Current time:0:00Total duration:12:36
0 energy points
Studying for a test? Prepare with these 5 lessons on Alternate coordinate systems (bases).
See 5 lessons

Alternate basis transformation matrix example part 2

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.