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Dönüşüm Matrisini Bulmak İçin Koordinat Sistemini Değiştirme

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Let's say we're in R2. Let me graph it. Let me draw my vertical axis, just like that. And then let me draw my horizontal axis, just like that. And let's say I have a vector 1, 2. So my vector looks like this, 1, and then we go up 2. So my vector looks like this. Let me draw it. That is my vector. It's the vector 1, 2. And I want to look at the line that is spanned by this vector. So let me just define some line L, and it's equal to the set of t times 1, 2, where t is any real number. So this is just a line with a slope of 2, right? You rise 2 for every 1 you go over. So let me draw my line just like this. There's nothing new at this point. Let me draw it a little bit better than that. There you go. And it keeps going, because obviously, you could get negative multiples of my vector, and it also goes here. If you look at all the points that are specified by all of the scaled up or scaled down or negative scaled versions of this vector, you're going to get all the points on this line if you draw those vectors in standard position. So this is my line right there. Now, I want to construct a linear transformation that reflects around this line. For example, if I have this vector-- let me draw it like this. Let's say I have that vector that specifies that point right there. Let me call that vector x. I want the transformation of x to be essentially the reflection of x around this line. So if I go in this direction right here, I want this to be the transformation. I want that to be the transformation of x, so it should look just like that. I'll do another example. Let's say I have this vector. Let me give you a particular vector. Let's say I have a vector that's orthogonal to my line. So let's say I have the vector, I'll call it v1. And let's see, something that's orthogonal to it, I can switch these two guys and make one of them negative. So if I go to the vector 2, minus 1, so if I have this vector right here, visually it looks orthogonal. So if I have v2-- let me call it v1. v1 is equal to 2 minus 1. Visually, it looks orthogonal, but then even if you actually take the dot product 2 times 1 plus minus 1 times 2, that's going to be equal to 0. So this is definitely orthogonal to my spanning vector, so it's definitely orthogonal to every point on that line. But if I apply the transformation to this guy right here, I'm just going to flip it over the line. So let me draw it like this. So it's going to be equal to-- it's just going to look like-- let me do it in a green color. So the transformation of this guy, I'm just flipping him over this line. I'm finding the reflection across line L. So this is going to be the transformation of v1, and it would be-- we don't know what the actual transformation matrix is and that's actually going to be the subject of this video, but if you just look at the transformation just for v1, it's going to be the negative of that, so it's going to be minus 2, 1. So I know I want to define this transformation. I know I want to define this transformation from R2 to R2. That is a reflection. So the transformation of some vector x is the reflection of x around or across, or however you want to describe it, around line L, around L. Now, in the past, if we wanted to find the transformation matrix-- we know this is a linear transformation. I don't have to go through all of the exercise of this is a linear transformation. But in the past, if we wanted to find the transformation matrix for a linear transformation, let's say we say T of x is equal to some 2-by-2 matrix, because it's a mapping from R2 to R2 times x. Before in the past to find A, we would say A is equal to the transformation applied to our first standard basis vector, so the transformation applied to 1, 0. And then our second column in A would be the transformation applied to 0, 1. And if we were dealing in Rn, we would do this n times and we would have the n basis vectors. We've seen this multiple, multiple times. Now, in the past, when we were constructing these transformation matrices, finding these things were pretty straightforward, but you'll see now it's not that straightforward. If I have 1, 0, if I have the vector 1, 0, finding its transformation, if I were to flip it over, it's going to look something like this. And we could do it if we were creative with our geometry and our trigonometry. We could actually figure this number out, and we could figure this number out. If I were to flip 0, 1 over, it might look something like that. We could figure it out, but it's not easy. So let me just write this down. It's not easy. And we want to at least-- I mean, you know, we could do this if we had to, but if there's an easier way, we should at least be exposed to that easier way. So you might have an insight here. We've been talking about alternate coordinate systems, and this transformation, it looks difficult in our standard coordinate system, because we're reflecting around this line right there. But what if we defined a coordinate system where reflection around this line was more natural? And you might already see an interesting coordinate system. What if we had a coordinate system with v1 as one of the basis vectors and then 1, 2 as the other basis vector? Then, all of a sudden, at least in that coordinate system, the reflection wouldn't be around this inclined line. The reflection would be around the second coordinate's axis. It seems like it would be a more natural transformation. So this leads us to a very interesting thing. Let me review what we've done in the last couple of videos. In the last couple of videos, we could say, look, if we're in standard coordinates, you multiply by some matrix A and you get the transformation of x. We've seen that multiple times. This is in standard coordinates. And we've also seen that we could change our coordinates. We can get coordinates with respect to some other basis. We can multiply x times C inverse and then we'll get the representation of x or in coordinates with respect to some other basis, and then we can multiply that times some other matrix D. And we've seen-- let me write this down-- that D is equal to C inverse times A times C. We showed this I think it was two or three videos ago, where C is just the change of basis matrix. C is just the matrix that has our new basis vectors as columns. And C inverse is obviously its inverse. So we can apply D. And then if we multiply D times this B coordinate version of x, we will get the B coordinate version of the transformation of x, so the transformation of x represented in B coordinates. And then, you know, you could convert back and forth between these guys. You could go this way if you multiply by C inverse. You could go that way if you multiply C. We've seen this in multiple videos. Now, in this particular case that we're dealing with, I want to find A. I've kind of defined my transformation in kind of touchy-feely terms. I've just verbally defined it, but I would like to have a transformation matrix with respect to my standard basis. But I just showed you that it's not easy. I'll have to break out some serious geometry and trigonometry to figure out what happens to 1, 0 when I reflect around the line spanned by 1, 2. It's not easy. But what if we change basis? What if we change basis to some new basis? Let's say we define some new basis, where our spanning vectors or our basis vectors are this guy, are the vectors 2, 1, and the vectors 1, 2. So you can almost view that our new horizontal axis-- obviously, it's not horizontal, but if you want to think of it this way-- is going to be like that, and our new vertical axis is like this. And now in this world, our transformation is just a reflection around our new vertical axis. So maybe it'll be easier to find D, because this might be a simpler transformation in this other coordinate system. And then if we can find D, then we can solve for A. We've seen this. A is just equal to C times D times C inverse. I showed that to you a couple of videos ago. So let's see if it's easier to find D. So let's just experiment a little bit. Let's label our transformation vectors. Let's say this is-- or our basis vectors. Let's say that's v1. That's the same label I applied to it over here. Let's say that this spanning vector right here of our line that we're reflecting around, let's call that v2. So our new coordinate system has the basis vectors v1 and v2. Now, what is v1 represented in our new basis? Remember, when you transfer to a new basis, all it means is how much do your coordinates-- your first coordinate now is going to be-- how much of your first basis vector do you need to get a linear combination of v1 and how much of your second one? Well, v1-- is just equal to 1 times v1 plus 0 times v2. I mean, if I wanted to, I could've found C inverse and multiplied by it, but it's almost trivial to find the linear combination of your basis vectors that gets v1. In fact, it's not almost trivial, it is trivial. v1 is 1 times v1 plus 0 times v2. So its coordinates with respect to the basis B are going to be 1 times v1 plus 0 times v2. Fair enough. Now, what is D going to be? Let's see. Let's say we want to find D first, because if we can find D, then we can just apply this formula here to figure out A. Now, let's just say D is equal to-- it's going to be a 2-by-2 vector. It's still mapping from R2 to R2. This is still a two-dimensional space. This is going to map within this two-dimensional space. So let's say D has just two column vectors d1 and d2, just like that. So what is D times this going to give us? So let me write it this way. Let me scroll down a little bit. I'd like to show that little chart right there. But D times the B coordinate version of our first basis vector is going to be equal to d1, d2-- those are the two columns of D-- d1, d2 times v1 in B coordinates, right? So it's times 1, 0. Now, what is this going to be equal to? Well, this is just going to be a linear combination of these columns. So it's going to be 1 times d1 plus 0 times d2, which is just d1, so it's the first column of this. But what is that also going to equal? If I multiply D times the B coordinate version of x, I've just gotten the B coordinate version of the transformation of x. So this is going to be equal to the B coordinate version of the transformation of v1. If you just put a v1 here instead of an x, v1's B version, right there, times D is going to be equal to the transformation of v1's B version Or its coordinates with respect to B. So it's just that thing right there. So there's a couple of interesting things here. We just got that the first column, just by applying D times this has got to be equal to this, we just figured out that the first column of D, because if you take this times 1, 0, which is the first basis vector's representation in its own basis, and you're going to see in general that when you represent these basis vectors in their own basis, they're going to look just like your standard basis vectors in this new coordinate system. So when you represented 2, 1 in your new basis with respect to this, of course, it's 1 times this plus 0 times this. It's going to be 1, 0. If you want to represent v2 in your new basis, well, what is it? It's 0 times this guy plus 1 times this guy. It's 0 times v1 plus 1 times v2. So it's going to be 0, 1. And you're going to see this is generally the case. And when you think about it, it's almost ridiculously obvious that the first basis vector is going be 1 times the first basis vector plus 0 times the second basis vector. And if you had n basis vectors, it would be 1 times the first basis vector plus 0 times all of the other ones. And in general, if you're dealing with the nth basis vector and you want to express it in your basis, you're essentially going to have a bunch of the zeroes all the way-- because it's 1 times your nth basis vector, and then everything else is zeroes, so this'll be the nth term. So this applies generally, or you could called this en, if you want to call your standard basis vectors that way. This is a bit of an aside. I just want to show you that this idea is generalizable. Now, what's useful about this? Well, just applying this little logic, D times this guy is that guy, we figured out that the first column of D is equal to the transformation applied to our first basis vector in B coordinates. So if I wanted to rewrite D, I could rewrite D like this. The first column is d1, which is the same thing as the transformation applied to v1, our first basis vector, in B coordinates. Now, what is d2? Well, let's do the same thing to v2, to our second basis vector. So if I multiply D times the second basis vector v2 in B coordinates, in our new coordinate system, that is equal to d1, d2 times-- what is this guy's representation in our new coordinate system? It's 0 times v1, plus 1 times v2. It's 0, 1. I wrote it up here. It's 0, 1. So this is going to be equal to 0 times d1 plus 1 times d2, which is equal to d2. And what is that going to be equal to? If you multiply D times some B coordinate representation of x, you're going to get the B coordinate representation of the transformation of x. So this is going to be equal to the B coordinate representation of the transformation of v2. And just like that, we see that column 2 in D, the second column, so this was the first column, the second column in D is just this guy. It's the transformation of v2 in B coordinates. Now, is this any easier to find than this stuff up here? Remember, before, we said, hey, if we just apply the transformation to the standard basis vectors, we can get the standard transformation or we can find the transformation matrix with respect to the standard basis, and that's all we wanted to do. And we said that's not easy. Is this any easier? Well, what is the transformation of v1? Let's go back to our original transformation definition. This is v1 right here. The transformation of v1 ended up being minus v1. It equals minus v1. So if we wanted to write the transformation of v1 in B coordinates-- so let me write this right here. So the transformation of v1, if I wanted to reflect it around the line L, it just was equal to minus v1. That's because v1 was orthogonal to the line, right? That's why we picked it. If you just take the transformation of this guy, it just flips over and you get the minus of the vector. So what is this guy, if we want to apply it, if we wanted to write this in B coordinates? Well, minus v1 is just equal to minus 1 times v1 plus 0 times v2. So minus v1's coordinates, in B coordinates, they're just going to be equal to minus 1 times v1 plus 0 times v2. Remember, these are just the weights on your basis vector. So this thing right here is pretty straightforward. This is just going to be equal to minus 1 and 0, the first column of D. Now, what's the second column of D? It's the transformation of v2. v2 was the vector that spanned our line. This is v2 right here. So what happens to v2 when you transform it? Well, you take the reflection of something that's on the line, around that line, it's just going to be the same. Nothing's going to change. You're taking a reflection around this. You can kind of imagine spinning that vector, but it's not going to change the vector at all, right? So the transformation of this guy, of 1, 2, is just going to be 1, 2. The transformation of v2 is just going to be equal to v2. Let me write that down. The transformation of v2 is just going to be equal to v2. As you can see, we've picked these basis vectors because transforming them was very natural. It was very easy. So what about this guy, if I wanted to represent this guy in B coordinates? Well, v2 in B coordinates, we already figured out is 0, 1. So this is just going to be 0, 1. So now we have our transformation matrix D with respect to basis B. It is equal to the transformation of our first basis vector with respect to the B coordinates, which is minus 1, 0. And then its second column is the transformation of our second basis vector with respect to B coordinates, so it's 0, 1, or the B coordinate representation of the transformation of our second basis vector. The language can get convoluted sometimes. And that was neat. That was easy. We now have solved for D. And now that we've solved for D, you notice that changing the basis was actually very natural, and it was very easy to find what D is. And now that we know what D is, we can now solve for A. We can now solve for the transformation matrix with respect to the standard basis. So to do that, we have to figure out C and C inverse. So C, remember, C is just the change of basis matrix. And all that is is the basis vectors. It's just a matrix with the basis vectors in the column. So we have our basis right here. So C is just 2, 1 and 1, 2. And then what is C inverse? C inverse is equal to-- well, we have to find that this guy's determinant, so it's 2 times 2, which is 4. 4 minus 1 times 1, so it's 3. So we do one over the determinant, 1/3, times-- you swap these dudes right here. So this becomes a 2 and a 2, and then you make these two guys negative. Negative 1, negative 1. That's C inverse. So now we're ready to solve for A. So A is going to be equal-- let me pick a nice color here. A is going to be equal to C, which is 2, 1; 1, 2 times D. D is minus 1, 0; 0, 1 times C inverse, which is 1/3, 2 minus 1; minus 1, 2. So let's solve this. Let's just do some matrix-matrix products. So let's do these two guys first. And so what do we get? We're going to get a 2-by-2 matrix. We get 2 times minus 1 plus 1 times 0, so this is just going to be minus 2. And then here, you get 2 times 0 plus 1 times 1, so you're just going to get 1, right? 2 times 0 plus 1 times 1. And then here, 1 times minus 1 plus 2 times 0. That's just minus 1. And then you get 1 times 0 plus 2 times 1. That's just 2. And now we're going to have to multiply this times this guy over here. Let me put the 1/3 out front, so we can worry about that later. And then let me multiply times this, so times 2, minus 1; minus 1, 2. And all of this is going to be equal to A. I'll do it in yellow. So this is going to be another 2-by-2 matrix. And so we get negative 2 times 2, which is minus 4. Let me write out, so I don't make careless mistakes Minus 4. Minus 2 times 2 plus 1 times minus 1, so minus 1. That's that term right there. The next term is minus 2 times minus 1, which is 2 plus 1 times 2, so plus 2. And then let's do this term. Minus 1 times 2 is minus 2 plus 2 times minus 1, which is minus 2. And then finally we have minus 1 times minus 1, which is 1, plus 2 times 2. 1 plus 4. Let me make sure I did that right. Minus 2 times 2 as minus 4, 1 times minus 1-- yeah, I think I got it. And, of course, we have a 1/3 out front, so this is going to be equal to 1/3 times the matrix. Minus 4 minus 1 is minus 5, 4; minus 4, and 5. Or it is equal to minus 5/3, 4/3, minus 4/3, and 5/3. That is our transformation matrix A with respect to the standard basis. Now, let me check something, because I have a suspicion because I did this before, and I don't have what I did before in front of me, but I think I got a different answer. So let me just make sure I did this correctly. So C, these are our basis vectors right there. Let's see, our basis vectors were the vectors-- oh, well, our first basis vector was the vector 2 minus 1. So that's where my error came from. My first basis vector was the vector 2 minus 1 and my second basis vector is the vector 2, 1. So that obviously is going to change my C. It won't change anything else, because we really didn't use this information anywhere else. So it's going to be 2, minus 1, so C is 2, minus 1. So C inverse, let's see, the determinant is going to be 2 times 2, which is 4 minus 1 times 1, so it's going to be 4 plus 1. So this is going to be a 5. Sorry for the error. And then we're going to switch these two guys. And then these two guys are going to become negative. So this guy's going to become negative. This guy's going to be a plus. Let me rewrite that. I apologize for the error. C inverse is going to be 2, 1; minus 1, 2 times 1/5, not 1/3. That's what that negative 1 changes. And so what are we going to get? There's going to be 2 minus-- this is C inverse right here, so it's going to be 2 minus 1. This is a plus 1 and this is going to be a 5. And then so this is going to be a 5 and this is going to be a plus 1. This thing wouldn't change, or actually, this will change, because C is now 2 minus 1. Actually, let me just redo this part, because I realize that I made it a bit of a mess, and just to go through this and correct this is a bit hairy. It's easier just to redo the whole multiplication. I apologize about that. I apologize for wasting your time with my error. So A is equal to-- let me rewrite them. 2, 1; minus 1, 2, right? 2, minus 1; 1, 2-- these are our basis vectors. I forgot that minus sign there-- times our vector D, minus 1, 0; 0, 1-- our matrix D-- times 1/5-- let me just write the 1/5 out here-- times 1/5. This is C inverse, so we're going to have 2, 1; minus 1, 2. And what is this going to be equal to? So first we can do these two guys right there. So it's going to be 1/5 times-- we have 2 times minus 1 plus 1 times 0, so it's minus 2. 2 times 0 plus 1 times 1 is 1, right? You have minus 1 times minus 1 plus 2 times 0, so that's just 1. And then minus 1 times 0, which is 0, plus 2 times 1, which is just 2. And then we multiply it times this guy, 2, minus 1; 1, 2. And what is this going to be equal to? This is going to be equal to 1/5 times minus 2 times 2 is minus 4 plus 1 times 1: plus 1. And then your next term is going to be minus 2 times minus 1, which is 2, plus 1 times 2, so it's plus 2. And then we have 1 times 2, which is 2, plus 2 times 1, which is 2, right? 1 times 2, plus 2 times 1. And then you have 1 times minus 1. You have 1 times minus 1, which is minus 1, plus 2 times 2, which is 4. So this is going to be equal to 1/5 times-- we have a minus 3, we have a 4, a 4, and a minus 3. Sorry, a plus 3. Minus 4, a minus 3, and then a plus 3. You can imagine when you deal with matrices, arithmetic is what will trip you up. So your matrix A, contrary to what I found out a few minutes ago, is equal to minus 3/5, 4/5, and then 4/5, 3/5, just like that. So barring my slight arithmetic error, which I got from writing down the basis vectors incorrectly, forgetting to put that negative number, now hopefully, you realize you can appreciate that we found an easier way to find this transformation matrix A with respect to the standard basis. You find the transformation matrix D first in a more natural basis coordinate system, and then you can solve for A from that. And you get this result down there, which is hopefully the right answer. If my memory serves me right, this is what I got the first time I did the problem.