importance of order of operations you could write it like this A'(A(BC))=A'((AB)C) and if you don't consider this as mistake then next step could bring you this mistake even more A'(A(BC))!=(A'A)(BC) before you prove associativity of this operation, which you want to prove using this property And actually it is often bad to prove something with something you learn in future, because this smth could be proved or invented only because of true of this thing that you want to prove And I know that you wrote this 8 months ago for me, but what if someone would see your prove as most simple, when it is wrong in a lot of different aspects

haters, poeple who don't believe in math (i assume there are some; i know it's nonsensical though), frustrated content creators and/or jealous people.

@@zokalyx bruh what? nowadays there are some people who don't believe in maths? ehhh the world's getting crazier

Same here bro

you mean right to left

Well... Manga readers are a bunch of weirdos, so...

Any person who knows a semitic language: ........

@@smolkafilip Omae wa mou shindeiru just kidding

I think someone said he uses python programing language

Like saleem khatib pointed out, he uses Python to generate his graphics and animate them. Search "manim github" in google, or go to github.com/3b1b/manim He has not documented how to use his code, and it has no interface. It would require prior knowledge of Python to work with.

How are you still alive? :(

I could help you, with this series from start

I apologize for my ignorance but what do you mean with "He has not documented how to use his code"? Isn't it used the same way by everyone? Anyway thanks for the info, I'll get to it.

The order is the same for (AB)C = A(BC), you are applying first transformation B, then transformation A. It's just that in (AB)C you are computing first the composite linear transformation AB and then applying it directly to C, and in A(BC) you are applying first transformation B and then A to the resulting space. It's the same order of transformations, just different order of doing the operations.

"Grant's New Etude" by Vincent Rubinetti

i too have this exact same question. previously he sheared j keeping i fixed and second time he sheared i keeping j fixed.

In the second example, the rotated i-hat is in the original j-hat position, so the shear now operates on whatever is in the original j-hat position (i.e. shear operates on the transformed i-hat). The author chose this example to show that, in general, matrix multiplication is not commutative. You could work up another example to show commutativity in a specific case; e.g. scaling by 3 with rotation by +π/2, followed by scaling by 1/3 with rotation by -π/2.

vk2ig By the example you gave, commutative property does work in this case, right?

I'm tripping on this too, I had to transform it myself to really understand what's going on. To anyone confused about that part, remember that the second transformation basically transforms the TRANSFORMED î and ĵ, so not the original î and ĵ, let's refer to the transformed vectors as î' and ĵ'. When you do shear first on the original basis vectors, only the ĵ part moved while î' stays the same, so in this particular case only vectors that have value on the y axis (I know it's not really interchangeable but because the basis vectors lie on the x and y axis I'll just use them for simplicity) is sheared, so ĵ is sheared into ĵ'. But when you do rotate first, ĵ' lies flat on the x axis, while î' now have a non zero scale on the y axis. Because the shear transformation in this case only affects those with y value, î' is sheared while the ĵ' stays the same because it has 0 y value.

I tripped over this one, too. The thing with transformation matrices is, that they are only an array of components not including a vector basis. So the transformation matrices in this example always transform with respect to the "x- and y-axis", because we chose to represent our vector space in this representation. If you would use a transformation tensor, you would have a transformation with respect to a basis. In our case î and ĵ. So it would be clear, that M1 and(!) M2 are transformations with respect to basis î and ĵ. This implies that M2 (the shear) would always "shear" the components of vectors in ĵ-direction no matter how many transformations were executed before.

Sir,Did you get it?

Brother,Did you understand it? I am unable to understand it?