So I realized that maybe in the last couple of lectures, I might have been going a little bit too fast. I noticed there was a little bit of uncertainty. We've been doing quite a bit of stuff with these structure of all these linear maps and the conjugacies. So I'm just going to take a few minutes to try to review a little bit to summarize what we've done and hopefully it will be helpful. So let's go back for the moment to the maps with real eigenvalues. So we have the two by two matrix with real eigenvalues lambda 1, lambda 2, and r distinct. So you remember we discussed the various topological conjugacy classes that you can have in this case. Did anyone work out how many topological conjugacy classes you can have? Because I did not write out explicitly the conjugacy class. I just wrote all the conditions. Did anyone kind of work out some diagram or something? How many? Three conjugacy classes. Which ones? Yes? That's right, but it also depends. You're right, those are the fundamental classes that determine if the fixed point is attracting or repelling or saddle. But is it true that every saddle, for example, any two saddles are topologically conjugate? Right. What distinguishes two different saddles that are not topologically conjugate? There was another aspect that needs to be taken into account. Sorry? If it's two saddles, we need to keep in mind the signs of the eigenvalues. Because that determines the orientation and whether we can construct the conjugacy or not, remember? So if we have two maps, for example, two saddles, so remember we can assume in this case, because these are linearly conjugate to a map in diagonal form. So we can assume that these are in diagonal form. This is A, and this is B. And if you remember, how do we construct the conjugacy? Well, in diagonal form, you have the fact that the horizontal and the vertical are the eigenspaces. And so we have a certain dynamics here in the horizontal, which depends on lambda 1. So this is the dynamics that depends on lambda 1. If lambda 1 is less than 1, then this is attracting. It's greater than 1, it's repelling. And here we have the dynamics that depends on lambda 1 prime. And so if you restrict yourself, remember, I'm just reviewing now, right? If you restrict yourself to the horizontal, these are just one-dimensional linear maps. And so there were some conditions that told us in which cases we can conjugate these linear maps, remember? And these linear maps, we can currently, they're topologically conjugate if the both, either both less than 1 and bigger than 1, and if the orientation is the same. In other words, if the sign is the same as well. So the same thing for the vertical, which corresponds to lambda 2, lambda 2 prime. And then we construct the conjugacy h is equal to a composition of these, is just the product of these two conjugacies. We construct one conjugacy h1 between the horizontal spaces, another conjugacy h2 between the vertical spaces. And then we get a conjugacy in the whole space by just taking the composition, these two as the coordinate functions of the homomorphism. That's how we constructed the conjugacy, right? So let's just briefly, again, try to explicitly write out all the various conjugacy classes that come in this case, using this construction. So we can assume, so because these are distinct, we can assume, so let's try to draw here the possibilities. This is lambda 1 and lambda 2, but this is this kind of parameter space. So this is the picture that I want to draw if, so each point here represents a diagonal matrix with these values of lambda 1 and lambda 2. And I want to study what the various possibilities for these maps, for these diagrams are. So let's suppose, let's draw this. This is 1. This is minus 1. Let me draw the diagonal here. OK, so for example, suppose we have a point here. What dynamics does this correspond to? So this here, a point here, has the property that lambda 1. So here we have this is lambda 1. So this is lambda 1 less than lambda 2 less than 1. What is the picture in this case? In this case, the fixed point is attracting. So now let's look at all the possibilities. How does this phase space divide? So this 1 is an important, so this is the coordinate 1. So for example, what about a point here? What does this point here? So this region here corresponds to a situation where lambda 1 is less than 1, but lambda 2 is bigger than 1. So what fixed point does that correspond to? That's a saddle. So this is a saddle. What about here? So here we have everything is bigger than 1. This is greater than 1, greater than lambda 1. Sorry, less than lambda 1, less than lambda 2. What kind of fixed point is that? Repelling. So I'm going to choose specifically everything above the diagonal because I'm going to assume here that lambda 1 is less than lambda 2 because everything else is completely symmetric. Because it doesn't make, since these two are different, one of them is bigger than the other. And just to simplify the diagram, I'm going to call lambda 1 the smallest one. It doesn't really make any difference. It just makes a difference of whether you look at the picture like this or like this. So it depends which one you call the horizontal and which one you call the vertical axis. So for simplicity, I'm just going to draw the picture in the case where lambda 1 is less than lambda 2, which means that I'm above this diagonal here. Lambda 2 is bigger than lambda 1. But you can fill in the diagram exactly in the same way, symmetric way below. So let's look at these other possibilities. So what is this possibility here? So here we have minus 1 less than lambda 1, less than lambda 2, less than 0 in this region here. You have both lambda 1 and lambda 2 are a negative because these are the positive quadrant. This is the negative quadrant. So everything is negative here. And then I have here I have that lambda 1 is less than minus 1, which is less than lambda 2, which is less than 0. And here, so this corresponds to attracting again. This corresponds to saddle. And this here, we have here that everything is less than 1. So here we have lambda 1, less than lambda 2, less than minus 1. And this is the pelling. Now here we have, so here we have everything is positive. So both eigenvalues are positive and we get these three cases. Here we have that both eigenvalues are negative and we get these three cases. In particular, what is the orientation of the map as a whole, in other words, the determinant. So this region here, the determinant is what? Positive or negative? The determinant is positive here because the determinant is the product of the two eigenvalues. Here also the determinant is positive, even though the two eigenvalues are both negative, their product is positive. So the determinant is also positive. But OK, so let's finish and then we will discuss a little bit this. So this case here, here you have one positive and one negative. So we have lambda 1 is negative. So we have minus 1, less than lambda 1, less than 0, less than lambda 2, less than 1. So lambda 1, so both of these have absolute value, less than 1. But one of them is negative, one of them is positive. So what's the fixed point here? Sorry? Sorry? It's a tactic because they both have absolute value, less than 1. So this is a tactic. What does it mean in terms of the dynamics? In terms of the dynamics, it just means that the eigenvalue in the lambda 2 direction is positive. In lambda 1 direction is negative. So points flip this way. So they are converging to 0. But in the vertical direction, it just gets closer to this. But in the horizontal direction, it flips. So if you take a point here, the orbit will go like this. It will converge to 0 like that, right? It will flip each time. You will still get these curves going like this. But this point will actually, if you look at the orbit of a single point, it will flip left and right, like that. Whereas what happens in this case where both are negative, what is the dynamics here? Then it flips both horizontally and vertically, right? Then you still have these curves that look like this. But where will this point go? This point starts with a point that has positive coordinates, and it maps to a point that has both negative coordinates. And then it comes back to here, and then it comes back to here, and then it comes back to here. So it converges to 0, but jumping back and forth between these two coordinates, right? Because both of these are negative. And so at each iterate, it flips horizontally and flips vertically and does this. If just one of them is negative, then it just flips like this. So you need to practice. OK, the reason why I'm doing this is just to make sure. So when you go home, you need to look at this diagram. And for each one, you need to draw yourself a little picture and try to understand what the dynamics is in each of those cases. OK, there's no point in me doing each of these on the board because it's not difficult. It's just a question of sitting down and trying each of these individual cases and getting a feel for that. OK, so let's complete here. So here we have lambda 1 is still. So here we have minus 1 less than lambda 1, less than 0. Less than 1, less than lambda 2. Lambda 2 is bigger than 1. So this is a saddle, OK? So again, the picture here will be a saddle and things will flip around, right? So this is a saddle, so the picture will be like this. So lambda 2 is the part that is greater than 1. So lambda 2 is the part that is expanding like this. So points will move like this. So you can see the different cases in which you have saddles. In this case, you have the negative, lambda 1 is negative. So lambda 1, again, is flipping each time. So if you start with a point here, it will not, the orbit will not just be along this curve, but it will flip from this curve to this curve. So it will flip from here to here to here to here and then it will keep flipping back and forth. Here we have, similarly, lambda 1 is less than minus 1. So this is lambda 1 less than minus 1, less than 0, less than lambda 2, less than 1, OK? So this is also a saddle. What is the difference between this saddle and this saddle in terms of the dynamics? Where does this point go? For this saddle here, what is this? So for one difference is the fact that in this case, it's lambda 1 that is expanding, right? So one difference is that in that case, everything is going in the other direction. So here we have lambda 1 is the expanding direction. And this is the contracting direction, OK? And then it's lambda 1 that is negative, OK? That is the same in both these cases, right? It's lambda 1 that's negative, so the flip goes like this. But it's just, if you start with a point here, it flips, but it moves away in this direction. And finally, we have this situation here in which everything is less than 1. So we have lambda 1 less than lambda 2 less than minus 1. And this is, what kind of point is this? Repelling, yeah, repelling. Sorry, no, this is positive, sorry. Lambda 1 less than minus 1, less than 0, less than 1, less than lambda 2. Sorry, lambda 2 is positive here, right? So this is, OK. So these are the conjugacy classes that you can get. Because if you think about the construction that we did, you'll see that inside each of these classes, you can do the conjugacy construction. Because once you fix this exactly in the statement of the theorem that I mentioned, I said two maps with distinctly eigenvalues are topologically conjugate if I gave a certain set of conditions. And if you look at those conditions, it corresponds exactly to distinguishing all these different conjugacy classes, OK? Because it distinguishes the signs of the various eigenvalues. So if in one case you have one eigenvalue that's positive and one that's negative, even though they're both contracting, you cannot conjugate them. Because you cannot conjugate one dimensional linear maps if one of them is orientation preserving and one of them is orientation. So we have 1, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 conjugacy classes. Any questions about this? Yes, because this is really, I mean, you don't even really need to reflect it. It's just that if we assume that we have real eigenvalues that are distinct, well, then, yeah, I can just call the smaller one lambda 1. And yes, it doesn't really matter which is the horizontal or which is the vertical axis. Yeah, I can rotate the whole thing. But that does not affect the topological conjugacy. So they're still topologically conjugate. Two maps that have the same eigenvalues. But if you switch lambda 1 and lambda 2 around, then obviously you can conjugate using the identity the horizontal here to the vertical here, the vertical to the horizontal. And then I define the same thing and I get the conjugacy. So if lambda 1 is greater than lambda 2, each map here is topologically conjugate to the corresponding one with the symmetric eigenvalues in a kind of obvious way. OK, so this is for real eigenvalues. I think we started getting a little bit confused with the complex eigenvalues. And it's possible actually when I thought about it, maybe there was a slight, is not completely correct, the statement I gave. Let's recall the situation with complex eigenvalues. So if you have complex eigenvalues, so suppose we have a map B, then remember that we discussed the normal form is of the form alpha beta minus beta alpha in the sense that if it has complex conjugate eigenvalues, it's always linearly conjugate to one of these. In the same way, the real eigenvalues are always conjugate to one of these. So in this case, the eigenvalues are alpha plus or minus i beta. And notice that in this case, the determinant of B is equal to the product of the eigenvalues, which is alpha squared plus beta squared, which is always positive in this case. So what we are really going to show, basically, is that any such map is always belong to one of these two conjugacy classes, this one or this one. So by introducing complex eigenvalues, you don't actually introduce any new topological conjugacy classes. If you now look at the space of all linear maps with distinct eigenvalues, hyperbolic invertible linear maps with distinct eigenvalues, you still have the same number of topological conjugacy class, the same number, because all the complex ones will be either depending on whether the eigenvalues are inside the unit circle or outside the unit circle, they will be topologically conjugate to one of these cases or one of these cases. So this is maybe a better way to formulate this statement. So let me write it maybe in the following way. Theorem, so B is topologically conjugate. So here we assume that beta is different from 0. Otherwise, we would not have real complex topologically conjugate to A equals lambda 1, 0, 0, lambda 2, with lambda 1 and lambda 2 positive, and lambda 1, lambda 2 in the interval 0, 1, if alpha plus i beta is in the interval 0, 1, and lambda 1, lambda 2 greater than 1, if alpha plus minus i beta is greater than 1. So I'm assuming here this is invertible and hyperbolic, so the fact that it's hyperbolic says that this is different from 1. So this is part of the assumption here that I have. This is different from 1. This is part of the hyperbolic assumption. And here also, I'm always assuming these are invertible and hyperbolic. So that's why all these conjugacy classes are the open domains here. They do not include the boundaries here, because I'm not including any situation which any of the eigenvalues is equal to 1. So hyperbolic means that lambda 1 means that this is different from 1, and this is different from 1. Different from 0, so different, so I don't know. They do not include 0 or plus or minus 1. We always look at those maps. OK, so I also realize that perhaps there's a more direct construction. How do we construct this conjugacy? So let's just recall here. If you remember, the structure of this map means that every point, so for all x0, y0 in R2, except for the origin, the orbit an of x0, y0 lies on the parameterized curve. t goes to lambda 1 t, x0, lambda 2 t, y0. So this is just to remind you, the structure what we have here is that if this is a point x0, y0, then the orbit, the images of each point are given simply by an x0, y0 equals lambda 1 and x0, lambda 2 and y0. And therefore, they lie on this parameterized curve. So here we have a curve. This is the point x0, y0. And this is the curve, the curve t. This is this curve here. Is this clear? Make sure ask me, OK, because this is going to be the basis of everything I'm going to say. So for every point, for any point we choose, the orbit of this point lies on a parameterized curve that has exactly this form. Is this clear to everyone? Yes. This is the orbit in forward and backward time. So you can see this is just choosing n as integer values of t. So in particular, this is a parameterized curve. We can call this gamma if you want. This is gamma. If I choose lambda 1, lambda 2 in 0, 1, then as t goes to plus infinity, these are going to 0. So this curve is converging towards the origin. As t goes to minus infinity, these are both going to infinity. And so we have every, including this, right? So this is also one of these curves. This is one of these curves. This is one of these curves. This is one of these curves. What about in the complex case? As we said last time, in the complex case, we have the same thing. It's just that the curves look a little bit different. So in the complex case, we also have this picture. And we pick a point here, which, as we saw, is useful to write in polar coordinates e to the i theta 0. And then as we saw, bn of this point, theta e to the i theta 0 is actually n e to the i theta 0 plus theta n. You all remember that? Yes, this is the orbit. Because we saw that the action of this is just like complex multiplication, which gives you exactly this, where r and theta depend on the eigenvalue here. And therefore here, this also lies on this curve. So lie on parameterized curve. t goes to r 0 r t e to the i theta 0 plus theta t for t in r. t in r. Except that in this case, the curve looks like a spiral. If r is less than 1, which corresponds exactly to this being inside the unit circle, then this will look like a spiral and will do this. OK, yes? So for different points, of course, we will have different spirals, right? If we choose a different initial condition here, well, it might be on the same spiral or it might be on a different spiral. So if we choose a different initial condition here, it will belong to a different spiral that is inside. So the whole space is foliated by the spiral. It's clear to see that any two such curves are disjoint. So remark. So in both cases, any two such curves either coincide or are disjoint. You agree with this? If I take another initial condition and I define this curve, if by chance I take the initial condition here, then I'm going to get exactly the same curve when I arrive at like this with this new initial condition. But if I take it somewhere else, then I will get a different curve. And it's the same thing here, right? Each initial condition defines a spiral, which in fact goes not only in forward time but in backward time as well. If I choose a different initial condition, if by chance I choose some place, some other point that happens to be on the spiral and I construct the spiral for this new point, I will get exactly the same spiral. If I construct it for a point that is different, I will get another spiral that never intersects the previous spiral. You can just see it just because they are explicitly defined these spirals in this way. So it's clearly that these two spirals, if you change the initial condition, either the initial condition is just for a different value of T and then, you know, geometrically they coincide or they're completely disjoint. So this applies to both cases. Okay, both this case and this case. Which means that the whole R2 is covered with these curves, is completely foliated by these curves.  Everywhere, okay? You can think, you can literally write R2 as a union, disjoint union of these curves. So this actually, there might be, using this idea, there might be a little bit simpler way of constructing this conjugacy, which also implicitly uses the fundamental domain idea, but in a way that is maybe a little bit simpler. So we're trying to construct the conjugacy, right? What did we do last time? Is we started with the unit circle on both sides. And what we want to do is to conjugate the dynamics. Now because every, what this means is you must send orbits to orbits. So if you take the orbit of this point, the orbit of this point always lies on this curve. So H maps this point, say, to some other point here, and therefore for this to be a conjugacy, the least that needs to happen is that at least H must map every point on this curve to the spiral at this point, right? Because any image of this point must map to the image of this point, the image of this point, or lie on this curve, the image of this point, or lie on the spiral. So basically what the conjugacy does is maps each one of these curves to one of these spirals. That's what we're trying to do. That's what we want to do for this conjugacy. So the question is, can we do that? And it's very easy because we just have two parameterized curves and we map these parameterized curves actually have a kind of natural mapping from one to the other, which is given by T, right? So let's try to write it down. So to construct the conjugacy H between NB, we need to map the invariant curves So the first thing we need to do is to figure out which curve we map to which spiral, right? So we need to kind of associate, have at least a bijection between all these invariant curves and all the invariant spirals here. It does not seem completely obvious how to do that to begin with. And then we need to make sure that, so we need to associate one curve to one spiral and then we need to make sure that the curve maps bijectively to the whole spiral, okay? So how are we going to do that? How are we going to map? How are we going to associate? First of all, once you've given a curve here, which spiral does this correspond to? There's an infinite number of spirals, right? Of course, because through every point there's a distinct spiral. How are we going to do that? Any suggestions? So Eila, yes, basically what we did. So we take the circle, we can take the unit circle here. So this is a unit circle, S1. And the key property, as I said last time, and that is why it's very useful to use with the diagonal form, is that each of these curves intersect the unit, the unit circle in a unique point. This is crucial property for this construction here. In other words, there exists a unique T such that this curve intersects the unit circle. And then we draw the unit circle here. Similarly, okay? So my picture is not perfect of the spirals here, but it follows clearly from this that also each one of these spirals intersects the unit circle in a unique point. Because this is monotonically, if I was different from one, then this distance from the origin is monotonically changing, right? So whether R is less than one or bigger than one, it doesn't matter where you start with, doesn't matter wherever you start with, you're monotonically either increasing or decreasing, which means the distance will be equal to one from the origin for a unique value of T, okay? So each of these spirals intersects this circle in a unique way. So how do we match up now these curves to these spirals? Well, all we need to do is define some bijection between the unit circle here and the unit circle here. And that will automatically match up all the invariant curves, because if we define a bijection that maps this point to this point, okay? Then automatically we have associated the unique curve, invariant curve that goes to this point with the unique spiral that goes to that point. So we can really, since we want everything to be a homomorphism, we just define a homomorphism between S1 and S1, right? So just define that H tilde from S1 to S1 be a homomorphism. EG, the identity is also a very good choice since we have S1. This associates to each invariant curve of A and invariant curve. So now I want to suggest a slightly different way that seems to me actually much simpler, I hadn't quite thought of. What we did, what we started doing last time was to use this invariant circle, look at the image of the invariant circle and define a fundamental domain and define a fundamental domain here and then map the fundamental domain and then extend it, which was exactly the same technique we used in the one-dimensional case. I thought that actually here you can also look at it a slightly different way and something I started mentioning at the end, because each of these curves is invariant, right? This curve is invariant, there is a dynamics on this curve, right? And just like on each spiral, there's a dynamics on the spiral. So on the curve, the points are moving towards the origin or towards infinity and essentially, even though this is a curve, it's just like the one-dimensional dynamics. What's the difference between looking at the eigenspaces and looking at this curve? There's no difference. Okay, just geometrically it's a curve. On the eigenspaces, which are straight, we understood that very well, that on the eigenspaces, you take a point here and you multiply by lambda one and you get a point here and you get a point here and you get a one-dimensional linear system, okay? If you look at this curve, it's not quite, the contraction rate is not quite calculated in the same way because it depends a little bit. It's not just multiplication by lambda one, but really you just take this curve and the dynamics on here is exactly one-dimensional converging. You can talk about the order between two points. You can talk about the fact that it's order preserving on this curve. Everything works in exactly the same way, right? So why don't we do this? Let's just, now that we have a correspondence between every invariant curve here and every spiral, let's first conjugate the dynamics on the corresponding curves and then join everything together and we will get a homomorphism of the whole thing. They're kind of equivalent ways of doing this, right? But the advantage is that there's a very explicit way of conjugating the dynamics on these two invariant curves, which is given by this parameterization. So the parameterization is a parameterization that is very, very tied to the dynamics, right? There's many different parameterizations of this curve as a geometric object in the plane, but this specific parameterization is a parameterization that is consistent with the dynamics because the dynamics corresponds to exactly just taking the integer values of t and you get the orbit. So this parameterization is natural in terms of the dynamics. And the same is true for this parameterization here. It's natural with respect to the dynamics for the same reason. So we can use this parameterization to map this invariant curve to this invariant curve in the sense that you just map the corresponding points for a given t, okay? You map it to the corresponding point on the spiral with the same t and then you will automatically conjugate the dynamics as well. So how do we map? So what we want to do, how do we write it? So for each, for each such curve, for each such curve, and for all t in R, we can define h of x zero lambda one. T, y zero. Sorry, so I can choose x zero, y zero here now on the unit circle. So for all t in R and for all x zero, y zero on the unit circle, let's say in S one, we can let h x zero lambda one, t, y zero lambda two, t to be equal to R zero to the t e to the i theta zero plus theta t R zero e to the i theta zero is the image of x zero, y zero on the unit circle. So you choose a point here, x zero, y zero, this gives a point on the unit circle here by this identity map. Okay, so this can even be an equality here without, if we take h tilde to be identity. So this maps this point on the unit circle and then what is the natural way to map this curve to the spiral? Well, any point on this curve now can be written in the form x zero, lambda one, t, y zero, lambda two, t and any point on this spiral here can be written as R zero, R t e to the i theta zero plus theta t. As t varies in R, we will get all the points on this curve. As t varies in R, we get all the points on this spiral. So you just associate the points with the same t and you get a bijection and a homeomorphism and explicit map from this to this, okay? Notice that this also makes it seem very natural to map a curve which instinctively looks finite in some sense, right? Because it just converges to zero in this way, whereas this curve looks infinite because it spirals round and round but in fact there's no real difference between the two, right? When you parameterize them by time as time goes to infinity, this converges to the origin. It's not really finite curve in this parameterization with respect to t. It's infinite, right? Because it converges to zero as t tends to infinity and so you get exactly the same property here. So this is how these two curves are really representing, they're both representing an infinite orbit in both directions, okay? So this is the map, right? And then this is it. This is the conjugacy. So you have now defined, this is really a conjugacy between the whole space because now for every point, we have for any point that we choose, we have a corresponding, any point that we choose is of this form and we map it to the corresponding point of this form and this gives a map between R2 to R2. So this gives, so this defines a bijection. It's clearly a bijection, H from R2 to R2, okay? We need to define also of course zero, let H of the origin equal the origin because that is the only one that you cannot apply this construction to and then we get a bijection of R2. So the only thing that's left to show is that it's a conjugacy and that it's a homomorphism, right? The conjugacy really follows immediately from the dip. I'm going to leave both of these as exercises because it's the best way for you to make sure that you really have understood this construction but they're relatively simple exercises. So check, exercise, H is a conjugacy. This follows directly from the, directly basically from the definition and H is a homomorphism and this is a little bit more subtle but also it's fairly straightforward because everything is very explicit, okay? So here of course you will need to use the fact for this construction for the definition of this we do not use the fact that both of these are attracting or both of these are repelling because the pictures look exactly the same, right? The only thing that changes is the direction of the arrows, okay? In fact, even for the fact that it's a conjugacy you're not going to need that, just that it's a conjugacy but for the fact that this conjugacy is continuous of course you will need the fact that in forward time that you preserve the omega limit so that in forward time you go to zero in forward time you go to zero. So it's just for that last point that you need to use the fact that they're both attracting or both repelling, okay? And so this concludes the proof and shows that as I said before when you include all which in some sense is the majority of linear maps with complex eigenvalues you're not actually introducing topologically anything new. It still belongs to one of the same topological conjugacy classes that we had before with the eigenvalues. Okay, any questions? So this is certainly the kind of most substantial part of this first part of this course of these first 10 lectures, right? So it's kind of the heart of what we're doing for this first part. So just to finish today, of course I want to finish with the structural stability which remember is one of our motivations for studying topological conjugacy. So once you have the topological conjugacy class you ask yourself, okay, is this other system structurally stable? In other words, if you perturb a little bit, what happens? And so we need topology. We need to know what it means to perturb a linear map. So to study structural stability, we need topology on LR2, LR2 which is equal to the space of all linear maps. Have you studied something like this before? Have you any ideas for how to compare to linear maps? So that's right. And then how do you compare the two matrices? We define a norm, that's what it means to define a topology, yes. So how do we define a norm, right? Of R2, Euclidean norm of R2, which what do you mean? So such a matrix has this form. So if a matrix A belongs to the space of linear maps, then A is of this form, right? So what's the natural parameter space for these matrices? R4, exactly. So there's a natural, so in fact it turns out there's many different norms we can put on this, okay? So I'm going to choose one, but in fact they're all essentially equivalent, so they all induce the same topology. But the most natural and the most elementary one is the one induced by R4, like you say. So this is parametrized, so LR2 can be parametrized by R4. So you can see every matrix is an element in R4, so there's a standard Euclidean metric in R4. So we can talk about the metric in R4, parametrized by R4. So natural norm is just the norm that says that the norm of A is the Euclidean norm, which means that it's A squared plus B squared plus C squared plus D squared. This is just the norm of the matrix. So you can see of course that if everything's zero, it's obviously the zero norm, and otherwise you'll have this norm, and this norm induces a metric and the metric induces a topology, so we have a topology. And clearly from this topology, what it means to change the matrix a little bit just means changing a little bit one or more of the entities of the matrix, okay? So it's a very natural topology to say basically two matrices are close if the entities are close, okay? This is natural. So one of the basic properties of this is that the eigenvalues, it's easy to see that the eigenvalues depend continuously on the matrix A because the eigenvalues are also just a basic algebraic formula of the entities. If you change, we have the explicit formula for the eigenvalues that I gave before. If you change a little bit these entities, you will get just a small change in the eigenvalue, so they depend continuously on A, okay? And this allows us, this immediately implies what we need about structural stability. So, proposition, suppose A in L R2, specifically in the maps, is invertible, hyperbolic with distinct eigenvalues, real or complex. Is it structurally stable with respect to topological conjugacy? If we perturb a little bit, is the perturbation still in the same topological conjugacy class? Why is that? That's right, because of basically of the picture we drew before. If you remember, the picture we drew before was a picture that you can think of in the space of eigenvalues, right? So we had this picture here, so one minus one. So we had all these different situations, right? So this was one conjugacy class. I didn't do this picture very well. Each of these was one conjugacy class. Not including, remember I emphasize of course that you're not including this because the fact that it's invertible and hyperbolic means I'm not including the zero axis and I'm not including these lines that are equal to one or plus or minus one, okay? So each of these legions is open because I'm not including the boundaries. And each of these legions, everything is topologically conjugate inside each of these legions. And so if you choose a map that is inside each of these legions, it'll have a certain eigenvalues that satisfy the corresponding properties. And if you change it a little bit, these eigenvalues depend continuously on A, which means that if you change it by small enough by some sufficiently small delta, then even if it's very close to the boundary, the eigenvalues will still belong to this, will still satisfy this condition. And therefore the new map will still be topologically conjugate because it will depend on the same topological conjugacy class, okay? That's in the real case. In the complex case is the same thing happens because we said that in the complex case, you're still topologically conjugate to one of these real maps in here. The complex case also, as long as your, if the complex conjugate eigenvalues are inside the unit circle, then you perturb a little bit. The eigenvalues depend continuously on A, so they remain inside the unit circle, so they remain still topologically conjugate inside the same conjugacy class. So then A is structurally stable with respect. Okay, so the proof is just what I said. I will not go into too much detail, but the proof is because each since eigenvalues depend continuously on A, the conjugacy classes all open in the topology. So the topological conjugacy classes are all open in the topology on L2. And that's all you need to say. If the top, if the conjugacy class is open, then the system is structurally stable because every system is in the interior of its conjugacy class, and therefore automatically it's structurally stable. Any questions? Okay, so I think this really completes what I wanted to say about linear maps. I've only looked at the two-dimensional maps, but you can easily generalize everything, and I think it should be fairly clear how to do it even, and if you're interested, it might be worth your while trying to think a little bit about the three-dimensional cases because as you saw, we kind of always reduce everything in one way or another to one-dimensional case. So the same in three dimensions. You can look at the diagonal case and the diagonal case. You can have one expanding, one contracting or two expanding and one contracting. You can have different possibilities, and that gives you different kind of dynamics, but also in that case, you can construct conjugacy just by conjugating the eigenspaces and getting a global conjugacy. So very similar technique and very similar ideas. So I think the two-dimensional is already not so trivial to understand the technicalities, so I think it's sufficient to do that example in full details. Okay, so what we're going to, we'll stop here for today. What we're going to do in the next couple of lectures is look at the case of circle maps, which arise, for example, in the case in which the map is not hyperbolic, as you remember, the linear map is not hyperbolic, then we get invariant circles, and then the dynamics on these circles is very interesting, and in fact, there is a whole field which is circle maps, just homomorphism of the circle and what can happen with homomorphism of the circle. So for the next couple of lectures, I'm just going to give you a little bit of introduction to that, and that will complete the first set of lectures. Okay.


Honestly, it feels like we do this every single year. Michael Vitt comes on Madden. Everybody freaks out, because he all sees the best card in the game. And then honestly, he's just not that good. Now don't get me wrong. Every time Michael Vitt comes out, he's crazy good. But this year, it is legitimately different. He really was not that good. In Madden 18, 19, 20, 21, when he came out, it was in the end of the world. Especially these last two years where we already had Lamar Jackson with escape artist. A lot of people like right handed quarterbacks better. Michael Vick wasn't all the buzz. This year, it's so cracked and I'm not even joking. Let me tell you why. EA wanted this year to make a little bit less reliance on abilities. They didn't want abilities to be so OP. They made quarterbacks without escape artist insanely fast already. So when the absolute fastest card in the game, Michael Vick with 93 speed, 95 agility and 93 acceleration is the only quarterback in the game who can get escape artist. It is comical how fast he is. On top of that, we got dashing dead ice. We get perfect pass accuracy up to 40 yards on the run. Every throw we make with Michael Vick is pretty much on the run. So, and honestly, the only reason he's not so broken that they took the servers down, he does get tired pretty fast. So recuperation is actually really clutch. As far as quarterbacks going, honestly don't even expect him to fumble that much. Especially because we put the ability trust on him. He cannot fumble as a result of a tackle. You have to rush for more than one yard like five or six times. It's actually kind of hard to activate, especially when people just send like three QB spies on you.  It does not matter how many QB spies they put on us. It's that crazy. Luna is joining us for today's video. My team is super random. It literally doesn't matter. Like, I don't know how to put it. Does not matter at all. It will be the Michael Vick show no matter what we do. And the beauty is house rules still exist when I'm recording this. I love doing gameplays with house rules.  We can see a bunch of different defenses. And yeah, Michael Vick's just gonna go off. Also a huge thank you to Vessie for sponsoring today's video. Work with Vessie a ton. They make incredible shoes. I've always worn the Vessie every day and weekend sneakers. They're stylish. They look good. They have awesome color ways. And they're weatherproof, which is what's crazy. Like hail, sleet, snow, mud, beach, whatever. Your feet stay dry. And even better, they just dropped a brand new shoe. I've got it with me right now. It's the everyday move. It's got that same awesome tech as the rest of Vessie's shoes, which is weatherproof. It's breathable. It's lightweight. It's honestly so clutched from Michigan right now. Been raining a ton. So I just keep these at the front door. Anytime I go to walk Luna, I just slide these puppies on. No matter what I encounter on that walk, I will be totally fine. And I do this every time because people don't believe me when I say they're everything proof. I'm gonna show you my Vessie test right now. Are you ready? Don't, I don't even know how to do it. So click the link below and use my code MMG to get $25 off your Vessie shoes. Vessie, thank you for sponsoring guys. Enjoy the rest of the video. So in this house rules, by the way, if you guys aren't familiar, touchdowns are worth 10 points. Interceptions give you three points and it's the first to score 15 points. So, bop, no way.   Wow, Karim Jackson just single-handedly made that guy rich, nobody else. I guess first game where hopefully I'll get to actually use Michael Vick. I'm starting out on defense. Oh, I still sacked him. Wow, that was a, that was impressive. Josh Allen looking like Michael Vick back there. Maybe I need to do Josh Allen video. B could be open in the back, but he's not. Because, who's Colin's brother? I have no idea who, oh, I'm the one now because he heard of honor. This is gonna be so sketchy. Will I run out of the back of my own end zone if I do this?  Oh no, dude, I could have stepped out. All right, that was the sketchy shit ever, but I'm really only willing to do that with Michael Vick. He's gonna try and use a rush. It's not gonna work. And you know what?  That was such a soft touch. Throw, oh no, he didn't throw, oh, he's getting to that. Shit. I thought that was made coverage. I thought Ryan Harris was gonna just keep going, but you know what? Michael Vick has a tackle. Um, A maybe? Oh, I was just behind that.  Inside zone maybe? Gotta get there, baby. Gotta get the, oh, he's playing extra, but we got him. Let's go. Game ain't over. First or 15 wins, so if he had gotten that, it would have been real bad. We can still win it right here. We marched down in score. He moved Edwards up, so you know what? Sammy Watkins. So you know what? Sammy Watkins! Go, go! Sammy Watkins, so how are you that slow? Are you kidding? You're Sammy Watkins, bro. A, easy. Kelsey's got it. I need the two-point here. Oh, let's run read options. Let's run read options, just keep it. Right, I probably fucking, I hate myself. I'm going straight for the QB.  Yup!  Almost got the seg, but you know what? There was nothing open. He could playmaker X. He's just about to throw. I don't actually know what happened, but he missed it. Now let's go win, baby. Does he QB spy? No. Well, then that's free yards, baby. It's literally free real estate. Why might have been open? Honestly, it doesn't even matter that he's QB spies. Oh! Oh! I just dropped everybody! Go, Vic! Go, Vic! Dude, the first stop and goal is the one you might not have seen there. We might have to run that one back. A little action replay for the boys. Oh, darn it. He's gassed, bro. Yeah, see what I mean?  Let's see if he actually gets to this. I highly doubt it. Yeah, it's so good. Oh no, but I'm gonna be on the one. The QB spy finally got me. There's a clutch stop. Clutch stop by him, but we just gotta stop one more time. Also, if we get an interception, it's game over. I'm gonna guess the, yup, yup, yup! His snap count is just so obvious. I know exactly where he's gonna snap.  We're gonna get the same thing! That's 15! God! Wow, Papa Meeks. It almost looks like it's a totally different day. And you're literally recording the same video. Yes. In the middle of me recording that last video, EA just took the servers down for like two hours. They just decided to patch Madden in the middle of the day. Problem! They took house drills out. And I can't play head to head seasons because that's my Wheel of Mud account. So, I hopped on Moshe Moshe Man. I've been on Moshe Moshe Man so long. I've been on X-Sex Hatter 69.  Like it's so mediocre. 78 overall, I have like base elites and golds at every position. It's kind of like a really balanced Wheel of Mud team. It's perfect because this Michael Vick is so cracked that he actually can carry a team all by himself.   I've got Michael Vick as my only really good player, really the only X-Factor. And I think we're still gonna go absolutely ballistic. That's the top three. Yes, I have an 80 overall in my top three. He's got 91, 93, 90. It doesn't matter. It doesn't matter. Ooh, this could be the run though. This could be the run. He wants that first down. Yup, yup, yup! Let's go! Fourth and one! Fourth and one! He is gonna throw!  Whoa, how was I not there? I totally thought I was gonna be there. I moved so much slower than I expected to. Oh, I guess this is not Devon White in the middle of the field here anymore, huh? Another run. Ooh, good grab. Another run, we're there? No! How was I not there? Yup, yup, we're here, we're here, we're here! Let's go! Do it again! Run it again! Oh, it's a pass, baby. It's a pass! Ooh, and he just barely gets the touchdown. All right, I respect it. I had a few chances to get that ball back, didn't get it. Let's just start calling plays, baby. I got base elite TJ Hawkins socks out here, and I have a really fast Michael Vick. Oh, I really could've thrown that. I could've thrown that to Y, I think that was open.  Why can't I streak Hawkinson? Are you kidding me?  Why does he run a wheel now? Fourth and 13! Marquis Brown! I'm fourth and 13, let's go! That was big. So sketchy, he's so open!  Weird anymore? I could've swore you just hold X.   That was not Vick's fault at all. That should've made a passing touchdown.  I'm so pissed off right now. Get down! What? That's my safety! That was my safety.  That was a bizarre, bizarre path. He just took fourth and seven from his own four, and he goes for it. Respect that. That is a true Madden player right there, baby. Ha ha ha! Yeah! There's my touchdown. I'm doing a fake PAT rollout pass. I'm pretty sure Michael Vick throws this. Is that, no, that guy looks a little too white. It's Ryan Tannehill. But you know what? He's open, we're gonna stop him going because he's mad bad! And we're not gonna get that. Nobody in my team knows how to dive. Oh my god. I swear to God, they just want people to win sometimes. Pick a guy to win and then they just hand it to him. What is that? Oh, he played that. Oh, why? Why'd you drop it?  That was just passing interference. This team is so sorry right now. No doubt in my mind that Vick is really good, but I think even Vick has a limit here. I just got to throw lasers, baby. And Vick can do that. Marquis Brown! Let's go! We're marching, baby. 4 for 9, 121, 200. Ooh. He will stat line for my boy. I feel bad about that, honestly. B might have been open, but I think he just made a mistake. This is the first time in the whole game he's not had a QB spy. And we got Mad Yard's offense, so I don't think he'll make that mistake again. I take it all back. He will make that mistake back to back. Now he's going to be so worried about the QB spy that I don't think he'll guard A. I think he's going to forget. No, he's just going to knock QB spy again. I'm going to get around that. Ah! No bubbles, man. Good work. Second and sixth. I'm literally just running this one play. Oh, why? He's so open. Marquis Brown in no man's land gets a spin. Oh, let's go. Oh, let's go, Vick. Let's go. Oh, I did get a yard. It's now fourth and one. I think we just hand this off right up the middle. We just give this a cream hunt and let him go, baby. Fourth and two. It's a short two and he's through. Let's go. That was huge because we have to score here. Go. Go, Vick. Let's go. Too easy. That one was just too easy. A second on the clock is actually nice. It's going to save us a little bit of this clock time. I'm going for two. And I should probably be able to hit random march. I just don't know if he can get off that block or if he can get off that press. Oh, I got in. I had to have. Let's go. Michael, Vick, finally taking over in the third quarter. That was huge. Why are Lyman blocking four of my dudes and Dawkins can't get off a block? That was smart. That was smart. You got to shoot the clock here. You better hope you can get in the end zone, sir. You can't get in the end zone. It won't matter. Yup. Big tackle. Wow. My defense is just too atrocious right now.  Not a man out there to stop me. And I should have made an actual juke. I just thought he'd switch off. This is activated. Let's go. Oh, wow. I tried to do too much. I thought I might have stayed on my feet there. Wait a second. I just got tackled for a loss. I'm so disappointed in myself. I just had two seconds. All right, we're going to leave Hawkinson in the middle of the field here so that we can playmaker him. Yes. What a play. Let's go. It's too bad. That's not there. Well, maybe because that's a touchdown if it is. Creamhunt out of the backfield. Easy. Let's get out of bounds, buddy. Hawkinson should be open. He is for the laser maybe. No. He toe dragged and caught that. Damn. I totally thought we had that angle. I had so much fun with Vick, but that's a tough fell. It's a little harder when your team is 78 overall in the whole line, but I shouldn't need a crutch. I should have been able to play well with Michael Vick without a good team. 11 for 56. Five yards per carry. I just hate that he had that fumble on the goal line. That's random. Marshall should have been in. All right. I will take mile A boys. Great time. As soon as there's an affordable escape artist QB, get one and try it out. It's insane. This is a blast. I'll see you boys in the next video. Peace.