 So, you have these definition of positive definiteness and I want to connect it to this matrix. I want to sort of claim in some sense that the matrix function x transpose Ax or basically if I just add a t plus 1 does not change much. This is also positive definite function in the sense of you know how we talk about positive definiteness. So, I want both definitions to align. So, I want to claim that x transpose Ax given that A is positive definite is a positive definite function. How do I do that? It is pretty straightforward. I can decompose A in this form because A is a symmetric matrix because I cannot talk about positive definiteness. So, A is symmetric. Therefore, you can decompose it in this form. This m transpose lambda m where lambda contains a diagonal matrix of eigenvalues and m is just the eigenvectors. So, I can do this sort of characterization here and then if I write my, if I write x transpose Ax in this form, basically I am just in a sense doing a similarity transformation if you think in terms of coordinates. So, if I write my x transpose Ax like this and I choose, so I can write this is mx transpose lambda mx and I choose y as mx. This is the new state. This is the similarity transformation. So, y is constructed out of the eigenvectors. Then this quadratic form can be written as this y transpose lambda y. All of this is very simple for symmetric matrices, real eigenvalues diagonalizable naturally and all that. If it is not, then you have more problem, but we are dealing with this nice symmetric matrices. So, this is essentially sum of lambda i yi squared and all the lambda i's are obviously strictly positive and so this has to be greater than equal to some lambda min norm y squared because I take just the smallest lambda i, it is a finite number of lambda i. So, I can take the min, so I can just write this as lambda min sum of yi squared. So, lambda min sum of yi squared is just lambda min times norm of y squared and so basically what I have is a class k function, it is lambda min times class k function of norm y because the norm itself is a class k function, it is like saying x square is a class k function. Similarly, norm of y square is a class k function, in fact it is a class k r function, it is a class k r function, not just a class k function. So, therefore, this and this is just a scaling, it is a positive scaling, it does not affect the class k nature of the function. So, this is just a positive scaling multiplied by phi norm y square, in fact I can take this with the norm y square and just say that it is, it is a class k function. So, what have we shown? We have just shown that x transpose Ax is, dominates a class k function, in fact it dominates a class k r function. Therefore, x transpose Ax itself is a positive definite function, ok, that is all we need, we need, it has to be 0 at 0 and all that, I mean it is 0 at 0, right, x transpose Ax is obviously 0 at 0, right and it has to be continuous which is also obviously the case, x transpose Ax is continuous in x, no problem, smooth in x in fact, yeah. So, this is a x transpose Ax has been proven to be a class, sorry, a positive definite function, ok and that is what we needed, we wanted to reconcile, yeah. So, even if I take my v tx with a time argument here, multiplied by x transpose Ax in this form, no problem, it is still greater than equal to x transpose Ax for all non-negative t. So, therefore, this is also a positive definite function, ok, does that make sense, all right, ok, great, all right. So, now that we have this sort of characterization for class k function, I will again say that it is not easy to verify this in general, yeah, because I do it all this, I sort of cheated, right, because I first constructed a phi and then constructed a v, in reality for most systems, you have to construct a v first and you do not think about constructing a class k function to dominate and all that, all right, but once you have a v, trying to find a class k function that it dominates is not very easy, all right and so here we have this nice and easier conditions which obviously have been sort of collected years from Vidya Sagar's book, ok and the characterization is very straightforward. If you have a function of only the state, ok, therefore I use a different notation, it is still the same v or whatever you want to call it, I have used wx because there is no time argument, I am using just a different notation to distinguish these cases. So, if you want to discuss positive definiteness of wx which is a function of the state only, then you require to check, you are required to check only two things, one that it is 0 at 0, this was anyway one of the conditions, same condition does not change anything, right and the second one is this guy, yeah, that wx has to be strictly positive for all x which is not 0 in this domain, yeah. So, this domain that is ball of radius r is fixed, we are somehow assuming that r states evolve in this ball, ok. So, you have to only verify two things, one that it is 0 at 0 value of the state and two that it is strictly positive whenever the states are non-zero, ok, again something that should remind you of the norm, right, norm also has such a property, right, it is 0 when the argument is 0 and it is strictly positive when argument is non-zero, ok. You can see that there are these, you know similarities between the two, ok, great. Now, the easy check does not remain so easy when you have a function of both time and state, in this case the only way to check is obviously you want the first condition which we cannot do away with in any case, alright. The only way to verify this is that you have to find a positive definite w to dominate which is only a function of the state norm, ok. So, instead of hunting for a class k function you are hunting for a positive definite wx, ok. So, I would not say this is significantly easier or anything like that, but still it is another characterization, ok. So, what this first characterization, more often than not we are dealing with time invariant or autonomous systems, we hardly talk about or we hardly see a lot of real examples where there is time varying quantities in the system, usually we do not, yeah. And even if we do more often than not the Lyapunov or the Lyapunov function means that we look at do not contain a time argument, ok. We use a time invariant v to analyze even systems that are time varying, at times, ok. So, this second one being not so useful does not impact us in a lot of scenarios, but it can also, yeah. So, these are the easy characterizations you can see all you have to check is that it is 0 for especially for this case where there is only the state, you just have to check that it is 0 at 0 and then it is positive for all non-zero values of the state, very easy, yeah. And the proof of this is obviously in Vidya Sagar's book, yeah. Please take a look at it, it is very interesting. What it basically says is that if you have this kind of a condition, then you can always find a class k function to dominate, ok. If this condition is satisfied you can always find a class k function to dominate, ok. So, this is rather nice, right, rather powerful. It is not a, it is not exactly constructive in the sense that the book is not actually showing you a construction of the class k function, but it just shows that there exists such a class k function, ok. So, this is pretty cool, ok, alright. The only thing is for v to be negative definite minus v needs to be positive definite, yeah, this guy, alright. And the notation we use is just the flipped over here, alright, ok, great. Acha, I do not know why I have repeated this because it is we already looked at this idea, ok. We already looked at this example, yeah. If you take v t x as t plus 1 x transpose Ax, ok, then it is, then in this case I know that it is greater than equal to Wx defined in this form for all non-negative t, right. And once I have Wx equal to this, I just need to verify this positive definiteness of Wx, right. It is not difficult at all, it is 0 at 0, no problem, yeah. And since, in fact, I do not have to even look at a lot of arguments because it is a positive definite matrix in between the quadratic form is always positive, right. By virtue of it being a positive definite matrix, x transpose Ax is always positive for non-zero x, yeah. That is what it means for matrix to be positive definite. So, even without looking at this eigenvalue decomposition, I can directly say this, right, because A is positive definite for all non-zero values of x, x transpose Ax has to be strictly positive. And that is all we need here, strictly positive for all non-zero states, ok, which exactly is satisfied in this case, ok. So, pretty straightforward. So, the idea is positive definite matrices lead to positive definite functions, ok. So, and please do not think of this as a trivial result. The point is we use in a lot of scenarios, we do use quadratically up, even for non-linear systems, ok. Especially when you have systems which are to a large extent feedback linearizable, ok. We have not talked about this obviously, we will come to this in the second half of the course, but a lot of systems can be linearized via feedback, yeah, a lot of aeromechanical system, ok, can be linearized via feedback, ok. And for most of those systems, we can use quadratic Lyapunov function, yeah, because once you have some kind, something linear appearing in your dynamics, then you can use the linear Lyapunov candidate, right, which is x transpose Ax or Px or whatever you want to call, ok, alright, excellent. What about this guy? This is this function, ok. Earlier we tried to verify that everything is a class k function and all that, but I know now that this is greater than equal to this guy, and I know that this is 0 at 0, and I know that for all non-zero values of the state which is norm x is non-zero. If norm x is non-zero, this is positive, right. So, therefore, v, v t, x is dominating a positive definite function, alright. So, I did not have to find any class k function to dominate again, yeah. Of course, in this case, this is also a class k function, but even if it is not the case, we do not have to worry about it, alright, ok. The next more stringent or more or a smaller class of functions is the radially unbounded function, ok. In this case, we can no longer take as argument states in a ball, but it has to take arguments which are all from all of R n, ok. So, no more Br ball of radius R n all that, yeah, because we are talking about radial unboundedness, which is a global property in some sense, ok. And again, everything else is the same, time and states maps to some real number, right. You require that the function is 0 at 0. So, the only difference now is that it has to dominate a class k R function, ok. The function has the function v t, x has to dominate a class k R function, ok, for all t in R plus and for all x in R n, ok. Now, slightly different picture, v is allowed to be oscillating, no problem, oscillating v is fine, again I should say, yeah, and oscillating v is fine, but it has to be above this class k R function, ok. Therefore, as you can imagine as x goes to infinity, v t x also goes to infinity, right, because it is always above this guy. So, if this guy is going to infinity, this also has to go to infinity, ok, alright. So, that is the difference. So, the property of going to infinity is inherited by the radially unbounded function also. Therefore, the word radially unbounded, ok. Why radially? Because you can think of somehow norm of x as some kind of, you know, radial direction, ok. So, it basically as norm of x goes unbounded, which is the same as saying state goes unbounded, v also has to go unbounded, ok. And as I mentioned, this is connected to global stability, ok, yeah. We will talk about why in some sense, I mean, may be later, but the idea is not that complicated. If your function sort of, I mean, if you have a function, say, if your function looks like this, yeah, I mean I am at 0, yeah, going to infinity on both sides, this is v and this is x, forget the time argument, ok. If the function looks like this, then great, I mean, if you say that, if you somehow say that in the y argument, that is in, if you somehow can claim that my function lies below this guy, yeah. If I can claim that my function all value always remains below this, right, then I can claim that my x lies within this, yes, no problem, ok. This is a radially unbounded function, ok. This is a radially unbounded function, yeah, it goes all the way to infinity. But now, if I have a different scenario, I have to make a different picture, sorry, I cannot fit it here. If I have a different scenario, which is that you have a function, which now does this, yeah, very much a class, I mean, positive definite function, right, because I am only concerned about, you know, say some domain, whatever, yeah, you can think of this as increasing even there, yeah. I mean, you can sort of imagine that this is also increasing, I mean, not actually flat, ok. But the point is, now if I say that my y, just look at this. Now, if I say that my y is restricted to this level, ok. Again, this level is not, whatever, I mean, this level is close to the upper bound in some sense. I cannot say much about x, right. x can be really large. The bound on x could be really, really, very large. Of course, if I give you smaller bound then, ok. But if I am actually giving you bound right on the boundary, then I cannot say anything much about x. x could be very large, ok. Therefore, you can see this is not the one on the left. This is not a radially unbounded function, ok. Even, let us assume that in both cases, the domain is Rn. Let us not worry about Brn all that. There is no Brn. Let us assume that the domain in both cases is Rn. But this function is this structure and this function is this structure. This function at this guy lets you actually conclude something about, yeah. For all possible values of v, you can claim some bound on x. Here you cannot. So, this invertibility sort of property is what makes radially unbounded functions amenable to global results, yeah. And here it will only give you local results. Why local? Only until some x you can, some levels you can work with. From this level, ok, this level, ok, this level, ok, here not ok. Beyond that, forget it. You cannot say anything. Beyond this level though, nothing, obviously. You cannot say anything about x. x could be anything, ok. Because we can never reach that level, all right. So, that keep that in mind. That is the idea. We can, we will of course prove some things so that these ideas are not just ideas and you see that it works in the math also. But it all depends. It is always using invertibility property. You will always think of using v inverse. Whenever you look at the proof, you will see. Entire proof goes by using v inverse and so on and that is what this is. This is v inverse, right, ok, all right, great, ok. So, examples. This function obviously class k r, just the norm in fact. This is the Euclidean norm. Yeah, this itself is a class k r function. So, if I take v as that, v has to be radially unbounded, because it is equal to a class k r function, all right, simple. This example, again obviously class k r. Why? Because this dominates this way, right. This dominates this for all non-negative time, right. So, again class k r, sorry, dominates a class k r function or a positive definite or a radially unbounded function, right. So, you are fine. So, this is radially unbounded, yeah. In fact, you can also think of it differently. You can say that this is dominating a class k function and goes to infinity as x goes to infinity in any direction. So, we are fine, ok, all right. I think that goes until like this. Yeah, let us look at this guy. What about this guy? 1 plus sin square t divided by 2 x 1 square plus x 2 square, ok. I am again claiming that this is greater than equal to half x 1 square plus x 2 square for all non-negative t. That means, yes, because sinusoid smallest value is 0 because I took a square deliberately. So, therefore, I get the half, ok. I have deliberately taken this example for one specific reason, ok. Easier conditions, ok. Unfortunately, for all the examples I am taking easier conditions and the normal conditions do not look too different, but I can promise you these easier conditions are what most people use. Yeah, they never try to find a lower bounding class k function and k r function, all right. What are the easier conditions? The first two conditions look exactly the same for radial unboundedness. The only thing additional is this going to infinity condition, right, because obviously this verifies that w is a class k function and radial unboundedness is just a class k function with going, it going to infinity, that is it, all right. Though again remember, it is not that simple. This is being verified for all Rn other than 0, not just x in a ball, ok, not just x in a ball. So, the first two conditions in Vidya Sagar's book especially, they tend to have additional notation, they use local positive definite LPDF and PDF. So, locally positive definite functions and positive definite functions. So, this is actually a PDF, a positive definite function in Vidya Sagar's notation. Why? Because if this is verifying this class k condition for all states, not just states within some ball around the origin, all right. So, that is the difference here. So, that is why I have written global positive definiteness plus the unboundedness condition, ok, great. Counter examples, ok, counter examples are very good because they tend to jog you. If you look at this function, this is positive definite radially unbounded, I have written an explanation. The function is x1 plus x2 whole squared divided by 2, yeah. What happens? There are only two states, x1 and x2 and I am giving you a function x1 plus x2 whole squared divided by 2 is irrelevant, but whatever. Is this class k, class kr, what is it? Is this class kr, ok? So, this is you are talking about class k, ok. So, what are those non-zero points? Right, as simple as that, yeah. This is not even class k. Why? Because if x2 and x1 are opposite signs, x2 is minus x1, all right. For all possible values of x2 equal to minus x1, basically this line, that is what I have drawn here, yeah. Along this line, v is 0. Along this line in state space, this is 0 and this does not satisfy our easier test. Easier test requires for all possible non-zero states, yeah. And these are obviously non-zero states, for all possible non-zero states except for this guy, v has to be strictly positive or w has to be strictly positive, which is not, ok. So, that is a problem, not positive definitely. What about this guy? x1 square plus x14, yes, yes. So, if I take points of the form this, where x2 is 0. So, basically along the, sorry, where x1 is 0, yeah. Basically along the y-axis, I take anything, this function is exactly 0, yeah. Again, 0 for non-zero values, so not positive definite, yeah. State being 0 means every element has to be 0, yeah. So, this is a non-zero state, 0, alpha and this is evaluating to be 0 at that, ok. So, these are counter. So, simplest thing to remember, if all the states do not appear in your w or v, whatever notation you want to use, if all the states do not appear, then it is immediately not positive definite, ok. And it is not a function we can do Lyapunov analysis with, ok. So, all states must appear, simple. This is the first key requirement, alright. Next, but this, this we have already done, right. This function, this is positive definite, right. We have already done that because it is strictly positive, yeah, for non-zero values of norm x, right. But this is not readily unbounded. That should also be very evident, why? It maxes out at 1, right. Maximum value this can take is 1 as x goes to infinity. Therefore, it is not unbounded, ok. So, this is something like a, I mean, I mean, maybe the shape will be different, but it is something like, you get, tapers off, does not go to infinity. So, this is again cannot be used for global stability analysis, only for local results you can use, ok, alright. Similarly, for the easier test for radial unboundedness or not necessarily easier anymore, if it is a function of time and state, you still need the zero condition and you need a Wx which is readily unbounded that you can dominate, ok. So, this is not an easier condition like I said because it is almost the same as trying to find a class Kr function, ok. So, for the time varying cases as you can see the easier conditions are not too easy, but that is about all you have, ok. So, but for the time invariant V or W, you have very easy regards to everything, ok.