I am new to linear algebra. Sal says several times that it is useful in computer science especially in handling graphics representing 3 dimensions on a 2 dimension screen. But isn't this what Renaissance painters did when they used perspective . linear algebra was developed in the 19th Century. My question is were Renaissance painters using a primitive or proto form of linear algebra which is simpler that full blown LA?

Renaissance painters learned about and portrayed perspective by carefully observing the world around them and replication what they saw when they stood in rooms or looked at buildings. Computers don't have this observational power, nor do they have senses with which to process the world, so they need mathematical algorithms to function. Linear algebra lends numerical rigor to principles that Renaissance painters first began to understand. The concepts are very similar, it's just that painters speak in the language of shading and color whereas computers speak in the language of numbers and matrices. So while the concept of representing 3 dimensions on a 2 dimensional screen is hundreds of years old, the method of doing it using matrices is fairly young.

I am not sure if this was asked before or if it is executed in another video but how does this logic apply to when you are dealing with two non standard bases?

The logic is exactly the same, but you must be very careful to identify on which basis you are working on. For example, let's suppose you have two non standard basis: ``` 𝗮 = { a⃗₁ , a⃗₂ , ⋯ , a⃗ᵤ } 𝗲 = { e⃗₁ , e⃗₂ , ⋯ , e⃗ᵤ } ``` If you want to create the change of basis matrix that goes from basis `𝗮` into basis `𝗲`, you would need to construct a matrix whose columns are the vectors of the basis `𝗮`, expressed in the basis `𝗲`: ``` 𝗗ₐ₋ₑ = [ [a⃗₁]ₑ [a⃗₂]ₑ ⋯ [a⃗ᵤ]ₑ ] ``` (Of course, in order to get to this, you would first need to create the change of base matrices between the standard basis and each of your basis, in order to be able to express the vectors in each of them).

At 12:25 Sal says "you want to pick the right basis". I wonder how to identify the right basis in order to get the optimization he refers to with computational processing. Can you provide some examples, please?

As Sal says, "Linear Algebra is the art of choosing the right basis"; as with any form of art, the only way to become proficient is with practice. Having said that, usually a good basis to work with a transformation is one where you can represent the transformation as a diagonal matrix. In the videos about eigenvalues and eigenvectors you will see some methods to achieve this: https://www.khanacademy.org/math/linear-algebra/alternate_bases/eigen_everything/v/linear-algebra-introduction-to-eigenvalues-and-eigenvectors

Thank Sal for a very clear tutorial. I have one question here. What is the difference between transformation matrix and change of basis matrix? To me they look similar.

the change of basis matrix is a kind of transformation matrix. Specifically it transforms the vector into an orthonormal basis. Does that make sense?

If D maps vector x (which is wrt some basis) on to a co-domain and with an image which is also wrt some basis. Can some transformation matrix map directly from x (standard basis) to x in co-domain ( with respect to B)?

Yes. You can map directly from x to T([x]b) using the matrix given by multiplying C^(-1)D or AC^(-1). You can also map directly from [x]b to T(x) with the matrix given by DC or the matrix given by CA. This also means that DC = CA and C^(-1)D = AC^(-1).

Does any of this relate to Tensors?

Loosely. Tensors are a more general mathematical object than vectors. Vectors are the simplest kind of tensor.

why A is equal to C^(traspose)*D*C ?

Just follow the box flowchart Sal drew and the formula writes itself, though it would've made more sense if he did the last step by multiplying the altered basis coords by C to basically create a closed loop rather than the other way around.

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Linear algebra

Course: Linear algebra > Unit 3

Lesson 3: Change of basis

Alternate basis transformation matrix example part 2

Name: Alternate basis transformation matrix example part 2
Uploaded: 2011-02-20T16:47:22Z
Description: Showing that the transformation matrix with respect to basis B actually works. Brief point on why someone would want to operate in a different basis to begin with.

Google Classroom

Showing that the transformation matrix with respect to basis B actually works. Brief point on why someone would want to operate in a different basis to begin with. Created by Sal Khan.

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Thomas Kember
Posted 10 years ago. Direct link to Thomas Kember's post “I am new to linear algebr...”
I am new to linear algebra. Sal says several times that it is useful in computer science especially in handling graphics representing 3 dimensions on a 2 dimension screen. But isn't this what Renaissance painters did when they used perspective . linear algebra was developed in the 19th Century. My question is were Renaissance painters using a primitive or proto form of linear algebra which is simpler that full blown LA?
Button navigates to signup pageComment on Thomas Kember's post “I am new to linear algebr...”
(12 votes)
Answer
- Erika Davidoff
  Posted 9 years ago. Direct link to Erika Davidoff's post “Renaissance painters lear...”
  Renaissance painters learned about and portrayed perspective by carefully observing the world around them and replication what they saw when they stood in rooms or looked at buildings. Computers don't have this observational power, nor do they have senses with which to process the world, so they need mathematical algorithms to function. Linear algebra lends numerical rigor to principles that Renaissance painters first began to understand. The concepts are very similar, it's just that painters speak in the language of shading and color whereas computers speak in the language of numbers and matrices. So while the concept of representing 3 dimensions on a 2 dimensional screen is hundreds of years old, the method of doing it using matrices is fairly young.
  Button navigates to signup page
  (15 votes)
Marcello Cruz
Posted 8 years ago. Direct link to Marcello Cruz's post “At 12:25 Sal says "you wa...”
At
12:25
Sal says "you want to pick the right basis". I wonder how to identify the right basis in order to get the optimization he refers to with computational processing. Can you provide some examples, please?
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
- Yamanqui García Rosales
  Posted 8 years ago. Direct link to Yamanqui García Rosales's post “As Sal says, "Linear Alge...”
  As Sal says, "Linear Algebra is the art of choosing the right basis"; as with any form of art, the only way to become proficient is with practice.
  
  Having said that, usually a good basis to work with a transformation is one where you can represent the transformation as a diagonal matrix. In the videos about eigenvalues and eigenvectors you will see some methods to achieve this: https://www.khanacademy.org/math/linear-algebra/alternate_bases/eigen_everything/v/linear-algebra-introduction-to-eigenvalues-and-eigenvectors
  Button navigates to signup page
  (8 votes)
damla.yengul
Posted 8 years ago. Direct link to damla.yengul's post “I am not sure if this was...”
I am not sure if this was asked before or if it is executed in another video but how does this logic apply to when you are dealing with two non standard bases?
Button navigates to signup pageButton navigates to signup page
(4 votes)
Answer
- Yamanqui García Rosales
  Posted 8 years ago. Direct link to Yamanqui García Rosales's post “The logic is exactly the ...”
  The logic is exactly the same, but you must be very careful to identify on which basis you are working on. For example, let's suppose you have two non standard basis:
  𝗮 = { a⃗₁ , a⃗₂ , ⋯ , a⃗ᵤ } 𝗲 = { e⃗₁ , e⃗₂ , ⋯ , e⃗ᵤ }
  
  If you want to create the change of basis matrix that goes from basis 𝗮 into basis 𝗲, you would need to construct a matrix whose columns are the vectors of the basis 𝗮, expressed in the basis 𝗲:
  𝗗ₐ₋ₑ = [ [a⃗₁]ₑ [a⃗₂]ₑ ⋯ [a⃗ᵤ]ₑ ]
  
  (Of course, in order to get to this, you would first need to create the change of base matrices between the standard basis and each of your basis, in order to be able to express the vectors in each of them).
  Comment on Yamanqui García Rosales's post “The logic is exactly the ...”
  (3 votes)
nhathieugc
Posted 3 years ago. Direct link to nhathieugc's post “Thank Sal for a very clea...”
Thank Sal for a very clear tutorial. I have one question here. What is the difference between transformation matrix and change of basis matrix? To me they look similar.
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(3 votes)
Answer
- loumast17
  Posted 3 years ago. Direct link to loumast17's post “the change of basis matri...”
  the change of basis matrix is a kind of transformation matrix. Specifically it transforms the vector into an orthonormal basis. Does that make sense?
  Button navigates to signup page
  (2 votes)
nasir_nasrat
Posted 8 years ago. Direct link to nasir_nasrat's post “If D maps vector x (which...”
If D maps vector x (which is wrt some basis) on to a co-domain and with an image which is also wrt some basis. Can some transformation matrix map directly from x (standard basis) to x in co-domain ( with respect to B)?
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(2 votes)
Answer
- Kyler Kathan
  Posted 8 years ago. Direct link to Kyler Kathan's post “Yes. You can map directly...”
  Yes. You can map directly from x to T([x]b) using the matrix given by multiplying C^(-1)D or AC^(-1). You can also map directly from [x]b to T(x) with the matrix given by DC or the matrix given by CA. This also means that DC = CA and C^(-1)D = AC^(-1).
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  (2 votes)
Jim Cunningham
Posted 5 years ago. Direct link to Jim Cunningham's post “Does any of this relate t...”
Does any of this relate to Tensors?
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(2 votes)
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- kubleeka
  Posted 5 years ago. Direct link to kubleeka's post “Loosely. Tensors are a mo...”
  Loosely. Tensors are a more general mathematical object than vectors. Vectors are the simplest kind of tensor.
  Button navigates to signup page
  (2 votes)
Teodor Chiaburu
Posted 8 years ago. Direct link to Teodor Chiaburu's post “An extra note: it may see...”
An extra note: it may seem that we can even use a shortcut to go from x in standard form straight to [T(x)]B, by multiplying x first with the inverse of C, then with D. So [T(x)]B = DC^-1 x (of course, provided that we know D, too).
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(2 votes)
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falcaodelcio
Posted 10 years ago. Direct link to falcaodelcio's post “what if instead of vector...”
what if instead of vector it's given a line and we are ask to find the standard matrix?
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(2 votes)
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Andrew Burns
Posted 8 years ago. Direct link to Andrew Burns's post “So D is the transform mat...”
So D is the transform matrix with respect to B?
D = [A]_B?
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(2 votes)
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Julius Minulescu
Posted 11 years ago. Direct link to Julius Minulescu's post “why A is equal to C^(tras...”
why A is equal to C^(traspose)*D*C ?
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(0 votes)
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- FishHead
  Posted 10 years ago. Direct link to FishHead's post “Just follow the box flowc...”
  Just follow the box flowchart Sal drew and the formula writes itself, though it would've made more sense if he did the last step by multiplying the altered basis coords by C to basically create a closed loop rather than the other way around.
  Button navigates to signup page
  (3 votes)

Video transcript

We've see now that we can apply linear transformations in different coordinate systems. The transformations that we've been performing before have all been with respect to the standard basis. So in the last video I said, look, in standard coordinates, if you have some vector x in your domain and you apply some transformation, then let's say that A is the transformation matrix with respect to the standard basis, then you're just going to have this mapping. You take x, you multiply it by A, you're going to get the transformation of x. Now, in the last video and in a couple of videos before that, or actually the one right before that, we said, well, look. You can do the same mapping, but just in an alternate coordinate system. You could do it in some coordinate system with respect to some basis B, and that should be the same thing. It should just be a different transformation matrix. And in the last video, we actually figured out what that different transformation matrix is. We had a change of basis, so let's say we had this basis right here. Let me actually copy and paste everything so that we understand what we did. So this was the example. Let me copy it. Let me paste it up here, put all of our takeaways from the last video up here and paste it right here. For the last video we said, OK, this is my basis right there. And then we said-- let me copy and paste. That was my alternate basis, and then I have my change of basis matrix and it's inverse. Those will be useful to deal with, so let me copy and paste that. OK, copy, And then I'm going to paste it. Edit, paste. Maybe I'll just write it over there. Not maybe the best order, maybe I should have written that first, but I think we get the idea. Then we want to write what our transformation matrix is with respect to the standard basis. And I wrote that write over here. This was all from the last problem, if you're wondering where I got all this stuff. Let me copy and then paste that. Edit, paste. So I'll paste that. And then the whole point of the last video is we figured out what the transformation matrix is with respect to this basis right here. So D, which was the big takeaway from the last video, was equal to this right here. Let me copy and paste that. Copy and paste. And so now we have all of our takeaways in one place. Edit, paste. What I want to do in this video is verify that D actually works, that I could start with some vector x-- let me write it up here. Let's take some example vector. So this transformation, it's entire domain is R2, so let's start with some vector x. Let's say that x is equal to 1 minus 1. Now, we could just apply the transformation in the traditional way and get the transformation of x. Let's just do that. The transformation of x is just this matrix times x. And so what is that going to equal? Let me see. Maybe I can just do it right here in this corner to save space. So it's going to be this matrix times x. So this first term right here is going to be 3 times 1 plus minus 2 times minus 1, or plus 2, right? Minus 2 times minus 1 is just 2, so it's going to be 3 plus 2, so it's going to be equal to 5. And then the second term right here is going to be 2 times 1 plus minus 2 times minus 1. Well, that's just positive 2, so it's going to be 2 plus 2, so that is 4. So that's just the transformation of x. Now, what is this vector x represented in coordinates, or I guess you could say, this alternate basis coordinates? So what is that vector x represented in coordinates with respect to this basis right here? Well, you saw it before. I wrote them out here. Maybe it'll be useful to do it right here. I'll copy this. Actually, let me copy both of these. These will both be useful. Edit, copy. As you can see, if you want to go from x to the x in an alternate basis or the alternate coordinate representations of x, you essentially multiply x times C inverse, so that's why I'm copying and pasting it. Let me copy, and then let me put it up here so that we can apply these, then paste it right there. So if we want to go from x to the B coordinates of x, I take my x and I multiply it times C inverse. C inverse is this thing right here. So if I take x and I multiply it times C inverse I'l get this version of x. So let's do that. So this times that. Let me just put the minus 1/3 out front. It's going to be equal to minus 1/3 times-- let's see if we can do this one in our head as well. So it's going to be 1 times 1 plus minus 2 times minus 1, which is just positive 2. So it's going to be 1 plus 2, so it's going to be equal to 3. And then it's minus 2 times 1, which is minus 2 plus 1 times minus 1, which is just minus 1. So it's minus 2 minus 1, so it's minus 3. So if we have minus 1/3 times this, the B coordinate representation of our vector x is going to be equal to minus 1 and then 1, just like that, which is actually interesting for this example. It just kind of swapped the first entry and the second entry. Now, let's see what happens when we apply D to x. So if we apply D to x, D should be our transformation matrix if we're dealing in the B coordinates. So let's see what happens. If we apply D to x-- let me scroll over a little bit, just so we get a little bit more real estate. So if we apply D to x, what do we get? And so this is going to be the transformation, or this should be the transformation of x in B coordinates. So what is this going to be equal to? You have to multiply this times D. So it's going to be minus 1 times minus 1, which is 1, plus 0 times 1. So it's just minus 1 times minus 1, which is 1. And then we're going to get 0 times minus 1 plus 2 times 1, so 2 times 1 is just 2. Now, for everything to work together and assuming I haven't made any careless mistakes, this thing, this is vector right here should be the same as this vector if I change my basis, so if I go from the standard basis to the basis B. When you go in that direction, you just multiply this guy times C inverse. And I'm just using this formula right here. If I'm in the standard basis and I multiply by C inverse, I'm going to get the B basis. So let's see what I get. So this guy, I'm going to multiply him times C inverse. Let me do it up here just to get some extra space. So I'm going to multiply the vector 5, 4. I'm going to multiply that by C inverse. We're going to have minus 1/3 times 1 minus 2 minus 2, 1, just like that. So this is going to be equal to-- Ill just write the minus 1/3 out front. We have 1 times 5, which is 5, plus minus 2 times 4, so 5 minus 8. And then we have minus 2 times 5, which is minus 10. And then we have plus 1 times 4. Minus 2 times 5, which is minus 10, plus 1 times 4: plus 4. So this is equal to minus 1/3 times minus 3, and this is what? This is minus 6. If you multiply the minus 1/3 times it, all the negatives cancel out, and you get 1 and 2, which is exactly what we needed to get. When you take this guy and you change it's basis to basis B or you change its coordinate system to the coordinate system with respect to B, you multiply it by C inverse, you get that right there. So literally is the B coordinate representation of the transformation of x. We just did it by multiplying it by C inverse, which is exactly what we got when we took the B coordinate version of x and we applied that matrix that we found, the transformation matrix with respect to the B coordinates, and you multiply it times this guy right here. We got the same answer. So it didn't matter whether we went this way around the little cycle or we went this way. We got the same answer. This isn't a proof, but it shows us that what we did in the last video at least works for this case, and I literally did pick a random x here. And you can verify it, if you like, for other things. Now, you should hopefully be reasonably convinced that we can do this, that you can change your basis and find a transformation matrix. We've shown how to do it, but the obvious question is why do you do it? And someone actually wrote a comment on the last video, which I think kind of captures the art of why you do it. I'm not looking at the comment right now, but if I remember correctly, they said their linear algebra teacher said that linear algebra is the art of choosing the right basis. Let me write that down. Or you could imagine, the right coordinate system. And why is there a right coordinate system? Maybe I'll put little quotes inside the quotation. What does it mean to have the right coordinate system? Well, if you look at your original transformation matrix with respect to the standard basis, it's fine. It's got this 2 by 2. But if you performed matrix operations with this, you've go to do some math. And if you had to perform it over and over, if you have to perform it on a bunch of vectors, it is what it is. But when you transfer your bases, when you go to a new basis, when you went to this basis right here, all of a sudden, you find that the transformation matrix is much simpler. It's a diagonal matrix. When you multiply a diagonal matrix times something, you're literally just taking its scaling factors of the first and second terms. When you multiply this guy times some vector-- we did it here. When you multiplied this guy times this vector, you literally just scaled the first term times minus 1 and you scaled the second term by 2. So it's a much simpler operation. And you might say, hey, but we had to do all of that work of multiplying by C inverse to get there, and then once you get this answer, you're going to multiply by C to go back into standard coordinates, you know, that's a lot more work than just what you save here. But imagine if you had to apply D multiple times. Imagine if you had to apply D times D times D times D to x. Let me say it this way. Imagine if you had to apply A times A times A or you have to apply A 100 times to some vector up here, then applying A 100 times to some vector would be much more computationally intensive than applying D 100 times to this vector, even though you had a little bit of overhead from converting in this direction and then converting back. So in a lot of problems, especially in computer science frankly, or some other applications you might be doing, you want to pick the right basis. The computations for many problems get a lot simpler if you pick the right coordinate system.